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

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

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

Figure 57 - Norms for Density of Defense

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.

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

Figure 60 - Diagram for Front Offensive Planning Factors

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.

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

Figure 63 - Aerial Reconnaissance

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

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

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

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

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

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

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

Figure 70 - Norms for Material Reserves in an army

Figure 71 - Requirements for material reserves in Offensive

Figure 72 - Operational Expenditure and Reserves

Figure 73 - Location of Rear Supply and Support

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

(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

Norms for Planning Combat Movement and Marches

Movement Planning Norms

Figure 76 - Movement data for tracked and wheeled vehicles

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

* 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

Figure 79 - Distance Between Column Elements

Figure 80 - March, Halt, and Rest Durations

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

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:

where:
t_Bt=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;
t_p=overall time for halts during march;
t_Bt=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

(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_
The formula is:

where:
D_n=distance to SL;
t_N=time column starts to move;
V_v=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)

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

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:

where:
T_d=time for deployment;
G_K=length of column;
G_R=depth of area;
V_d=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

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:

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;
G_K=length of column;
G_R=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.
t_p=duration of halts en route in hrs.

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

(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:

when G_K < G_r

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;
D_i=distance of start point from original assembly area in km;
G_K=depth of column in km;
G_r=depth of the new assembly area in km;
V=average rate of movement of column in km/hr;
t_p=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

(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:

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

(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:

where:

D=length of route in km;
L_i=length of each sector of different types of road each allowing V_i speed of movement of columns in km;
t_d=overall time on the route in hrs;
V_i=speed of movement on a type of sector of the route in kph;
V=average speed of movement in kph.
G_K=length of column in km;
G_P=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;
t_n=overall time of halts.

Figure 88 - Calculation of Duration of March over Complex Route

(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:

where:
G_k=formation depth in km;
N_m=number of vehicles;
d_m=distance between vehicles;
N_K=number of columns;
d_K=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

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

(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:

where:
N_m=number of vehicles;
d_m=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.

The formula for major obstacles:

where:
G_k=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. G_k=)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

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

(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:

where:
ti=SL crossing time of the head of the march column i in hours and minutes;
T_i-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;
d_i=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;
D_i=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

(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:

where:

t_e=expected tome of meeting in hours;
D=initial distance between opposing groupings in km;
V_f=movement rate of friendly troops in km per hr;
V_e=movement rate of enemy forces in km per hr;
d_l=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

(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:

where:
t_a=time for crossing final deployment into line of attack - "Che" hour -in minutes;
D_a=distance of line for going into attack formation from the forward edge of the enemy position in km;
V_a=rate of movement in attack formation in kph;
t_r=time for crossing the line of deployment into company columns at )"Che" - minutes(
D_r=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;
t_b=time for crossing line of deployment into battalion columns at )"Che" - minutes(;
D_b=distance of line for deployment into battalion columns from line of deployment into company columns in km;
t_rr=time for crossing the last regulation line or point prior to the deployment into battalion columns;
D_rr=distance of last regulation line from the line for deployment into battalion columns;
t_i=time for passing the start line at )"Che" - minutes(
D_i=distance of start line from last regulating line in km;
t_vit=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(;
G_k=depth of the mounted column of first echelon of units in km;
d_k=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

(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:
t_v=}D + ])t_n x V_n( + )t_p x V_p([{ ÷ )V_n + V_p(
T_n=t₁ + t₂ t₃; T_P=t₁' + t₂' + t₃'
where:
t_v=expected time if contact with enemy in hours;
D=distance between forces of the two sides in km;
t_n=total delay time for own force in hrs;
V_n=speed movement of own forces in km per hr;
t_p=total delay time of enemy in hrs;
V_p=speed movement of enemy in km per hr;
l_p=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

(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:

where:

t_o=time to overtake enemy in hours;
D=distance to the enemy in km;
V_n=friendly speed in pursuit in km per hr;
V_p=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: t_o=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: V_n=))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:

(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:

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;
V_e=rate of enemy advance kph;
t_p=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

(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:

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;
V_s=rate of movement of leading sub-units;
V_c=rate of movement of CP during its relocation;
60=coefficient for conversion of hr to min;
t_bd=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

(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:







and through the mathematical expectation )mean value( of damage:

where:
P_n=probability of mission accomplished by a homogenous group of weapons;
P₁=same by a single system;
M_n=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 P_n < P₁

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

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

where P_nis 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: P_n=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:

where:
P_n=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;
M_n=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: P_n=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

(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

Figure 101 - Expected Losses

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 )K_p( 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:

b( in defense:

where:
K_s1, K_s2, K_sp=coefficients of comensurability of combat means of units of country A;
K_b1, K_b2, K_bp=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;
K_p=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" K_s1=1.5, and quantity, i=3;
- APC, K_s2=0.8, and quantity, i=1.

Defenders:
- APC, K_b1=1.4, and quantity i=1;
- AT gun, K_b2=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:
- S_n=}])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 S_n is greater than or equal to 1, then the sub-unit will fulfill its mission, if )as in this example( the S_n 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:
- S_n1=}])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.
- S_n2=})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:

when:
Y_N )1 - P_N ( < )1 - P_N^K( and Y_o )1 - P_o( < P_o^K(
where:
N_N, N_o=number of units of attackers and defenders, measured in combat potentials )"utils"(;
P_N, P_o=expected losses of the units of each side up to the arrival at the immediate mission line;
P_N^K, P_o^K=critical loss of a unit of attacker and defender;
F_N=width of front of attack in km;
F_o, G_o=width of front and depth of the defender's area in km.
Y_N, Y_o=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 K_p 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:

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 N_H=800 for attacker and N_O=700 for defender and the following other values for variables:
Y_N=90%; Y_O=80%; P_N=0.1%; P_O=0.4%; K=1.5;
P_N^K=0.5%; P_O^K=0.7%; F_O=20 km; G_O=30 km; F_N=15km;

The resulting equation is as follows:

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:

where:
W_m=width of strike sector;
W_o=total width of zone;
C_o=overall correlation of forces;
C_m=correlation of forces in strike sector;
C_s=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: W_m=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 C_m=W_o ÷ W_m )C_o - C_s( + C_s ; 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

where:
M=required level of destruction of enemy in percent;
C_N=beginning correlation of forces;
C_T=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

(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, K_C, in an equation such as V=140 x K_C. 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 K_C 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=K_C or 40 ÷ 140=0.29. Enter 0.29 on the scale for K_C 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 ÷ KTV_max 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;
V_max=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)

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

(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

Figure 106 Force attrition nomogram front

(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:

M_i=C_i - A_i ÷ B_i ; A_i=)n_f P_f( x N_f ;


B_i=)n_e -N_e P_e( x N_e ; m_i=B_i x M_i

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;
C_i=required degree of superiority ratio for a particular type of the i type of weapon;
A_i=availability remaining on hand quantity of weapons of a particular type i of own forces;
B_i=availability of remaining number of weapons of a particular type i for enemy forces;
n_f )n_e(=initial full strength of own troops )enemy troops( by type i of weapons pers/ proportions;
P_f )P_e (=losses of own troops )enemy troops( pers/proportions;
N_f )N_e (=TOE number of weapons of a particular type i of own forces )enemy forces(;
m_i=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

(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:

where:
D=the expected radiation dose of personnel in roentgens;
L_i=the length of the route within the i zone of radioactive contamination in km;
N_i=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(;
R_i=mean radiation level on the travel route in roentgens per hour;
V_i=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.