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SOVIET VOROSHILOV ACADEMY LECTURES
HANDBOOK

 

CHAPTER FIVE

NORMS AND CALCULATIONS



 

TABLE OF CONTENTS

 
 

I SOVIET PLANNING NORMS 5-1
Introduction 5-1
Norms for Accomplishing Tasks 5-3
Rate of Advance 5-6
Average Density of Forces in Defense 5-6
Location Distances 5-8
Norms in Mission Assignment Theory 5-9
Norms for Depth of Missions and Widths of Sectors for Offensive Operations 5-11
Reconnaissance Norms 5-14
Combat reconnaissance unit 5-14
Division Reconnaissance Norms 5-14
Security 5-15
Aerial Reconnaissance 5-15
Space Reconnaissance 5-16
Artillery Reconnaissance 5-18
Air Defense Reconnaissance 5-18
Electronic Reconnaissance 5-19
Planning for Reconnaissance 5-20
Information for the Estimate of the Situation 5-20
For the Commander's Decision 5-20
For Detailed Planning 5-20
Norms in Planning Air Army Operations 5-21
Planning Factors for Airborne Operations 5-22
Norms in Air Defense Planning 5-24
Air Defense Army 5-25
Factors Used in Calculating Air Defense Capabilities 5-25
Rear Service Norms and Planning Factors 5-26
Army Rear Services: 5-26
Front Rear Services 5-28
Some Fundamentals of Troop Supply 5-31
Medical Norms and Planning Factors 5-36
Casualty estimates 5-36
Protection against nuclear radiation 5-37
Norms for Planning Combat Movement and Marches 5-40
Movement Planning Norms 5-40
Road Marches 5-41
Rail and ship movement 5-43
Calculations in the Decision Process 5-43
Clarification of the Mission 5-43
Estimate of the Situation 5-43
Enemy 5-43
Friendly 5-44
Terrain 5-45
Calculations Essential for Correct Decisions 5-45
Mathematical Modeling - the Basis of Calculation Techniques 5-46

II CALCULATIONS FOR PLANNING COMBAT 5-52
Operational and Tactical Calculations 5-52
Basic Time and Distance Calculation 5-52
Calculation of Time to Begin Move to Start Line 5-54
Calculation of Time to Deploy into a New Assembly Area 5-55
Calculation of Time a Unit will be in a New Area 5-56
Calculation of the Duration of a March from one Area to Another 5-57
Determine the Required Movement Rate for a Unit to Regroup in a New Area 5-58
Calculation of Length of Route, Average speed and Duration of Movement of Moving Column 5-59
Calculation of Overall Depth of Column Consisting of Several Sub-columns 5-61
Calculation of Duration of Passage of Narrow Points and Difficult Segments 5-63
Calculation for Passage Times Across Start Point (SL) by the Head and Tail of the Column 5-67
Calculation of Expected Time and Distance of Probable Point of Contact with Advancing Enemy 5-68
Calculation of Time Required for Advancing and Deploying Sub-units to Change From Line of March into the Attack 5-70
Calculation of the Time and Distance to the Line of Contact 5-73
Calculation of Expected Time and Rate of Overtaking when Pursuing the Enemy 5-75
Calculation of the Work Time Available to the Commander and Staff for Organizing Repulse of Advancing Enemy Forces 5-77
Calculation of Length of Time to Operate Command Post in a Single Location 5-79
Determination of Quantity of Various Weapons, Reconnaissance, Support, Communications etc. for Task Performance 5-80
Modeling battle 5-83
Calculation of Strike Capability of Sub-units 5-87
Calculation of the Width of Main Attack Sector 5-89
Calculation of Required Destruction of Enemy 5-89
Calculation of Rate of Advance in Relation to Correlation of Forces 5-91
Determine the Possible Friendly and Enemy Losses in Relation to the Correlation of Forces and Rate of Advance 5-93
Determine the Required Amount of Manpower and Weapons for Bringing Sub-Units back up to Sufficient Strength to Restore their Combat Capability 5-96
Form for the Calculation of the Required Amount of Personnel 5-97
Determine the Expected Radiation Dose 5-98
Calculation to Select the Optimal Travel Route 5-102
Calculation to Determine Optimal Distribution of Weapons 5-104
Calculation to Determine the Effectiveness of Fire Destruction Means 5-108
Reconnaissance Planning 5-109
Calculation of Effectiveness of Reconnaissance and Required Duration for a Reconnaissance Mission 5-109
Calculation of Detection of Targets by Reconnaissance 5-111
Sample Calculations for Division and Army Staff 5-113
Calculations for Front Offensive Planning 5-122
Calculating Operational Scale Rates of Movement 5-125
Methods for Calculating Marches in Complex Situations 5-131

 
 

LIST OF FIGURES - CHAPTER FIVE

 
 

Figure 55 Norms for times to accomplish tasks 5-5
Figure 56 Norms for average rate of advance 5-6
Figure 57 Norms for density of defense 5-6
Figure 58 Norms for locations 5-8
Figure 59 Diagram of division offensive norms 5-12
Figure 60 Diagram of front offensive planning factors 5-13
Figure 61 Combat reconnaissance units 5-14
Figure 62 Limits of visual observation 5-15
Figure 63 Aerial Reconnaissance 5-16
Figure 64 Drone reconnaissance aircraft 5-16
Figure 65 Visual reconnaissance range 5-17
Figure 66 Artillery reconnaissance 5-18
Figure 67 Preparation time for reconnaissance report 5-21
Figure 68 Tactical-technical characteristics transport aircraft 5-22
Figure 69 Echeloning material reserves of the front 5-33
Figure 70 Norms for material reserves in army 5-33
Figure 71 Requirements for material reserves in offensive 5-34
Figure 72 Operational expenditure and reserves 5-34
Figure 73 Location of rear supply and support elements 5-35
Figure 74 Protection against radiation 5-38
Figure 75 Coefficients of reduction of radiation 5-39
Figure 76 Movement data for tracked and wheeled vehicles 5-40
Figure 77 Average speed of march and length of daily march 5-40
Figure 78 Times to form unit column 5-41
Figure 79 Distances between column elements 5-42
Figure 80 March, halt, and rest durations 5-42
Figure 81 General character of tactical calculations 5-48
Figure 82 Nomogram for calculating duration of a march 5-53
Figure 83 Nomogram for calculating duration of movement from assembly area to start line 5-55
Figure 84 Nomogram for calculating time required for mobile column to deploy into new area 5-56
Figure 85 Form to calculate time unit is in new area 5-57
Figure 86 Form for calculating duration of march 5-58
Figure 87 Form to calculate required rate of travel 5-59
Figure 88 Calculation of duration of march over complex route 5-60
Figure 89 Nomogram for calculating length of mobile formation consisting of several columns 5-62
Figure 90 Nomogram to calculate time required to pass narrow spot on route 5-65
Figure 91 Nomogram to calculate time required to pass difficult section of route 5-66
Figure 92 Nomogram to calculate time head and tail of column will pass regulation point 5-68
Figure 93 Nomogram to calculate time and distance to point of meeting engagement 5-69
Figure 94 Calculation of time to advance and deploy into the attack from line of march 5-72
Figure 95 Form for calculation of expected time and location of meeting engagement 5-74
Figure 96 Nomogram for calculating expected time and speed to overtake retreating enemy 5-76
Figure 97 Nomogram to calculate time available to plan fire on advancing enemy 5-78
Figure 98 Calculation of duration of operation in one location 5-79
Figure 99 Nomogram for calculating the combined effectiveness of several systems 5-82
Figure 100 Coefficients of comensurability 5-84
Figure 101 Expected losses 5-85
Figure 102 Nomogram to determine required losses to achieve correlation of forces 5-90
Figure 103 Nomogram relating correlation of forces to rate of advance (1) 5-92
Figure 104 Nomogram relating correlation of forces and rate of advance (2) 5-92
Figure 105 Force attrition nomogram - army 5-94
Figure 106 Force attrition nomogram - front 5-95
Figure 107 Table for calculating restoration of combat effectiveness 5-97
Figure 108 Form for calculating expected radiation doses 5-100
Figure 109 Nomogram to calculate expected radiation dose 5-101
Figure 110 Table to enter description of travel routes 5-102
Figure 111 Graph of effectiveness of alternate routes 5-103
Figure 112 Table of effectiveness indicators for march routes 5-104
Figure 113 Table of effectiveness of artillery fire on targets 5-105
Figure 114 Table of effectiveness of artillery fire on targets 5-106
Figure 115 Table of effectiveness of artillery fire on targets 5-107
Figure 116 Form for calculating weapons effectiveness 5-108
Figure 117 Form for calculating probability of target detection 5-110
Figure 118 Form for calculating search length 5-110
Figure 119 Nomogram for calculating probability of target detection 5-112
Figure 120 Value of coefficient 5-130
Figure 121 Nomogram to determine quantity of march routes 5-130
Figure 122 Conditions of formation shift (a) and deployment (b) of columns on a march 5-132
Figure 123 Formation change of columns in movement 5-132
Figure 124 Relationship between speed, time, and distance required for columns to move from sequence head-to-tail into line abreast 5-133
Figure 125 Values of coefficient K 5-134

 

 

I SOVIET PLANNING NORMS

 
 

Introduction

It is impossible to overestimate the role and importance of "norms" in every aspect of Soviet life. The Soviet officer has been brought up thinking in terms of fulfilling plans and accomplishing tasks in accordance with established norms. We will not discuss all the ramifications of this phenomena here. Suffice it to say that thinking in materialist terms that can be quantified about all aspects of life is part of the essence of Marxism-Leninism. In addition to his need to quantify in order to believe he understands, the Soviet officer obtains a measure of self-confidence by believing he is doing things "po normu". The confidence stems from believing he has the weight of history and science behind him, and, in a more practical vein, it allows him to justify his decisions by "scientific substantiation" when questioned by superiors. In addition the use of norms provides readily understandable measures for success and failure which may be used during operations to determine if things are going well.

Soviet military planners use a variety of standard figures or norms in the preparation of plans. There are several ways to classify them. One way is to consider norms for time required to accomplish tasks, norms for space allocation (width and depth or location) required for units or tasks, and norms for quantities of resources provided or required to accomplish a mission.

In the following tables and discussion we have grouped the norms in this way for use by interested individuals. In addition they are grouped by functional area, such as operations, reconnaissance, artillery, air defense, and air. A somewhat different scheme for classification is employed by The Soviet Studies Research Centre at RMA Sandhurst, whose excellent work, The Sustainability of the Soviet Army in Battle, we have used freely in compiling these norms. The following description is quoted from that book.

"The standards, which have the force and authority of regulations, i.e. law, are known as "NORMATIVY": or "NORMY", in English, simply "Norms".

The Great Soviet Encyclopedia describes a norm as the minimum of something, as established by a rule or plan, for example, a time norm or a norm for sowing grain. However the definition in the Soviet Military Encyclopedia is more detailed. The word norm originates from the Latin "Normatio" meaning "to regularize" and for military purposes, norms are subdivided into a number of groups operational and tactical (spatial and temporal) norms, norms of expenditure, and norms of supply.
A. The first type of norms are those scales which characterize the spatial and temporal factors of operations and tactical tasks of combat forces and the terrain of which they act.
- Spatial operational-tactical norms take the form of:
--- depth of battle tasks;
--- dimensions of zones, areas, and sectors of combat operations;
--- areas of grouping of forces; battle order, formations, and groupings along the front and in depth;
--- scale of re-deployments and regrouping.

Temporal norms are concerned with the time taken to complete a task, march or maneuver. They are worked out taking into account the fighting strengths and capabilities of ones own forces and those of the enemy, battle experience, experience of operations and tactical exercises, degree of preparedness and training of personnel, the results of special research, terrain conditions, time of the year, and time of day.

B. The second group of norms, which are relevant to the military sphere are "NORMY RASKHODA" - norms of expenditure of material resources. These norms are concerned with the accounting of supplies in units of mass or volume or as individual items in their expenditure by servicemen, weapon systems, sub-units, units, formations, and armies. Again, the norms are laid down by the Soviet Ministry of Defense on the basis of researched and calculated data. For instance, the basic norm of consumption of fuel (diesel, gasoline, oil and lubricants (POL)) is laid down in liters or kilograms for each vehicle, usually for 100 km of movement or for one hour of operations or of static running. When special conditions prevail (difficult terrain, bad weather, etc.), a supplement is added to the basic norm. Norms of expenditure are laid down for; ammunition in "BOYEVOY KOMPLEKT" (BK) "Units of Fire"; fuel in "ZAPRAVKIY" refills; rockets in batches. Norms of expenditure for each type of ammunition and fuel for battle are worked out well beforehand on the basis of the action or operation envisaged. For example, the norms of fuel consumption for tanks in an offensive battle are calculated according to the planned depth of the operation taking into consideration terrain conditions, weather, and coefficients of maneuverability. As a rule, norms of expenditure also take into account the availability of material resources.

C. The third type of norms are "NORMY SNABZHENIYA", norms of supply. These are the amount of material resources laid down for supply to servicemen, sub-units, units of formations and designated for use in a specific period of time. Under this category are included: spare parts, types of instruments, material stores: ammunition, POL, and rations. They are closely linked to norms of expenditure. So in essence the three types of norms covered, (including temporal with the operational_tactical group), are:
- operational-tactical norms, i.e. norms of performance;
- norms of expenditure;
- norms of supply.

Norms for exploitation of equipment are not included."

The classification in this handbook also has three categories, but lumps the norms for time and space together as operational norms while dividing the norms for resources in to separate categories for expenditure and supply.

Another way of looking at norms is to consider that some are meant to stand alone as specific criteria in themselves for use as direct measuring or reference points immediately understandable to the user. Other norms are not too meaningful as bits of data, but are meant to be used as part of the process of calculating requirements or capabilities. With this in mind it should be noted also that quite a bit of "normative" prescription is also contained in the formulas to be used in the calculations. Thus it is impossible to separate the Soviet norms from the calculations in which many are used. In support of the calculation process the Soviet headquarters is now increasingly equipped with computers. However, when these are not available, the officer may still use the well-established methods. These are three in number: direct calculation from a formula ( with perhaps a hand calculator), drawing lines on a nomogram, and filling out the blanks in a table. Samples of all three are found in this handbook.

A critical set of norms for Soviet planners, which are not included, is composed of data on enemy forces. In principle, Soviet perceptions of U.S. and NATO force characteristics should be added, since they are crucial to the Soviet perception of combat.

 
 

Norms for Accomplishing Tasks

The following table shows estimated times required for Soviet headquarters to complete the planning indicated, or for the units and formations to accomplish the activities specified.

 

Figure 55 - Norms For Times to Accomplish Tasks

NORMS - TIMES FOR DECISIONS OR ACTIONS

These are average times required to perform the indicated tasks:
Unit Regiment Division Army Front
Clarification of mission 20 min 30 min 40 min 60 min
Estimate of situation 1.5-2 hrs 3-4 hrs 4-6 hrs 6-12 hrs
Make decision 30-40 min 40-60 min 1-1.5 hrs 1.5-2 hrs
Write directive or order 30 min 40 min 60 min 1.5 hr
"Recognosirovka" 1.5-2 hrs 2-3 hrs 4-5 hrs 5-6 hrs
Prepare and write map plan 1-1.5 hrs 3-4 hrs 12 hrs 24 hrs
Preparation of planning recon 30-60 min 1.5 -2 hrs 6-8 hrs 12-18 hrs
Preparation and plan - artillery and rocket 1-2 hrs 3-4 hrs 6-8 hrs 12-18 hrs
Plan of PVO (air defense) 30-60 min 3-4 hrs 6-8 hrs 12-18 hrs
Plan air army -- -- -- 12-18 hrs
Plan engineer force 1-2 hrs 3-4 hrs 6-8 hrs 12-18 hrs
Plan chemical support 30-60 min 1.5-2 hrs 6-8 hrs 12-18 hrs
Plan rear services 1-1.5 hr 4-6 hrs 8-10 hrs 18-24 hrs
Plan signal 1.5-2 hr 2-3 hrs 6-8 hrs 12-18 hrs
Plan REC (radio electronic combat) - 6-8 hrs- 12-18 hrs
Prepare plan of first nuc strike - 1-2 hrs 4-6 hrs 8-12 hrs
Prepare plan to occupy the FUP (forming-up area) 1-2 2-3 hrs 4-6 hrs 8-12 hrs
Conduct occupation of FUP 1 night 1-2 nights 1-2 nights 2-3 nights
Prepare plan for regrouping 30-60 min 1.5-2 hrs 4-6 hrs 8-12 hrs
Conduct regrouping 1 night 1-2 nights 1-2 nights 2-3 nights
Prepare plan of maskirovka 30-60 min 1.5-2 hrs 4-8 hrs 8-12 hrs
Prepare plan of defense against enemy mass wpns and maintain combat readiness afterwards 30-60 min 1.5-2 hrs 4-8 hrs 8-12 hrs
Prepare plan deployment and move CP's 30-60 min 1.5-2 hrs 2-3 hrs 4-6 hrs
Prepare plan for morale and political work 30-60 min 1.5-2 hrs 6-8 hrs 12-18 hrs
Prepare report 20 min 30 min 40-60 min 1 hr
Prepare combat instructions 30 min 30-60 min 1.5-2 hrs 2-3 hrs
Prepare operational summary 30 min 60 min 1.5-2 hrs 2-3 hrs
Prepare plan for commitment to battle for 2nd echelon 30-60 min 1.5-2 hrs 4-6 hrs 8-12 hrs
Change plan for location of unit 30 min 2-3 hrs 4-6 hrs 6-8 hrs
Prepare new decision during battle and planing commitment of 2nd echelon and issue mission 1.5-2 hrs 2-3 hrs 6-8 hrs 12-18 hrs
Commitment into combat of 2nd echelon in three phases
1 assembly 10 min 20 min 30 45-60 min
2 movement depends on distance - 30 km/hr 2 hrs 1 night
3 deployment 15 min 30 min 2-3 hrs 1 day
Crossing water without defense by enemy 1-2 hrs 2-4 hrs 6-8 hrs 3-4 days
Crossing water against defending enemy on march 3-4 hr 5-7 hrs 1-2 day 3-4 days
Passing water by force against prepared enemy with stop for preparation 4-6 hrs 6-10 hrs 2-3 days 4-5 days
Forcing river by battalion acting as forward detachment on the march 2-2.5 hrs
Time mobile obstacle detachment of one battalion can lay mine field on a front of 5 kms is one hour, maybe 15 min to prepare to move, then move 20 km/hr, then lay mine field 5 km wide 1 hr
Preparation plan airborne 2-3 hrs 4-6 hrs 6-8 hrs 8-12 hrs
Conduct airborne operation - This is divided into three phases:
1. assembly: Depends on distance to move to assembly area, the norm is 30 km/hr; it also depends on size of unit.
2. move in air: depends on distance and type of aircraft:
3. landing: depends on size of unit:
For example, a battalion or company takes 30-40 min to move to assembly area and load. To move 100 km by helicopter takes 30-60 min. Landing takes another 20-30 min to assembly and move off.
For a regiment to load on aircraft at three airfields, one for each bn, depends on location and distance from assembly area to airfields, but might be about 40-60 min. For a landing at a distance of 200 km, movement by aircraft takes 30-60 min. The landing and assembly takes another 30-40 min.

Vayner, Ivanov, Altukov, and other Soviet authors discuss the use of PERT Chart methods for determining the optimum organization of planning and other preparations for combat. See especially Troop Control Through PERT Methods by Pavel G Skachko, V. M. Kulikov, and G. K. Volkov, Voyenizdat, Moscow, 1974. The above estimates of times required to accomplish tasks could be used to develop PERT diagrams, especially of the planning activities at each level headquarters. Evidently the Soviet practice is to have a series of representative PERT diagrams prepared ahead of time, one for each basic contingency. See the discussion of the commander's estimate of time available for decision making and planning in Chapters Two, Three, and Four and the sample Calendar Plans shown in Chapter Six.

 

Rate of Advance

Soviet norms for average rates of advance in various conditions are shown in the following table. These may be use as very general planning factors when considering what is a desirable achievement. For a more sophisticated approach use the methods for calculating the advance rate as a function of the correlation of forces and situational factors. This technique is discussed in the section of calculations below.

 
 

Figure 56 - Norms for Average Rate of Advance

AVERAGE RATE OF ADVANCE DURING ACTION INDICTED

ACTION RATE IN KM/DAY
Average rate during breakthrough of fortified region 20-25
Breakthrough of prepared position
25-30
Breakthrough occupied position 30-40
Offensive in depth of defense 60-70
Offensive in mtns, forests, swamp 30-50
Offensive flat and steppe 70-80
 
 

Figure 57 - Norms for Density of Defense

 
 
Average Density of Forces per KM Width in Defense
Personnel 0.2-0.3 Bns
Tanks 7-10
Guns and mortars 5-6 units
 
 

The modern defense is very dispersed. On the direction of the main effort multiply by 1.5 to 2.0. Densities in the attack depend on the nature of the defense and the correlation of forces required to achieve stated missions. These are discussed in the chapters on division, army and front planning and in the section on Norms in Mission Assignment, below

 
 

Figure 58 - Norms for Location Distances

The following table indicates the standard Soviet norms for the distances of various command posts, second-echelons, and other units from the forward edge of combat.

 
 

DISTANCES FROM FEBA IN KM

Unit Battalion Regiment Division Army Front
1st echelon CP - attack 1-2 4-6 10-12 30-50 100-150
in defense 2-3 5-7 12-15 40-60 100-150
2nd echelon CP - attack 2.5-3 5-6 10-12 40-60 200-300
2nd echelon CP- defense 1.5-2 5-7 15-20 80-100 150-225
1st echelon rear service CP in attack 4-5 12-15 30-40 40-60 120-170
1st echelon rear service CP in defense 5-6 15-20 40-60 100-120 150-200
2nd echelon rear service CP in attack
2nd echelon rear service CP in attack
Regt Arty Group - 3-4 - - -
Div Arty Group - - 4-5 - -
Army Arty Group - - - 5-6 -
Arty in defense - 3-4 4-6 8-10 -
PVO - 2-3 6-10 10-20 10-30
Depth of recon

(behind enemy)

- 6-20 100 400-500 1000-1500
Depth of air reconnaissance Tactical - 400 - 500 Operational - 800 - 1000
Width of attack total zone 1.5 3-5 8-12 - 15-20 60-100 250-400
Width of breakthrough 1 2 4 8-12 27-30
Width defense 3-5 10-15 30-40+ 100-150 350-500
Depth defense 2-3 9-10 20 100-150 250-300
 
 

Norms in Mission Assignment Theory

The Soviets initiate a combat operation normally in accordance with a detailed plan worked out well in advance; however, once the combat action starts, many factors begin to influence the course of the combat and thus the outcome of the operation. (See Chapter One for more discussion of this issue.) Military science suggests that the very bilateral nature of armed struggle creates an inevitable dynamism in combat operations with every action meeting a reaction while the reaction becomes an action for another reaction from the opposing side. This process could continue forever. Therefore no plan could be fully implemented as worked out but timely adjustments and even changes become necessary in the course of combat operations. The commanders are required to make necessary adjustments and changes in the plans in such a way which may not compromise the aim of the operations. However, sometimes and in certain areas of the operational zone the commander may be forced to change the missions and even the aim of combat operations, once it is known that the initial plan can not serve the actual combat situation.

Therefore in planning the combat operation, the preplanned missions are assigned to different levels of command in accordance with their combat capabilities to deal with possible reactions of the enemy in a framework of specified time and space. For example, preplanned missions for a motorized rifle regiment in attack encompass a dimension of 3-4 km x 8-10 km and 3-4 hours. It is assumed that under normal conditions when the regiment is reasonably reinforced with supporting arms and means and when it attacks on the main axis in a normal terrain, the regiment can accomplish its assigned preplanned missions on time with accepted level of support by the division. After the accomplishment of the preplanned mission, the regiment is assigned a new mission (normally one mission each time) in the course of the combat according to the situation.

Based on the concept discussed above the preplanned missions for different levels of command are as follows.

- squad / platoon and company - immediate mission;
- first-echelon battalion - immediate and subsequent missions normally in the depth of the enemy's battalion;
- first-echelon regiment - immediate and subsequent missions normally in the depth of the enemy's brigade;
- first-echelon division - immediate, and long range (daily) missions normally in the depth of enemy's division and corps blocking positions;
- first-echelon army - immediate and long range operational missions normally in the depth of the enemy's field army or force;
- first-echelon front - immediate and long range operational missions normally in the depth of the enemy's theater reserves;
- second echelons are given an immediate mission and an axis for further advance.

In the course of combat operations the enemy can react by in various ways using different means, particularly by the commitment of his reserves. In an offensive operation the following enemy counterattacks are expected to interfere with the accomplishment of the offensive missions of different levels of command.

1. Counter attack by battalion reserves can slow the attacking battalion attack and can stop the attack of at least one company. Courses open: commitment of the attacking battalion's second echelon at a preplanned time or early commitment of the regiment's second echelon.

2. Counter attack by the defending brigade reserve can slow the regiment attack and can stop the attack of at least one first echelon battalion of the regiment. Courses open: commitment of the regiment's second echelon on time or early commitment of divisional second echelon.

3. Counter attack by division's reserve can slow the attack of the division and can stop the attack of at least one regiment. Courses open: timely commitment of division's second echelon and use of mobile anti-tank reserve.

4. Counter attack by corps reserve can stop the attack of at least one division. Courses open: commitment of uncommitted reserves of first echelon divisions or early commitment of part of the army reserve.

5. Counter attack by army group reserve can stop the attack of at least one operational axis or the army operation. Courses open: opening another axis of attack by commitment of army's reserve or early introduction of the front second echelon.

Time and space for dealing with the enemy's counterattack(1):
- Enemy battalion reserve in depth of 1 km, one hour after the beginning of the attack;
- Enemy brigade reserve in depth of 2-3 km, 3 hours after the beginning of the attack;
- Enemy division reserve in depth of 6-7 km, 4-5 hours after the beginning of the attack;
- Enemy corps reserve in depth of 8-12 km, 7-8 hours after beginning of the attack.

At the army level the correlation of forces shows an estimated balance as follows:
- up to 8 battalion counterattacks versus 20 battalion second echelons;
- up to 4 brigade counterattacks versus 10 regimental second echelons;
- up to 2 divisional counterattacks versus 4-5 divisional second echelons;
- up to 1 corps counterattack versus 1 army second echelon and 1 army reserve.

This calculation suggests that battalion, brigade, and divisional counter-attacks can not interfere with the accomplishment of the army's immediate mission, but corps counter-attack, if launched on the first day, could force the army to change it's plan for the first day of offensive operations.

Other factors affecting the implementation of the initial plan of the army's operation include:
- air attacks;
- nuclear attacks;
- chemical attacks;
- destruction of communication routes;
- training of the troops;
- troop morale ( fear of isolation);
- uprising of local population;
- failure in other directions or operational axis;
- lack of initiative.

 
 

Adjustment and Change of Plans:

Even a small adjustment or change in initial plan requires a large amount of reorientation and coordination of procedures. It will take normally 2 hours delay in the advance to change the line of commitment of the divisional second echelon, provided the decision is made in the shortest possible time.

During combat operations, the army staff closely follows the combat action and makes adjustments in plans for interaction and support. The major changes are reported to the commander. The division commanders send situation reports after each 1-2 hours and on important occasions. The situation reports contain the situation of opposing sides, the commander's decisions and the recommendations.(2) The army staff processes the reports and submits the summary with suggestions to the commander. Based on these reports and suggestions the army commander makes his decision, which is conveyed by the staff to the division commanders through combat instructions. The army commander uses the following to influence the situation.
- supporting air force;
- SSM and artillery;
- mobile antitank reserve and mobile obstacle detachment;
- second echelon and reserves;
- airborne (seaborne) assault landings;
- assistance from higher echelon or adjacent units.



Norms for Depth of Missions and Widths of Sectors for Offensive Operations

The norms for offensive operations generally consider that there will be two successive front operations and each front operation was to consist of two successive army operations. The army has initial and subsequent missions. The army frontage is 60-80 kilometers, some may have frontages to 100 km. A tank army has a frontage of 45-60 km. The depth of the first operation is 250-350 km. The subsequent operation is 300-500 km more. In each of these the depth of the initial mission is 100-150 and the depth of the subsequent mission is 150-200 km. The final goal for the two front operations in the Western TVD may be deep into France, perhaps to the Spanish frontier. The duration of each operation is 6-9 days. The rate of advance during breakthrough with conventional weapons is 30-35 km a day. The penetration of a fortified area is 20-25 km a day. The attack in the depth of the enemy position is 60-70 km a day. The average rate of advance is 40-50 km a day. In mountains the rate is 30-50 km a day. The width of the army penetration is 6-8 km. Sometimes a reinforced army might have a frontage of 8-10 km.

 
 

Figure 59 - Diagram for Division Offensive Norms


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Figure 60 - Diagram for Front Offensive Planning Factors


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Reconnaissance Norms

The following are some standard planning factors for the organization and capabilities of Soviet reconnaissance forces.


FIGURE 61 - Combat reconnaissance unit

The following table indicates the organic troop reconnaissance elements in the division, army and front. These are long range patrol units.

COMBAT RECONNAISSANCE UNITS

Unit Size Operating location
Recon Co. division 9 groups 100 km
Recon Co army 8 groups 450 - 500
Recon Bn of front 35-40 groups 1000 - 1500
Total in front 250 - 300 groups
 
 

Each group in the course of a 24 hour period can perform reconnaissance two objectives, each in an area 40-50 square kilometers. Each group receives 1-2 missions and in 5-6 hours after insertion they begin to report information about the enemy.

 
 

Division Reconnaissance Norms

Reconnaissance detachments composed of a platoon to company+ are detached from division or regiment. They operate in a sector 4-5 km wide. The distance from own troops is 30-40 km (for div) 20-25 km for regiment.

Combat reconnaissance patrol (up to a platoon) is detached from battalion or company at a distance of 6-8 km.

The number of reconnaissance observation posts in a division and regiment=2-3 posts (3 observers in each post). The battalion has 1-2 posts. The company and platoon have 1-2 observers. "A reconnaissance in force is composed of a company size unit with attack mission to the depth of 1.5-2 km to seize an enemy strong point.
Deep reconnaissance is conducted by teams of 5-7 men (9 teams in the division) at a distance of up to 50 km. The team is assigned a specific target or area for cover. The dimension of the area can be 25-50 km sq.

Each of the division's radio interception post can monitor 3-4 radio nets. The division has 2-3 radio direction posts, which perform 30 verified fixes an hour. The division's radio technical posts can take a bearing of an average of 20-30 radar stations of the enemy in one hour and can determine their technical specifications. The division's 1-2 radar reconnaissance stations (SNAR-2) operate with a range of 14-16 km for ground targets and up to 38 km for water targets.

The division artillery sound ranging battery deploys 4-6 km from the front and conducts reconnaissance within a zone of 6-8 km. The range depends on the caliber of enemy weapons and can reach 18 km.

The artillery flash spotting battery deploys on a 5 km front. It conducts reconnaissance on a 6-8 km front with a range under favorable conditions of 12-15 km.

 
 

Security:

Norms for the locations of security elements:
- advance guard is 15-30 km from main body.
- rear guard is 10-20 km. behind the main body.
- interval between battalions in the column is 3-5 km.
- interval between regiments is 10 km.
- distance of GPZ, BPZ, TPZ, are 3-5 km.
- bivouac security detachment (bn) has a sector 10 km wide.
- bivouac security detachment (co) has a sector 5 km wide.
- bivouac security patrol (platoon) sector is 2 km wide.



 
 

Aerial Reconnaissance

Depending on the conditions of observation and the terrain nature, the line of bivouac security elements can be 4-6 km on the average away from the troops being protected.

 
 

Figure 62 Limits of Visual Observation

LIMITS OF RANGE OF VISUAL OBSERVATION OF OBJECTS

Type of Object (Target) and Conditions of Observation Flight Altitude meters Slant Range, of Observation, km
Column of Troops and Combat Equipment Moving Along a Road. - on a dark night 200-400 0.5
- on a moonlit night 600-1000 1-1.5
- when illuminating terrain with aid of a light bomb 2000-4000 2-3
Headlights of Automobiles, Uncamouflaged 2000-4000 5-15
Large Populated Points (Illuminated Cities) 2000-4000 60-100
Highways and Dirt Roads in Summer: -- on dark nights 500 0.5
-- on moonlit night 1000-2000 1.5
 
 

Figure 63 - Aerial Reconnaissance

 
 

AERIAL RECONNAISSANCE UNITS IN THE FRONT AIR ARMY

UNIT NUMBER AIRCRAFT DEPTH OPERATION
Operational Recon Regt 30 bomber type 800 - 1000 km
Tactical Recon Regt 40 fighter type 400 - 500 km
Combat Regiments 10 - 12 outfitted for recon also
 
 

An air army, which has five divisions and three reconnaissance aviation regiments, can have 287 reconnaissance aircraft and 377 pilots, including 60 tactical reconnaissance aircraft and 30 operational reconnaissance aircraft. Fifteen percent of the non-organic reconnaissance aviation is designated for carrying out reconnaissance. In this case the reconnaissance aviation on the first day will carry out 500-700 flights. Reconnaissance crews in an area of 10-15 kilometers can reconnoiter 2-3 targets in one flight.

In the front there are 3-4 tactical pilotless aviation squadrons. Each squadron has 4 aviation flights. The squadron can conduct in all 36 pilotless reconnaissance sorties. Each squadron is attached for reconnaissance in support of a first echelon army. The time of flight is 45 minutes, the width of photography is 5 kilometers, the overall range of photography is 75 kilometers. The squadron can conduct reconnaissance as follows:

 
 

Figure 64 - Drone reconnaissance aircraft

 
 

Drone reconnaissance aircraft

Flight altitude Flight Radius
600 meters 80-100 km
3000 meters 150-170 km
7000 meters 200- 250 km
 
 

Space Reconnaissance

The Soviet Union uses space reconnaissance means at a distance of 200-250 kilometers above the earth. A single pass can photograph a 40-50 kilometer zone.

 
 

Figure 65 - Visual Reconnaissance Range

 
 

VISUAL RECONNAISSANCE RANGES FROM AIRCRAFT

Limits of Range of Detection and Distinguishing Objects (Targets) Daylight in Good Visibility
Object Flight Altitude Slant Range, km Detection Distinguish
Artillery in fire positions, uncamouflaged 100 2-3 1.-1.5
300 3-3.5 2-3
600 3-4 2-3
1000 3-4 2-3
4000 5 undistinguish-able
Rocket launcher, artillery guns, automobiles, tanks, armored transporters, radar stations on the ground and in movement in columns along highways and on dirt roads, uncamouflaged 100 4-5 2-3
300 6-7 3-4
600 6-7 3-4
1000 6-9 4-5
6-7000 10-12 Undistinguish-able
8-10,000 12-15 "
Tanks, automobiles, armored transporters under cover in trenches, camouflaged 100 3-4 1.5-2
300 6-7 3-4
600 6-7 3-4
1000 6-7 4-5
4000 7-8 5
Tanks, automobiles, armored transporters, uncamouflaged 4-500 3-4 1-2
2-3000 4-5 -
3-4000 5-6 -
Artillery and mortar trenches, camouflaged against terrain background 4-5000 5-6 Undistinguish-able
5-6000 8-10 "
 
 

Artillery Reconnaissance

An army artillery reconnaissance regiment has 2 battalions: the first battalion has a sound ranging battery (zvukovaia razvedka), one radio-technical reconnaissance battery, and one optical reconnaissance battery; the second battalion has a photography battery, a topo-geodetic battery, and a meteorological service battery.

 
 

Figure 66 Artillery Reconnaissance

ARTILLERY RECONNAISSANCE EXPECTED LOCATION ERRORS

Reconnaissance System Range Error - % Range from Observer Deflection Error - % Rg from Observer or m
Visual Recon Theodolite 0.5-0.8 0.1
Stereoscopic Telescope 0.8-1.1 0.2
Range finder 1.0-1.5 0.2-0.3
(Dependent on range finder base) (1.0-1.2) (0.1)
Flash ranging Large (0.5-0.8) Large (0.1)
Timing flashes Large -
Sound Ranging 1.0 (1.0) 0.4 (0.4)
Helicopter Recon (1.4) (0.9)
Radar A few meters
Visual Detection from an Aircraft (No instruments) 100-150 meters CEP (50-70 meters)
Aerial Photo 15-20 meters CEP
 
 

Air Defense Reconnaissance

An army PVO radio-technical battalion can conduct reconnaissance to a depth of 80-100 kilometers on a front of 160 kilometers (at great heights and average heights) and on a front of 80 kilometers, (at low and very low heights). An army PVO radio-technical battalion has 4 companies. The battalion is deployed in two echelons. In the first echelon are two companies at a distance of 10 kilometers from the forward edge. The distance between the two companies is 50 kilometers. On the width of the army zone, two companies are situated in the first echelon, and two companies are in reserve to build up the radar posts in the course of the offensive. The radio-technical battalion will relocate once in 24-36 hours at a distance of 30-50 kilometers during the operation.


 
 

Electronic Reconnaissance

The radio and radio-technical company of a division has 3 platoons of interception and radio direction. The company can establish 5 interception posts, 3 direction finding posts, 3 radio-technical posts.
- depth of radio reconnaissance 25-30 km; Radio-technical - 60 km;
- radio interception numbers 3-4 radio nets;
- radio direction net of 2-3 posts; 30 verified fixes an hour;
- RT can take a bearing of an average of 20-30 radar stations per hour and can determine their technical specifications;
- SNAR - 2 ground targets up to 16 km; water targets up to 38 km;
- sound ranging battery deploys 4-6 km; covers 6-8 km, depth of reconnaissance 18 km.;
- flash spotting battery deploys 5 km covers 6-8 km, depth of recon 12-15 km;
- reconnaissance detachment (Co) 4-5 km sector, Div 30-40 km; Regt 20-25 km;
- recon group of 4-5 platoons;
- RCP 6-8 km from Bn.


A separate radio battalion (osnaz) at army level has the following:
- 3 radio intercept companies - 23 posts;
- 1 radio direction finder company - 10 posts;
- 1 reconnaissance helicopter flight.


A separate radio-technical battalion (osnaz) has the following;
- 3 radio-technical companies - 24 radio-technical and 11 radio intercept posts;
- 1 reconnaissance helicopter.

The battalion can carry out reconnaissance on ground targets to a depth of up to 60-120 kilometers and reconnaissance on aerial to a depth of 300-350 kilometers.

A separate front radio regiment (osnaz) has the following;
- 2 radio intercept battalions;
- 5 radio direction finder companies;
- 1 radio relay reconnaissance company;
- 1 reconnaissance helicopter;
- 1 laboratory.

The regiment can set up 125 posts, including 97 radio intercept and 28 radio direction finder posts. Radio direction finder posts can conduct reconnaissance at a depth of 1000 kilometers with a width of 500 kilometers. The radio interceptor post can conduct reconnaissance to a depth up to 2000 kilometers.

A separate front radio-technical regiment has the following:
- 2 radio-technical battalions;
- 1 radio interceptor battalion
- 1 aviation reconnaissance flight;
- 1 laboratory.

The regiment can organize 97 posts, including 46 radio-technical and 50 radio interceptor posts. The aviation reconnaissance has 2 airplanes, which have radio-technical means. The regiment can conduct reconnaissance at a depth of 500-2000 kilometers.

During the organization of the first offensive operation, the radar means of the army do not operate, but the reconnaissance of the enemy is conducted with front equipment. Two to four radar posts of the front will be located in the army zone.

The second echelon of radar is organized by forces of the front. It is situated 50 kilometers from the forward edge.


 
 

Planning for Reconnaissance

Soviet combat doctrine places a strong emphasis on extensive reconnaissance. Soviet planners consider detailed knowledge of the defender's unit locations (front line strong-points, reserve assembly areas, artillery positions, and command posts) to be essential. They expect to have such information available in timely fashion.

To obtain this information the Soviet commander is served by an extensive reconnaissance establishment that includes agent networks, SPETZNAZ units, and long range reconnaissance units under army, front, and TVD control; air reconnaissance from front and TVD; combat patrols from divisions; radio and radio-electronic combat units at army and front; artillery sound ranging and counter battery target acquisition units; engineer and chemical warfare reconnaissance units from divisions, army, and front; and national technical intelligence collection platforms which provide reports through front intelligence headquarters.

The commander of an army or front in the Western TVD, for instance, has a priceless intelligence file containing information built up over 40 years on every nook and cranny in his operating area. One might speculate that the potential volume of information available to the Soviet commander is so massive that the principle limitation on its effective exploitation is his own finite human ability to master it.

The Soviet planner expects to have the information needed to serve his purposes available when he wants it. He organizes the collection effort and flow of information through the analysis process accordingly.

The following is the phased sequence of information flow related to the decision /planing process. This shows that the level of detail required in the information increases as the planning process moves from general outline of the concept to specific targeting of force on force. It shows that the Soviet norm for timeliness allows more time for the acquisition of the more detailed information.



 
 

Information for the Estimate of the Situation

This information is required within 2-3 hours of the start of planning:
- grouping, disposition, and nuclear weapons;
- movement and likely intentions;
- strength and composition;
- weak points in dispositions;
- meteorological situation.

For the Commander's Decision

The following information is required within 4-6 hours of start of planning:
- changes in disposition;
- nuclear delivery means;
- density of troops and means in one km of front;
- density on each axes and along the entire front;
- disposition down to regiment or brigade level;
- location of artillery;
- exact location of command posts;
- support aircraft locations

For Detailed Planning

The following information is required within 12-24 hrs of start of planning:
- exact location of battalions on the ground;
- exact location of support arms;
- location of mine fields;
- location of nuclear delivery means and artillery bns;
- location of reserves and logistic support elements;
- fortifications;
- air defenses and anti-tank weapons systems locations;
- airfield composition and support aircraft.

The time to prepare the reconnaissance report of the chief of reconnaissance for his commander is shown in this table.



 
 

Figure 67 Preparation time for Reconnaissance

TIME REQUIRED TO PREPARE RECONNAISSANCE REPORT

for regiment commander 15-20 min.
for division commander 30-60 min
for army commander 1-1.5 hr
for front commander 2-3 hr
for commander TVD 4-5 hr

 
 

Norms in Planning Air Army Operations

The average time to prepare the air army daily plan of operation is 2-3 hrs. During battle, to launch a squadron of fighters or fighter bombers requires 10 min. and a regiment requires 20 min., if the unit is in category 1 readiness.

In an air operation, when bomber and fighter-bomber aviation strike against enemy airfields, 20-30% of fighter aviation will escort and cover them during the action.

During a front offensive operation, the main effort of the front aviation is concentrated at a depth of 30-40 kilometers from the front line and sometimes up to 80 kilometers.

During actions to support a front offensive operation, the aviation can operate initially with a capability of 3-4 sorties per day tapering to 1.5-2 sorties toward the end of the operation. The movement of one aviation squadron from one airfield to another deployment airfield will take 30-60 minutes (including take-off, travel time, landing, and replenishment).

A fighter-bomber and a fighter regiment each consists of 43 planes. If the division has in its composition 3 regiments of each, this division will have in its composition 130 fighter-bombers and 130 fighters.

Only one aviation regiment is located on one airfield. If the conditions are good, one regiment can be deployed at two airfields. A squadron is located at a distance far from other squadrons, so that one medium nuclear blast will not destroy two squadrons. The distance between planes is 200 meters. A caponier (an earth parapet made of dirt piles around the current position) is built around each aircraft to protect it from fragments of enemy bombs.

At readiness # 1 for fighter aviation and fighter-bomber aviation units: the crew is located in the airplanes, and with the reception of a signal, the aviation regiment becomes airborne in 8-10 minutes. At readiness #2 for fighter aviation and fighter bomber aviation units: the crew is located near the airplanes, and with the reception of a signal, the regiment becomes airborne in about 20 minutes.



 
 

Figure 68 Tactical Technical Characteristrics of Transport Aircraft

BASIC TACTICAL-TECHNICAL CHARACTERISTICS OF TRANSPORT AIRCRAFT

Main Features AN-12 AN-22
Crew 7 7
Maximum speed (km/hr) 683 710
Average speed 500 --
Flight Altitude (meters) 12000 11000
Weight in Air: Minimum 54 196.5
Maximum 61 225.0
Maximum range (km) 6450 8800
Load Capacity: Maximum 20 --
Normal 10.6 50
Passenger Capacity: Troops 93 295
Paratroops 60 152
Wounded 80 177
Tactical Radius (km) 900 --

 
 

Planning Factors for Airborne Operations

The depth of an air drop or an air landing depends on its scale and type as follows:
- a strategic operational desant - 500-600 kilometers;
- an operational desant in conventional war - 150-300 kilometers;
- an operational desant in nuclear war - 300-400 kilometers;
- an operational-tactical desant in conventional means is 100-150 kilometers;
- an operational-tactical desant in nuclear war - 250-300 kilometers;
- a tactical desant - 50-100 kilometers or more.

An airborne division can continue its operation independently for 6-7 days. The division is dropped at a distance so that, after 2-3 days, the main forces reach the area of the division landing. This distance will be 150-300 kilometers from the front line.

A transport aviation regiment consists of 32 airplanes. If the division has 4 regiments, then the division will have 130 planes. A transport aviation division of four regiments can make an air landing or air drop of one parachute regiment. A parachute regiment of 1600 people will its weigh 700-800 tons. For the landing of one desant division by means of parachutes, 455 AN-12 planes are required, broken down as follows: for the transport of the parachute dropping group 415 planes are needed, and for the landing group 40 planes are needed. For a parachute regiment to be dropped 80 AN-12 airplanes are needed, or one transport aviation division.

A transport aviation regiment of the type AN-12 can become airborne in the following times according to readiness status:
- readiness #1 in 15 minutes;
- readiness #2 in 1 hour;
- readiness #3 in 4 hours.

According to the experience of exercises bringing the military transport aviation from routine combat readiness to full combat readiness requires 25-27 hours. Included in this, to raise by an alarm requires 2 hours; to prepare the troop control of transport aviation )organization of airborne operations( requires 81 hours. For loading the airplanes and completing the preparation of military transport aviation, 2-3 hours are required.

If military transport aviation received the mission beforehand, and it is located at the permanent or deployment airfields in the first or second degree of readiness for take-off, the time for preparation significantly decreases, in this case the time will be 5-7 or 5-8 hours.

One airborne )desant( division is air dropped on one flight in 4-5 hours; by landing )air landing( method, the process requires 20-28 hours.

For each military transport division composed of 3-4 regiments, 4 main airfields and 1-2 reserve airfields are needed. Up to 20 airfields are needed to deploy 4 military transport divisions in the staging area; included in this, 16 of the airfields will be main airfields and 4-6 will be reserve airfields.

The same number of airfields are designated for strategic aviation, front aviation, and front and national air defense and civil aviation.

For resupply of one AN-12 aircraft, 3- 15 tons of fuel are needed. At one airfield, at which one transport regiment is concentrated, up to 500 tons of fuel are required. At all 20 airfields, for 4 transport divisions, from 10,000-12,000 tons of fuel are needed. The assembly area/waiting area of an airborne division is designated at a distance of 5-10 kilometers from the airfield. The area of the staging area for an airborne division can be 300 by 400 kilometers.

Fuel resupply areas for military transport aviation are designated at a distance of 200-300 kilometers from the front.

The time for landing, fuel resupply, and take-off of one military transport regiment will be 2-3 hours. The flight of aviation to the desant area, the boarding of the desant on the aircraft, fuel resupply and take-off of aviation from the staging area to the objective of the air drop requires 7-9 hours. To load military equipment requires 2-3 hours, and it must be finished 30-50 minutes before the take-off. At the same time the fuel resupply is being carried out and the military transport regiment is brought up to readiness #2. The boarding of personnel is concluded 10-15 minutes before starting the engines of the aircraft and the regiment is brought to combat readiness #1. The boarding of personnel requires 10- 20 minutes. The speed of the flight of transport aviation is 500 km/hour.

The depth of the formation of a transport regiment during flight will be 32 kilometers during the day and 110 kilometers at night. The time of the air drop of one regiment during the day will be 4 minutes, and at night it will be 13 minutes. The depth of a military transport division formation during flight will be 180 kilometers during the day and up to 900 kilometers at night.

The time of the air drop of one division will be 25 minutes during the day and 1 hour 47 minutes at night.

An airborne division can carry with it material reserves for two full days. Each day an airborne division expends in their combat activities 250 tons of material. To supply such a load of material, 20-25 AN-12s are required.

Air transport of the troops in great distances is more advantageous, because at a distance of 500 kilometers, the average speed of movement is 50 km/hour; at a distance of 3000 kilometers, the average speed of movement will be 215 km/hour; at a distance of 6000 kilometers, the average velocity of movement will be 250 km/hour. The conditions of the above calculations are the following: transport and movement of troops, two hours are needed; for loading, 4 hours are needed; for unloading, 3 hours are required. The average speed of the flight is 500 km/hr.

The requirement of military transport aviation for transport of a motorized rifle division without heavy equipment is 803 AN-12s; the transport of a motorized rifle regiment without heavy equipment requires 138 AN-12s; the transport of an artillery regiment requires 122 AN-12s.

For the supply of one basic ammunition load and one refueling of POL for a motorized rifle regiment, 30 AN-12s, or almost one transport regiment, is needed.



 
 

Norms in Air Defense Planning

The time to assemble a battalion, move, deploy, and take a new position depends on distance, with 10 min to assemble and 20-30 min to deploy. A regiment takes 15 min to assemble. The time to move varies with the distance. At the new location it requires 30-40 min to deploy and be ready to fire.

Capability:

The front air defense is capable of destroying 15-25 percent of the enemy aviation, which participates in a massive strike. Losses of PVO in the enemy first nuclear strike will be 20-35 percent, but with the use of only conventional weapons the losses will be 5-10 percent.

Location:

The depth of the air defense belts in a strategic operation will be:
- First Echelon Second Echelon 300-500 kilometers. 700-1000 km

The SA-2 rocket regiments of an army and front are located at a distance of 10-20 kilometers from the FEBA. The distance between battalions is 10-30 kilometers. The regimental command point is located in the center where its distance should not be more than 20 kilometers from the battalions. The technical battalion deploys in the rear of the rocket regiment. The distance to the furthest rocket battalion should not exceed 30 kilometers. The distance between rocket battalions should not be less than 5 kilometers, so that one nuclear blast will not destroy two battalions.

In crossing a wide water obstacle )rivers(, the position of the forward air defense rocket battalions will be as follows:
- SA-2 rocket battalions - 10-11 kilometers from the bank of the river;
-SA-3 rocket battalions - 6 kilometers from the bank.

Composition:

Front

Forces and means of the air defense of a front which has in its composition 3 combined arms and one tank army and one air army will be the following )without consideration of the division PVO forces and means(:
- 6-11 anti-air rockets of the type SA-2
- 1-2 PVO divisions;
- 4-8 separate antiaircraft artillery regiments of the type S-60;
- 1-2 anti-air rocket regiments of the type SA-3;
- 1 PVO radio-technical regiment;
- 4 PVO radio-technical battalions;
- 2-3 fighter aviation divisions including 6-9 regiments.

The front - level PVO forces and means are the following:
- 2-3 PVO rocket regiments of the type SA-2;
- 1-2 PVO rocket regiments of the type SA-3;
- 1 antiaircraft artillery division;
- 1 radio-technical regiment.

Army

The organic air defense means and forces of the army are the following:
- 1-2 PVO rocket regiments of the type SA-2;
- 1-2 antiaircraft rocket regiments of the type S-60;
- 1 radio-technical battalion.



 
 

Air Defense Army

The composition of an army of the national PVO is not constant. It depends on the mission, importance, direction, and nature of the theater, and on the number of targets which are being defended. The PVO army, which has in its composition 1-2 corps and 2-4 PVO divisions, will have the following forces and means:
- 5-7 PVO rocket brigades;
- 15-20 PVO rocket regiments;
- 6-12 fighter aviation regiments;
- 3-6 radio-technical regiments or brigades;
- 1 separate radio-technical regiment of spetsnaz;
- 2-3 separate radio battalions of spetsnaz;
- 1 signal center and engineer, chemical, rear services and other units

In a continental theater of military operations )TVD( the following PVO forces and means participate:
- 100-150 formations and units of rocket troops;
- 30-40 fighter aviation regiments;
- 50-70 antiaircraft artillery units;
- 30-40 radio-technical formations and units;
- 3-4 radar patrol boats;
- 60-80 air defense artillery ships.

Radio-technical means of the front conduct the reconnaissance of the enemy at a depth of 300 kilometers at great and average heights and at a depth of 150 kilometers and also at very low heights, along a width of 120 kilometers. The command post of the PVO of a front simultaneously reports 40-45 targets, while the command post of the PVO of an army reports 30 targets.



 
 

Factors Used in Calculating Air Defense Capabilities

There are several types of factors, weapons effective numbers for individual weapons, and expressions of general capability associated with SAM and air defense artillery units. Air defense has been one of the most dramatically improving branches of the Soviet armed forces during the past 15 years. Thus the technical characteristics and associated capabilities factors for air defense weapons and radar systems are continually changing. The following factors were official as of the mid-1970's and may be taken as characteristic examples of the type of factors used in Soviet planning methods. However, for assessments of current capabilities they should be significantly increased.

According to the experience of Soviet exercises the attrition on Soviet forces from the enemy first massive air strike may be:
- in nuclear conflict: 20-35%
- in conventional conflict: 5-l0%



 
 

Destruction Coefficients of Selected SAM Weapons

WEAPONS WITH JAMMING WITHOUT JAMMING
S-75M (SA-2( 0.45-0.72 0.66-0.77
S-125 )SA-3( 0.4 0.7
STRELA-2M )SA-7( 0.22-0.26 0.72-0.76

 
 

Fire Capabilities of Selected Units

- S-75 )SA-2( regiment can engage one target at a time and shift after two minutes
- S-60 AAA regiment of division can engage one target at a time ) 24 guns(
- S-60 AAA regiment of army can engage two targets at a time ) 36 guns(
- infantry and tank regiment air defense battery can engage one target at a time and shift after one minute
- STRELA-2M squad )three launchers( can engage one target at a time with destruction probability 0.53-0.60:
- formula used to convert the weapons effectiveness of individual weapons into that for a unit or group of weapons is as follows: P=1 - )1 - P(n
- using the effectiveness of the individual Strella to compute the effectiveness of the squad gives the indicated value. P=1 - )1 - 0.22( )1 - 0.22( )1 - 0.22(=0.53

The formula is generally used to determine how many weapons are required in order to reach a certain desired effectiveness. Thus P is given and the equation is solved for n. A table giving the values for P achieved by each n can also be constructed.



 
 

Rear Service Norms and Planning Factors

The following information on the organization and capabilities of army and front level rear service organizations has been taken from the lectures delivered at the Soviet General Staff Academy in the mid-1970's.

Army Rear Services:

The army material support brigade )MSB( can maintain up to 7,000 tons of material. It can have up to 7,000 men and 2,500 transport vehicles. The army MSB has as its components the following units, subunits, and establishments:
- base headquarters;
- signal platoon;
- separate chemical defense company;
- separate rear services signal battalion;
- separate gas decontamination battalion.

A state bank branch is one of the components of a rear mobile army base branch, as is a mobile bakery plant )with a capacity of 18 tons in a 24 hour period( and a military post office, a military shopping )trade( facility with a capacity of 40 tons, personnel of all types of supply )artillery, armor, motor, engineering, chemical, signal, POL, foodstuffs, clothing, medical, furniture( and a service company, which can load or unload 2500 tons in a 24 hour period.

The Army MSB has various supply storage facilities )depots(. The capacity of each is as follows:
- artillery depot - 2000 tons;
- fuel depot - 3000 cubic meters;
- food depot - 400 tons;
- tank depot - 1000 tons;
- motor tractor depot - 150 tons;
- engineer depot -250 tons;
- signal depot - 80 tons;
- chemical depot - 300 tons;
- goods and property depot - 40 tons;
- medical depot - 60 tons;
- quarters depot - 25 tons;
- trade depot - 40 tons.

A motor transportation regiment in the brigade has 4 battalions and one battalion for transporting POL. The regiment can transport up to 5,000 tons, including 630 tons of POL.

Two separate road construction and traffic battalions are designated for preparing the army's main vehicle roads. These battalions are not part of the mobile support brigade.

The rear services of an army are also composed of the following rear units:
- separate battalion for the recovery of tanks;
- separate company for the recovery of motor transport equipment;
- separate engineer company for recovery and repair.

An army can receive from the front several rear services units and establishments, such as a tank repair battalion, and a separate automobile repair battalion.

The air army also has as one of its components a rear service. In the rear service of an air army there are the following units and establishments:
- the rear services headquarters;
- 1-2 air army mobile bases;
- one separate technical support regiment )each division has aviation technical support battalions; each regiment also has technical support battalions(;
- separate companies for technical support of the airfields.

In the staging position for an offensive, the army material support brigade and rocket technical base are located 40-60 kilometers from the front, and reserve positions are 15- 20 kilometers from the main locations. If several formations of an army operate on separate axes, a branch of army's material support brigade is designated to resupply and reinforce them. The rear units and establishments move forward in accordance with the rate of advance. As a rule, the medical support detachment, motor ambulance company, and units for recovery and repair of damaged vehicles move behind the first echelon divisions.

An army material support brigade is located so that its distance will not be more than 125 kilometers, i.e., one half of the daily march, from the division material support battalions. At a rate of advance of 30-50 kilometers per 24 hour period, the army material support brigade relocates once every 2-3 days. The rocket technical base, as a rule, moves behind the units of rocket troops.

Some of the rear services of the army, such as the army mobile base, army rocket technical base, mobile mechanical bakery, separate recovery and repair battalion for armored vehicles, and recovery company for motor transport, relocate in one move, while other elements shift locations in increments.

The organization of the rear services in a defensive operation:

In an offensive, the depth of the rear services of a division will be up to 40 kilometers, and in defense, the depth of the rear services of a division will be up to 60 kilometers. Division medical battalion, separate medical detachments which the division has received as reinforcement, and the repair battalion, are located between the first and second echelons of the division, or in the depth of the division defense. The remaining units of the division are situated at a distance of 30-40 kilometers from the FEBA, behind the combat formation of the division. The rear of second echelon divisions of the army, and the rear of other, which are located in the units, as a rule, are located in the region of the deployment of such divisions and units. The depth of the army rear services area in defense will be 100-150 kilometers.

The rear services of an army is deployed as follows:
- the army rocket technical base is at a distance of 30-40 kilometers from the positions of the army rocket troops, or 60-80 kilometers from the positions of the rocket battalions of the divisions;
- the army material support brigade is situated at a depth of 100-120 kilometers from the forward edge, or 60-80 kilometers from the divisions' material support battalions;
- a branch army material support brigade on a separate direction is situated at a distance of 60-80 kilometers from FEBA;
- the separate recovery and repair battalion for tanks, and a the separate recovery and repair company for motor transport vehicles, and the ambulance company are situated behind the first echelon divisions for the recovery of equipment and vehicles, and the sick and wounded;
- the separate medical detachments which are in the reserve of the army are situated behind the second echelon divisions near the rear command post of the army.

Their relocation to alternate positions is conducted, in accordance with the order of from the chief of the rear services of the army, in case of threat to their position or use of chemical weapons.


 
 

Technical support of the army:

During World War II, 8-9 percent of the tanks were knocked out of operation every twenty four hours. With the use of nuclear weapons, these losses will be 12-15 percent every 24 hours. The technical repair means of the troops can handle 100% of the running light repair and 15-20 % of the medium repair. During a nuclear war, in the course of an operation losses will be as follows: 50-80%; 30-40 percent of the APCs; 40-50 percent of wheeled vehicles.



 
 

Front Rear Services:

The composition of the rear services of a front is not constant; it depends on the composition of the front, the mission, and the conditions of the theater of military operations. The rear services of the front are made up of 250-300 rear services units and installations, consisting of 160,000-170,000 men, up to 25,000-27,000 vehicles. The following rear services formations, units and installations make up the rear of a front:
- 1-2 rear service bases of the front;
- 2-3 forward bases of the front;
- 4-6 forward hospital bases of the front;
- 2-3 hospital bases of the front;
- 2-3 automobile transportation brigades )each brigade can load 6600 tons, including 1440 tons of POL(;
- 2-3 railroad brigades (each brigade can lay 20-25 kilometers of railroad in a 24 hour period, or up to 9 kilometers of railroad when it has been completely destroyed);
- 2-3 road construction and commandant service brigades )each brigade can reconstruct up to 900 kilometers of road; it can reconstruct a bridge with a load capacity of 60 tons, length of 110 kilometers(, can establish 16 traffic regulation posts, can construct a 400 meter floating bridge of 16 tons;
- 2-3 pipeline brigades )each brigade can put together 600 kilometers of pipeline; in a 24 hour period the brigade can refuel, using pipe with a diameter of 100 millimeters, 800 tons of fuel, and with pipe with a 150 millimeter diameter, 2000 tons of fuel over a length of 75-150 kilometers. In a 24 hour period, it can lay 65-75 kilometers of pipe(;
- separate battalions for transport of rocket fuel )each battalion can transport 640 tons of rocket fuel(;
- rocket fuel depots of the front )each depot can store 500 cubic meters of rocket fuel(;
- separate medical ambulance battalions )in one trip each can evacuate up to 0003 sick and wounded(;
- separate air transport medical regiments )including 23 AN-2 airplanes; in one flight a regiment can evacuate 180 seriously ill men(.

A front has the following units and installations for recovery and repair, which are subordinate to the chiefs of services: ) 17 repair plants and evacuation units(:
- separate railroad exploitation regiments;
- separate rear services signal regiments of the front;
- guard division for the protection of the rear services;
- bridge construction brigade, )it is a central reserve and establishes crossing sites on wide rivers(;
- railroad bridge construction brigade.

The following is the organization of some of the rear installations: As a rule the front mobile base is designated to resupply armies of the first echelon with material requirements. Its capacity is 8000 tons of materials. Which meets 3-4 day requirements of the respective grouping of forces to be resupplied. A front forward base is assigned to one or two armies. In its composition are the following elements: base headquarters, storage facilities )each type of supply has 1 control element(. The capacity of each facility is as follows:
- artillery depot - 250 wagon loads 38 cubic feet per load(;
- fuel depot - 4000 cubic meters;
- food depot - up to 250 wagon loads;
- tank depot - 250 wagons;
- motor and tractor depot - 25 wagons;
- engineer troop depot -200 wagons;
- communications depot - up to 70 wagons;
- chemical depot - up to 150 wagons;
- light goods and property depot - up to 150 wagons;
- medical depot - 300-350 wagons;
- topographic depot - up to 500 wagons.

A front forward base also has as part of its composition the following:
- 2 separate service companies )each company can load or unload up to 2500 tons(;
- an engineer company;
- a separate chemical defense company;
- 3 mobile field bakeries;
- repair plants for repair of equipment organic to POL, food, and clothing services;
- lubricant reprocessing station;
- two field mobile laundry detachments;
- laboratory;
- military post office;
- automobile transport regiment with a load capacity of 3300 tons.

The headquarters of the base includes a chief of the base, his deputy, political assistant, planning and organization section, movement section, six assistant chiefs for supply of ammunition and armament, POL, motor-tractor, tank, food-clothing, and combat technical equipment. The front rear base can maintain material reserves for 10 days. As a rule it is situated on a railroad. The load capacity is 57,000 tons. The organizational structure is as follows: base headquarters depots with capacities as follows:
- artillery depot - 250 wagons;
- artillery weapons - 250 wagons;
- POL depot - 8000 cubic meters;
- food depot - up to 350 wagons;
- tank depot - up to 250 wagons;
- motor tractor depot - up to 150 wagons;
- engineer depot - up to 200 wagons;
- signal depot -up to 100 wagons;
- goods and clothing depot - up to 150 wagons;
- chemical depot - 500 wagons;
- medical depot - up to 800 tons;
- veterinary depot - up to 7 wagons;
- household goods depot - up to 50 tons.

Besides this, the front rear base has in its composition a separate transport battalion with a load capacity of 1100 tons, a separate engineer company, a separate service battalion (the battalion can load and unload 7500 tons in a 24 hour period(; a separate chemical defense company, 3 mechanical field bakeries and other repair and other units including, repair plants to repair equipment used by POL, food and clothing, supply services, two mobile POL reprocessing stations, 7 field laundry detachments, centers for unloading and distribution of front transportation, testing laboratory, field post office. The composition of the rear base headquarters is the same as forward base headquarters.

The deployment and relocation of the rear services of the front:

The boundaries of the front rear service area are determined by the rear services directive from the supreme commander-in-chief. In the FUP area it is located up to 400 kilometers from the forward edge. As the leading divisions in the offensive move forward, the depth of its location increases and may reach 1000 kilometers. The rear services deploys in two echelons. In the first echelon are the following elements: the forward base of the front is located behind the armies of the first echelon at a distance of 80-100 kilometers, near the railroads in an area of 150 square kilometers; rocket technical units, which are located at a distance of 30-40 kilometers from the positions of the rocket troops; rocket fuel depots and pipeline units and large units; the forward hospital base of the front, which is located 50-70 kilometers from the FEBA; recovery and repair units of the front, as a rule, are located in the area of the armies or are given to the armies as reinforcement. The second echelon is situated in the following manner: the rear base of the front is situated on a railroad; it is deployed in a large area. The zone covers an area of 200 kilometers wide by 50-100 kilometers in depth. Storage facilities are located along the railroad in the depth. As a rule the rear hospital base of a front is situated in 2-3 regions by a railroad. Its distance from the forward edge will be 50-70 kilometers, or 200-300 kilometers. When there is a railroad, its distance will be the greater one, i.e. 200-300 kilometers.

Repair plants are situated, as a rule, near the front rear bases, and use local plants if possible. If the front has two rear bases, each base will be situated on separate axes to resupply the armies. One branch of the rear base of a front is located at 120-150 kilometers from the FEBA; and the second branch remains in the reserve to be used during the operation.

Relocation of the front rear during an offensive operation:

The first echelon of the front rear services should not fall behind the armies rear services more than 150 kilometers, i.e., not more than one half of the length of supply movement over a 24 hour period. If the speed of the offensive is 45-50 kilometers per 24 hour period, the first echelon will be relocated once in three days. If the speed of the offensive is 80-100 kilometers, the first echelon will be relocated once every two days.

The front forward base, as a rule, relocates at full strength. If conditions require, it can send forward a branch.

The mobile rocket technical base of the front relocates behind the advancing troops each time after 150-200 kilometers, or they relocate once after 220-250 kilometers, so that their distance will not be more than 340 kilometers from the positions of the rocket troops. Rocket fuel depots simultaneously relocate with the rocket technical bases.

Mobile repair units of the front approach the sectors where a large number of damaged combat equipment is massed, and conduct repair work. Formations, units, and installations of the second echelon of the rear services of a front move forward with the preparation of the railroad. The front rear base, in the course of an offensive operation, sends forward, in turns, its branch. It relocates in full strength only at the end of an operation.

Organization of supply of material in the offensive operation of the front: The principle of resupply is from above downward. Resupply is carried out continually with the use of the sum total of all types of transport. Under conditions of using nuclear weapons, the volume of overall resupply of a front will be 300,000 tons; with the use of conventional weapons, the overall volume of resupply of the front will be 450,000 tons, in which one third of the resupply will be the air army, PVO troops, various reserves, and the rear services of the front. Resupply of the first echelon armies will be 20,000 tons per 24 hour period.

The average daily (24-hour) haul of transport means of a front will be 300 kilometers; for transport means of an army this will be 250 kilometers; for troop transport, this will be 200 kilometers.



 
 

Some Fundamentals of Troop Supply:

Material reserves are replenished up to the norms every day. The first priority for delivery of resupply is those troops which are successful in the operation. All types of transport )front, army, troops( are used together. Front second echelon troops organize resupply by means of their own transport. As a rule, when the troops are moving forward very fast or an airborne assault is being air landed, resupply and transport is organized by air, and airfields are prepared. Superfluous loading is prohibited, that is, loading from one vehicle to another; this wastes time. It is better when the materials and ammunition are resupplied on one vehicle, for example, from a front depot and unloaded at a division depot.


The role of the types of transport:

Resupply from the front forward base to the army material support brigade amounts to 90% by motor transport means and 5% by air transport means. From the front rear base to the front forward base, 15% of the transport is by railroad, 70% is by motor transport means, 10% is by pipeline, and 5% is by air transport. To the front rear base, 75 % of the transport is by rail, 15 % is by motor transport, and 10% is by pipeline.

The preparation, content, and use of roads for a front offensive operation: as a rule, in an offensive operation all types of transport routes are used, such as railroad, waterways, air lanes, vehicular roads, and pipelines. In the area of the front there will be 2-3 front railroads and 2-3 lateral railroads. Their capacity will be 70 pairs of trains in a 24 hour period. In the course of the operation, 1-2 railroad axes are prepared with a capacity of 30 pairs or trains in a 24 hour period. 40-45 kilometers of rail can be laid by two railroad brigades in a 24 hour period; under conditions of mass destruction, 20-25 kilometers can be laid in this period. As a rule, for an army 2-3 distribution stations and 1-2 reserve stations are allocated.

Unloading stations are allocated to an army as follows: one per division, 2-3 for the army material support brigade. One or two temporary loading areas are organized for a front.

Waterways: A distribution port is designated for a front; an unloading port is designated for an army.

Front military vehicular roads:

These roads join the front bases with their branches, and with the army mobile bases. Behind each first echelon army of the front there is one frontal military vehicular road. The capacity of this road will be up to 01,000 vehicles in a 24 hour period.

Main field pipeline:

It is designated for the transport of fuel from the permanent and front storage facilities )depot( to the troops. As a rule, it is laid in the direction of the main attack of the front.

Military air transport:
- 7-8 percent of the material resupply is carried out by military air transport.

Requirements/expenditures on an Operation and Creation of Reserves at the end of the Operation

The material requirements at the front will be up to 700,000 tons. Included in this will be the following:
- small arms ammunition: 4-4.5 basic loads;
- artillery and mortar ammunition: 7.5-9 basic loads;
- tank ammunition: 7.5-9 basic loads;
- antiaircraft ammunition: 8.5-9.5 basic loads;
- aviation ammunition: 22-32 basic loads;
- motor gasoline: 8-9 refuelings;
- diesel: 11-31 refuelings;
- aviation gasoline: 26-27 refuelings;
- food: 30 daily rations.



 
 

Figure 69 - Echeloning Reserve Material of the Front

ECHELONING RESERVE MATERIAL AT THE FRONT

Ammunition )basic load Fuel Refueling Food

daily rations

Small Arms Arty & Mortars Tanks Anti Air Aviation Gas Diesel
Troops 1.0 1.0 2.25 2.0 1.7 2.4 13
Army )2 days( 0.15 0.3 0.4 0.5 0.46 0.7 2
Air army reserves 1.75 17.5 3.0 3.5 21
Forward base of the front 0.22 0.45 0.6 0.7 0.6 1.0 3-4
Rear base of the front 0.78 1.5 2.0 2.55 0.3 3.5 01
Total 2.15 3.25 5.25 5.75 17.5 5.15 7.65 28-29

 
 

Figure 70 - Norms for Material Reserves in an army

NORMS OF RESERVE MATERIAL AND ITS ECHELONING DURING WAR

Ammunition )basic Loads( Fuel )refueling( Food )daily ration(
Small arms Arty and mortars Rockets Tanks Antiair Gas Diesel
Army storage 0.15 0.3 0.3 0.4 0.5 0.46 0.7 2
Troops storage 1.0 1.0 1.0 2.25 2.0 1.7 2.4 13
Division storage 0.2 0.2 0.2 0.75 0.5 0.4 0.75 2
Units 0.8 0.8 0.8 1.5 1.5 1.3 1.65 11
Total 6-7 days 1.15 1.3 1.3 2.65 2.5 2.16 3.1 15

 
 

Figure 71 - Requirements for material reserves in Offensive

ARMY MATERIAL REQUIREMENT IN A OFFENSIVE OPERATION

Ammunition (basic load) Fuel (refueling) Food daily ration
Small arms Arty and mortars Tanks Antiair Gas Diesel
Use in operation(nuclear) 1.0-1.6 2.1-3.2 2.4-3.2 3.5-5.6 1.4-2.4 2.8-4.0 7-8
Use in operation (non-nuclear) 3.5-5.6 1.4-2.4 2.8-4.0 7-8
Creating reserves toward the end of the operation 1.15 1.3 2.65 2.5 2.15 3.1 15
Overall need on an operation )nuclear( 2.15-2.75 3.4-4.5 5.5-5.85 6.0-8.6 3.56-4.56 5.9-7.1 22-23
Overall need on an operation )non-nuclear( 6.0-8.6 3.56-4.56 5.9-7.1 22-23

 
 

Figure 72 - Operational Expenditure and Reserves

OPERATIONAL EXPENDITURE AND RESERVES

Echelons Ammunition (basic loads) Fuel (refueling)

Food )daily rations(
Infantry weapons Arty and mortar Tank Air defense Air Auto fuel Diesel Aviation fuel
Total in front 2.15 3.25 5.25 5.75 17.5 5.15 7.65 15.0 28-29
Troops 1.0 1.0 2.52 2.0 -- 1.7 2.4 -- 13
Army (2 days) 0.51 0.3 0.4 0.5 -- 0.64 0.7 -- 2
Air army 1.57 -- -- -- 17.5 3.0 3.5 7.5 21
Forward rear base 0.22 0.45 0.6 0.7 -- 0.6 1.0 -- 3-4
Front rear base 0.78 1.5 2.0- 2.55 -- 2.3 3.5 7.5 10
Total require- ment 4-4.5 7.5-9 7.5-8 8.5-9.5 22-32 8-9 11-31 26-27 30

 
 

Figure 73 - Location of Rear Supply and Support

LOCATION OF REAR SUPPLY AND SUPPORT ELEMENTS

1st Echelon sub-units/units Approximate depth from battle zone
Offensive Defensive
COMPANY: First aid post --- 50-100 m
Ammo distribution points --- 100-150 m
Rations & water point --- up to 1 km
BATTALION: Medical post 1-2 km 1.5-3 km
Ammo distribution points 3 km 2-3 km
Rations and field kitchen 3 km 2-4 km
Technical observation post 1-2 km 2-4 km
REGIMENT: Medical post 5-7 km 6-10km
Transport platoon 5-7km up to 20km
POL point 10-15km 10-20km
Ammo Distribution points 10-15km 10-20km
Rations 10-15km 10-20km
TK & MT repair & evacuation group up to 8km up on 10km
Damaged vehicle collection pound 5-7km 6-10km
2nd echelon regiments - All rear service elements 16-20km
DIVISION: Medical post 10-14km up to 20km
Truck repair & depot 10-14km up to 20km
Tank repair & depot 20-40km 35-60km
Arty & small arms repair/depot 20-40km 35-60km
Transport Bn 20-30km 20-40km
Engineer Bn 20-40km 35-60km
Ammo dump 25-30km 35-50km
POL dump 25-30km 35-50km
Rations dump & field bakery 25-30km 35-50km
Baths/laundry/water 25-30km 40-45km
2nd echelon divisions - All depots 40-70km

 
 

Medical Norms and Planning Factors


Army medical support:

The following medical units are components of the rear services of an army:
- 10-12 separate medical detachments in a combined arms army )OMO(;
- 6 separate medical detachments in a tank army )OMO(;
- a separate ambulance company )ASR(;
- a separate anti-epidemic medical detachment )ASPEO(;
- an army medical reinforcement detachment )OMU(;
- a veterinary-epizootic detachment )VEO(.

Each medical battalion of a division and separate medical detachment can take in a 24 hour period up to 500 men. The evacuation capability of a medical battalion of a division is 80 sick and wounded per trip; a separate medical detachment can evacuate 160 men on one trip; the medical ambulance company of the army can evacuate up to 1000 sick and wounded per trip.

The army medical anti-epidemic detachment and reserve medical detachment, and army medical reinforcement detachment are located between the first and second echelon of the army near the army command post. Separate medical detachments and recovery and repair subunits, which are attached to reinforce first echelon divisions operate in the division zone.


Front medical support:

There are up to six forward hospital bases in the front. The capacity of each base is 6500 beds. Each base has 13 separate hospitals, such as a mobile surgical unit, mobile therapeutic unit, and a mobile neuropathological unit, as well as others. There are also various sub-units, units, and installations. A front has 2-3 rear hospital bases. The overall capacity of a rear hospital base of a front is 20,000 beds, including 5900 beds in mobile hospitals and 14,100 beds in evacuation hospitals. The base is situated in 2-3 regions near the railroad. It has 48 various hospitals/units, including a medical ambulance company, a special medical support battalion, and others. The forward hospital bases of the front are deployed near the centers of large personnel losses, usually about 40-50 kilometers from the front line.



 
 

Casualty estimates

During World War II losses of personnel were 0.8-1.0 percent in a 24 hour period. With the use of nuclear weapons, this loss of personnel would be 3.70-5.30 percent per day. In all operation there will be 27-42 percent losses. With the use of conventional weapons, losses will be 1.10-1.30 percent in a 24 hour period, while in the entire operation this will be 7.40-01.40 percent.

At the front level with the use of nuclear weapons, losses of personnel for the entire operation will be 35-40%; in a 24 hour period losses will be 2-2.06%. In accordance with the type of weapons, losses will be as follows:
- nuclear weapons - 16-18 %;
- conventional weapons )artillery fire, rifle, and aviation( - 6-7 %;
- chemical weapons - 5-6%;
- biological weapons - 1.5-2%;
- disease - 1.5-2%.

The greatest losses will be during the first nuclear strike: losses will be up to 33% of all losses. With the use of conventional weapons, losses for the entire operation will be 12-13%; on the average, in a 24 hour period losses will be 0.7-0.9%. With respect to the above mentioned losses, 120,000-130,000 hospital beds, including 40,000-50,000 at the beginning of the operation will be needed. The fact that a front does not have such a quantity of beds means that 2 men will have to be put on one bed to provide for the treatment of the sick and wounded )each bed is multiplied by two(.

The following are the allowable doses of radiation from nuclear blasts for personnel:
- once in the course of every four days - up to 50 roentgens;
- several times in the course of 10-30 days - 100 roentgens;
- several times in the course of 3 months - up to 200 roentgens;
- several times in the course of one year - 300 roentgens.

The contamination from nuclear blasts is measured and affected areas are divided into zones. The outermost zone ) A ( varies from 40 to 400 roentgens, the next, ) B( is from 400 to 1200, the next )C( is from 1200 to 4000, and the innermost zone )D( is from 4000 to 10000:
- the average zone of destruction )zone A(: contamination doses on an external border - 40 roentgens, and on an internal border - 400 roentgens;
- the zone of strong contamination )zone B(: on an external border - 400 roentgens, and on an internal border - 1200 roentgens;
- the a zone of dangerous contamination )zone C(: for an external border - 1200 roentgens, and for an internal border - 4000 roentgens;
- the zone of extremely dangerous contamination )zone D(: for an external border - 4000 roentgens, and for an internal border - 10,000 roentgens.



 
 

Protection against nuclear radiation

The most effective means for reducing the destructive action of penetrating radiation are various types of coverings. The cover may be layers of iron, concrete, brick, or earth, as well as combat equipment found in sub-units.

The protection effectiveness of a material used for a cover depends on the density and thickness of the layer. Heavy materials such as lead, steel, concrete, and ferro-concrete have the greatest absorption capability, and, thus, the greatest capability for reducing gamma radiation. Walls of concrete or brick protect to a greater degree than wooden ones or ones built of foamed concrete. Dirt has good protection properties and, practically speaking, the greatest value as a covering material. This is illustrated by data included in the table.

The use of coverings as protection against neutrons is a very complex problem. Materials with high density, which are good protection against gamma radiation, do not provide sufficient protection against neutrons. And it should be remembered that emission of secondary gamma radiation accompanies the capture of neutrons by the nuclei of the atoms. In this respect it is necessary to use materials which weaken gamma radiation. Materials used as protection against neutrons can be divided into three groups:
- the first is materials which slow down fast neutrons;
- the second is materials which absorb neutrons which have been slowed down;
- the third is materials which weaken secondary gamma radiation.

Light materials containing much hydrogen )paraffin, water, some plastics( are used to slow neutrons down, and materials such as cadmium, boron, and others are used to absorb neutrons. Some of the above-mentioned materials are employed as elements of anti-neutron finishes used in new generations of combat vehicles.

Concrete and moist earth are also good protection against neutrons and gamma radiation. Although they do not contain elements with a high atomic mass, they do, however, have a sufficiently high amount of hydrogen so as to slow down and capture neutrons, and a sufficient amount of limestone and silicon so as to absorb gamma radiation. In connection with this, a sufficiently thick layer of concrete or moist earth can be used to assure simultaneous protection against neutrons and gamma radiation.

The factors for reducing the penetrating radiation of a neutron explosion by combat vehicles are, respectively:
- armored personnel carrier - 1.1-1.2
- BMP - 1.5
- tanks without an anti-neutron cover - 1.5-2.0
- tanks with anti-neutron cover - as much as 10 times.

In protection activities field fortification installations can be used for protection against penetrating radiation. The multiplying factor of reducing the dosage of the penetrating radiation from atomic, fission, thermonuclear, and neutron explosions is given in the table.



 
 

Figure 74 - Protection against radiation

Layers for Partial Reduction )in centimeters( of Penetrating Radiation for Several Protection Materials

Type of material Density of material kg per cu decimeter Radiation
Gamma Neutron
Fission Synthesis Fission Synthesis
Armor 7.8 3.5 3.5 11.0 12.0
Concrete 2.3 9.5 12.5 8.2 9.8
Brick wall 1.6 13.0 18.0 9.0 11.0
Earth 1.6 13.0 18.0 9.0 11.0
Water 1.0 20.4 28.0 2.7 4.9
Polyethylene 0.9 21.8 31.0 2.7 4.9
Wood 0.7 30.5 40.0 9.7 14.0


 
 

(A layer for partial reduction is a layer of material which reduces by half the strength of the dosage of radiation)

 
 

Figure 75 - Coefficients of reduction of Radiation

Coefficients of Reduction of the Total Dosage of Penetrating Radiation From a Nuclear Explosion by Field Fortification Objects

Type of object Protection layer )in centimeters( Type of explosion
Surface Air
Atomic charge: q >/=10 thousand tons
Field Shelter 140 600 500
Combat shelter 90 90 70
Covered fissure 60 20 3
Uncovered trench -- 4 3
Thermonuclear charge: q=50-200 thousand tons
Field shelter 140 1400 820
Combat shelter 90 180 100
Covered fissure

30

24

12

17

6.2

Neutron charge
Field shelter 140 600 430
Combat shelter 90 90 65
Covered fissure 30 8 5
Uncovered trench -- 4 3
Note: the coefficient of reduction for a trench is determined for a depth of 75 centimeters
 
 

Norms for Planning Combat Movement and Marches

Movement Planning Norms

Figure 76 - Movement data for tracked and wheeled vehicles

MOVEMENT CAPABILITY OF TRACKED AND WHEELED VEHICLES

Vehicle type Speed )km/hr Range in POL - )km( )1 refill( Range in length of vehicle life )km(
Average paved roads Dirt roads Max on paved roads Engine Track
Medium tanks 35 27 50 440-550 6000 - 9000 3000
Light tank 35 25 44 260 4500 - 7500 3000
BMP up to 50 25 70-85 260 -- 2000
APC's 50 25 80 500-600 -- --
Towing vehicle 25-30 15-20 40-55 250-300 -- 2000 - 5200
Wheeled vehicles 50 25 60-90 500-650 -- --


 
 

Figure 77 - Average speed of march and length of daily march

AVERAGE SPEED OF MARCH OF MARCHING COLUMNS AND LENGTH OF DAILY MARCH

Conditions of movement Speed of movement Length of daily march
Day Night March hours Kilometers
Motorized march columns:

Paved roads

30-40 25-30 10-12* 250-350
Dry dirt roads 20-25 18-20 10-12* 180-300
Muddy dirt roads, cities 10-15 8-10 10-12* 80-180
Tank & mixed columns:

Paved roads

20-30 15-20 10-12* 200-250
Dry dirt roads 11-20 12-15 10-12* 120-170
Muddy dirt roads, cities 10-12 8-12 10-12* 80-120

* The rest of the 12-14 hours per day are spent on the following activities:
- technical maintenance and services 3-4 hours;
- serving hot meals 1-1.5 hours;
- assembly or formation of column and concealment 1.5 hours;
- movement to march starting line 1-1.5 hours;
- rest 4-8 hours.

The average rate for movement motor rifle or tank units during deployment into company and platoon columns in close proximity to enemy when he may give artillery fire is as follows:
- motor rifle unit 15 km/hr;
- tank unit 15 km/hr.

The average rate for battalion or regiment columns is 20-30 km/hr when not under enemy artillery fire. Time for refill tank battalion with fuel using mechanical means 20 -30 min. Time for reload tank with ammunition 1 - 2 hrs.


 
 

Road Marches

A combined arms and tank army consisting of 4-5 divisions, as a rule, can perform a march on 4-5 or 6-7 routes; the operational formation of the march will be in two echelons. The distance of the first operational echelon of the march from the second operational echelon will be 80-100 kilometers. The depth of the march formation of the division on 2 routes will be 80-100 kilometers. The depth of the march formation of a division on 3-4 routes will be 40-50 kilometers.

The depth of the first operational march formation of an army on 8 routes will be 130 kilometers. The depth of the first operational march formation of an army on 9 -10 routes which normally will be in the last day of movement 60 kilometers and the depth of the second operational formation of the army will be 1851-300 kilometers. The overall depth of a march formation of both echelons of the army while moving on 7 routes will be up to 300 kilometers and more. The overall depth with movement on 5 routes will be 500-600 kilometers. The overall depth of a march formation of a motorized rifle division on 3 routes, without march security echelon, can be 70-80 kilometers, while the distance between vehicles will be 25 meters and the distance between battalion columns will be 5 kilometers and the distance between regiments will be 10 kilometers. To perform a march a distance of 1500-1700 kilometers, 5.5- 6 refuelings of gasoline are required and 8.5 -9.5 refuelings of diesel. Overall this will be 62,000 to 92,000 tons of fuel. One refueling of fuel for a combined arms army consists of 5300-5000 tons of fuel. The capacity of the motor transportation units of the army is 6000-7000 tons of fuel.


 
 

Figure 78 - Times to form unit column

TIMES TO PASS FROM ASSEMBLY AREA TO MARCH COLUMN

Unit Minutes
Motorized Rifle Company 5
Motorized Rifle Battalion 10 to 15
Artillery Battalion 15 to 20
Artillery Regiment 40 to 50
Motorized Rifle Regt )Reinforced( 60 to

 
 

Figure 79 - Distance Between Column Elements

MARCH INTERVALS

Combat Reconnaissance Patrol - Advance Security Elements up to 10 km
Advance Security Elements

Advance Guard Main Body

5-15 km
Patrol Vehicle-Unit Sending It Out 1.5-2 km
Advance Guard-Main Body 5-30 km
Vehicle - Vehicle 25-50 m

up to 100 m in nuclear threat area - 25 m at night or possible less

Co - Co )in a Bn Column( 25-50 m

Up to 300 in nuclear threat area

Bn - Bn ) in a column( 3-5 km
Regt - Regt )in a column( 5-01 km
Regt Main Body-Regt Rear Svc 3-5 km
Div Main Body-Div Rear Svc 15-20 km

 
 

Figure 80 - March, Halt, and Rest Durations

MARCH, HALT, AND REST DURATIONS

Average Day's March Duration 10-12 hours
Maximum March Per Day, Emergency 16-18 hours
Normal March Between Halts 2-3 hours
Northern Areas Between Halts 1-1.5 hours
Heavy Ice Between Halts 1 hour
Poor Roads - Favorable Weather 1.5-2 hours
Foot March Between Halts 50 minutes
Short Halt After 2-3 Hours of March 20-30 minutes
Short Halt After 3-4 Hours of March 45 minutes
Long Halt Mid-Day 2-4 hours
Foot March Minutes per Hour 10 minutes

 
 

Air movement

As the experience of exercises has shown, a motorized rifle division without heavy equipment is transported by air transport aviation at a distance of 1700 kilometers in the course of 8 hours. Four aviation transport divisions ) 800 AN-12's( are required for two flights to transport a motorized rifle division without heavy equipment.


 
 

Rail and ship movement

The width of an army movement sector is 150-200 kilometers; with one or two railroad routes, which gives the overall capability of the railroad 50-60 trains in a 24 hour period. 70% or 80% of these trains will be used for military purposes, i.e., 35-50 trains can be used for military purposes in a 24-hour period. These 35-50 trains can provide for the transport of heavy equipment of up to two divisions in a 24 hour period.
For the transport of one motorized rifle division, 50-60 trains are needed. For the transport of a tank division, 48 steam trains are needed. For an army composed of four motorized rifle divisions and one tank division, 400-450 trains are needed.

The average speed of the train in the territory of the Soviet Union can be 600 kilometers in a 24 hour period, and in certain directions up to 1000 kilometers per 24 hour period.

For the transport of one motorized rifle or tank division, the following naval ships are needed:
- 30-35 ships, each 2.5-3 thousand tons, or
- 16-20 ships, each 4-5 thousand tons, or
- 7-8 ships, each 12-13 thousand tons.



 
 

Calculations in the Decision Process

The Soviet planning process includes a great deal of time and effort in making detailed calculations. These calculations are done at every stage of the decision process. The decision process for division, army, and front commanders and their calculations are discussed in Chapters Two, Three, and Four, respectively.


 
 

Clarification of the Mission

While clarifying his missions, the commander must calculate the depth and width of the missions, the time to achieve them, and the required rate of advance. From these calculations he derives his general idea of the number of forces required and their echeloning and formation. He and/or the chief of staff must calculate the time available for planning and preparation of the troops for combat. From this they develop a time schedule for accomplishing all these needed actions. This time is used in preparing the calendar plan. )See Chapter Five for samples of calendar plans for army and front.(



 
 

Estimate of the Situation

Enemy

During this estimate the commander first calculates the density of enemy forces in each different area and for various depths. He calculates the enemy nuclear capability in terms of the number of targets and kilotons it is possible for the enemy to deliver. He also calculates enemy artillery capabilities in terms of hectares of target per salvo, aircraft capabilities in terms of numbers of sorties per day and enemy air defense in terms of the theoretical number of aircraft that could be shot down at one time, if the enemy launched a massive air strike. There are also more sophisticated models used to compute the expected value of damage to own forces averted if a number of enemy forces are destroyed.

The commander calculates the time and space factors, first those related to mobilizing and preparing the units for combat and then those that show when units can reach their combat starting locations. For these he includes the enemy operational and strategic reserves in order to establish how soon they will reach the areas he believes the enemy will want to assign to them. These calculations make use of simple rate of movement formulas and established norms for movement over various roads as well as norms for accomplishing each activity such as debarking, dismounting, deploying and etc. Next the calculations take into consideration the disturbances to the time schedules that might be introduced by disruptions to the line of communications, blocking of ports, destruction of air fields and other similar events.

The probable enemy concept of operations is assessed by estimating the length of delay actions he can achieve on each line based on the calculation of the density of forces and means. If the density is one company per kilometer then a division can hold for a day or so within its 21 km deep position. At this time the possible locations at which the enemy reserves can intervene in the battle are noted from the calculation of when and from where they can launch counterattacks.

If the initial enemy position is to the rear of his preferred battle position, calculations are made to find out if a meeting engagement between the large units is to be expected.


Friendly

When the commander turns to the estimate of friendly forces, he makes many of the same calculations. First there is the movement from garrison including time to mobilize and bring the forces to full combat readiness and time to establish the unit attack groupings. These calculations are mostly reconfirmation of existing planned activities. The commander can turn to the staff all of whom know what will be asked of them ahead of time to insure that units can arrive on time. The calculations require information on the status of units, and where they draw supplies or how the supplies will be delivered.

The combat capability of friendly forces includes calculation of nuclear capability in kilotons and numbers of warheads, artillery firepower in hectares of target per salvo, air defense capability in numbers of aircraft destroyed and aircraft in squadron sorties per day. The air defense calculations are especially complex since they involve detailed numerical factors for each type of weapon and target acquisition.

The commander must next establish the correlation of troops and means. This is shown in a table titled COMPOSITION OF FORCES AND DENSITY. The friendly and enemy forces are shown in terms of nuclear rounds, nuclear delivery means, divisions, artillery, tanks, antitank missiles, air defense weapons, and aircraft. The ratios are calculated using quantitative and qualitative factors and are figured for the sector as a whole and for each individual axis and for each relevant depth of mission. They are calculated for; the start of the operation, after the initial nuclear strike, at the end of the first day, at the end of the immediate mission of the armies, at the end of the formation's immediate mission, and at the end of the subsequent mission.

The calculations for correlation and density for each of the points in time after the start involve calculations of the estimated losses that each side will have incurred. The calculation for losses in the initial nuclear strike is made by taking the total number of rounds allocated )or estimated for the enemy( and from this the number and yield that will be targeted against divisions to get a number of rounds per division. Then norms are applied to estimate losses. One norm is that if a division is hit by more than 6 -7 nuclear rounds it suffers medium damage and is incapacitated. If it is hit by more than 21 rounds it is destroyed. The effect of losses is estimated and 03% is considered heavy casualties while 05 - 06% will equal destruction. Losses for each day of combat are calculated according to norms for conventional and nuclear warfare. The correlation at the end of the first day would include loss norms of about 5% for personnel and 8% for tanks and lesser numbers for other equipment. One norm is that in 7 days of fighting a loss of 05 - 08% for tanks is expected. Some other norms are for army level in conventional war 1.1 - 1.3 % personnel per day; for nuclear 3.8 - 5.3 % per day; and 7.7 - 01.4 for the entire operation in conventional war and 72 - 24 % for the entire operation in nuclear war. Equipment loss norms include conventional of 8-9% per day and 04 - 06 % for vehicles and 05 - 08 % for tanks All these norms are used to calculate the correlation of remaining forces for the various subsequent times. For instance, at the end of 5 days in an operation it might be expected that the attacker will have suffered losses of 7% in personnel, 04% in tanks, 52% in APC and 53% in other vehicles while the defender will have suffered losses of 5% in personnel, 53% in tanks, 02% in APC and 03% in vehicles.

After he calculates both sides, he is able to make a deduction on the proper distribution of troops to the several axes and then to distribute the combat support arms and reserves, naval and airborne assaults and other support.

An important set of calculations is for electronic warfare, used in determining how many communications links above division can be neutralized by the available REC assets. Each radio electronic warfare battalion has a capability based on its means to jam a certain number of radio nets of a certain power or type. The enemy can also jam certain links.

Once the commander has determined the missions, there are then calculations related to the interaction between forces. These are to establish how groupings will be created and what times will be involved. A table showing who will do what at each time is prepared.

The calculation for the locations and times for movement of the command posts are based on the planned course of the offensive. The commander then considers the role of adjacent forces. He makes calculations to see how the missions of adjacent forces might involve the formation and vice versa. For instance, the time an airborne division can sustain itself before linkup with ground forces is used in calculating when the airborne operation should take place. One of the adjacent forces is the strategic rocket and air force. The timing of their strategic nuclear strike, if any, or the strategic air operation is considered in calculations on when to launch the operational air strike.



Terrain

The commander then considers terrain in calculations to refine the plans. The capacity of routes, ports, airfields, bridges, etc. is considered to insure that the forces can move as planned. The economic situation in the theater is the basis for calculations on the availability of local resources such as supplies and transportation means.



Calculations Essential for Correct Decisions(3)

The following sections on the theory of using mathematics to forecast battle outcomes as part of the commander's decision making process is translated from Vayner's book Tactical Calculations. it is included here because it is one of the clearest expositions of the Soviet approach to mathematical modeling combat as a "real time" decision tool.

Each commander in his practical activity constantly encounters the necessity to make decisions. A characteristic feature of any situation which requires a decision is the presence of several possible variants of action, from which one, that is, the optimal, which ensures the successful fulfillment of the combat mission, must be selected.

In modern conditions, it is impossible to make an optimal decision for a battle without comprehensive evaluation of the situation and determination of the combat capabilities and without having learned to correctly foresee the consequences of a particular action variant.

As a result of the dynamic opposition of the two sides, on the field of battle a complex situation is created in which without foreknowledge and without forecasting it is impossible to command the forces in a goal oriented manner and to achieve success.

Calculations are an important instrument for giving the commander the required quantitative data which make it possible to evaluate the effectiveness of actions. The various composition of forces and the methods and means for their application relative to the situation ensure the achievement of the goal of the battle to a varying degree. The commander, making a decision, attempts to achieve the maximal effectiveness of actions, to inflict the greatest damage to the enemy with the least losses of his own forces and to successfully fulfill the assigned mission.

The complexity of evaluating the situation and selecting the best action variant, besides enormous expenditures of creative energy and efforts of will, makes it necessary to use the tools of quantitative methods. Using calculation techniques, it is possible to determine the various indicators of effectiveness of combat actions: the effectiveness of fire strikes, the length of a march, the required number of combat equipment for inflicting the assigned damage to the enemy and so on.



Mathematical Modeling - the Basis of Calculation Techniques

Effectiveness means the degree of correspondence of the actions to the intended goal )the formulated mission(. The more successfully the goal is achieved )the mission fulfilled(, the more effective the action. When there are several possible action variants, they must be evaluated in some way and the best selected. To do this, an effectiveness "measure" is required. The strict scientific measure of effectiveness is the criterion of effectiveness. How effective a particular action variant is may be determined from the size of the criterion.

Selecting the criterion of effectiveness is an important logical problem. An incorrectly selected criterion leads to inaccurate evaluation of effectiveness with all the attendant negative results. The correct criterion must first strictly conform to the goal of the actions, be understandable, be expressed by a number and be calculated. The value of the effectiveness criterion must be changed relative to a change in the substantive factors which impact on the course and outcome of the operations. For instance, if the main mission is to march from one region to another in the shortest time, then the effectiveness criterion of the possible movement variants will be the expected length of the march )time(. To evaluate the effectiveness of actions under a particular variant, the specific value must be calculated with consideration of all the basic factors which effect the travel speed, i.e., a calculation must be made. However, it is impossible to accomplish such a calculation without the appropriate mathematical model of travel.

As a strict scientific method for cognition of activity, modeling has quite ancient roots and was always associated with practice. Throughout history, physical and mental models - samples of actual subjects, processes and phenomena - were used for predicting in decision making. Since models have a number of similar features with the originals, they act as a "Substitute" for the object being studied. The presence of a communality between the model and the object being modeled is the objective basis of the modeling.

On one hand, the capabilities of mathematical modeling are caused by the level of development of mathematical methods and on the other, by the degree of study of the processes and phenomena being modeled. The rapid development of military science, the theory of operations research and computer equipment had great meaning for the success in the development and wide use of mathematical modeling in force command and control. All of this transformed the abstract capabilities for mathematical modeling of the processes of armed combat into reality, into a practical instrument of troop control.

There are two kinds of parameters; parameters of the conditions and parameters of management.

Parameters of conditions include those whose values in a given specific situation are not a function of the decision the commander makes. These may include, for instance, the time of the year, the meteorological conditions and others. The values of the management parameters, on the other hand, are a function of the commander' decision and are caused by how the commander intends to act in the particular situation. For instance, how many and what kinds of forces and means and where and when he decided to use them for the fulfillment of the assigned mission. In completion of a march, the management parameters may be the number of route columns on the routes, the distances between them, the movement start time )the passage of the initial line( and so on. In repelling enemy tanks, the management parameters may include the expended number and types of anti-tank weapons, the munitions for them, their disposition in the terrain, the time fire is opened and so on. From the above, it is easy to see that the classification of parameters to a particular group depends on the goal of the actions and the situation. One parameter in one case will be a condition parameter and in another, a management parameter.

The general order of the construction of a mathematical model can be briefly reduced to the following. First, the following must be determined: what results must be achieved, the size )unit of measure and the arithmetic precision of their calculation. Then the factor )parameters( are exposed on which the size of the required indicator depends. For instance, if a march length must be calculated, then there can be many such factors )march length, the composition of the columns, the road characteristics, the presence of bottlenecks, the time of the year and the day and so on(. Naturally, to make the model more precise )adequate to reality(, as many of the substantial factors as possible must be taken into consideration. However, it is not always possible to quantitatively express a particular factor. Therefore, from the totality of exposed factors, only those which may be expressed by a number )measured, calculated( are selected. Then, if possible, the selected factors such as the road condition, the weather conditions, the characteristics of the transport equipment and others, may be expressed by a single combining indicator: the travel speed. Here, the model is sort of rounded off, but then, it is significantly simplified, which is quite important for speed and convenience of calculation. As is evident, on the one hand, the necessity for building a more precise model is present and on the other, the necessity to have the most weighty, but smallest possible model.

After final determination of the factors which effect the value of the required indicator, the relationship between them is established. In the simplest case this means that the physical essence of the action being modeled is described by an analytical relationship, i.e., a mathematical formula or several formulas are deduced, in which the factors will be variable initial data for calculating the required indicator.

Many various calculations are conducted in headquarters and other troop control bodies. All of them have their own purpose and target orientation and are fulfilled by their own specific techniques.

Tactical calculations may be fulfilled by different responsible individuals and their results must serve as the basis for a more profound evaluation of the situation, the making of the most expedient decision, justified planning, and comprehensive battle support.

In the nature of the questions being solved, all tactical calculations are subdivided into direct, inverse and optimization calculations )See table below(.

Direct calculations normally make it possible to obtain quantitative data for determining the expected result of using specific on hand forces and means in a plan, with a planned action variant.

For instance, having a specific number of anti-tank weapons and having planned a variant for their use, direct calculation is conducted through a previously developed technique and data are obtained about the expected volume of damaged enemy tanks, i.e., the effectiveness of the plan variant is evaluated.

Having obtained data on several planned variants and using the calculated indicators, the best, the most expedient variant in the particular situation is selected. Having a specific number of crossing means at one's disposal, it is possible through direct calculation to obtain data about the expected length of time for crossing a water obstacle and so on.

In direct calculations, information about the obtained or earmarked forces and means is used as the initial date, as well as information which characterizes the conditions for using these means, i.e., the utilization plan. Using the calculation, this plan is analyzed and its effectiveness is evaluated.



 
 

Figure 81 - General Characteristics of Tactical calculations

GENERAL CHARACTERISTICS OF THE BASIC TYPES OF TACTICAL CALCULATIONS

Calculation types Initial data for calculations Calculation results
Direct calculations Number of forces and means - Variant of plan for using forces and means Expected effectiveness of the variant of the plan for using the forces and means
Inverse calculations Required result - Plan variant for using forces and means Required number of forces and means for achieving the required result
Optimization calculations Number of forces and means - Conditions for their use Most beneficial plan variant for using the on hand forces and means which provides the best effectiveness


 
 

Inverse calculations are conducted in those cases when during evaluation of the situation and decision making, it is required to determine what number of forces and means is required for achieving an assigned result of operations through a projected plan variant. For instance, in evaluating the predicted variant of a fire strike by destructive means, inverse calculation is used to obtain data about how many of the means )guns and mortars( should be used and how much ammunition must be expended in order to provide the required degree of damage of a particular enemy target. In creating a mixed mine field on a specific line with an assigned density, data about the required number of mines can be obtained using the inverse calculation technique.

Inverse calculations are conducted on the basis of the initial data about the required result of the actions and information which characterizes the projected variant for using particular forces and means.

Optimization calculations make it possible to obtain quantitative data for determining the most beneficial variant of the predicted actions, i.e., how, using the on-hand forces and means, to achieve the greatest effectiveness and the best result, since the same number and quality of forces and means may be used with different effectiveness depending on their distribution on the target, the location and the time.

Optimization calculations are the most difficult and the techniques for their accomplishment are the most complex. Optimization calculations as a rule require the use of complex, modern mathematical equipment and are realized during high speed electronic computer equipment.

For instance, optimization calculations may be used to solve the problem of the optimal target distribution, when it must be determined with which variant of distribution of the on-hand destructive means the maximal damage will be inflicted on the enemy targets in the particular conditions. It follows from the essence of optimization calculations that they produce the most useful quantitative indicators of the predicted actions for decision making.

In each specific case, the goal of the calculation is determined, along with the indicators )the quantitative data( and by what time the calculations are required. In accordance with this, the technique of tactical calculation is selected, which may be used to calculate the actions in the designated variant. However, in all cases, calculations are primarily conducted which are required for the complete and comprehensive explanation of the assigned mission. Then, quantitative data are calculated for evaluating the situation and for determining the forces, methods, means and terms of mission fulfillment. Based on the obtained quantitative data, the most expedient variant of the actions is selected. And finally, the indicators are calculated which are required for justification of the combat missions of subordinates, for the planning and comprehensive support of combat operations. Normally, the direct nd inverse calculations are first fulfilled and then, the optimization calculations based on them.

The great significance of tactical calculations for decision making causes the great requirement placed on them. The essence of these requirements can be reduced to timeliness and precision.

The requirement for timeliness of calculations is caused by the necessity for urgently making a justified decision, to plan force operations in a timely manner and to organize comprehensive battle support. Therefore, the results of the calculation must be obtained in times dictated by the situation. *untimely fulfillment of calculations, even if they are the most correct, lose all sense when their results because of the delay may not be used as designated in the interest of decision making and battle planning.

The maximally possible speed of calculations is sometimes determined as work efficiency. At the same time, work efficiency, besides speed, includes another important requirement - calculation precision. Only a combination of calculation speed and objective correctness and precision may be identified as the efficiency of calculation support.

The precision of calculations is determined by three basic factors: the precision of the technique used to calculate particular values, which, in turn, is a function of the mathematical model of the predicted actions used and the degree of its correspondence to the actually occurring process; the reliability of the initial data, on the basis of which the calculation is conducted and the error free manner in which the calculations are fulfilled by the executor.

Often, the concept of precision of calculations is confused with the concept of the arithmetic precision of the calculated result of the calculation, which hinders a correct understanding and evaluation of the quality of the calculation techniques and the means for their realization.

When greater or lesser calculation precision is discussed, the point is not the number of significant digits obtained as a result of the calculations, but the degree of correspondence of the calculation result to reality.

Consequently, a factor is significant or insignificant not of itself, but with respect to the specific conditions.

In principle, any mathematical model and, consequently, calculation technique, has some sort of assumptions. It is important that these assumptions be as few as possible. The fewer the assumptions, the more precise the calculation technique and, consequently, the calculation itself.

Various models of computer equipment are being used more and more often in order to further reduce the calculation times in the forces. They not only reduce the calculation times, but simultaneously increase their precision and reduce the labor expenditures of responsible individuals. Even the simplest computers )calculation rulers, nomograms( and keyboard equipped computers increase the efficiency of calculations in the tactical link by 2-5 times.

The commander must creatively consider not only the calculated data, but other unconsidered factors of the situation which impact on the course and outcome of the predicted actions. This may only be done by an officer who has a good knowledge of the essence of mathematical modeling, the role and meaning of quantitative methods and the specific means and methods of calculations.

Among the basic measures accomplished by commander and the headquarters in organizing a battle are making the decision for the battle, planning combat operations and their comprehensive support.

The decision making of the commander is the most complex and critical act in troop control.

As a rule, the decision making process begins with an explanation of the received combat mission. Here, it is necessary to understand the role and place of one's own subunit )unit( in the upcoming combat operations, the concept of the next senior commander )superior(, the action of the adjacent units in the fulfillment of the received mission, the times for preparing for fulfilling the assigned mission, and other questions.

Different calculated data may be required for a more profound and comprehensive explanation of the mission. For instance, in organizing an offensive battle, it is important to calculate the effectiveness of the action of forces and means used by the senior commander )superior(, the capabilities of assigned and supporting subunits. The results of calculations of the effectiveness of the weapons of the senior commander )superior( in the axis of actions of the subunit )unit( allow the commander to have a more precise idea about the nature of the mission and the conditions of its fulfillment.

After the clarification of the assigned mission, the commander calculates the time, organizes reconnaissance, and gives the required preliminary orders to prepare for combat operations.

A modified program evaluation and review technique )PERT( method may be successfully used for rapid and correct calculation of the time for fulfilling all measures to organize a battle. Using it, preserving the logic, significance and sequence of the measures to be conducted, the distribution of the time expended for each operation is optimized so that all the organization of the battle will be finished in the established time.

In organizing reconnaissance, the capabilities of the available forces and means to find targets and the expected effectiveness of mission fulfillment in reconnaissance must be calculated )calculations are conducted of the optimal distribution of forces and means(. The duration of the operations of reconnaissance forces and means to fulfill individual missions in the established times and others may also be calculated.

An important element in the situation is the action of the enemy. Information about the enemy will often be incomplete and insufficiently accurate. Hence, the necessity for probabilistic calculations for predictions.

Both reconnaissance data, as well as information about enemy operational tactics, his organizational structure, the tactical and technical characteristics of his armaments and equipment, and other data known ahead of time may be used as the initial data for calculating the actions of the enemy. Evaluating the enemy, calculations are conducted of his personnel, and calculations to evaluate his combat capabilities )fire, strike, maneuver(. The very same techniques may be used as were used in the evaluation of friendly forces, using the corresponding initial data.

For instance, calculating the maneuvering capabilities of the enemy, techniques for calculating the advance of forces, such as calculation of the duration of regrouping, the length of the break out to a line, deployment time, and others may be used. In evaluating fire, strike, and maneuver capabilities of an enemy, it is expedient to conduct calculations through several designated variants. The obtained calculation results are quantitative data for justified conclusions about the most probable nature of the enemy actions and his combat capabilities. These data make it possible to compare different variants of his actions and to more justifiably judge just how much particular enemy forces and means and actions will hinder the successful fulfillment of the assigned mission and what should be done to reduce the combat capabilities of the enemy and to defeat him.

Evaluating friendly forces, the commander determines their combat capabilities in the existing situation from the point of view of the assigned combat mission.

The combat capabilities of friendly forces are evaluated as a whole, by branch of service and even by individual types of combat equipment. To evaluate the capabilities of defeating the enemy, calculations are required with are associated with the use of nuclear weapons, calculations of the capabilities of artillery which is firing from protected fire positions, direct aim field guns and anti-tank guided missiles )PTUR(. They include calculations of the degree of enemy defeat by the assigned )available( number of guns and shells or calculations of the required number of guns and shells for solving the fire assignments and the optimal distribution of the weapons against targets, by objects and by missions, calculations to evaluate the effectiveness of the means ear marked for battle against armored targets and others. For instance, to determine the combat capabilities of PVO weaponry, their effectiveness in defeating aerial targets in different variants of action of the air enemy is calculated.

Evaluating the capabilities of motorized rifle subunits, their strike power, maneuverability, the effectiveness of their fire and the calculated. In comparing friendly and enemy forces, different variants of calculations of the correlation and the densities of the forces and means of the sides are conducted.

Maneuverability is evaluated on the basis of data from calculations of the expected duration of advancement and deployment of forces, change in regions, positions and so on.

In evaluating the situation, calculations are required for determining the capabilities of special force subunits, for instance, engineering forces in fortification of the terrain, ion providing for force crossings of water obstacles, equipping routes, and in setting up obstacles and in removing them. These include calculations of the time it takes to erect engineering structures, to equip the terrain in an engineering sense, calculations of the required amount of forces and means to fulfill the missions of engineering support in the established times, calculations of the time it takes to put forces across a water obstacle and the required amount of crossing equipment.

Besides calculations to evaluate the capabilities of organic, assigned and supporting forces and means, the combat capabilities of adjacent units, forces operating in advance and other subunits, with which interaction is required in the fulfillment of the assigned mission, are also calculated. Here, the very same calculation techniques are used as in evaluating the capabilities of one's own subunit. In evaluating the capabilities of friendly forces, calculations with are associated with command and control organization are pertinent. These include, for example, calculations to determine the numerical characteristics of a communications system, its reliability, the required number of forces and means for communications must be fulfilled in planning comprehensive battle support. These include calculations to determine the requirement for different types of materials and the supply of them and calculations to determine the distribution of means, transport, repair, evacuation and many others.

Evaluating the terrain, the commander determines how it impacts on the fulfillment of the assigned mission, what the conditions for observation, cover and camouflage are, where the suitable lines are, and what the passability of the terrain and traffic ability of the roads are and the possible changes in the terrain as a result of combat operations. For instance, calculations of the fields of blind spots, the capacity of shelters, the traffic capacity of routes, the dimensions of zones of destruction, obstructions, fires and floodings may be conducted.

The characteristics of calculations during battle planning are that they are fulfilled according to data which correspond to the idea of the operations in the decision made. In other wards, the multiple variant nature of the calculations in this particular stage is replaced by greater detail and specificity of each calculation. Despite the fact that in basic content, the calculations fulfilled during decision making and battle planning may be identical, their techniques may differ. As a rule, planning calculations will be more detailed. Moreover, there are calculations which may be conducted only on the basis of a decision made. These calculations include, for example, calculation of the march schedule, calculation of the artillery preparation and detailed calculations of the types of support for combat operations.

Many data obtained as a result of calculations conducted during explanation of the mission and evaluation of the situation, may be used in justifying a battle plan. But often, the decision made requires the very same kind of calculations again, on the basis of data, determined by the commander's decision. A decision cannot be considered fully formed if the planning does not include all the basic indicators for organizing the battle, for interaction and for comprehensive support of combat operations.

The results of the basic calculations may be set up in the form of tables, in a workbook, on a chart or in planning documents. These data may be used in briefing and justifying the decision to the senior commander )superior(, in the delivery of the missions to subordinates, in organizing interaction and supporting combat operations and in troop control of sub-units during the battle.



 
 

II CALCULATIONS FOR PLANNING COMBAT


 
 

This section describes some of the mathematical calculations and procedures used by Soviet commanders and staff while making decisions, predicting the future course of combat and preparing plans. The nomograms and practical examples have been taken from several Soviet books and articles, most notably the two editions Tactical Calculations by Vayner and a series of articles in the formerly classified journal, Military Thought. Many of the same examples were also used in the book Sustainability of Soviet Army in Battle, prepared by the staff of the Soviet Studies Research Centre, Royal Military Academy,Sandhurst, England. The Soviet sources discuss exclusively tactical level planning and calculations. However, with a little imagination one can extrapolate to the operational level. Moreover, the authors have drawn on their direct experience in making operational level calculations to prepare the sections on division, army, and front calculations included in this chapter.


 
 

Operational and Tactical Calculations

The first group of calculations are those made by the commander and operations department for predicting the course of combat and planning how to control it. Probably the most pervasive and characteristic calculation is determining the time and distance required for troop movements of various kinds. Soviet operational and tactical planning places great stress on the troops arriving at the right place at the right time in a carefully orchestrated sequence to apply maximum combat power at the chosen "decisive" point. Therefore they have developed many simple or elaborate variations on the basic algebraic equation that distance equals time times velocity. Other calculations relate to the time units can remain in one place between moves. Other calculations are used to determine the probability of accomplishing complex tasks, given the experimentally determined probabilities of accomplishing individual sub-parts of the task. In this class are the calculations on probabilities for destroying the enemy with a given set of weapons of known effectiveness. However, the Soviet literature does not contain nearly as many examples of these equations as it does those for unit movements.


 
 

(1) Basic Time and Distance Calculation

This simple formula is used for determining the approximate time required to move a unit from one area to another, not counting the time required to move out of the initial area and reach the start line. The information required is the length of the march as measured from the initial starting line )SL( )at a distance outside the original assembly area( to the nearest point of the new assembly area; the average rate of march of the column, the length of time spent in halts, and the time required to deploy from the road into the new area.

 
 

The formula is:

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where:
tBt=is computed if the depth of new area is less than depth of the march formation.
t=total time of march;
D=length of route;
V=average speed of column on march in kph;
tp=overall time for halts during march;
tBt=time required to deploy into new area;

 
 

Example problem using nomogram: Calculate the duration of a move along a 80 km route with an average speed of 35 km/hr, duration of halts total 1 hr & 30 min, and time taken to deploy into new area is 30 min.

Solution: Start at the 80 point on the bottom scale "Length of March" go up to the "Speed of movement -35 kph" line then horizontally across to the I line. Draw a line from that point to the II line passing through the .5 point on the "Pulling in" line, then another line downwards from the II line passing through 1.5 on the "Duration of halts" line. This intersects the "Duration of march" line at 4 hrs and 20 min.


 
 

Figure 82 - Nomogram for Calculating Duration of March


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(2) Calculation of Time to Begin Move to Start Line:

This calculation is used to determine when a unit should begin moving out of its assembly area in order for the head of the column to cross the start line )SL( at the prescribed time. The given data are the distance from the assembly area to the start line and the rate of march.
(NOTE: in these forumlae the06 should be 6o to convert time to minutes_
{short description of image}The formula is:








where:
Dn=distance to SL;
tN=time column starts to move;
Vv=rate of movement to SL;
06=conversion factor Hr to min;
T=time head of column passes SL;


Example problem: Determine the starting time for a column when the time for the head of the column to pass the start line is planned for 2100 hrs, the distance to the start line is 9 km, and the rate of march while moving out is 15 kph.

Solution:

The 12:00 should be21:00)

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Figure 83 - Nomogram for Calculating Duration of Movement from Assembly area to start line


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Example using nomogram (Figure 83):

Using the same initial data as the previous example enter the nomogram on the X axis at 9 km move up to 15 kph line then across to the 36 min on the Y axis.

Example 2: Calculate the required speed of movement for the column to reach the start line with a distance of 7.5 km and a time of 45 min.

Solution: Draw lines from the 7.5 km and 45 min points on the scale. These intersect on the 10 kph line.



 
 

(3) Calculation of Time to Deploy into a New Assembly Area:

As noted in formula (1), this only must be calculated when the depth of the new area is less than the length of the mobile column. This is because in this case the head of the column will have stopped at the far end before the tail reaches the near side of the area. The formula gives the time it takes to deploy, once the head of the column has reached the new area.

The formula is:

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where:
Td=time for deployment;
GK=length of column;
GR=depth of area;
Vd=speed during deployment;
60=convert hr to min;

Example problem: Calculate the time required for a column to occupy a new area if the length of the column is 7 km, the depth of the new area is 3.5 km and the speed of movement during deployment is 10 kph.

Solution:=(7 - 3.5)/10 ( x 60=0.35 x 60=21 Min


 
 

Figure 84 - Nomogram for Calculating Time Required for Mobile column to deploy into new area


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Using the nomogram (Figure 84) provides the same answer. Enter at 7 on the length of column scale cross 3.5 on the depth of area scale then horizontally to 10 kph and then down to 21 min on the duration of movement scale.


 
 

(4) Calculation of Time a Unit will be in a New Area

This calculation combines the previous formulas in order to determine the clock time a unit will be deployed in the new area. It takes into consideration the time required for a unit to deploy into an area when the depth of that area is less than the length of the marching column. It also includes time for halts en route.

Formula for calculation:

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where:

t=clock time a unit will be regrouped in new area in hrs;
T=time )astronomical( of passing start point )line( by front of column in hrs/mins;
D=length of route and distance away of new concentration area in km;
V=average speed of movement;
GK=length of column;
GR=depth of new concentration area in km )considered only when the depth is less than the length of column(;
0.6=coefficient, which takes into account the lowering of the average march speed while deploying into the new assembly area.
tp=duration of halts en route in hrs.

This calculation may be conveniently performed by entering the data into the table provided.

 
 

Figure 85 - Form to Calculate Time unit is in new area

CALCULATION OF TIME UNIT IS ASSEMBLED IN NEW AREA

No Initial data and values to be calculated Units Calculation variant Remarks
Example 2 3 4
1 Length of march route km 167
2 Average rate of movement 18
3 Length of moving column 7.5
4 Depth of new assembly area 4
5 Duration of halts 1.5
6 Time of passing start line )SL( 10.0
7 )1( ÷ )2( 9.3
8 )3( - )4( 0.1 3.5
9 )2( x 0.6 0.1 10.8
01 )8( ÷ )9( 0.1 0.3
11 Overall duration of march

)5( + )7( + )10(

0.1 11.1
21 Time unit is concentrated in new area )6( + )11( hrs 21:60

 
 

(5) Calculation of the Duration of a March from one Area to Another

This is a more sophisticated version of the basic march formula to take account of possible reductions in the capacity of the road or other influences on the achievable rate of movement of the columns. The formula is:

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when GK < Gr

where:
t=duration of march in hours;
D=length of march in km;
K=coefficient for reduction in average rate of march of moving columns during entering and leaving the route of march;
Di=distance of start point from original assembly area in km;
GK=depth of column in km;
Gr=depth of the new assembly area in km;
V=average rate of movement of column in km/hr;
tp=duration of halts or delays during movement in hrs.

Example problem: Determine the duration of march for a column 11.2 km long to a new area at a distance of 87 km. The start point is 4.5 km from the original assembly area and the depth of the new area is 7 km. The average rate of march is 18 kph with a coefficient of reduction of speed of 0.6. There will be a total of 1 hr of halts.

Answer is 6 hr 38 min.


 
 

Figure 86 - Form for Calculating Duration of march

TABLE FOR CALCULATING DURATION OF A MARCH

No Initial data and values to be determined Units Calculation variant Remarks
Example 2 3
1 Route length km 87
2 Speed reduction factor -- 0.6 )1( x )2(
3 Distance to start point km 4.5 + )3(
4 Column depth km 11.2 + )4(
5 Depth of new concentration region km 7 - )5(
6 Rate of march km/hr 18 ÷ )6(
7 Speed reduction factor -- 0.6 ÷ )7(
8 Length of halts hr 1 + )8(
9 March duration hr 6.64 =ans

 
 

(6) Determine the Required Movement Rate for a Unit to Regroup in a New Area.


This is a more elaborate version of the basic movement formulas to take into account more variables and possible interactions during the movement.

The formula is:

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Example problem: Determine the required rate of march if a column has a depth of 8.7 km and the time allowed to assemble in the new area 5.5 km deep is 6 hours. The distance to the new area is 128 km and to the start point is 6 km. The coefficient for reduction of rate is .7 and the duration of planned delays is 45 min.

Answer is 27 kph.


 
 

Figure 87 - Form to Calculate Required Rate of Travel

FORM FOR CALCULATING REQUIRED TRAVEL SPEED

No Initial data and values to be determined Units Calculation variant Remarks
Example 2 3
1 Route length km 128
2 Speed reduction factor -- 0.7
3 Distance to start point km 4.5
4 Column depth km 8.7
5 Depth of concentration area km 5.5
6 Maximum travel time allowed hr 6
7 Halt time hr 0.75
8 Speed reduction factor -- 0.7
9 Required travel speed km/hr 27

 
 

(7) Calculation of Length of Route, Average speed and Duration of Movement of Moving Column


This calculation combines the basic equations. It is used when the total distance to be traveled is composed of segments having different route characteristics. The different characteristics result in different possible movement rates over the individual sectors. The initial data is the length of each sector, the movement rate over each sector, the length of the column and the depth of the new assembly area. Formulas for calculation:

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where:

D=length of route in km;
Li=length of each sector of different types of road each allowing Vi speed of movement of columns in km;
td=overall time on the route in hrs;
Vi=speed of movement on a type of sector of the route in kph;
V=average speed of movement in kph.
GK=length of column in km;
GP=depth of concentration area in km;
0.6=coefficient of reduction in speed of the column while deploying into the new area or depending on local conditions;
tn=overall time of halts.


 
 

Figure 88 - Calculation of Duration of March over Complex Route

CALCULATION OF TRANSIT TIME OVER MULTI-SEGMENT ROUTE

No Initial data - form of calculations Units Calculation variants Remarks
Example 2 3
1 Length of paved roads km 42
2 Speed of movement on 1 km/hr 35
3 Length of improved dirt roads km 18
4 Speed of movement on 3 km/hr 25
5 Length of dirt roads km 21
6 Speed of movement on 5 km/hr 15
7 Length of field tracks km 8
8 Speed of movement on 7 km/hr 10
9 Length of moving column km 6.8
10 Depth of new assembly area km 3
11 Overall time for halts hr 1.5
12 Total length of route

)1( + )2( + )5( + )7(

km 89
13 )1( ÷ )2( 1.2
14 )3( ÷ ))4( .7
15 )5( ÷ )6( 1.4
16 )7( ÷ )8( 0.8
17 Time of movement

)13( + )11( + )15( + )16(

hr 4.1
18 Average speed )12( ÷ )17( km/hr 22
19 )9( - )10( ÷ ]0.6 x )18([ 3.8
20 Time of deploy - new area

)19( ÷ ]0.6 x )18([

hr .3
21 Duration of move

)11( + )17( + )20(

5.9

 
 

(8) Calculation of Overall Depth of Column Consisting of Several Sub-columns:


This technique is for calculating the total length of a moving formation on a route given the number of vehicles in the moving columns and the distances between them is known. It is also used to determine the required distances between vehicles. The formula is:

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where:
Gk=formation depth in km;
Nm=number of vehicles;
dm=distance between vehicles;
NK=number of columns;
dK=distance between columns;
1000=convert meter to km.

Example problem: Determine the length of a moving formation consisting of four columns, if the overall number of vehicles is 169, distance between columns is 600 meters, and distance between vehicles is 40 meters. Solution: (Note the 006 should be 600 and the 0001 should be 1000


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Example using the nomogram Figure 89:

Determine the length of a moving formation if there are 3 moving columns and the distances between them is 400 meters, the overall number of vehicles is 65, distance between vehicles is 25 meters )variant a(.

Solution: First find the column depth without considering the distances between col use the right side of the nomogram and draw a perpendicular line from the "65" mark on the "Total number of vehicles" scale to the intersection with the "distances between vehicles- 25" line; from this point draw a horizontal line to the intersection with the "Column depth" scale. In the left part of the nomogram from the "3" mark on the "Number of columns in route formation" scale draw a perpendicular line to the intersection with the "Distances between columns- 400" line, from this point draw a horizontal line to the unnamed scale. Then connect the two obtained marks and find the calculation result on the "Depth of marching formation scale.

Answer: is 2.5 km.



Example variant b: Determine the required distances in a column consisting of 83 vehicles given that the length of the column must not exceed 2.5 km.

Solution: On the nomogram draw a horizontal line from the "2.5" mark on the "Depth of column" scale. Then from the "83" mark on the overall number of vehicles" scale draw a perpendicular line and at the intersection of these lines read the required distances between the vehicles. The answer is 30 meters.


 
 

Figure 89 - Nomogram for Calculating Length of Mobile Formation consisting of several Columns


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(9) Calculation of Duration of Passage of Narrow Points and Difficult Segments

This calculation is the basic one for determining the time it will take a column of given length to pass through a constriction in the route. It does not take into consideration the issue of vehicle bunching up at the halt before the constriction nor the time to regain column vehicle separation distances after the passage. Therefore more elaborate formulas are used to determine the overall effect of a constriction on a full march. The formula for short obstacles is:

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where:
Nm=number of vehicles;
dm=distance between vehicles;
t=time required to overcome obstacle in minutes;
V=speed of column through obstacle;
0.06=conversion factor km/hr to meters/min.

 
 

Example Calculation )A(: Calculate the time required to cross an obstacle by a column of 54 vehicles with distance between vehicles of 75 meters and a maximum speed of 10 kph.
t=)54 x 77( x 0.06 ÷ 10=24 Min

There are two types of difficult sections on routes; the first is minor ones whose length is less than the marching column, and the second is major obstacles with length greater than the length of the column. The main factor for shorter obstacles is the number of vehicles in the column, the distances between them and their speed of movement while passing the obstacle. The main data for the larger obstacles are the length of the column, the length of the sector and the speed of movement.


 
 

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The formula for major obstacles:

where:
Gk=length of column;
D=length negotiated segment;
t=duration to overcome obstacle in hours;
v=speed through obstacle.


Example calculation )B(: Determine the time for a column 2.5 km long to pass through an obstacle 5.5 km long at a movement rate of 15 km per hr.

)2.5 + 5.5( ÷ 15=8 ÷ 15=0.53=32 min

Example calculation )C(: Determine what length of column can negotiate a pass 2.5 km long at a speed of 8 kph in a 45 min. Gk=)V x t( - d=)8 x 0.75( - 2.5=3.5 Km
Solution: 3.5 km


Example using nomogram Figure 90

Example calculation: Using data from example )A(, start at 54 on "Number of vehicles" line and draw a perpendicular to the intersection with the "Distances between vehicles - 74" line. From that point draw a horizontal line to the intersection with the "Travel speed - 10" line. From this point drop a perpendicular line to the "Duration of surmounting obstacle" scale at which point the result shows 24 minutes.

Example calculation variant B: Determine the number of vehicles able to cross an obstacle within 30 min, if the allowable movement speed is not more than 15 km per hr and the distance between vehicles is 100 m.

Solution: Start at 30 on "Duration of surmounting obstacle" scale, move vertically to "15 km per hr on speed" scale, then horizontally to "Distance between vehicles 1-00 m" and down to "Number of vehicles" scale where the result shows 75 vehicles.

Example calculation variant C: Calculate the distance between vehicles in a column of 80 vehicles in order that the column crosses a bridge within 36 min at rate not more than 10 kph.

Solution: Starting at 80 on the "Number of vehicles" scale and at 35 min on the "Duration of surmounting obstacle" scale draw perpendicular lines. From the intersection of the perpendicular with the "Speed of movement -10" scale draw a horizontal line to intersect with the first perpendicular. The point of intersection is on the "distance between vehicles- 75" line. This means that the distance between vehicles must be no more than 75 meters.

Example using nomogram Figure 91
Example calculation variant A: Determine the time required to pass through a damaged strip of road, if the length of the sector is 5.5 km, the length of the column is 2.5 km, and the average speed while crossing the sector is 15 kph.
Solution: Mark on the "Depth of column" scale at 2.5 and the "Length of sector" scale at 5.5 and then draw a line through these points to the intersection with the Y or vertical axis. From this point draw a horizontal line to the "Speed of movement - 15" line and draw a perpendicular down to the "Time of surmounting" line to read the result of 32 min.

Example calculation variant B: Determine what length of column can negotiate a pass 2.5 km long at speed of 8 km per hr in a given time.
Solution: From the 45 mark on the "Time of surmounting" scale draw a perpendicular to the intersection with the "Speed of movement- 8" line. From this point draw a horizontal line to the Y axis. Connect this point with the 2.5 mark on the "Length of sector" line and continue it to intersect with the "Depth of column" line. This shows the result is 3.5 km. This means that a column of 3.5 km length may negotiate the passage in the given time.


 
 

Figure 90 nomogram to calculate time required to pass narrow spot on route

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Figure 91Nomogram to Calculate Time Required to Pass a Difficult Section of Route


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(10) Calculation for Passage Times Across Start Point )SL( by the Head and Tail of the Column

This calculation also determines the time for a column to pass a given point, however, since there is no delay as with an obstacle, the time interval is governed by the length of the column and its velocity. Since we are not interested in the length of time the column requires, but the clock times the head and tail cross, the equations yield time in military time.

The formula is:

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where:
ti=SL crossing time of the head of the march column i in hours and minutes;
Ti-1=the time for passing the line by the tail of the leading march column in hours and minutes; for the head of the first column this time is the time specified for passing the line by the head of the entire formation, for instance, by the column of the leading unit on a route;
di=the specified distance between the lead and the i march column in km;
06=factor for converting hours into minutes;
V=average speed in km per hr;
t'i=the time for passing the line by the tail of the i route column in hours and minutes;
Di=depth of the i column in km.

Example: to determine the passage time of the starting line or other regulation point by the head and tail of the third column in a formation, when the passage time of the cited point by the tail of the previous column is 21:15, the established distance between the columns in 1.5 km, the depth of the column is 1.8 km and the movement speed is 25 km per hr.

solution:
t3=21:15 + ])1.5( x )60({ ÷ 25=21:15 + 0.04=21:19;
t'3=21:19=1.8 x 60 ÷ 25=21:19=0.04=21:23;
this means the third column in the march formation will pass the regulation point with its lead at 21:19 and its tail at 21:23 hrs.

Example calculation using the nomogram Figure 92: The nomogram may be used to speed up the calculation of the passage of a line by the head and tail of the column. To determine the passage time of an initial line by the head and tail of a march column 7 km line with the condition that the time for crossing the line by the tail of the lead column is 20:20, the distance between the columns is 5.5 km and the travel speed is 25 km per hr.
Solution: Draw a perpendicular line up from the horizontal axis, "Depth of column or distance between columns" scale from the 5.5 mark to the intersection with the "Average speed of columns -25" line. From this point draw horizontal line to the "Time of passing point" scale and read the result=13.3 or approximately 13 minutes. This is the time in which the head of the stated column must pass the point after it is passed by the previous column )at 20:00(. The time for passing a point by the tail of a stated column is solved in a similar manner. For this draw a perpendicular line from the 7 mark on the "Depth of column axis" to the intersection with the "Average travel speed line - 25". From this point draw a horizontal line to the "Time of passing point" scale and read the result of 17 minutes. This means that this column must pass the control point with its tail at 20:30.

 
 

Figure 92 - Nomogram to Calculate Time Head and Tail of Column will Pass Regulation Point


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(11) Calculation of Expected Time and Distance of Probable Point of Contact with Advancing Enemy

Clearly this is one of the most important calculations Soviet commander's make regularly during the course of combat. As the discussion of meeting engagements in Chapters One and Two indicates, the commander's effort to control the flow of battle focuses heavily on the relative times and place of introduction of his second echelon versus the enemy's reserves, both of which are moving forward.

The formula is:

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where:

te=expected tome of meeting in hours;
D=initial distance between opposing groupings in km;
Vf=movement rate of friendly troops in km per hr;
Ve=movement rate of enemy forces in km per hr;
dl=distance from friendly initial position to expected point of contact in km;


Example calculation: Determine the expected time of meeting and distance to probable encounter line when enemy is located 63 km away, his average forward speed is 25 km per hr, and average speed of friendly troops is 20 km per hr.
Solution: the equation yields distance of 28 km and time of 1 hr and 24 min.


Example using nomogram Figure 93: Determine the expected time of meeting and distance to line of contact with enemy if at 18:00 the advancing enemy is located at a distance of 64 km, his average speed is 15 km per hr, and friendly troops are moving at 20 km per hr.

To use nomogram )variant a( find the marks "20" and "15" on the "Speed of movement of own forces" and "Speed of movement of enemy" lines respectively, draw a line through these points to intersection with horizontal scale and read mark of 35. Then move downward along the line shown by the dots to the intersection with the perpendicular established from the "64" mark on the "Distance between our own and enemy forces" scale. From this point of intersection draw a horizontal line to the "Anticipated time of meeting" scale and read the calculation result of 1 hr and 50 min. This will be the length of time from the start time to meet the enemy. That means 18:00 plus 1:50 gives 19:50.

To determine the distance of the probable meeting line with the enemy (variant b) from the result of 1 hr 50 min, draw a horizontal line to the intersection with the speed line (ie at mark 20), which corresponds to the travel speed of the friendly troops. From the obtained point, drop a perpendicular line to the "Distance between friendly and enemy forces" scale and read the calculation result of 36 km.

 
 

Figure 93 - Nomogram to Calculate Time and Disance to Point of Meeting Engagement


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(12) Calculation of Time Required for Advancing and Deploying Sub-units to Change From Line of March into the Attack

Determination of the time a unit should begin to move for the advance from its assembly area to the line of commitment into battle and assault on the enemy position is a complex application of the basic time and distance formula. All times are measured backwards from "CHE" hour, the moment the troops hit the first line of the defending enemy's position. The total time from beginning of movement in the assembly area is composed of the segments of time while moving in each type of deployment, that is: line of attack, company column, battalion column, and regimental column. It also includes the time it takes to shift from one formation to the other and any time for halts and delays en route. This is one of the most important and fundamental of tactical calculations. The times for sub-unit movement are tied exactly into the times for the artillery preparatory fire and air strikes.

The required given data are the distances between each of the deployment and regulating lines, distance of the attack line from the enemy's forward line of defense, distance of the start line from the unit assembly area, the average speed of movement while mounted in the columns, the coefficient for speed reduction during deployment actions, the speed of movement in attack formation, the depth of the columns, and distance between first and second echelons of the units. The set of formulas are:

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where:
ta=time for crossing final deployment into line of attack - "Che" hour -in minutes;
Da=distance of line for going into attack formation from the forward edge of the enemy position in km;
Va=rate of movement in attack formation in kph;
tr=time for crossing the line of deployment into company columns at )"Che" - minutes(
Dr=distance of line of deployment into company columns from the line of deployment into attack formation in km;
V=average rate of movement of subunits in mounted formation during march;
tb=time for crossing line of deployment into battalion columns at )"Che" - minutes(;
Db=distance of line for deployment into battalion columns from line of deployment into company columns in km;
trr=time for crossing the last regulation line or point prior to the deployment into battalion columns;
Drr=distance of last regulation line from the line for deployment into battalion columns;
ti=time for passing the start line at )"Che" - minutes(
Di=distance of start line from last regulating line in km;
tvit=time to begin movement of subunits in assembly area;
D=distance of start line from the last unit assembly area;
t'i=time for crossing the start line by the 2nd echelon of units at )"Che" - minutes(;
Gk=depth of the mounted column of first echelon of units in km;
dk=distance between the tail of the first echelon and head of the 2nd echelon of the units in km;
The numbers 60 and 90 are coefficients for conversion of time in minutes for the average speed of movement during the deployment of the units at each of the lines. For example, the 90 shows that the average maneuver speed of the unit during the actual deployment on a line decreases by a factor of 1.5 in comparison with the speed of forward movement between the lines. Depending on the concrete conditions and situation for movement on the terrain, this could be some other factor.

To perform the calculation the required data may be entered into the table. The result will be the planning data for start of forward movement and deployment times for each shift of sub-unit columns. Since all times are measured backward from "Che", the planner must remember to subtract the values indicated in lines 12, 15, 18, 21, 23, and 27 from Che; and add the value for time in line 30 to the time in 24 to obtain line 31.



 
 

Figure 94 - Calculation of time to advance and Deploy into the Attack from Line of March

CALCULATION OF TIME TO ADVANCE AND DEPLOY SUB-UNITS FOR SHIFT INTO ATTACK FROM LINE OF MARCH

No Initial data, values and calculations to perform Units Precision Calculation variant
1 2 3
1 2 3 4 5 6
1 Distance of attack line from enemy front line
2 Distance of line to deploy into company columns from attack line
3 Distance of line to deploy into battalion columns from line of company columns
4 Distance of regulation line from line to deploy into battalion columns
5 Distance of regulation line from start line
6 Distance of start line from the unit assembly )FUP( area
7 Movement speed into the attack
8 Average speed during march
9 Dept of first echelon march column
10 Interval between first and second echelon columns
11 )1( x 60
12 Time to cross line for shifting to attack )Che - min(

)11( ÷ )7(

13 )2( x 90
14 )13( ÷ )8(
15 Time to cross line of deployment into company columns: )Ch - ( )12( + )14(
16 )3( x 60
17 )16( ÷ )8(
18 Time to cross line of deployment into battalion columns: )Ch - ( )15( + )17(
19 )4( x 60
20 )19 ( ÷ )8(
21 Time to cross regulation line )Ch - ( )18( + 20(
22 )5( x 06
23 )22( ÷ )8(
24 Time to cross start line )Ch - ( )21( + )23(
25 )6( x 90
26 )25( ÷ )8(
27 Time to start moving out of assembly area - 1st echelon )Ch - ( )24( + )26(
28 )9( + )10(
29 )28( x 90
30 )29( ÷ )8(
31 Time for second echelon to pass start line )SL( )Ch - ( )24( - )30(

 
 

(13) Calculation of the Time and Distance to the Line of Contact


This method takes into consideration the many situational factors that are ignored in the simpler formula and nomogram. In fact, there are so many possible influences on the time the two sides will meet and hence the location of meeting that a nomogram can only give a crude approximation of the answer. With the use of computers or even hand calculators and an established procedure such as that shown in this table it is possible to assess the influence of many more factors.
The set of formulas is:
tv=}D + ])tn x Vn( + )tp x Vp([{ ÷ )Vn + Vp(
Tn=t1 + t2 t3; TP=t1' + t2' + t3'
where:
tv=expected time if contact with enemy in hours;
D=distance between forces of the two sides in km;
tn=total delay time for own force in hrs;
Vn=speed movement of own forces in km per hr;
tp=total delay time of enemy in hrs;
Vp=speed movement of enemy in km per hr;
lp=distance to expected line of meeting with enemy;
t1 )t'1(=delay - )time difference( start of movement of one side versus other in hrs;
t2 )t'2=duration of halts of forces of each side in hrs;
t3 )t'3(=duration of delay of forces due to strikes by opponent en route in hrs;



Astronomical time of meeting enemy depends on complex conditions. Determine it by adding the relative time obtained as a result of the calculation with the astronomical time of the beginning of the approach of the opposing forces.

For example if the time to start march of own forces is 9PM the starting time for the enemy is 10PM then astronomical time to start the approach is 9PM. If the calculated time for meeting is about 2.5 hours then the time for meeting would be 11:30.

Example problem: determine the expected time of meeting and the distance to likely line of contact with the enemy and the duration of movement to that line under following conditions:
- start time of own forces - 20:00 hrs;
- start time of enemy forces - 21:00 hrs;
- distance to enemy - 105 km;

The commander decides there will be a break of 20 minutes ).3( hr during the advance. The plan is to delay enemy forces 30 -40 minutes ).6( hr. It is assumed that during movement enemy will be required to take halts of 30 minutes )0.5( hrs. The speed of movement of own forces is 28 km per hr. The speed of movement of the enemy is 19 km per hr.

The answer is that contact will be at 11:20 at a distance of 84 km. Duration of movement to meeting line is 3 hrs.



 
 

Figure 95 - Form for Calculation of Expected time and location of Meeting Engagement

CALCULATION OF EXPECTED TIME AND DISTANCE TO PROBABLY LINE OF MEETING ENGAGEMENT

No Initial data, values and calculations Units Precision Calculation variant
Example 2 3
1 2 3 4 5 6
1 Friendly force starts advance hr, min 20:00
2 Enemy force starts advance hr, min 21:00
3 Distance between opponents at start km )1.0( 105
4 Delay in advance of friendly relative to enemy hr )0.1( if 1 < 2 --
5 Total duration of halts of friendly hr )0.1( 0.3
6 Total duration of delays of friendly by enemy hr )0.1( --
7 Friendly force travel speed km/hr )1.0( 28
8 Delay of enemy start relative to friendly hr )0.1( if 2 < 1 1
9 Total duration of halts of enemy hr )0.1( 0.5
10 Total duration of delays of enemy due to friendly hr )0.1( 0.6
11 Enemy force travel speed km/hr )1.0( 19
12 )4( + )5( + )6( )0.1( 0.3
13 )12( x )7( )0.1( 8.4
14 )8( + )9( + )10( )1.0( 2.1
15 )14( x )11( )1.0( 40
16 )3( + )13( + )15( )1.0( 153
17 )7( + )11( )0.1( 47
18 Expected time of meeting )relative( )11( ÷ )17( hr )0.1( 3.3
19 Duration of travel time to meeting line: )18( - )12( hr )0.1( 3
20 Distance to meeting line: )19( x )7( km )1.0( 84

 
 

(14) Calculations of Expected Time and Rate of Overtaking when Pursuing the Enemy

Naturally the commander hopes to use this calculation often! The initial data are the distance between the enemy and friendly forces, the average travel speed of friendly and enemy forces, or the ordered time to overtake him. The formula is:

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where:

to=time to overtake enemy in hours;
D=distance to the enemy in km;
Vn=friendly speed in pursuit in km per hr;
Vp=enemy speed in retreat in km per hr;


Example: Determine how much time it will take for forces to overtake a retreating enemy when the distance to him is 20 km, his rate of retreat is 10 km per hr, and the rate of advance of friendly forces is 25 km per hr.
Solution: to=20 ÷ )25 - 10(=1.3 hrs ;

Determine the pursuit speed required to enable friendly forces to overtake enemy in 45 min, when he is 15 km away and his travel speed is 12 km per hr.

Solution: Vn=))15 + 0.75( ( 12) ( ÷0.75=33 km per hr.


Examples of calculations using the nomogram Figure 96:

Determine the expected time to overtake enemy when his distance is 30 km, his travel speed is 20 km per hr, and the speed of pursuit is 28 km per hr. In the calculation )variant a( establish a perpendicular line from the "30" mark on the "Distance between friendly and enemy forces" scale. Then draw a line through the "28" mark on the "Friendly forces travel speed" and the "20" mark on the "Enemy travel speed" scale to its intersection with the horizontal upper axis. From this point draw a line down as shown by the dots to the intersection with the previously set perpendicular line. From the point of intersection draw a horizontal line to the right and read off the result of 3 hrs and 45 min on the "Expected encounter time" scale.

Determine the required pursuit speed to intercept the enemy in 1 hr and 20 min, when the enemy is at a distance of 40 km and is traveling at a speed of 5 km per hr.

In calculation )variant b( establish a perpendicular line from the "40" mark on the "Distance between friendly and enemy forces" scale to the intersection with the horizontal line, drawn from the 1 hr and 20 min mark on the "Expected encounter time" scale. From the meeting point draw a line, as shown by the dots and dashes, to the horizontal scale. Draw a line through the point of intersection of this scale and the 5 mark on the "Enemy travel speed" scale and continue it to the intersection with the "Friendly force travel speed" scale, where the result is 35 km per hr. This means that the pursuit speed must he at least 35 km per hr.


 
 

Figure 96 Calculation of Expected Time and Rate of Overtaking when Pursuing the Enemy:

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(15) Calculation of the Work Time Available to the Commander and Staff for Organizing Repulse of Advancing Enemy Forces:

This is obviously a critical issue in meeting engagements and encounter battle. It is also relevant to defensive situations when preparing fire on the attacker. The method is designed to determine the time the commander and staff will have to organize the enemy's defeat by firing on the advancing forces relative to the distance to the enemy, the speed of his advance, the effective range of friendly weapons and the time required to prepare the sub-units to fire. In this example it is assumed that the enemy will be taken under fire beginning at the maximum range of the firing weapons. The formula is:

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where:

t=time available for commander and his staff to organize repelling the advancing force in minutes;
D=distance to advancing enemy km;
d=max range of friendly weapons km;
60=conversion factor hours to minutes;
Ve=rate of enemy advance kph;
tp=time required to prepare subunits to destroy enemy with fire in minutes;

Example: Determine how much time is available for the commander and staff to organize destruction of the advancing enemy if it is located at a distance of 25 km, his average speed is 15 km per hr, the effective range of the weapons is 12 km., and the time for preparing the sub-units is 03 min.

Solution: t=} (25 - 12) x 60 ÷ 15{ - 30=22 min.

Example calculation using the nomogram Figure 97: Determine the time the commander and staff spend in organizing the destruction of an advancing enemy if he is 15 km away, his rate of advance is 12 kph, the effective range of friendly weapons is 6 km, and the sub-unit preparation time is 20 min.

Solution: Mark the points "15" and "6" on the "Enemy distance" and "Effective range of friendly fire" scales, respectively. Draw a straight line through them to intersection with the vertical axis. From this point draw a horizontal line to the "Enemy advance speed - 12" line. From that point drop a perpendicular line to the horizontal axis and make a mark on it. Then draw a line through that mark to the "20" point on the "Time for preparing the sub-units for combat" scale. At the point of intersection of this line with the "Commander and staff work time" scale read the result of 25 min.



 
 

Figure 97 - Nomogram to Calculate Time Available to plan fire on advancing Enemy

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(16) Calculation of Length of Time to Operate Command Post in a Single Location

One of the important calculations performed by the operations staff is to determine the duration of time a command post can remain at one location during an offensive and still keep up with the advancing troops. Obviously, the longer the CP can remain in place the more work that can be done and less time lost to transfers, however, the CP must not lag too far behind the forward edge of troops, if the commander is to exert direct supervision and control. The required data are the established maximum and minimum distance norms of the command post from the forward edge of battle, the rate of movement for the leading troops, the movement rate for the command post, and the time required for loading up and deploying the command post for each move. The calculation would be similar for any other entity that should remain between maximum and minimum distance from another, for instance rear service installations. The formula is:

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when D < d

t=length of time available for work at the CP in one location;
D=maximum distance of CP from forward edge in km;
d=minimum distance of CP;
Vs=rate of movement of leading sub-units;
Vc=rate of movement of CP during its relocation;
60=coefficient for conversion of hr to min;
tbd=duration of loading and unloading time for CP.

Example problem: Determine the working time for a CP when maximum distance is 7 km, minimum distance is 1.5 km, rate of movement for units is 4 kph, rate of movement for CP is 25 kph, and time to load and deploy is 15 min.
Answer is 54 min


 
 

Figure 98 - Calculation of Duration of Operation of CP in one Location

CALCULATION OF DURATION OF OPERATION OF CP IN ONE LOCATION

No Initial data and values to be determined Units Calculation variant Remarks
Example 2 3
1 Permissible distance of CP from front km 7
2 Minimum distance of CP from front km 1.5
3 Sub-unit rate of advance km/hr 4
4 Command post travel rate km/hr 25
5 Time to deploy and break down CP min 15
6 Duration of operation in one location min 54

 

(17) Determination of Quantity of Various Weapons, Reconnaissance, Support, Communications etc. for Task Performance

This general technique is applicable to a wide variety of activities in which it is desired to know what the overall effectiveness of a group of agents will be when the probability of success of an individual agent is known. First one calculates the action of the number of identical equipments, and then determines the total effectiveness or total requirement for the systems. the initial data for the calculations are information about the number of available systems, the assigned degree of success for each, the effectiveness of the systems )which is expressed in probability of mission fulfillment or by the mean value of the applied damage to a particular target.( A single system also means a complex of systems combined into a whole unit. Such data, for instance, are the probability of target destruction, the man damage applied to an enemy target, the reliability of a communications channel, the probability of enemy target detection, the probability of the flawless operation of a water crossing for a specific time interval, the probability of overcoming the anti-air forces of the enemy and so on. These data may be obtained on the basis of the results of studies, from statistical data nd from tactical and technical characteristics.

The formulas for calculating the degree of mission fulfillment by a designated number of systems is expressed through mission fulfillment probability:{short description of image}







and through the mathematical expectation )mean value( of damage:
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where:
Pn=probability of mission accomplished by a homogenous group of weapons;
P1=same by a single system;
Mn=average )mean( value of damage inflicted on the enemy by a group of systems;
M1=average value of damage inflicted on the enemy by one system;
n=quantity of available systems.

The formulas for computing the required number of systems when the effectiveness of the systems is expressed by probability of task performance is:

when Pn < P1

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when it is expressed by the mathematical expectation )the mean value( of the inflicted damage, the formula is: when Mn < M1

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The formula for calculating the effectiveness of different systems fulfilling a common mission is :

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where Pnis the overall effectiveness of systems used or in other words the likelihood of fulfilling the mission, and the individual "P"'s are the effectiveness measures of each system being summed for the task.

Example: Determine probability of spotting an enemy target with combined use of 3 reconnaissance systems, if their effectiveness expressed as the probability of detection of enemy target is:

P1=0.4; P2=0.6; and P3=0.8

Solution: Pn=1 - )1 - 0.4( x )1 - 0.6( x )1 - 0.8(=0.95 or approximately 1.

Example calculations using nomogram Figure 69: Determine the probability of enemy target destruction with a strike against it by three systems when the probability of target destruction by a single system of the particular type is 0.4.

Solution: )variant a( Draw a perpendicular from the "3" mark on the "Required number of systems" scale to the intersection with the "Probability of mission fulfillment by one system - 4" curve. From this point draw a horizontal line and on the "Probability of mission fulfillment by a group of systems" scale read that the probability of target destruction by three systems is 0.78.

It is possible to determine the reliability of communications in a link which consists of two channels )n=2(, when the reliability of each channel is 0.6. From the nomogram, the probability of faultless operation of communications in this instance will be 0.84.

It is possible to evaluate the effectiveness of using four similar reconnaissance systems to detect an enemy target in an assigned region when the probability of detection by one system is 0.5. According to the nomogram, the probability of target detection by four systems will be close to one (.94.

Determine how many weapons systems must be assigned to inflict no less than 90% damage an enemy target, when the average damage inflicted byu a single system is 70%.

Solution: Draw a horizontal line from the "0.9" mark on the "Probability of mission fulfillment by a group of systems" scale to the intersection with the "Probability of mission fulfillment by a single system - 0.7" curve. From the intersection point drop a perpendicular line and on the "Required number of systems" scale find the result - 2. This means two systems must be used to achieve the assigned damage.

It is possible to determine the required number of communication channels to provide for 90% reliability when the reliability of a single channel is 0.6. From the nomogram it appears that at least three channels are required to ensure this reliability.

From these examples it is evident that the technique may be used for calculating a wide range of direct and inverse problems associated with the use of various forces and means. With proper additional factors this method may also be used to calculate rapidly the effectiveness and required number of forces and systems when taking into account also probable enemy countermeasures.

The formulas for calculating the effectiveness of forces and means with consideration for probable enemy countermeasures are:

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where:
Pn=the probability of mission fulfillment by a group of systems;
P1=the probability of mission fulfillment by a single system of this type;
Q=the probability indicator of the enemy countermeasure ;
n=the number of forces and means of the given type;
Mn=the mathematical expectation of the damage inflicted by a single system of this type'

Example calculation: To determine the probability of mission fulfillment by five weapons systems when the probability of target destruction by a single system is 60%, while the probability of successful enemy countermeasures is 50%.

Solution: Pn=1 - )1 - 0.6 x 0.5(5=1 - )1 - 0.3(5=0.83.


 
 

Figure 99 - Nomogram for Calculating the Combined Effectiveness of several Systems

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(18) Modeling battle(4)

There is much discussion in the Soviet literature about the theory of development of coefficients of comensurability. The coefficient is designed to make it possible to aggregate the effective contribution of various weapons to a combined arms battle into a composite score for each side. This makes the measurement of the correlation of forces relatively simple, ie. only one number for each side rather than the long list of individual types of weapons found in the standard tables of correlation of forces. However, what sounds simple in theory is quite difficult in practice. Soviet authors point out that even within types of weapons such as tanks or artillery the "proving ground scores" assigned on the basis of technical/tactical characteristics may not be true reflections of the actual value of the weapon to the commander in all the diverse conditions of real combat. When it comes to establishing a uniform score that will place tanks, artillery, aircraft, and small arms, etc on one scale; a lot of judgment is involved. There is some argument for two different methods, one would develop a "average" score for each weapon reflecting its varied value in different circumstances, and the other method would establish a single theoretical value number and then provide situation modifiers to be applied to account for each actual set of combat conditions. We do not have available an actual set of Soviet "utils" as they call their weapons scores. The following table is taken from a Soviet article, but the scores are purely hypothetical for educational purposes. Nevertheless, the following method represents a typical Soviet approach which may be used by American officers with an appropriate set of "utils".



 
 

Figure 100 - Coefficients of Comensurability

COEFFICIENTS OF COMENSURABILITY

Nationality Type of combat equipment Coefficient of comensurability
Country A Tank 5 1.0
Tank 6 1.12
Tank 7 1.5
APC 1.6
AT gun 0.3
Country B Tank 06 1.02
APC 1.4
PTRK 0.95
AT gun 0.3
Country V Tank 1 1.09
APC 0.45
PTRK 0.78
RPG 0.12

 
 

Figure 101 - Expected Losses

EXPECTED LOSS IN COMBAT EFFECTIVENESS

Nationality of forces Defense Offense Meeting battle
Prepared Non-prepared On prepared line On unprepared line
Country B 55 45 30 35 40
Country V 60 50 35 40 45

 
 

The mathematical expectation of the level of destruction of a side from fire and strikes of artillery and aviation )M( in the experience of the Great Fatherland War could reach on defense to 04-06% and on the offense to 02-03%. The losses of combat effectiveness due to destruction in various types of combat is shown in the table of expected losses. Another Soviet source gives 03% loss as the critical break point for attackers and 04% loss as the typical break point for defenders. Of course the effect of losses on unit combat effectiveness depends heavily on the rate of loss and the size of the unit. One more indicator is the coefficient of superiority )Kp( of the defenders over the attackers, which in the experience of combat typically is three times.

The possibility of sub-units for destruction of the enemy during the course of accomplishing their combat missions is calculated according to the following formulas:

 
 

a( in offense:

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b( in defense:

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where:
Ks1, Ks2, Ksp=coefficients of comensurability of combat means of units of country A;
Kb1, Kb2, Kbp=coefficients of comensurability of combat means of units of country B or V;
i=quantity of combat means of given type;
Z1, Z2=level of destruction at which the unit loses combat effectiveness;
M=mathematical expectation of the level of destruction due to artillery and aviation fire;
Kp=coefficient of superiority of defender over attacker;
S=combat capability of the sub-unit.

For calculations of combat capability during meeting engagement )battle( use formula )1( but for the coefficient of superiority of defender over attacker use 1.


Example calculation: The tactical situation; a tank platoon of country A )with tanks 7( with motor rifle section in a APC has the mission of attacking from the march to destroy defending section of country B.
Initial data for the calculation:
- a. type of combat action is attack;
- b. model of combat means of the sides, their quantity and coefficient of comensurability calculated qualitative indices, are shown in table of comensurability.

The attackers:
- tanks "7" Ks1=1.5, and quantity, i=3;
- APC, Ks2=0.8, and quantity, i=1.

Defenders:
- APC, Kb1=1.4, and quantity i=1;
- AT gun, Kb2=0.3, and quantity i=3;

c. the mathematical expectation of the level of destruction of the defenders in the time of artillery preparatory fire and attack support fire is M=0.4;

d. losses of the sub-unit during the time of approach to the line of going over to the attack must not exceed Z1=0.3;

e. level of destruction of the defenders at which they loose combat effectiveness is found in the table of expected loss of effectiveness, Table 2, Z2=0.55;

Taking the initial data and using formula 1 to create a mathematical model of the battle gives the following:
- Sn=}])1.5 x 3( + )0.8 x 1([ x ]1 - )0.3 + 0.55 - 0.4([{ ÷ }3 ])1.4 x 1( + )0.3 x 3([ x )1 - 0.4(=0.7


Analysis: If Sn is greater than or equal to 1, then the sub-unit will fulfill its mission, if )as in this example( the Sn is less than 1, then the defender with his forces will be superior. This means that it is necessary to raise the capability if the force to achieve likelihood of accomplishing the mission.

Decision: Reenforce by fire the platoon on the march from a range of 0002 meters to destroy the enemy APC, and after that conduct fire on the AT gun.

The problem may be decided by means of mathematical modeling of battle:
- Sn1=}])1.5 x 3( + )0.8 x 1([ x )0.3 + 0.55 - 0.4({ ÷ )3 x 1.4 x )1 - 0.4((=1.19;

In other words, the APC will be destroyed.
- Sn2=})1.5 x 3 + 0.8 x 1( x )1 - )0.3 + 0.55 - 0.4({ ÷ }3 x 0.3 x 3 x )1 - 0.4({=1.9.

Outcome: Attacking from the march with forces of three tanks and APC to destroy an APC and then a AT gun. The line of going over to the attack is designated in accordance with the characteristics of the terrain, but not closer than 0002 meters from the forward edge of the enemy defenses.

As is evident from the example, the conduct of the calculation is difficult work demanding a significant amount of time and corresponding conditions. If the commander has an electronic calculator he could work out the answer quickly.

What is not evident is to what level of aggregation this method and these formulas may be taken without loosing validity. The given example illustrates duels and very small unit combat. With the use of computers it would be possible to calculate combat at the level of a full regiment or even division but so many other factors enter in and effect combat outcomes at that level that a simple formula is no longer valid.


 
 

(19) Calculation of Strike Capability of Sub-units(5)

This method is designed to calculate the expected depth of penetration of a unit on the offensive as a measure of its strike capability, or, in reverse, to forecast the ability of a unit of given combat capability to achieve the depth of penetration required in its mission. The initial data for the calculation includes the composition of the attacking and defending units, the conditions of combat and combat potentials with calculated losses of the sides and also the designated critical level of losses necessary to cause unit failure.

The critical figure is considered to be the loss at which the sub-unit losses its combat capability and cannot continue the kind of action it was performing without re-enforcement. The calculation is made over the width of the front of the attacker and the designated area of the defense. See the section on modeling battle for more discussion of the theory and use of the critical level of loss as an indicator in forecasting and modeling combat. Together these formulas may be used to determine the initial data on the three measures of effectiveness described in Chapter One as those deemed critical for assessing and forecasting battle outcomes.
The formula for calculating tbe expected depth of penetration follows:

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when:
YN )1 - PN ( < )1 - PNK( and Yo )1 - Po( < PoK(
where:
NN, No=number of units of attackers and defenders, measured in combat potentials )"utils"(;
PN, Po=expected losses of the units of each side up to the arrival at the immediate mission line;
PNK, PoK=critical loss of a unit of attacker and defender;
FN=width of front of attack in km;
Fo, Go=width of front and depth of the defender's area in km.
YN, Yo=composition of the units of each side as a percentage of unity 001%.
K=coefficient of combat effectiveness of the defending side. )This is the same as Kp in the formulas for modeling battle.

Sample calculation: Determine the expected depth of penetration of attacking unit whose combat potential at 100% measures 150, during an attack on a enemy having a combat potential of 194 and a manning of 85%, if the attackers losses in the immediate battle reach 10% and the defender's losses reach 40%. The critical loss for the attacker is 50% and for the defender is 70%. The attack is conducted on a front of 5 km. The defending unit occupies an area of 8 km wide and 10 km deep, and the coefficient of effectiveness of the defender is 2.5.

Solution:

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This should be 150 (1.0 (1-0.1) - (1 - 0.5)) x 8 x 10
over 194 (0,85 (1 -.04) - (1 - 0.7) ) x 5 x 25=9.5


This means the attacking unit can be expected to succeed in reaching a depth of 9.5 km in the defender's position. The Soviet author does not indicate the scale of this combat, and in fact uses the term for sub-unit )ie. a battalion or smaller(, but from the dimensions of the combat area )8 km by 01 km( we may assume this is a regiment attacking of a brigade frontage to the full depth.

If we use the same formula to evaluate battle at division level we may assume "utils" of NH=800 for attacker and NO=700 for defender and the following other values for variables:
YN=90%; YO=80%; PN=0.1%; PO=0.4%; K=1.5;
PNK=0.5%; POK=0.7%; FO=20 km; GO=30 km; FN=15km;

The resulting equation is as follows:

{short description of image}







Again, the numbers should be 800, 20, 30,700, and 15 - the computer reverses many numbers

With these assumptions the attackers may be expected to achieve a depth of penetration of 53 km.

Using the same formula to evaluate battalion scale combat we may assume both sides have "utils" of 40, the defender has a superiority coefficient of 2.5 and the widths are 4, 2, and 1.5 km. Then the formula yields the result as a penetration of 3.2 km. These are not unreasonable depths for the types of units being considered.



 
 

(20 ) Calculation of the Width of Main Attack Sector

There are several different ways to determine the possible, desired, or required width of the main attack sector depending on the purposes and criteria. The following is one quick way to calculate the relationship between the width of a strike sector and the total width of zone in relation to the correlation of forces in the strike sector and the total correlation. For these calculations the correlation of forces may be calculated in terms of "utils" or in terms of some other aggregate measure based on the quantities of forces and means of the two sides. The following formula applies:

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where:
Wm=width of strike sector;
Wo=total width of zone;
Co=overall correlation of forces;
Cm=correlation of forces in strike sector;
Cs=minimum desired correlation of forces in rest of zone excluding strike sector.

Example calculation: Given the width of full zone of action is 120 km, the overall correlation of forces is 1:1, the correlation of forces in the strike sector is to be 3:1, and the minimum allowable correlation of forces in the rest of the area is 0.5:1; determine the possible width of the strike sector.

Solution: Wm=120 )1 - 0.5( ÷ )3 - 0.5(=24 km.

Example calculation: Determine the correlation of forces possible in the strike sector given that the total width of zone is 400 km, the overall correlation of forces is 0.8:1, width of strike sector is desired to be 120 km, and the minimum correlation of forces allowable in the rest of the area is 0.5:1.

Solution: The formula may be restated as Cm=Wo ÷ Wm )Co - Cs( + Cs ; inserting the given data produces a result of 1.5 as the possible correlation of forces for the strike sector. The commander may then decide on several alternatives such as narrow the strike sector, weaken the non-strike sector even more, bring in reinforcements, or weaken the defenders with preliminary artillery and air bombardments. Having done the initial calculation the commander may use the same formula in various ways while deciding on alternatives. For instance, to determine the required reenforcement to obtain a desired higher correlation the relationship is C=) 1.25 - 0.8( ÷ 0.8 x 100=56.25%, that is by reinforcing by 56% the correlation in the overall area may be raised from 0.8 to 1.25 to 1.



 
 

(21) Calculation of Required Destruction of Enemy

Another calculation related to the preceding is to determine what damage should be inflicted on the defender prior to the attack in order to bring the correlation of forces into the desired range, considering that there will be a given amount of loss on the friendly side as well. The formula is as follows:

100 not 001
{short description of image}






where:
M=required level of destruction of enemy in percent;
CN=beginning correlation of forces;
CT=required correlation of forces;
P=forecast friendly losses from enemy action in percent.

Example calculation: In the break through sector regrouping of forces may bring the correlation of forces to 2:1, however for successful break through we need to bring the correlation up to 4:1 by means of initial artillery and air bombardment inflicting losses on the defenders. Determine what level of damage must be inflicted, if it is expected that the enemy will also inflict 30% loss on the attacking strike grouping.

Solution: M=100 - 2 ÷ 4 ) 100 - 30 (=65%. Assume that the initial "utils" of the attackers are 2000 and the defenders are 1000 for a correlation of 2:1. The enemy will inflict 30% losses bringing the attacking force to 1400 while the attackers must inflict 65% losses to bring the defender to 350 "utils" to achieve the 1400/350=4:1 ratio. For more rapid calculation of this relationship to determine the required losses to inflict one may use the following nomogram )Figure (. The nomogram is based on the above formula. The example shown is for the same situation. Enter with the correlation of forces 2 on the bottom left and draw vertical line to the correlation of forces 4 line, then a horizontal line across to the 30% loss line and again vertical down to read the required enemy loss of 65%.

 
 

Figure102 - Nomogram to Determine Required Losses to Achieve Correlation of Forces


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(22) Calculation of Rate of Advance in Relation to Correlation of Forces(6)

Various Soviet articles have discussed the concept of the relation of advance rates to the correlation of forces of the sides. The relationship is complex and also depends on several other variables. Evaluation and analysis of the actual correlations of forces achieved in various operations of World War II and recent local wars shows that a single norm for the correlation of forces and means does not exist, but it all depends on the concrete conditions. Nevertheless sufficient figures to make approximate calculations can be developed, especially for academic and training purposes. The actual rate of advance, V, in km per day may be related to a coefficient reflecting the influence of the correlation of forces, KC, in an equation such as V=140 x KC. In this the number 140 represents the collected resulting effect of variables acting in the area, for instance the Western TVD to give a practical maximum technically possible speed of movement for mechanized forces in km per day and also an indicator characterized by the influence of terrain, engineer obstacles, time of year, length of daylight, etc. The factor KC is a coefficient depending on the correlation of forces of the sides. This approximate relationship can be shown in a graph. Analogous graphs can be made for other theaters and conditions. The formula together with the graph is used to evaluate the formation and determine the correlation of forces required to achieve a required rate of advance or in reverse the expected rate of advance for a given correlation of forces.

Example calculation: A given strike group has a planned average rate of advance of 40 km per day. Determine what the required correlation of forces superiority is to achieve this rate.

Solution: Taking the formula V ÷ 140=KC or 40 ÷ 140=0.29. Enter 0.29 on the scale for KC in the nomogram and draw a horizontal line to the curve and then a vertical line down to the scale of correlation of forces scale. This shows a required correlation of about 3.4:1 for the strike sector.

Example calculation: In the zone of action of the strike group we have created a 2.5:1 superiority in correlation of forces. Determine the approximate tempo of advance.

Solution: From the nomogram read that the correlation of 2.5:1 corresponds to a K factor of 0.13. Then using the formula V=140 x 0.13=18.2 km per day advance rate.

Another approach is to use the following formula to calculate a factor corresponding to the K factor in the previous method. In the following nomogram the rate of advance is related also to movement distance, terrain type, length of the operation and a theoretical maximum movement speed. These variables are joined into a single factor according to the equation: F=D ÷ KTVmax where:
D=distance )depth( of operation;
K=terrain coefficient )1.25 - level; 1.00 - rough-level; .75 - rugged hills; .75 - urban; .50 - mtns.(
T=time required for operation in days and fractions of days;
Vmax=theoretical speed in km per day.

According to this nomogram, to achieve a depth of advance of 03 km on level terrain in one day with a theoretical maximum speed of 06 km per day requires a superiority in correlation of forces of 4:1. With a correlation of forces of 3:1 on rough-level terrain and a theoretical maximum speed of 50 km per day in two days the force may advance 20 km.



 
 

Figure 103 Nomogram relating correlation of forces to rate of advance (1)


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Figure 104 Nomogram relating correlation of forces and rate of advance (2)


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(23) Determine the Possible Friendly and Enemy Losses in Relation to the Correlation of Forces and Rate of Advance

A simplified expression of the relationship between attrition and the correlation of forces may be shown with a nomogram. In actuality many factors influence this relation, however, for general planning at front level the averages shown in the nomogram may be sufficient. This nomogram is built with the following assumptions.

1. The attacker needs a 3:1 advantage represented by the line at 45 degrees to break even and have the same loss rate as the defender.
2. The losses to each side are related to the force ratio in a relatively simple and direct way, but the greater the force ratio advantage and the longer the time interval the larger the difference in attrition will be. So losses accumulate faster for the side with the disadvantage in forces;
3. For an army size force the losses are about .5% per day when the correlation of forces in 1:1



 
 

Figure 105 Force attrition nomogram - army


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Figure 106 Force attrition nomogram front


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(24) Determine the Required Amount of Manpower and Weapons for Bringing Sub-Units back up to Sufficient Strength to Restore their Combat Capability

This calculation is made to determine how many men and weapons are needed to restore the sub-unit or unit to combat capability sufficient to accomplish the mission. It is not expected that the replacements will restore the sub-unit to its original full strength, but just to the necessary ratio of correlation of forces and means between the two sides. The mission was determined in the first place on the basis of a calculated correlation of forces and means deemed sufficient )such as 3:1(. Since both sides have suffered losses the quantity required to restore the original favorable correlation will not equal the loss.

The basic data needed for the calculation are information on the original full strength of the troops of the two sides, the losses sustained and the superiority ratio of correlation of forces and means required for carrying out the task. The results of the calculation represent the quantities necessary to restore the force to its combat capability. Combat capability in this case means sufficient strength to accomplish the mission. Obviously this is one of the most important calculations Soviet commanders must make during combat. The formulas for the calculation are:

Mi=Ci - Ai ÷ Bi ; Ai=)nf Pf( x Nf ;


Bi=)ne -Ne Pe( x Ne ; mi=Bi x Mi

where.
M=index of the reduction of combat capability in terms of the required ratio of manpower and weapons for the execution of the mission;
Ci=required degree of superiority ratio for a particular type of the i type of weapon;
Ai=availability remaining on hand quantity of weapons of a particular type i of own forces;
Bi=availability of remaining number of weapons of a particular type i for enemy forces;
nf )ne(=initial full strength of own troops )enemy troops( by type i of weapons pers/ proportions;
Pf )Pe (=losses of own troops )enemy troops( pers/proportions;
Nf )Ne (=TOE number of weapons of a particular type i of own forces )enemy forces(;
mi=number of weapons of a particular type which are required for the restoration of own troops combat capability.
If M is less than 0 then the calculation is no longer performed since the filling out of the forces and means is not required to restore combat capability. This method may be used both in calculating the number of various types of combat equipment, as well as in calculating the correlated quantity/quality indicators.

What these formulas mean essentially is that M is an index showing the reduced combat capability of friendly forces in terms of the change in the correlation of forces resulting from the differential losses by the two sides. It is equal to C )the required correlation of forces ( minus the actual ratio )correlation( of forces.

The term "A" is the remaining quantity of forces or means of the particular type for friendly forces and it is equal to the initial strength minus the product of the friendly initial strength times the losses to friendly forces all times the TOE strength.

The term "B" is the same thing for the enemy forces.

Then m is the quantity of the forces or means required to restore the initial favorable )required( correlation and it is equal to the remaining strength of the enemy )B( times the M index of reduced combat capability )deficit in correlation(

Example calculation using the table: Determine the required number of tanks and guns to re-equip a sub-unit if the required level of superiority is 2.5:1, the initial full strength of own troops was 88% )0.85(, the losses of own forces in tanks is 40% )0.4(, the initial full strength of the enemy was 60% )0.6(, his losses in tanks are 40% )0.4(.

Solution: Entering the data in the table we find that to resupply the sub-unit and restore its combat capability requires 2 tanks. Replacements of guns are not required since, given the losses on both sides, the required level of superiority necessary to fulfill the mission is preserved.


 
 

Figure 107 Form for the Calculation of the Required Amount of Personnel

CALCULATION OF FORCES AND MEANS TO RESTORE COMBAT EFFECTIVENESS
No Initial data, values and operations to be determined Unit (Accuracy) Forces and equipment
Tanks Guns Mortars BTR Etc
1 Initial full strength of friendly forces percent 0.85 0.7
2 Friendly forces losses percent 0.4 0.2
3 Tabular volume of friendly force equipment units 25 32
4 )1( x )2( )0.01( 0.34 0.14
5 )1( - )4( )0.01( 0.51 0.56
6 )3( x )5( )1.0( 13 18
7 Initial full strength of enemy forces percent 0.6 0.6
8 Enemy forces losses percent 0.4 0.3
9 Tabular number of enemy force equipment units 18 25
10 )7( x )8( )0.01( 0.24 0.18
11 )7( - )10( )0.01( 0.36 0.42
12 )9( x )11( )1.0( 6 10
13 Established degree of superiority of friendly forces )proceeding from the ratio of forces and equip( 2.5 1.5
14 )6( ÷ )12( )0.01( 2.17 1.8
15 )13( - )14(* )0.01( 0.33 -0.3
16 Number of equipment required to fill out: )12( x )15( units )1.0( 2 -
* At )15( is less than 0, there is no need to fill out the forces and equipment, the subunit is combat ready and may successfully fulfill the established mission.

 
 

(25) Determine the Expected Radiation Dose

The initial data for this calculation are the length of travel route within the zone of radioactive contamination, the average radiation level on the travel route and the speed and direction of travel of the sub-units. The degree of shielding of the personnel is assigned by a coefficient of radiation reduction by the transport equipment. The formula is:

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where:
D=the expected radiation dose of personnel in roentgens;
Li=the length of the route within the i zone of radioactive contamination in km;
Ni=factor which considers the direction of zone relative to the axis of the radioactive pattern. )If travel is along the axis of the pattern N equals 1, and if perpendicular to the axis N equals 0.52, and at an angle to the axis N equals 0.375(;
Ri=mean radiation level on the travel route in roentgens per hour;
Vi=the speed of travel in km per hr;
K=the coefficient of reduction of radiation by the protection of the transport equipment;
n=the number of zones to be crossed.

Example calculation using the form: Determine the possible radiation doses of personnel of a subunit while they are negotiating two zones of radioactive contamination with mean radiation levels of 187 and 165 roentgens per hour. The length of the route within the first zone is 12 km and the second is 10 km. The travel speed of the sub-unit is 25 and 18 km per hr respectively. The first zone is crossed along a route perpendicular to the pattern axis and the second at an angle to the pattern axis. The reduction coefficient is 4.

Solution: performing the calculations using the table we find that the total radiation dose received by personnel may reach 15 roentgens.

The formula and method for calculation may be simplified by considering the entire zone as a whole and using average values for the above variables. In this case the expected dose of radiation may be expressed as D=)R x L( ÷ )K x V(;
and the required speed of movement is V=)R x L( ÷ )K x D(;
This relationship is shown on the following nomogram.

Example calculations using the formulas: Determine the anticipated dose of radiation of personnel crossing a sector of radio-active contamination with a level of 78 R/hr, if the length of route is 18 km, the sub-unit speed is 25 km/hr, and the coefficient of protection is 7. The solution is 8 roentgens.

Determine what is the required speed to negotiate a contaminated zone 17 km deep when the mean radiation level is 95 R/hr, the protection coefficient is 4 and the allowable radiation level for personnel is less than 20 R. Solution is approximately 20 km/hr.

Example calculations using the nomogram: Determine the expected personnel radiation dose from passing through a sector of radio-active contamination of 60 R/hr, when the length of route is 12 km, the unit travel speed is 15 km/hr and the protection coefficient is 2. Solution: )Variant A( From the "12" mark on the "Length of sector of contaminated route" scale draw a perpendicular line to the intersection with the "Average level of radiation - 60" line. From that point draw a horizontal line through the obtained point to the intersection with the "Speed of movement 15" line and then draw a vertical line to the "Coefficient of degradation 2" line." From that point draw a horizontal line to the "Dose of radiation of personnel scale, on which read the result=24 roentgens.

)Variant B( is to calculate the speed for negotiating a contaminated zone so that the personnel will receive less than 48 roentgens when the route length is 18 km, the radiation level is 80 R/hr, and the coefficient for protection is 2. Draw a perpendicular line from the 18 mark on the "Length of sector" line to intersect with the "Average level of radiation - 80" line and draw a horizontal line to the left. Then draw a horizontal line from the "48" mark on the "Dose of radiation of personnel" line to the intersection with the "Coefficient of degradation - 2" line and from that point draw a vertical line to intersect with the previously drawn horizontal line. The intersection point on the "Speed of Movement" lines at the 15 line shows that 15 km/hr is the minimum acceptable speed for crossing the contamination.

Another method for calculating the expected radiation dose for personnel may be used with a nomogram by considering only averages over the entire width of the radiation zone. The initial data for this calculation are the length of the contaminated section of the route, the mean radiation level over the route, the unit travel speed during passage of the contaminated area, the degree of shielding of personnel by the vehicles. This method is used to calculate the speed required to pass the contaminated area so that personnel do not receive a larger radiation dose than allowed. The formulas are: D=RL ÷ KV; and V=RL ÷ DK
where:
D=the expected radiation dose of personnel in roentgens;
L=the length of the route in km;
R=mean radiation level on the travel route in roentgens per hour;
V=the speed of travel in km per hr;
K=the coefficient of reduction of radiation by the protection of the transport equipment;

The mean radiation level is determined on the basis of radiation reconnaissance data by averaging the readings at various points along the route.

Example calculations using the formula: Determine the expected radiation dose for personnel in crossing a radioactive zone 18 km deep when the mean radiation level is 78 roentgens per hr, the sub-unit travel speed is 25 km/hr, and the reduction coefficient is 7.

The solution is D=78 x 18 ÷ 7 x 25=8 roentgens.

To determine the speed required to cross a contaminated area 17 km deep when the mean radiation level is 95 roentgens/hr, the reduction coefficient is 4, and the allowable radiation level is less than 20 roentgens. The solution is: V=95 x 17 ÷ 4 x 20=20 km per hr.

Example calculations using the nomogram Figure 109: Variant A: Determine the expected personnel radiation dose in crossing a contaminated zone with radiation level of 60 Roentgens/hr when the length of route is 12 km, the travel speed is 15km/hr and the reduction coefficient is 2.

Solution: From the 112 mark on the "contaminated route length" scale draw a perpendicular line to the intersection with the "Mean radiation level - 60" line. From this point draw a horizontal line to the intersection with the "Travel speed - 15" line. Draw a vertical line through this point to the intersection with the "Reduction coefficient - 2" line and then draw a horizontal line to the "Personnel radiation dose" scale. At that point read the result=24 roentgens.

Variant B: Determine the speed for crossing a contaminated zone so that the personnel receive less than 48 roentgens, when the route length is 18 km, the mean radiation level is 80 roentgens per hr, and the radiation reduction coefficient is 2.

Solution: Draw a perpendicular line from the 18 mark on the "contaminated route length" scale to the intersection with the "Mean radiation level - 80" line and then draw a horizontal line to the left. Draw a horizontal line from the 48 mark on the "Personnel radiation dose" scale to the intersection with the "Reduction coefficient - 2" line and through this point draw a vertical line to the intersection with the previously draw horizontal line. The point of their intersection on the "Travel speed - 15" line shows the result is 15 km/hr. This is the minimum speed required.



 
 

Figure 108 - Form for Calculating Expected Radiation Dose

FORM FOR CALCULATING EXPECTED RADIATION DOSES

No Initial data, calculations to be performed Units and accuracy Calculation variant
Example 2 3
1 Length of route within the first zone of radioactive contamination km )1.0( 12
2 Travel speed km/hr )1.0( 25
3 Factor for travel direction )0.001( 0.25
4 Average radiation level R/hr )1.0( 187
5 Radiation reduction factor for transport equipment )0.1( 4
6 )1( ÷ )2( )0.01( 0.48
7 )6( x )3( )0.01( 0.12
8 )7( x )4( )1.0( 22
9 Radiation dose of personnel: )8( ÷ )5( R )1.0( 6
10 Route length in second radiation zone km )1.0( 10
11 Travel speed km/hr )1.0( 18
12 Factor for travel direction )0.10( 0.375
13 Average radiation level R )1.0( 165
14 Radiation reduction factor for transport equipment )0.1( 4
15 )10( ÷ )11( )0.01( 0.56
16 )15( x )12( )0.01( 0.56
17 )16( x )13( )1.0( 35
18 Personnel radiation dose: )17( ÷ )14( R )1.0( 9
19 Total dose: )9( + )81( R )1.0( 15

 
 

Figure 109 - Nomogram to Calculate Expected Radiation Dose


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(26) Calculation to Select the Optimal Travel Route

This is an extremely important and varied class of calculation. The example given here is only illustrative of the possibilities for employing the given method for solving problems of optimization from among a small set of possible variants.

The commander of a sub-unit has a mission to break out to an assigned line and assume the defense in the shortest possible time. There are five possible travel routes to reach the assigned position. All routes pass through a zone of radioactive contamination. Therefore, an additional condition of the advance is the expected dose of radiation received by sub-unit personnel. It is apparent that in this example there are five decision variants. The main criterion for evaluating their relative effectiveness is the travel time. The shorter the better. However, the commander must also strive to keep the received radiation dose at a minimum, or at least not allow it to exceed a set maximum per norms, for instance 05 roentgens.

The conditions for the sub-unit to advance to the assigned mission line may be expressed in a table showing quantitative characteristics. From the data it becomes obvious that the possible movement routes differ from each other in length, road quality, and mean level of radiation.

From the table we see that on route 1 in advance variant 1 there are four segments of 16 km of highway, )travel speed 40 km/hr( 10 km of improved dirt road )travel speed 25 km/hr(, 9 km of dirt road )travel speed of 15 km/hr (, and 5 km of damaged, nearly impassable road )travel speed > 5 km/hr(. The mean radiation level on the entire first route is 100 roentgens per hr. From this information we can calculate the expected duration of travel time to reach the mission line, as follows:

t1=)16 ÷ 40( + )10 ÷ 25( + )9 ÷ 15( + )5 ÷ 5(=2.4 hrs or 2 hrs, 24 min. The first route passes through a radiation zone of 100 roentgens per hr. Using the previous method for calculating radiation with a reduction factor of K=4 for equipment D=) 100( )2.4( ÷ 4=60 roentgens.

After calculating the same data for the other four routes the information needed for the decision may be shown in another table of movement time and expected radiation dose.



 
 

Figure 110 Table to enter description of travel routes

QUANTITATIVE CHARACTERISTICS OF ALTERNATE ROUTES
Advance route number Length, km, of route segments of different road quality Mean radiation level on route, R/hr
1st 2nd 3rd 4th
1 16 10 9 5 100
2 -- 20 12 4 120
3 29 12 6 2 120
4 -- 15 20 5 60
5 16 15 3 4 80
Travel speed along each segment km/hr 40 25 15 5

 
 

Figure 111 Graph of effectiveness of alternate routes

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Figure 112 Table of effectieness indicators for march routes

EFFECTIVENESS INDICATORS FOR TRAVEL ON VARIOUS ROUTES
Route )variant( number Time to move forward )hr( Expected radiation dose, R
1 2.4 60
2 2.4 72
3 2 60
4 3 45
5 2 40

 
 

These results may now be compared to determine the optimum decision of which route to choose. Displaying this data in the table makes it easy to see that the optimum solution is route number 5, having both the shortest travel time and lowest radiation dosage, however this might not always be the case. We can also see that the shortest route )no 2( is actually worse than the longest ) no 3(. However, a more general technique for solution, when the answer is not evident is to prepare a graph to plot the information. In this example plot the radiation dose on the y axis and the travel time on the x axis. Then enter the coordinates for the given mission requirements, such as maximum travel time of 3 hrs and maximum radiation dose of 50 roentgens. Draw vertical and horizontal lines from the axes to the selected coordinates. Plot the data for all the possible variants. Those variants whose position falls within the rectangle formed by the axes and the lines to the decision point.



 
 

(27) Calculation to Determine Optimal Distribution of Weapons

One of the most important classes od decision for the commander is the optimal distribution of weapons against various targets. This is a more complex calculation that the previous example of optimum route given two decision criteria. As an example consider the distribution of weapons against a group of targets. The calculation technique shows the course and result of the calculation for optimization. Of course, the ideal solution technique is to use computers, and Soviet headquarters are certainly using this method. In the field, however, the commander may be required to make a rough calculation without having a computer available. The following simple method produces sufficiently reliable results when a small number of forces and means and targets are being considered. The illustration, nevertheless, indicates the type of problem for which computer solutions may be developed.

In this example the commander must find the most effective distribution of five different groups of weapons against five strong points in order to inflict the maximum casualties on the enemy. having calculated the effectiveness of fire of each of the groups of weapons against each target, for instance, using the technique with the nomogram ) ( the commander may create a table of effectiveness of damage to each enemy target by each weapon.


 
 

Figure 113 Table of effectiveness of artillery fire on targets

EFFECTIVENESS OF ARTILLERY FIRE DAMAGE ON TARGETS
Weapons group Damage effectiveness in hectares for target number
1 2 3 4 5
1 3 4 3 5 9
2 4 7 5 4 7
3 6 8 3 5 9
4 2 6 9 8 5
5 9 4 2 2 4

 
 

The damage achieved by different weapons against various types of targets is not identical because of the difference in the quantity and quality of the weapons, the quality of the shells, and the relationship of the ballistic characteristics of the weapons )dispersion pattern etc( to the dimensions of the targets, and the degree of shelter of the enemy personnel.

According to the data in the table, the first group of weapons when hitting the first enemy target inflicts casualties to the personnel in an area of 8 hectares, the second weapons group against the same target damages 4 hectares, and so on.

If the commander assigns the first weapons group against the first target, the second group to the second target, etc., then the effectiveness of the distribution will be the total area damaged or 03 hectares. However, this solution will not be optimal, since the actual combat capabilities of the available weapons in these conditions are not used in the best way. In order to find the best distribution and solve the problem quickly, another method for calculation must be used. The initial data for this are shown in the next table; the weapons groups, the number of target groups and the firing effectiveness.

The calculation of the optimal distribution is performed directly on the table. First, select the two greatest values for the effectiveness values in each line. In the first line these are 9 and 8, and in the second they are 7 and 7, etc. Then subtract the lesser value from the greater and write the results in the right hand column, as shown. Perform the calculation first by each line and then by each column. Examine the differences by lines and columns, select the greatest value and mark it )*(. In this example the result is the fifth line, but the maximum may also be in a column. Then in the line or column in the fields for which the greatest value shows, find the maximum indicator and underline it. )This is the 9 in line 5 column 1.( This indicates that the fifth weapons group should be targeted against the first target group. Eliminate this pair from further consideration.
The course of the further calculation is shown in the next table. The remaining weapons are in a similar manner distributed against the remaining target groups until all groups of weapons are distributed.. This shows that the first group should fire on the fourth target, the second on the second, the third against the fifth, and the fourth against the third. This results in the greatest damage to enemy personnel, a total area of 42 hectares. This is the maximum value of damage from the 120 possible combinations of weapons and targets in the example.

 
 

Figure 114 Table of effectiveness of artillery fire on targets

EFFECTIVENESS OF ARTILLERY FIRE DAMAGE ON TARGETS )COURSE I( Course 1
Weapons group Damage effectiveness in hectares for target number
1 2 3 4 5
1 8 4 3 5 9 1 )9-8(
2 4 7 5 4 7 0 )7-7(
3 6 8 3 5 9 1 )9-8(
4 2 6 9 8 5 1 )9-8(
5 9 4 2 2 4 5* )9-4(
C-1 1 )9-8( 1 )8-7) 4 )9-5( 0 )8-8( 0 )9-9(

 
 

Figure 115 Table of effectiveness of artilleyr fire on targets

EFFECTIVENESS OF ARTILLERY FIRE DAMAGE ON TARGETS )COURSE I( Course 1 C 2 C 3 C 4
Weapons group Damage effectiveness in hectares for target number
1 2 3 4 5
1 8 4 3 5 9 1 )9-8( 1 1 -
2 4 7 5 4 7 0 )7-7( 0 0 0
3 6 8 3 5 9 1 )9-8( 1 1 1
4 2 6 9 8 5 1 )9-8( 1 - -
5 9 4 2 2 4 5* )9-4( - - -
C-1 1 )9-8( 1 )8-7) 4 )9-5( 0 )8-8( 0 )9-9(
C-2 - 1 4* 0 0
C-3 - 1 - 3* 0
C-4 - 1 - - 2*

In this example the total area of damage was considered the indicator of effectiveness. Sometimes one must minimize the effectiveness indicator instead of maximizing it. For instance, the indicator might be the time required for fulfilling the mission, the expenditure of munitions or other materials, etc. Top optimize the distribution according to a minimum indicator value, use the same technique, only in calculating the differences in the lines and columns, select two minimum values instead of the two maximum ones and subtract the least from the greatest. Then find the minimum indicator in the appropriate line or column and fix it first.
This comparatively simple and practical technique for determining the optimum distribution of forces and means is useable in the field. The materials being distributed might be weapons, reconnaissance means, communications systems, water crossing equipment, etc.




 
 

(29) Calculation to Determine the Effectiveness of Fire Destruction Means

This is a more elaborate version of the basic formula designed for use with calculators. The initial data is the number of fire targets, the number of weapons, rate of fire, and coefficient of effectiveness of fire, the length of time for conducting fire, and the coefficient of resistance )counteraction( of the enemy. The formula is:

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where:
M=quantity of targets to be destroyed;
Nt=total number of targets;
Pi=probability of destruction of target from one weapon by one shot;
q=probability of destruction of weapon by fire of enemy;
Nw=number of weapons;
=rate of fire )number of rounds per min(;
t=duration of fire in min.

Example problem: Determine the expected number of destroyed targets from 30 observed targets, if 12 weapons with a rate of fire of 3 rounds a minute are used for their destruction. The probability of destruction by 1 shot is 0.2 and firing time is 5 min. The probability of destruction of our weapon by enemy in the time of one shot is .3


 
 

Figure 116 Form for calculating weapons effectiveness

FORM FOR CALCULATING WEAPONS EFFECTIVENESS CONSIDERING ENEMY COUNTERMEASURES
No Initial data and values to be determined Units Calculation variant Remarks
Example 2 3
1 Number of targets units 30
2 Probability of destruction in one shot -- 0.2
3 Probability of destruction by enemy during one shot -- 0.3
4 Number of weapons units 12
5 Firing rate rounds / min 3
6 Firing time min 5
7 Number of destroyed targets units 18

 
 

Reconnaissance Planning

Reconnaissance planning makes use of many of the same calculations and formulas used for operational planning, such as time and distance for movements and optimization of resources. The following are some calculations specifically designed for reconnaissance planning.


30) Calculation of Effectiveness of Reconnaissance and Required Duration for a Reconnaissance Mission

The given data for the calculation is the speed of movement of the reconnaissance means, the actual range of observation of reconnaissance, the dimensions of the area being covered and the duration of the patrol. The formula is:

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P=probability of observing an object )target(;
e=natural logarithm 2.72;
R=radius of observation of targets by reconnaissance in km;
V=rate of movement of reconnaissance means during the mission in km/hr;
t=duration of patrol;
S=area of the region being observed in sq km.

If it is required to observe several targets in one area, then the mathematical expectation of coverage )average number of targets seen( for observed targets can be calculated as M=N x P
where M=average number of observed targets; N=total number of targets in the area;
and R=probability of detection of one target.

To determine the duration of patrol in order to observe a target with a pre-determined likelihood for observation the formula is:

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Example problem: Determine the likelihood of observation of targets after 2 hours of reconnaissance in an area of 58 sq km if the effective observation distance is 1.5 km and rate of movement of the patrol is 12 kph. The table shows the answer that the probability of observation coefficient is .71. Answer is that if there are 15 targets in the area then they can find 10 to 11 of them.



 
 

Figure 117 Form for calculating probability of target detection

FORM FOR CALCULATING PROBABILITY OF DETECTING TARGETS

No Initial data and values to be determined Units Calculation variant Remarks
Example 2 3
1 Effective target detection range km 1.5
2 Reconnaissance system speed km/hr 12
3 Search duration hr 2
4 Search region area sq km 56
5 Target detection probability -- 0.71

Example problem 2: Determine the duration of time to perform the reconnaissance mission to observe a target in an area of 50 sq km to a likelihood percentage of 80%, if observation distance is 1.2 km and rate of movement is 15 kph. Answer is 4.5 hr


 
 

Figure 118 Form for calculating length of search

FORM FOR CALCULATING LENGTH OF RECONNAISSANCE SEARCH
No Initial data and values to be determined Units Calculation variant Remarks
Example 2 3
1 Target detection probability -- 0.8
2 Search region area sq km 50
3 Effective target detection range km 1.2
4 Reconnaissance system speed km/hr 15
5 Search duration hr 4.47


 
   

(31) Calculation of Detection of Targets by Reconnaissance

If the probability of detection of targets at a given range may be considered as approaching 1, then the formula for calculating the probability of detecting a target in an area during a patrol of given length can be simplified to eliminate the logarithm. The simplified formula may in turn be structured into a nomogram for even faster calculations. In this case the initial data for the calculation are the length and speed of the search, and the range of reliable target detection ) probability near 1.0( of the reconnaissance systems. The formula is:

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where:
2RVt is less than or equal to S;
P=probability of target detection;
R=effective range of reliable observation in km;
V=search speed in km per hr;
t=search time in hours;
S=surface area of the search region in square km.

Example calculation using formula: Determine the probability of target detection in 2.5 hours in an area of 37 sq km when the search speed is 4 km/hr.

Solution is P=}2 x 1.3 x 4 x 2.5{ ÷ 37=0.7


 
 

Example calculation using the nomogram Figure 119: Determine the probability of detection of a target in 3 hrs in an area of 50 sq km, when the range of detection is 1.2 km and the search speed is 6 km/hr.

Solution: For this calculation )variant a( draw a perpendicular line from the "1.200" mark on the "Reliable detection range" scale to the intersection with the "Search speed - 6" line. From the point draw a horizontal line to the right to intersect the "Search length - 3" line, then from this point draw a horizontal line to the left to the "Probability of detection" scale. Read the answer at 0.86.

Example 2: Determine the search length in an area of 60 sq km, when the target detection probability is 0.9, the reliable detection range is 1.3 km and the search speed is 8 km/hr.

Solution: For this calculation )variant b( draw a horizontal line to the right from the "0.9" mark on the "Probability of detection" scale to the intersection with the "Search region area - 60" line. From this point draw a vertical line. Then draw a perpendicular line from the "1.300" mark on the "Reliable detection range" scale to the intersection with the "Search speed - 8" line. From this point draw a horizontal line to the right to the intersection with the previously draw vertical line. They intersect along the "Search length - 2.5" line. The answer then is approximately 2.5 hours required to complete the assigned search.


 
 

Figure 119 Nomogram for calculating probability of target detection {short description of image}


 
 

Sample Calculations for Division and Army Staff

 
 

Example calculation using formulas: On the basis of army's initial instruction the division staff calculates the time its lead regiment requires to move to the border area and deploy there when:
depth of regiment column - 30 km
depth of the area of deployment - 6 km
average speed of march - 20 km
distance to the border area - 120 km

Solution: Use the formula:

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where
t=time the regiment needs to deploy in new area
D=distance to border area
V=average speed of march
GK=depth of regiment column
GR=depth of the area of deployment
tp=time spent for halts

t=(120 ÷ 20) ( + (30 - 6) ÷ (0.6 x 20) + 1=9 hrs

Example calculation: On the basis of the same instructions the division determines the time for the division to deploy and occupy the designated departure area when:
- depth of division columns on 3 routes - 80 km
- depth of the area of deployment - 35 km
- average speed of march - 25 km
- distance to departure area from start line - 125 km
- time to complete engineer work - 8 hours

Solution: To solve the problem the staff applies a combination of calculations and norms. Calculations are done both in general terms )time to complete the deployment for the entire division( and calculations to determine the time of deployment of different echelons such as first and second echelons, rear service troops, air defense troops, attack helicopters etc. The method is the same for all categories. In each case normatives are applied to determine the time required for specific tasks such as engineer work, establishment of fire system, delivery of supplies, loading and unloading, establishment of security, etc.

A. Calculations in general terms:
T=)D ÷ V( + tp + )Gk - GR( ÷ 0.6V + tengr
T=125 ÷ 25 + 1 hr + )80 - 35( ÷ )0.06 x 25( + 8 hrs
T=5 + 1 + 3 + 8=17 hrs.

B. Calculations in specific terms

1. For the first echelon regiments )depth of column 30 km, depth of area of deployment 10 km(:{short description of image}








T=(150 ÷ 25) + 1 + (30 - 10) ÷ (0.6 x 25) + 8 hrs ....(1)
T=6 + 1=1:20 + 8 hrs=16:20 hrs

2. For second echelon regiments use the same formula:
T=170 ÷ 25 + 1 + (30 - 11) ÷ (0.6 x 25) + 8 .....(2)
T=6:48 + 1 + 1:20 + 8=17.08

3. For the rear service use the same formula:
T=195 ÷ 25 + 1 + (10 - 5) ÷ (0.6 x 25) + 8 ......(3)
T=7:48 + 1 + 0.20 + 8=17:08


 
 

Notes: )1( Movement distance for first echelon regiments is assured to be 150 km on the basis that the regiments should move another 25 km from the division deployment line to reach there designated position in the first echelon 125 + 25=150

)2( The distance for the second echelon is assumed to be 0`7 km on the basis that the regiment moves 35 km from the head of the column )depth of first echelon regiment plus 5 km interval( and to reach its designated area it should move another 01 km from the divisions to reach its area of deployment in the second echelon (125 + 35 + 10=170 km)

)3( The distance for the rear service is assumed to be 195 km to include its distance from the head of the column )65 km( and the 5 km interval ) 125 + 65 + 5=195 km(

Example problem: The division commander must clarify his mission and calculate the following on the basis of the army commander's instructions.
- depth and width of the division missions;
- width of the area of penetration;
- required rate of advance;
- number of regiments required in first and second echelon.

Solution:
1. The depth and width of the division mission is measured on the map in the following manner:
- depth of the immediate mission 17 km with a width of 8 km
- depth of long range mission is 21 km and a width of 12 km
- the width of penetration area is 3.5 km, requiring the forces and means of two regiments )2 km and 1.5 km of penetration area assigned to them(, the remaining 4.5 km of the front should be covered by part of one of the regiments of the penetration area )the right flank regiment( and forces and means of another regiments.

2. The immediate mission to be accomplished in ------- hours )assume 7 hours( therefore the average rate of advance is 17 ÷ 7=2.5 km/h. The long range mission is to be accomplished in ------ hours )assume 5 hours(, therefore the average rate of advance should be 21 ÷ 5=4 km/h

3. On the basis of the width of penetration area which is 3.5 km )the norm is 2 km per regiment( and the overall width of the division sector i.e. 8 km there can be two alternatives for the echelonment of the troops:

a. Two regiments in the first echelon, one BMP regiment covering the front and one regiment in the second echelon to be committed after the penetration of the enemy's brigade defensive position, while the BMP regiment will then constitute the second echelon to be committed after the penetration of enemy's division defenses.

b. Three regiments in first echelon, with the BMP regiment coming to the second echelon after the attack is begun.

4. Therefore the commander can tentatively determine the following:
- rate of advance
- direction of the main attack and the width of penetration area
- combat formation of the division for the attack

5. Issues mentioned in point 4 are further examined , elaborated and confirmed during the estimate of the situation and "recognasirovka".

Example problem: As the chief of operation section prepare for the commander calculations to determine the required correlation of forces and means to support the assigned rate of advance.

Solution: To solve the problem use the rate of advance nomogram. To use the nomogram first find the F factor
F factor=D ÷ KTVmax;
D=distance )depth( of the mission
K=terrain coefficient:
-- 1.25 level
--1.00 rough-level
--.75 rugged hills
--.75 urban sprawl
--.50 mountainous
T=time required for action in days and fraction of days
Vmax=theoretical speed in km/day
F=38 ÷ (1.25 x 1 x 60 )=0.5

Now see in the nomogram what correlation of forces and means is required when the F factor is 0.5 The answer is 4.3:1

Example problem: Determine the width of the main sector on the basis of the following facts:
- width of the overall area of the division is 8 km
- overall correlation of forces and means is 3: 1
- required correlation of forces in the main sector is 4.3:1
- correlation of forces and means below which we can not drop in the rest of the division area is 2:1

Solution: Use the size of the sectors formula which is as follows:

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Wm=width of the main sector
Wo=width of the overall area
Co=overall correlation of force
Cm=required correlation of force
Cs=correlation of force below which one can not drop in the rest of the action area.

Wm=8 (3 - 2) ÷ (4.3 - 2)=3.5 km

Note: You can increase the width of the main sector area by accepting a lower correlation of forces and means in the rest of the area of division attack, for example:
Wm =8 (3 - 1.3) ÷ (4.3 - 1.3)=13.6 ÷ 3=4.5 km

Example problem: Calculate the time for a division to advance from the assembly area )departure area( and deploy for shift into attack from the line of march when:
- distance of attack line from enemy forward line=1 km:
- distance of line to deploy into company column=4 km:
-distance of line to deploy into battalion column=12 km:
-distance of regulating line to line of deployment into battalion column=20 km:
- distance of regulating line from start line=40 km:
- distance of start line from assembly area=5 km:
- depth of first echelon regiments=30 km:
- interval between 1st and 2nd echelon regiments=10 km:
- movement speed into attack=8 km:
average speed during march=24 km/h:

Solution: This calculation can be done either by using several formulas to calculate each section of the advance and finally to combine them together or by filling in prepared tables.
a. Using formulas: 06 should be 60 and 09 should be 90:

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ta=time for crossing final deployment into line of attack minus )H( in minutes:
Da=distance of line for going into attack formation from the forward edge of the enemy position in km:
Va=rate of movement in attack formation in km/h:
tr=time for crossing the line of deployment into company column )H -minutes(:
Dr=distance of line of deployment into company columns from the line of deployment into attack formation in km:
v=average rate of movement of units in mounted formation during march:
tb=time for crossing line of deployment into battalion column )H -minutes(:
Db=distance of line of deployment into battalion columns from line of deployment into company columns in km:
trr=time for crossing the last regulation line )LRL( prior to deployment into battalion columns )BC(:
Drr=distance of LRL to BC:
ti=time for passing the start line )SL(:
Di=distance of SL from LRL:
tvit=time to begin movement from assembly area
D=distance of SL from assembly area:
ti'=time for crossing the SL by second echelon:
Gk=depth of the mounted column of first echelon in km:
Dk=distance between the tail of the first echelon and the head of the second echelon in km:
60 and 90=coefficient for conversion of time in minutes for the average speed during the deployment on each line:

ta=)1 x 60 ÷ 8=7.5 ) H - 7.5 min(
tr=7.5 + )4 x 90( ÷ 24=7.5 + 15 min=)H - 22.5 min(
tb=22.5 + )12 x 60 ÷ 24=22.5 + 30 min=)H - 53 min(
trr=52.5 + )20 x 60 ( ÷ 24=25.5 + 50=)H - 1hr,43 min(
ti=01:43 + )40 x 60( ÷ 24=01:43 + 01:40=)H - 03:23(
tvit=03:23 + )5 x 99( ÷ 24=03:23 + 00:19=)H - 03:42(
ti'=03:42 + )30 + 11( x 90 ÷ 24=03:42 + 02:30=) H - 06:12

b. Using the tables:
The same calculations can be done by using the tables given above with the individual calculations and filling in the numbers in each line. Such tables are pre-prepared in advance in blanks and the operation staff can use them to do calculations taking different options into consideration.

Sample calculations: Determine the time and distance to the line of meeting with a counter-attacking enemy reserve when:
- the enemy reserve )up to 2 mech and 2 tank battalions( is sighted 28 km from the forward line of division's attacking troops:
- enemy's speed of advance is about 15 km/hr:
- a delay of 30 minutes is expected in the enemy movement due to a narrow area along the road:
- planned air strikes and artillery fire's are expected to delay the enemy for another 40 minutes:
- the speed of own attacking forces in the first echelon is 4 km/hr due to isolated enemy's strong points across the front:
- the attack on enemy's position 4 km further in the depth )the troops are expected to reach there within an hour( is expected to delay 45 minutes for minor regroupment.


 
 

Solution: use the formula:

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tv=expected time of meeting the enemy in hours:
D=distance between opposing forces in km
tn=total delay time for own forces in hours
Vn=speed of movement of own forces
tp=total delay time for enemy forces
Vp=speed of movement of enemy forces

Results:
tv=}28 + ])0.75 x 4( + )1.15 x 15 ([{ ÷ )4 + 15(
tv=]28 + )3 + 17.35({ ÷ 19
tv=)28 + 02.25( ÷ 19 : tv=2.54 or 2 hrs and 33 minutes

Now to determine the distance of meeting with the enemy use this formula:
lp=Vn )tv - tn (
lp=4 )2.54 - 0.75(
lp=7.16 km

This means that the first echelon forces will be able to destroy the enemy in this intermediate defensive position before it can launch its counter-attack, provided the enemy's reserve is delayed by air strikes and artillery fire for not less than 40 minutes and the enemy does have to slow down 30 minutes delay to cross the narrow pass and own forces do not take more than 45 minutes to regroup in order to continue the attack. If the division commander determines that the line of meeting with the enemy is not convenient, he can chose to repel the counter-attack from a line further in back or he might want to further delay the enemy so that the first echelon troops can move further than 7.16 km before the enemy launches his counter-attack. This calculation can also be conducted by filling in the pre-prepared form.

Example calculation: On the basis of the assumptions which were mentioned in exercise 01 determine the number of anti-tank weapons )ATGM and AT guns( to repel the enemy tanks when:

- the number of enemy tanks are estimated to be 80:
- no less than 50% of enemy tanks must be destroyed:
- the probability of destruction of a single tank by one weapon with one shot is 0.2:
-up to 8 rounds may be fired by each weapon in the time the tanks are located in the effective zone of fire:

Solution: To determine the required number of anti-tank weapons use the formulas or the nomogram.

Formulas:

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Mn=degree of destruction of tanks by artillery )in percentage(;
P1=probability of destruction by one weapon in one shot;
N=required number of anti-tank weapons to accomplish the mission;
m=weapon rate of fire or number of shots one weapon can shoot during the time the target is in within range;
M=expected number of attacking tanks.


This method is a lengthy one and difficult to use under field conditions.
The nomogram can be used for up to 40 tanks. However, for over 40 tanks divide the number into pieces less than 40 and calculate on the basis of the nomogram and add the results: 60=40 + 20, or 77=40 + 37 etc.

In this example the following calculation can be done on the nomogram: Draw a perpendicular line from 0.5 mark on the "Required amount of destruction of targets" scale to intersect with the "Probability of target destruction by one round 0.2" curve. From this point draw a horizontal line to the intersection with the "Number of attacking ground targets - 40" line and then a vertical line up to the "Number of firing by one weapon - 8" line. From this point go along the horizontal to "Required number of antitank weapons" scale and read 17 then multiply it by 2 to get the required number of AT weapons for 80 tanks: 17 x 2=34 AT weapons.

Example calculation: The division commander decided to use the divisional AT reserve at the line for repulsing the enemy's counter-attack, which previous exercise set at 7 km up from the current forward line of the first echelon troops. The AT reserve is now 8 km from the forward line. The enemy is 12 km away from the line of repulsion of counter-attack with an expected delay of 1 hour and 10 minutes on the way )due to planned air strikes and artillery's fire(. The speed of advance of the enemy is 15 km/hr the effective range of AT weapons is 3 km. The AT reserve will need 30 minutes to deploy on the fire line and prepare for action. It's speed of movement 12 km/hr.


 
 

Determine how much time is available for the division commander and staff to assign mission to the AT reserve.

Solution: Use the formula:

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where:
t=time available for the commander and his staff to assign mission;
D=distance of the enemy to the line of contact;
tp=total delay time for enemy forces;
Vp=speed of movement of enemy forces;
d=effective range of AT weapons;
tn=total time required for AT reserve to move to the line of the repulsion of the enemy's counter-attack and time to prepare for action.
1. First determine the tn
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tn=time for AT reserve to move and prepare for action;
Vn=speed of movement of AT reserve;
tr=time to prepare for action;
tn=)8 + 7( ÷ 21 + 0.5=1.57.

2. Now determine the t:
t=}]D + )tp x Vp([ - d{ ÷ Vp - tn
t=}]21 + )1.15 x 15 ([ 2{ ÷ 15 - 1.75
t=])21 + 17.5( - 2[ ÷ 15 - 1.75
t=)38 - 2( ÷ 15 - 1.75
t=36 ÷ 15 - 1.75
t=2.25 - 1.75=0.5
t=30 minutes.
This means that the division commander and staff should assign mission to the AT reserve within 30 minutes )not later( so that the AT reserve will arrive and get prepared on the line of repulsion before the enemy tanks reach the effective range of AT weapons at the line.

Example calculation: In planning the commitment of divisions's second echelon the area within 4 km of the enemy intermediate defensive position, which is to be attacked by the second-echelon, is open and when the regiment moves and deploys to company and platoon columns and assumes the combat formation in this area, it should be covered by artillery strikes conducted on enemy strong points at the line of commitment and on the flanks. The line of attack is 1 km and the line of fire safety is 400 m from enemy position. The speed of movement is 20 km/hr and speed of attack is 6 km/hr. Determine the duration of artillery strike to cover the deployment and attack of the second-echelon regiment.

Solution: To determine the duration of artillery strike calculate the time it takes the regiment to deploy and move to the line of attack and then to the safety line of fire in front of enemy position (400 m).

Use equations:
t=ta + tr ...........)1(

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t=time of artillery strike;
ta=time for crossing the attack line up to the fire safety line;
tr=time spent from deployment in the company columns up to the line of attack;
Da=distance of attack line from enemy position;
d=distance of fire safety line to the enemy position;
Va=speed of movement in attack;
Dr=distance of deployment into company columns;
V=speed of movement of troops;

Results:
ta=)1 -0.4( x 60 ÷ 6=6 minutes
tr=3 x 90 ÷ 20=27 ÷ 2=13.5 minutes
t=ta + tr=6 + 13.5 approximately 20 minutes

This means that the artillery strike should be conducted for 20 minutes to cover the commitment of the second echelon regiment.

Example calculation: The division has accomplished its immediate mission and continues the attack in depth to complete the destruction of the enemy forces in its tactical zone and accomplish the long range mission by the end of the day. The first echelon regiments are fighting with the enemy forces, which conduct delaying action and cover the withdrawal of its main forces across the Schmalin River. Two enemy battalion size columns 15 km from the river are withdrawing to the river apparently to establish defense on the river. The division commander has decided to assign a forward detachment to prevent the arrival of these enemy battalions on the river. The distance to the enemy columns from the head of the assigned forward detachment is 15 km. A 20 minute delay is expected in the movement of enemy columns due to planned friendly air strikes. The speed of movement of enemy's columns is 15 km/hr.

Determine the expected time and rate of overtaking of the withdrawing enemy by division's forward detachment.

Solution: Here the crucial issue is to overtake the enemy before he is able to cross and establish defense at the river.

1. First determine how long does it take the enemy to reach the river:
t=D + )tp x Vp( ÷ Vp
t=time it takes the enemy to reach the river;
D=distance of the enemy to the river;
tp=expected delay in enemy's movement;
Vp=speed of enemy's movement;

Results:
t=]15 + )0.5 x 15 ([ ÷ 15
t=(15 + 7.5) ÷ 15=22.5 ÷ 15=1.5
t=1 hr and 30 minutes

Therefore the enemy's columns must be overtaken within less than 1.5 hours.

2. Assume that the speed of movement of the forward detachment is 20 km/hr:
Now to=]D - )tp x Vp ([ ÷ )Vn - Vp (
to=time to overtake the enemy )hours(;
D=distance to the enemy;
tp and Vp=same as in 1;
Vn=speed of movement of own forces )forward detachment(;
to=15 - (0.5 x 15) ÷ (02 - 15)
to=)15 - 7.5 ) ÷ 5
to=7.5 ÷ 5=1.5
to=1 hour and 30 minutes

This means that at a speed of 20 km/hr the forward detachment can not catch the enemy columns before they reach the river. In order to overcome this, either the speed of movement should be increased or the enemy should be further delayed by air strikes, airborne assault troop, artillery fires, mines etc.

3. In order to find the required speed of movement of the forward detachment to overtake the enemy in one hour perform the following calculation, using the formulas:

Vn=}]D - tp x Vp([ + )to x Vp({ ÷ to
Vn=]15 - )0.5 x 15 ([ + ])1 x 15 ([ ÷ 1
Vn=(7.5 + 15 ) ÷ 1=22.5 ÷ 1=22.5
Vn=22.5 km/hr.

Therefore the speed of movement of the forward detachment should be at least 22.5 km per hour to ensure the enemy's interception within an hour. At that time the enemy will be 7.5 km from the river.
D=)t x Vp( - )tp x Vp(
D=)1 + 15( - )0.5 x 15(
D=15 - 7.5=7.5 km


 
 

Calculations for Front Offensive Planning

One of the most important preliminary calculations made by the front commander is for the allocation of his forces to first and second-echelons and to main and secondary attack axes )directions(. This is done on the basis of the correlations of forces required to achieve assigned results in the attack sectors and the minimum correlations allowable in other )holding attack( sectors. In addition the quantity of artillery available to meet density norms will also be a governing factor.

Example calculation: Presume the initial instructions received by the front commander establish the following: the front is assigned a mission for offensive operation with the scope:
- overall depth: 640 km
- depth of immediate mission: 280 km
- depth of long range mission: 360 km
- width of the frontage: 340 km
- duration of operations 14 days

The front is composed of four combined arms armies on D day, a tank army will join the front on D + 2.
During the clarification of the mission the commander determines the following:
- number of attack directions and breakthrough areas;
- number of armies in the first echelon;
- rate of advance, and required correlation of forces for such rate of advance.

Answer:

Since the direction of the main attack is determined by the superior commander, the front will facilitate the establishment of the appropriate grouping of forces and means and support of the axis. In this case the front also has a choice to determine the number of supporting attacks and forces allocated to them.

1. On the basis of initial data the front has one direction of main attack. According to theory a major part of the forces should be allocated to this direction: two armies with a frontage of not more than 60 km in a European type of terrain ) 60 x 2=120 km of front). Now what is left is 340 - 120=220 km. Therefore this 220 km is to be covered by two more armies.

If the front launches a supporting attack with one army, it cannot give the army a sector of more than 08 km.

)therefore: 220 - 88=140 km(.
Now if one army is assigned this 140 km front, it cannot attack; but only will hold the line or it may attack in a narrow sector merely to support the main attack of the supporting attack directions.

These are preliminary deductions based on the clarification of the mission. These variants can be further developed during the estimate of the situation.

The determination of the number and overall width of the breakthrough areas depends on the number of armies and divisions in the first echelon. This is again a tentative and rough assessment, while the more detailed calculation can be conducted later, on the basis of artillery capability and number and type of enemy targets.

Penetration areas:
- main direction: 2 armies each 12 km=24 km
- supporting direction: 1 army 10 km - 10=total of 34 km

2. Rate of advance:

a. 640 ÷ 14=45 km/day

This is the average rate of daily advance which should be maintained. In order to see if this rate of advance can be maintained, make an overall and summary correlation of forces and means. A 3:1 correlation supports an average rate of advance of 40 to 60 km/day.
b. To find out the required correlation of forces and means )basically on the main direction( use the formula/nomogram of )"Correlation of forces needed for rates of advance"(:
f )factor(=D ÷ KTVmax;
f=correlation factor;
k=terrain factor )1.25 for open terrain(;
Vmax=theoretical speed in km/day.

On the nomogram read that a correlation of 4.6 to 1 is required to achieve the accomplishment of the mission in 14 days. This is a rough calculation and the enemy's detailed capabilities are not taken into consideration. This correlation mostly applies to the axes of main attack, while on axes with holding attacks a lower correlation can be accepted.
Assume that the four divisions attacking in the right flank army and the three divisions attacking in the army adjacent to it to the south are the main direction.

On the supporting direction three divisions of the left flank army attack in the first-echelon with one division of the adjacent army to the north participating in the breakthrough. Determine the overall breakthrough area and number of artillery pieces required.

Answer:

1. Main direction: 4 divisions + 3 divisions=7 divisions: the norm for the width of the breakthrough area per division is 4 km. Therefore 7 x 4=28 km. General norm for number of artillery pieces required per km of breakthrough area is 100 (90-110). Therefore 28 x 100=2800 artillery pieces.

2. Supporting direction: 3 divisions + 1 division=4 divisions: 4 x 4=16 km (width of breakthrough area): 16 x 100=1600 artillery pieces.


Exercise calculation: Determine the correlation of forces and means in the holding area when overall correlation of forces and means is 3:1 and as discussed in the above exercise, the width of the main sector is 120 km, of supporting attack sector 80 km, and overall width of front'soperation area is 340 km. The required correlation of forces and means on the main direction as discussed above is 4.6 to 1 and in the supporting direction is 4 to 1.

Answer: {short description of image}Use the following formula:






1. Main sector versus the entire front:

120=340 )3 - Cs( ÷ )4.6 - Cs(
552 - 120 Cs=1020 - 340 Cs
220 Cs=468 ÷ 220
Cs=2

Therefore in the rest of the frontage )out of the main sector and overall correlation of 2:1 is required.

2. Supporting sector versus the rest of the front )340 - 120=220 km(
80=220 )2 - Cs( ÷ )4 - Cs(
320 - 80 Cs=440 - 220 Cs
140 Cs=120; Cs=120 ÷ 140=.85
Cs=.85

It means that if we establish a 4:1 correlation of forces and means in the supporting attack sector, the overall correlation of forces and means in the rest of the front )excluding the main and supporting attacking sectors( cannot be more than 0.8:1 which will support only defensive action.

3. Suppose that the army which will be assigned in this 041 km sector between the main sector )021 km( and supporting attack sector )08 km( decides to launch attack by one division in a 02 km sector with a 3:1 correlation to support the flank of the main or supporting attack sectors. In this case the overall correlation of forces and means in this sector will further drop from its original 0.58:1. Here is how it calculates using the same formula above.
20=14 )0.58 - Cs( ÷ )3 - Cs
60 -20 Cs=119 - 140 Cs
120 Cs=59; Cs=59 ÷ 120
Cs=0.5

This means that if the army launches a division size attack in part of its sector ( 20 km ) then the correlation of forces and means in the rest of the sector will drop to 0.5:1 or the enemy will have a superiority of 2:1 in this sector. If a temporary defensive action can be acceptable to the front commander in this sector, then the army can choose this course of action.


 
 

Calculating Operational Scale Rates of Movement

The necessity for evaluating terrain passability in a zone of prospective operations arises equally with the study and analysis of its protective and maskirovka qualities every time an offensive operation is planned and conducted or when troops are regrouping or maneuvering. This is natural since passability of terrain is an important operational factor for successful troop movements and actions in an operation.(7)

Passability of terrain in the operational plan depends on the relief features, the presence of large forests, the density and condition of the road network, hydrologic and soil conditions as well as on natural obstacles, barriers, and the amount of destruction in the zone of troop operations. These qualities of terrain have a significant effect on determining the overall operational plan, its scale, the selection of main axes of operations, the depth of mission, the forms and methods of deployment, logistical organization, engineer effort, and other types of operational support.

It is know that the speed of troop movement and the most efficient organization of their combat formation depend greatly on the quantity and type of obstacles on the routes of march. When planning a march or regrouping on an operational scale calculations are also complicated by the fact that troop movements must be carried out in a concentrated manner along several parallel routes in strictly specified periods of time with minimum expenditure of personnel and equipment for the preparation of march routes and for keeping them in passable condition. In order to fulfill all these requirements it is necessary to consider some conditions determining a subsequent operational decision.

The most important of these conditions are as follows:
- sufficient distance between routes of march to insure the safety of troops moving on any one of them in case of an enemy nuclear attack on the adjacent route;
- the quantity and location of routes of march which are most consistent with the operational plan;
- efficient traffic loads on the routes of march to achieve not only maximum rate of traffic on each of them, but also the highest possible total rate of traffic on routes of march to insure organized movement within prescribed periods of time.


In effect this means that the greater traffic carrying capability on the route of march the more troops should be moved over it. It is important to apply such a procedure not only when considering the operational plan, but also when calculating the simultaneous movement of the main body of forces to an assigned area or to the deployment position.

The most convenient criterion determining traffic carrying capability of a given route of march is the average speed of movement over it. It takes into account not only the march speed of the columns, based on the technical condition of the road and the tactical-technical characteristics of the equipment involved, but also various delays en route connected with restoring the route of march or maintaining it in passable condition. Such an approach to evaluating passability of routes of march takes into account changing conditions in the conduct of operations when the factor of time and high speeds of movement become decisive.

The traffic carrying capability of a road is understood as the quantity of combat or transport vehicles passing over it in a prescribed interval of time. Traffic carrying capability is determined according to the following formula (1):

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Where:
N=the number of motor vehicles and other combat vehicles passing through in one hour;
Sav=the average speed of movement )meters/hours(;
ll=the minimum distance between vehicles )meters(;
lo=the overall length of the motor vehicle )meters(

This formula reflects the tactical and technical sides of the problem. However, for operational calculations it cannot be used in this form for two reasons. First, this concept of traffic carrying capability does not take into account the peculiarity of troop movement in march formation, when a definite distance is established between separate columns. Second, according to the computed value for traffic carrying capability it is difficult to draw the correct conclusions required for solution of operational problems.

Obviously, for operational calculations it is more expedient to introduce the concept of operational-tactical traffic carrying capability, which can be determined by the quantity of formations and units passing through in a prescribed time period over a given route of march. Operational-tactical traffic carrying capability of a route of march is determined according to the following formula )2(:

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where:

P=the operational-tactical traffic carrying capability of a route of march )the quantity of formations and units passing through in one hour(. When calculating major operational regroupings consideration should be given to specific features of the column for each branch of troops, for the amount of available equipment of losses suffered and other data which make possible determination of approximately equal depth in route order of large units or units for application during actual conditions.(;

Sav=average speed of column movement, km/hr;
K=the number of large units or units moving along the route of march;
De=depth of march formation of troops )km(.

As a result the traffic carrying capability as well as the speed of movement computed according to the given formula can describe the passability of a route of march in the form of a numerical indicator.

When planning a major maneuver of troops in any particular zone it is important to determine not only the passability of individual routes of march, but also the overall passability of terrain in the zone. The characteristic passability of a zone can be its traffic carrying capability, which will depend on the quantity of routes of march, the speeds of movement over them by large columns, and on the organization of the march formation. Traffic carrying capability in a particular zone )Pz( can be defined as the sum of operational-tactical traffic carrying capabilities available on its routes of march )P(, i.e.:

Pz=/P (3)


Thus, the criterion for traffic carrying capability of a zone can be taken as the quantity of large units and units, with a typical length of march formation for the given conditions, passing through the entire zone in the given period of time.

For example: In the zone there are four routes of march. On the basis of operational and tactical requirements and the evaluation of conditions on the routes of march the average rates of speed were established as follows: on route 1 -- 25 km/hr; on route 2 -- 19 km/hr; on route 3 -- 28 km/hr and on route 4 -- 20 km/hr.

On each of them moves a column of troops consisting of five troops consisting of five troop units with a given composition and an overall depth of march formation of )De( equaling 200 km. The traffic carrying capacity of each route is determined according to formula )2( and the following results are obtained:

for route 1 -- P1=25.5 ÷ 200=0.6; for route 2 -- P2=19.5 ÷ 200=0.5;

for route 3 -- P3=28.5 ÷ 200=0.7; for route 4 -- P4=20.5 ÷ 200=0.5.

Operational tactical traffic carrying capability for the entire zone is determined according to formula )3( -- Pz=0.6 + 0.5 + 0.7 + 0.5=2.3.

This means that over the given zone in the period of one hour there can pass through two troop units with a typical or characteristic depth of march formation for the determined conditions as provided in the given instance.

The operational/tactical traffic carrying capacity of march routes, as seen from formula )2(, depends on the speed of movement, the overall depth of the march formation, and the quantity of large units and units traveling over the route of march.

For this reason it is most important to use all means to increase speed of movement and decrease the length of the troop column. In addition to careful reconnaissance of routes of march and engineer support on each route, it is also expedient to make up a march graph. Such planning increases the independence of columns during regrouping and will promote increased speed of troop movement and, consequently, traffic carrying capacity on the routes of march. Where possible as many troop routes as possible should be designated to reduce the overall depth of the march formation. It is advisable to assign the best routes of march to troops forming a substantially long column.

When preparing for an operation or during its progress the quantity of routes of march for troops will be determined on the basis of the operational disposition of the troops, available roads in the theater of military operations, total traffic capacity of the roads, availability of time for movement and the distance involved, as well as the expected enemy action with weapons of mass destruction. To take all these factors into consideration and make a decision quickly on the allocation of the required number of routes of march for any particular large unit will be, undoubtedly, quite difficult. After proper analysis and research the mathematical function of those factors have been determined and expressed by the following formula
(4) :{short description of image}







where:
K=the quantity of routes of march;
a=the coefficient determining the relationship between the average speed of movement, Sav, to the actual speed during forward movement or deployment, Sfm, of the troop column.
a=Sav ÷ Sfm;
De=depth of the march formation )km( during movement along one route of march;
T=the time allotted for the movement )hours(;
Sav=the average speed of movement )km/hr(;
D=the distance on the route of march )km(.


Assume that the troops are required to complete a march over a distance where D=200 km at an average speed of Sav=25 km/hr in a time of T=12 hrs; the depth of the march formation during movement over only one route of march, De, is 200 km, and the speed during forward movement Sfm=12.5 km/hr. According to the given formula it is possible to calculate how many routes should be prepared or assigned, with the computation as follows:



(25 ÷ 12.5 ) x 200 [ ÷ ] (12 x 25 ) - 200 [=400 ÷ 100=4

For rapid calculations under field conditions it is more convenient to use graphs or nomograms. In this connection a proposed graphic variant in the form of a nomogram is provided for determining any particular unknown parameter when evaluating the passability of terrain over the routes of march.

Shown on the nomogram as an example is the graphic determination of the quantity of march routes based on the data contained in the above formula. From the point T=12 hours, located on axis O-T, a perpendicular is drawn to the intersection of the slant line corresponding to the speed of movement V=25 km/hr. From this point a horizontal line is extended to the left to intersect with the line where D=200 km. Then a perpendicular is dropped to the curve when De=2100 km and then a horizontal line is extended to the right. The required quantity of march routes is indicated on the axis O/K.

The nomogram provided is constructed on the assumption that:

a=Sav ÷ Sfm=2.

In this case if the relationship between the speeds of movement of all the route of march and during deployment between the speeds of movement of all the route of march and during deployment are different from the prescribed, then the value of K, determined graphically, of necessity will increase on the corrected coefficient a1, based on the change in speeds.

In formula )4( and in the nomogram the same speed of movement is used over the entire route of march. In practice, even in preliminary evaluation of march routes, the speed can vary. In case the difference in speed of the march does not increase more than 01-51 percent, then during operational calculations )by formula or on the Nomogram( the average speed of movement over all selected routes is used. To assure the simultaneous appearance of troops in the designated area it is advisable to have the troop columns proportional to the possible speeds of movement. If the difference in speeds of movement turns out to be substantial, then initially a determination is made on troop groupings which can complete the march over the best roads in the allotted period of time. Mathematically this means that according to formula )4( with a given value K the depth of the march formation De is determined. After this the remaining quantity of troops is distributed over the routes of march which permit slower speeds of movement and calculations are made according to the same formula.

With the aid of formula )4( and the nomogram, analysis of interrelated parameters for passability of terrain leads to certain conclusions. First, establishing the degree of terrain passability makes it possible to determine the optimal quantity of march routes in a given operational tactical situation. Second, with an increase in the number of march routes in the zone of troop operations during constant speed of troop movement over the roads, the periods of time for regrouping are decreased as a result of the reduction in time for deployment of forward movement of columns consisting of shorter march formations on each route of march. Third, the possibility is provided for the rapid and correct determination of the quantity of routes of march for troop movements in anticipation of a meeting engagement or an attack from the march, and also when completing a march in compressed periods of time over short distances. Fourth, the main portion of time for troop movements over large distances is lost on their deployment. Therefore in given cases it is advisable to increase the number of march routes by a maximum use of dirt roads and cross-country routes. And, conversely, if troops are displaced over considerable distances, then to reduce march time it will probably be more advantageous to designate fewer roads providing they permit a higher speed of movement.


 
 

Table for value of coefficient a:

Value of Coefficient a

a1
Given value: a=Sav ÷ Sfm Value of coefficient a1
1 ½
1
1 ½ ¾
2 1
3 1 ½

 
 

Figure 121 Nomogram to determine quantity of march routes

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Methods for Calculating Marches in Complex Situations(8)

March decision-making and planning, change in the order of march with an abrupt situation change, as well as deployment of columns, all involve laborious computations. On one hand it is necessary to compute different variants in order to compare and select the optimal variant; on the other hand the selected variant must be worked out in detail.

A march computation is performed usually in stages. First one determines the approximate values of the basic march indices required by the commander for a comparative analysis of various possible variants. The commander performs these calculations personally or jointly with his chief of staff )staff officers(. The calculation variant selected by the commander can be adopted by his staff officers as a basis for detailed elaboration )on the commander's map and their working maps( prior to the final march decision. This portion of the calculations will be the final stage of the computation. Obviously the faster it is completed the sooner the commander will be able to reach a thoroughly-substantiated march decision covering all elements.

The success of a march also depends on how quickly the requisite calculations are performed and decisions made by the commander pertaining to continuation of the march under conditions of abrupt situation changes or when it is necessary to deploy for combat directly from the march columns. The principal problems which must be solved in deployment of troops from march formation include determination of the distance at which columns following one another could line up with column heads abreast, and the time required for this -- depending on speed, length of columns and column lead. )Shown in Figure 15(.

Calculations for drawing abreast or deploying columns during a march, manual determination of requisite distances, time, or speeds are performed according to the following formulas.

For example, the distance )1( at which the head of column B will be abreast of the head of column A can be computed with formulas: l=) V1 x S( ÷ )V2 - V1 ( )1( or l=S ÷ )K - l( )2(

Where l=distance at which column B reaches the head of column A; V1 and V2 -- speeds of columns A and B )respectively(; S=lead between columns A and B )or vehicles A and B(; K -- Coefficient indicating the ratio of column speeds --V2 ÷ V1.

The time )t( it takes for the heads of columns A and B to stand abreast is determined with formulas

t=l ÷ V1 )3( or t=l + S ÷V )4(


If the distance l was not determined )or does not need to be determined(, and it is necessary to determine only the time required for the columns to stand abreast, it may be obtained with formula:

l=S ÷ )V1 )K - l((. )5(


Example problem: at what distance )l( and how much time will it take for the head of column B to reach a point abreast of the head of column A if the first column )A( proceeds at a speed of V1=16 km/h, while the second column )B( can proceed at a speed of V2=24 km/h and if the lead )S( between column B is 8 km )Fig 122(.


 
 

Figure 122 Conditions of formation shift

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Figure 123 Formation change of columns in movement

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Figure 124 nomogram to determine relationships between speed comumns and distance

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Figure 125 Values for coefficient K

Values of Coefficient K=V2 ÷ V11
17 20 22 25 27 30 32 35 37 40 42 45 47 50 60
10 1.7 2 2.2 2.5 2.7 3.0 3.2 3.5 3.7 4.0 4.2 4.5 4.7 5.0 6.0
12 1.4 1.7 1.8 2.1 2.2 2.5 2.7 2.8 3.0 3.3 3.5 3.7 3.9 4.2 5.0
15 1.1 1.3 1.5 1.7 1.8 2.0 2.1 2.3 2.5 2.7 2.8 3.0 3.1 3.3 4.0
17 1.0 1.2 1.3 1.5 1.6 1.8 1.9 2.0 2.2 2.3 2.5 2.6 2.8 3.0 3.5
20 - 1.0 1.1 1.2 1.3 1.5 1.6 1.7 1.8 2.0 2.1 2.2 2.3 2.5 3.0
22 - - 1.0 1.1 1.2 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.7
25 - - - 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.4
27 - - - - 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.2
30 - - - - - 1.0 1.1 1.2 1.2 1.3 1.4 1.5 1.6 1.7 2.0

 
 

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Substituting specific data in formula )1(: we find that: l=)16 x 8( ÷ )21 - 16(=16 km

According to formula )3(: t=l ÷ V1we find that t=16 ÷ 16=1 hr
But in this example the question can also be stated differently. For example, we must determine the speed )V2( at which column B must move in order to reach a position abreast of column A in 1 )2, 3, etc( hour, or by covering a distance of 16 )20, 30, etc ( km. This problem is solved with the above formulas by substituting known quantities.

These formulas can be used to solve any of the above problems fairly rapidly. They can be solved even faster and more simply, however, by utilizing the nomogram )Figure 3(.

With the aid of the graph one can quickly find such quantities as 1 and t with various values of S and column or vehicle speeds )V(. The sequence of use is shown in this example. Assume that we wish to determine at what distance, 1, and in what time, t, column B )following behind column A( will reach a position abreast of column A if the column lead )S( is 30 km, the first column )A( is traveling at 20 km/h, and the second column )B( is proceeding at 30 km/h.

First find index K, that is the ratio V2 ÷ V1. In this case K=30: 20=1.5. In order to speed up determination of index K with various column speeds V1 and V2, use a table compiled in advance.

Then find the values of 1 and t on the graph in Figure 3. For this, draw a horizontal line from the mark "30 km" on line S to its point of intersection with the diagonal line corresponding to index K=1.5, and from the point of intersection draw a perpendicular line down to line 1. At the point of intersection between the perpendicular line and line 1 find 1=60 km. This means that column a will travel a distance of 60 km before column B reaches a position abreast.

Time )t( is found on the same graph by drawing a perpendicular line from point 1=06 km to the point of intersection with the horizontal line designating speed V1 )in this case equal to 20 km/h(. The point of intersection of the perpendicular line and this line will show the time it will take for column B to reach a position abreast of column a with the given ratio and specific speeds; the time is found on diagonal line t at which the point of intersection occurs. In our example t=3 hrs.
We can also use the graph top solve inverse problems such as determination of the ratio of speeds V1 and V2 at which the columns can reach a position abreast )or deploy for combat( in a give time or at a given distance 1. For example, assume that we must determine at what ratio of speeds )V1 and V2( two columns can be brought to a position abreast if this must be accomplished in a distance 1=20 km, and the column lead is S=10 km. In order to solve this problem draw a horizontal line from point " 10 km" on line S and a vertical line from point "20 km" on line 1 to their point of intersection. Find in the region of this point of intersection the closest diagonal line

K=V2 ÷ V1 .

In this example this point was on diagonal line K=1.5. This means that the two columns can reach an abreast position with a ratio of column speeds of 1.5:1, that is V2=5 V1.

We must stipulate that the graph can be utilized with sufficient accuracy in all cases where the speed of the forward columns is at least 5 km/h, and the column lead )S( is more than 5 km.

One typical problem which must be solved quickly is determination of time and speeds for deployment of troops from march columns. In this case two indices are taken for the basis of computation -- the distance )S( from the head of the first column to the tail of the following column, if it is necessary for all vehicles in these two )and more( columns to deploy, and the possible speed of vehicles at the head and tail of these columns )that basis of road conditions or distance 1 and available travel time )t(. The graph enables one quickly to determine all possible values of the sought quantities )1,V, and t( and to determine the best variant on the basis of situation demands.

For example, determine the time )t( and speed )V1( at which the lead vehicles of column A must travel if the vehicles at the tail of column B can move at a speed )V2( not exceeding 30 km/h under given road conditions; the length of each column is 8 km, and the column gap is 4 km )resulting in a total column length of S=20 km(.

To solve this problem draw from the "20 km" point on the vertical line of values s a horizontal line, intersecting all diagonal lines indicating values of ratio

K=V2 ÷ V1.

From the points of intersection sequentially draw perpendicular lines to line 1 and continue them to the point of intersection with the horizontal lines of various values of V1. At the points of intersection of the perpendicular lines and lines 1 and V1 we find the specific values of these quantities )1 and V1( for the corresponding ratio of K.

For example, for ratio K=3 )V2=30 km/h, V1=10 km/h( 1=10 km, t=1 hour; for ratio K=2 )V2=30 km/h, V1=15 km/h(, 1=20 km, t=1 hour 20 minutes; for ratio K=1.5 )V2=30 km/h, V1=20 km/h( 1=40 km, t=2 hours; for ratio K=1.2 )V2=30 km/h, V1=25 km/h( 1=100 km, t=4 hours. All these figures are obtained on the graph within a few seconds, which makes it possible quickly to asses different variants, selecting from them that which most corresponds to the situation conditions.

Having determined the average speed, one must estimate capability to achieve it in relation to the given conditions. One should proceed thereby from the features of individual route sections, determining V average for each of these sections. It may turn out that on some stretches speed is below the average figure for a larger segment of the route, and possible for the entire march. In this in the general form average speed is determined by formula

V average=Stot ÷ ttot.

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where Stot is the total length of the route or individual )of significant length( route section, while ttot is the time distance on that section, in calculating particular values of V average one must proceed from road conditions:
This, differentiated approach to determining average speeds on route sections, taking into consideration the required average value for the entire route, will be the most correct. The fact is that in crossing obstacles all vehicles in the column travel at the same speed; the lead vehicles )subunits( cannot increase speed until the entire column has crossed the given obstacle. In order to reduce the time required for this operation it is advisable to cross obstacles with compressed columns, reducing the vehicle gap when approaching the obstacle )1.7 km(, while with a compressed column the obstacle can be crossed in almost half the time. In march computations it is important to determine not how much time will be required to cross a given obstacle but rather how much time will be lost thereby. This approach simplifies calculations, since it is sufficient to add the total time lost in crossing obstacles to the average travel time index for the section )or for the entire route(.

 
 

Index

 
 

Adjustment and change of plans 5-01
Basic time and distance calculation 5-25
Bilateral nature of armed struggle 5-9
Calculations
- in decision process 5-34
Calendar plan 5-34
Command post
Time to operate 5-97
Commander
- Calculates time available 5-77
- Makes estimate of situation 5-34
Division advance
- Calculation of time required 5-115
Formula
- to calculate anti-tank requirements 5-118
- to calculate CP operating time 5-79
- to calculate degree of mission fulfillment 5-80
- to calculate division advance 5-116
- to calculate duration of march 5-57
- to calculate effectiveness of fire destruction means 5-108
- to calculate effectiveness of multiple systems fulfilling missio 5-80
- to calculate effectiveness of reconnaissance 5-109
- to calculate effectiveness with consideration of countermeasures 5-81
- to calculate expected radiation dose 5-98
- to calculate expected radiation versus travel speed 5-99
- to calculate length of route, average speed, and duration of mov 5-59
- to calculate mathematical expectation of inflicted damage 5-80
- to calculate operational carrying capacity of route 5-126
- to calculate overtaking enemy during pursuit 5-75
- to calculate passage accross short obstacle 5-63
- to calculate passage through major obstacle 5-63
- to calculate probability of target detection 5-111
- to calculate replacements required to restore combat capability 5-96
- to calculate required damage to inflict on defenders 5-89
- to calculate required movement rate to regroup in new area 5-58
- to calculate required number of systems to achieve results 5-80
- to calculate required speed to overtake enemy 5-112
- to calculate route traffic factors 5-127
- to calculate time and distance to contact 5-73
- to calculate time and location of division meeting engagement 5-117
- to calculate time and place of meeting engagement 5-68
- to calculate time available to organize fire on enemy 5-77
- to calculate time for movement 5-55
- to calculate time for unit move 5-52
- to calculate time head and tail of column cross start line 5-67
- to calculate time required to deploy to new area 5-56
- to calculate time to advance and deploy into attack from march 5-70
- to calculate time to start move 5-54
- to calculate total depth of column 5-61
- to calculate traffic capacity of roads 5-125
- to calculate width of breakthrough sector 5-115
- to calculate width of strike sector versus correlation of forces 5-89
- to determine time and place of meeting counterattack 5-117

Meeting engagement
Organizing fire for 5-77
Nomogram to calculate time for movement 5-55
Norms
- Classifications of 5-1
- Distances from FEBA 5-8 estimated times 5-3
- Role of 5-1

Optimization
- mathematical techniques for 5-102
Reconnaissance planning
- Calculations and formulas for 5-109
Width of the main sector
- Calculations to determine 5-115


1. See Chapter One for full discussion of counterattacks.
2. See Chapters Two and Three for discussion of reporting
3. The following section is translated from Vayner Tactical Calculations. The author indicates the critical importance Soviet officers give to mathematical calculations as part of "scientific" decision making.
4. This discussion is taken from a recent Soviet article "Modeling Battle" by Colonel P. Ulugbenikov
5. This very interesting calculation is the only major new formula found in the new book Informatics in military affairs by A. Vainer, published by DOSAAF in 1989. The publication of the book itself, devoted to the use of computers and other methods for the practical application of mathematical analysis by commanders to combat as a text for high school students is an indication of the unrelenting Soviet effort to educate a new generation in the scientific study of leading forces in combat.
6. Among recent discussions of this topic is the article "Toward the question of the creation of strike groupings in offensive operations" by General Major A. E. Tatarchenko in Military Thought May, 1982. In addition to interesting theoretical issues the article contains several useful formulas, one for determining the ratio of widths of strike sector and total width of front in relation to the ratio of correlation of forces in the strike sector and across the total front (see this section here); the second for determining the relationship between initial and final correlations and the losses of two sides; and this formula for the relation of correlation of forces and advance rates.
7. This section is taken from an article by Colonels Korneychuk, Lapshin, and Galitskiy which appeared in Military Thought in April 1967.
8. This section is from an article "On the Question of Methods of March Calculation", by Colonel N. Shishkin, which appeared in Voyennaya Mysl', in 1973.