SOVIET VOROSHILOV ACADEMY LECTURES
OPERATIONS CALCULATIONS

This section describes basic calculations performed during operational planning

The same set of calculations, nomograms and tables are in Handbook Chapter 5 Norms and Calculations where the nomograms are numbered from the 80's up.

Operational and Tactical Calculations 1

Basic Time and Distance Calculation 1

Calculation of Time to Begin Move to Start Line 3

Calculation of Time to Deploy into a New Assembly Area 4

Calculation of Time a Unit will be in a New Area 5

Calculation of the Duration of a March from one Area to Another 6

Determine the Required Movement Rate for a Unit to Regroup in a New Area 7

Calculation of Length of Route, Average speed and Duration of Movement of Moving Column 8

Calculation of Overall Depth of Column Consisting of Several Sub-columns 10

Calculation of Duration of Passage of Narrow Points and Difficult Segments 12

Calculation for Passage Times Across Start Point (SL) by the Head and Tail of the Column 16

Calculation of Expected Time and Distance of Probable Point of Contact with Advancing Enemy 17

Calculation of Time Required for Advancing and Deploying Sub-units to Change From Line of March into the Attack 19

Calculation of the Time and Distance to the Line of Contact 22

Calculation of Expected Time and Rate of Overtaking when Pursuing the Enemy 24

Calculation of the Work Time Available to the Commander and Staff for Organizing Repulse of Advancing Enemy Forces 26

Calculation of Length of Time to Operate Command Post in a Single Location 28

Determination of Quantity of Various Weapons, Reconnaissance, Support, Communications etc. for Task Performance 29

Modeling battle 32

Calculation of Strike Capability of Sub-units 36

Calculation of the Width of Main Attack Sector 38

Calculation of Required Destruction of Enemy 38

Calculation of Rate of Advance in Relation to Correlation of Forces 40

Determine the Possible Friendly and Enemy Losses in Relation to the Correlation of Forces and Rate of Advance 42

Determine the Required Amount of Manpower and Weapons for Bringing Sub-Units back up to Sufficient Strength to Restore their Combat Capability 45

Form for the Calculation of the Required Amount of Personnel 46

Determine the Expected Radiation Dose 47

Calculation to Select the Optimal Travel Route 51

Calculation to Determine Optimal Distribution of Weapons 53

Calculation to Determine the Effectiveness of Fire Destruction Means 57

Reconnaissance Planning 58

Calculation of Effectiveness of Reconnaissance and Required Duration for a Reconnaissance Mission 58

Calculation of Detection of Targets by Reconnaissance 60

Sample Calculations for Division and Army Staff 62

Calculations for Front Offensive Planning 71

Calculating Operational Scale Rates of Movement 73

Methods for Calculating Marches in Complex Situations 79

Figure 1 Nomogram for calculating duration of a march 2

Figure 2 Nomogram for calculating duration of movement from assembly area to start line 4

Figure 3 Nomogram for calculating time required for mobile column to deploy into new area 5

Figure 4 Form to calculate time unit is in new area 6

Figure 5 Form for calculating duration of march 7

Figure 6 Form to calculate required rate of travel 8

Figure 7 Calculation of duration of march over complex route 9

Figure 8 Nomogram for calculating length of mobile formation consisting of several columns 11

Figure 9 Nomogram to calculate time required to pass narrow spot on route 14

Figure 10 Nomogram to calculate time required to pass difficult section of route 15

Figure 11 Nomogram to calculate time head and tail of column will pass regulation point 17

Figure 12 Nomogram to calculate time and distance to point of meeting engagement 18

Figure 13 Calculation of time to advance and deploy into the attack from line of march 21

Figure 14 Form for calculation of expected time and location of meeting engagement 23

Figure 15 Nomogram for calculating expected time and speed to overtake retreating enemy 25

Figure 16 Nomogram to calculate time available to plan fire on advancing enemy 27

Figure 17 Calculation of duration of operation in one location 28

Figure 18 Nomogram for calculating the combined effectiveness of several systems 31

Figure 19 Coefficients of comensurability 33

Figure 20 Expected losses 34

Figure 21 Nomogram to determine required losses to achieve correlation of forces 39

Figure 22 Nomogram relating correlation of forces to rate of advance (1) 41

Figure 23 Nomogram relating correlation of forces and rate of advance (2) 41

Figure 24 Force attrition nomogram - army 43

Figure 25 Force attrition nomogram - front 44

Figure 26 Table for calculating restoration of combat effectiveness 46

Figure 27 Form for calculating expected radiation doses 49

Figure 28 Nomogram to calculate expected radiation dose 50

Figure 29 Table to enter description of travel routes 51

Figure 30 Graph of effectiveness of alternate routes 52

Figure 31 Table of effectiveness indicators for march routes 53

Figure 32 Table of effectiveness of artillery fire on targets 54

Figure 33 Table of effectiveness of artillery fire on targets 55

Figure 34 Table of effectiveness of artillery fire on targets 56

Figure 35 Form for calculating weapons effectiveness 57

Figure 36 Form for calculating probability of target detection 59

Figure 37 Form for calculating search length 59

Figure 38 Nomogram for calculating probability of target detection 61

Figure 39 Value of coefficient 77

Figure 40 Nomogram to determine quantity of march routes 78

Figure 41 Conditions of formation shift (a) and deployment (b) of columns on a march 80

Figure 42 Formation change of columns in movement 80

Figure 43 Relationship between speed, time, and distance required for columns to move from sequence head-to-tail into line abreast 81

Figure 44 Values of coefficient K 82

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 a 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.
The formula is:

where:
D_n=distance to SL; \
t_N=time column starts to move;
V_v=rate of movement to SL;
60=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:

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:

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) x 60=0.35 x 60=21 Min 10

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.

Figure 84 - Nomogram for calculating time required for moble column to deploy into new area

(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 of 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.

Figure 85 - Form to calculate time unit is in new area

(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 to calculate 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 - Form for 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:

Figure 89 - Example using the nomogram :

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 to calculate the lenfth of mobile formation consisting of several columns on march

(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 75) 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 -100 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.

Figure 91 - Nomogram to calculate time required to pass difficult section of rooute:

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.

(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;
60=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 the front and tail of a column will pass a specific location.

(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 the time anddistance to point of a 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:

t'i=ti - (Gk + dk) x (90)/ V

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, 24, 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 adevance and deploy sub-units for shift into 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) Calculation 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.

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=32 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 Nomogram for calculating expected time and speed to overtake retreeating 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 30 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 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_n is 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 96: 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 to calculate combined effectiveness of several system

(18) Modeling battle⁽¹⁾

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 Table for coefficients of comensurability

Figure 101 Table for expected loss in combat effectiveness

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 40-60% and on the offense to 20-30%. 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 30% loss as the critical break point for attackers and 40% 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 2000 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 2000 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⁽²⁾

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 the 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 100%.
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 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 10 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=20km; G_O=30km; F_N=15km;

The resulting equation is as follows:

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:

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%.

Figure 102 Nomogram to deermine required enemy loss to achieve favorable correlation of forces

(22) Calculation of Rate of Advance in Relation to Correlation of Forces⁽³⁾

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_Cis 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 30 km on level terrain in one day with a theoretical maximum speed of 60 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 that relates correlation of force to advance rate (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 frontlevel 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 104 Nomogram for calculation of force attrition Army

Figure 106 nomogram for calculation of attrition - 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 85% (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 0.25, 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 105: Variant A: Determine the expected personnel radiation dose in crossing a contaminated zone with radiation level of 60 Roentgens/hr when the length of route is 12 km, the travel speed is 15km/hr and the reduction coefficient is 2.

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

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

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

Figure 108 Form for calculating expected radiation doses

Figure 109 Nomogram to calculate expected radiation doses

(26) Calculation to Select the Optimal Travel Route

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

The commander of a sub-unit has a mission to break out to an assigned line and assume the defense in the shortest possible time. There are five possible travel routes to reach the assigned position. All routes pass through a zone of radioactive contamination. Therefore, an additional condition of the advance is the expected dose of radiation received by sub-unit personnel. It is apparent that in this example there are five decision variants. The main criterion for evaluating their relative effectiveness is the travel time. The shorter the better. However, the commander must also strive to keep the received radiation dose at a minimum, or at least not allow it to exceed a set maximum per norms, for instance 50 roentgens.

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

From the table we see that on route 1 in advance variant 1 there are four segments of 16 km of highway, (travel speed 40 km/hr) 10 km of improved dirt road (travel speed 25 km/hr), 9 km of dirt road (travel speed of 15 km/hr), and 5 km of damaged, nearly impassable road (travel speed < 5 km/hr). The mean radiation level on the entire first route is 100 roentgens per hr. From this information we can calculate the expected duration of travel time to reach the mission line, as follows:
-----t₁=(16 ÷ 40) + (10 ÷ 25) + (9 ÷ 15) + (5 ÷ 5)=2.4 hrs or 2 hrs, 24 min. The first route passes through a radiation zone of 100 roentgens per hr. Using the previous method for calculating radiation with a reduction factor of K=4 for equipment D=(100) (2.4) ÷ 4=60 roentgens.

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

Figure 110 Table to enter description of travel routes

Figure 112 Table of effectiveness of alternate routes

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

(27) Calculation to Determine Optimal Distribution of Weapons

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

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

Figure 113 Table of effectiveness of artillery fire on targets

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

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

If the commander assigns the first weapons group against the first target, the second group to the second target, etc., then the effectiveness of the distribution will be the total area damaged or 30 hectares. However, this solution will not be optimal, since the actual combat capabilities of the available weapons in these conditions are not used in the best way. In order to find the best distribution and solve the problem quickly, another method for calculation must be used. The initial data for this are shown in the next table; the weapons groups, the number of target groups and the firing effectiveness.

The calculation of the optimal distribution is performed directly on the table. First, select the two greatest values for the effectiveness values in each line. In the first line these are 9 and 8, and in the second they are 7 and 7, etc. Then subtract the lesser value from the greater and write the results in the right hand column, as shown. Perform the calculation first by each line and then by each column. Examine the differences by lines and columns, select the greatest value and mark it (*). In this example the result is the fifth line, but the maximum may also be in a column. Then in the line or column in the fields for which the greatest value shows, find the maximum indicator and underline it. (This is the 9 in line 5 column 1.) This indicates that the fifth weapons group should be targeted against the first target group. Eliminate this pair from further consideration.


The course of the further calculation is shown in the next table. The remaining weapons are in a similar manner distributed against the remaining target groups until all groups of weapons are distributed.. This shows that the first group should fire on the fourth target, the second on the second, the third against the fifth, and the fourth against the third. This results in the greatest damage to enemy personnel, a total area of 42 hectares. This is the maximum value of damage from the 120 possible combinations of weapons and targets in the example.

Figure 114 table of effectiveness of artillery fire on targets

Figure 115 Table of effectiveness of artillery fire on targets

In this example the total area of damage was considered the indicator of effectiveness. Sometimes one must minimize the effectiveness indicator instead of maximizing it. For instance, the indicator might be the time required for fulfilling the mission, the expenditure of munitions or other materials, etc. Top optimize the distribution according to a minimum indicator value, use the same technique, only in calculating the differences in the lines and columns, select two minimum values instead of the two maximum ones and subtract the least from the greatest. Then find the minimum indicator in the appropriate line or column and fix it first.

This comparatively simple and practical technique for determining the optimum distribution of forces and means is useable in the field. The materials being distributed might be weapons, reconnaissance means, communications systems, water crossing equipment, etc.

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

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

Formula is:

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

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

Figure 116 Form for calculating weapons effectiveness

Reconnaissance Planning

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

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

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

The formula is:

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

If it is required to observe several targets in one area, then the mathematical expectation of coverage (average number of targets seen) for observed targets can be calculated as M=N x P

where M=average number of observed targets; N=total number of targets in the area;
and R=probability of detection of one target.

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

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

Figure 117 Form for calculating probability of target detection

Figure 117 Form for calculating probability of target detection

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

Answer is 4.5 hr

Figure 118 Form for calculating search length

(31) Calculation of Detection of Targets by Reconnaissance

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

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

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

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


Example calculation using the nomogram Figure 112: Determine the probability of detection of a target in 3 hrs in an area of 50 sq km, when the range of detection is 1.2 km and the search speed is 6 km/hr.

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


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

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

Figure 119 Nomogram to calculate probability for target detection

Sample Calculations for Division and Army Staff

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

Solution:

Use the formula:

where:
t=time the regiment needs to deploy in new area
D=distance to border area
V=average speed of march
G_K=depth of regiment column
G_R=depth of the area of deployment
t_p=time spent for halts
t=(120 ÷ 20) + (30 - 6) ÷ (0.6 x 20) + 1=9 hrs

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

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

A. Calculations in general terms:
T=(D ÷ V) + t_p + (G_k - G_R) ÷ 0.6V + t_engr
T=125 ÷ 25 + 1 hr + (80 - 35) ÷ (0.06 x 25) + 8 hrs
T=5 + 1 + 3 + 8=17 hrs.

B. Calculations in specific terms
1. For the first echelon regiments (depth of column 30 km, depth of area of deployment 10 km):

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


2. For second echelon regiments use the same formula:
T=170 ÷ 25 + 1 + (30 - 10) ÷ (0.6 x 25) + 8 .....(2)
T=6:48 + 1 + 1:20 + 8=17.08


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

Notes:
(1) Movement distance for first echelon regiments is assured to be 150 km on the basis that the regiments should move another 25 km from the division deployment line to reach there designated position in the first echelon (125 + 25=150)
(2) The distance for the second echelon is assumed to be 170 km on the basis that the regiment moves 35 km from the head of the column (depth of first echelon regiment plus 5 km interval) and to reach its designated area it should move another 10 km from the divisions to reach its area of deployment in the second echelon (125 + 35 + 10=170 km)
(3) The distance for the rear service is assumed to be 195 km to include its distance from the head of the column (65 km) and the 5 km interval (125 + 65 + 5=195 km)

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

Solution:

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

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

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

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

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

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


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

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

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


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

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

where:
W_m=width of the main sector
W_o=width of the overall area
C_o=overall correlation of force
C_m=required correlation of force
C_s=correlation of force below which one can not drop in the rest of the action area.
W_m=8 (3 - 2) ÷ (4.3 - 2)=3.5 km


Note:
You can increase the width of the main sector area by accepting a lower correlation of forces and means in the rest of the area of division attack, for example:

W_m =8 (3 - 1.3) ÷ (4.3 - 1.3=13.6 ÷ 3=4.5 km

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


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

a. Using formulas:

t_a=time for crossing final deployment into line of attack minus (H) 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 km/h:
t_r=time for crossing the line of deployment into company column (H -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 units in mounted formation during march:
t_b=time for crossing line of deployment into battalion column (H -minutes):
D_b=distance of line of deployment into battalion columns from line of deployment into company columns in km:
t_rr=time for crossing the last regulation line (LRL) prior to deployment into battalion columns (BC):
D_rr=distance of LRL to BC:
t_i=time for passing the start line (SL):
D_i=distance of SL from LRL:
t_vit=time to begin movement from assembly area
D=distance of SL from assembly area:
t_i'=time for crossing the SL by second echelon:
G_k=depth of the mounted column of first echelon in km:
D_k=distance between the tail of the first echelon and the head of the second echelon in km:
60 and 90=coefficient for conversion of time in minutes for the average speed during the deployment on each line:

t_a=(1 x 60 ÷ 8=7.5 ( H - 7.5 min)
t_r=7.5 + (4 x 90) ÷ 24=7.5 + 15 min=(H - 22.5 min)
t_b=22.5 + (12 x 60 ÷ 24=22.5 + 30 min=(H - 53 min)
t_rr=52.5 + (20 x 60) ÷ 24=52.5 + 50=(H - 1hr,43 min)
t_i=01:43 + (40 x 60) ÷ 24=01:43 + 01:40=(H - 03:23)
t_vit=03:23 + (5 x 90) ÷ 24=03:23 + 00:19=(H - 03:42)
t_i^'=03:42 + (30 + 10) x 90 ÷ 24=03:42 + 02:30=(H - 06:12

b. Using the tables:

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

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

Solution:

Use the following formula:

where:
t_v=expected time of meeting the enemy in hours:
D=distance between opposing forces in km
t_n=total delay time for own forces in hours
V_n=speed of movement of own forces
t_p=total delay time for enemy forces
V_p=speed of movement of enemy forces

Results:
t_v={28 + [(0.75 x 4) + (1.15 x 15)]} ÷ (4 + 15)
t_v=[28 + (3 + 17.35)} ÷ 19
t_v=(28 + 20.25) ÷ 19 : t_v=2.54 or 2 hrs and 33 minutes


Now to determine the distance of meeting with the enemy use this formula:
l_p=V_n (t_v - t_n )
l_p=4 (2.54 - 0.75)
l_p=7.16 km


This means that the first echelon forces will be able to destroy the enemy in this intermediate defensive position before it can launch its counter-attack, provided the enemy's reserve is delayed by air strikes and artillery fire for not less than 40 minutes and the enemy does have to slow down 30 minutes delay to cross the narrow pass and own forces do not take more than 45 minutes to regroup in order to continue the attack.

If the division commander determines that the line of meeting with the enemy is not convenient, he can chose to repel the counter-attack from a line further in back or he might want to further delay the enemy so that the first echelon troops can move further than 7.16 km before the enemy launches his counter-attack. This calculation can also be conducted by filling in the pre-prepared form.

Example calculation: On the basis of the assumptions which were mentioned in exercise 10 determine the number of anti-tank weapons (ATGM and AT guns) to repel the enemy tanks when:
----- the number of enemy tanks are estimated to be 80:
----- no less than 50% of enemy tanks must be destroyed:
----- the probability of destruction of a single tank by one weapon with one shot is 0.2:
----- up to 8 rounds may be fired by each weapon in the time the tanks are located in the effective zone of fire:

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

Formulas;

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

This method is a lengthy one and difficult to use under field conditions.

The nomogram can be used for up to 40 tanks. However, for over 40 tanks divide the number into pieces less than 40 and calculate on the basis of the nomogram and add the results: 60=40 + 20, or 77=40 + 37 etc.

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


Example calculation: The division commander decided to use the divisional AT reserve at the line for repulsing the enemy's counter-attack, which previous exercise set at 7 km up from the current forward line of the first echelon troops. The AT reserve is now 8 km from the forward line. The enemy is 21 km away from the line of repulsion of counter-attack with an expected delay of 1 hour and 10 minutes on the way (due to planned air strikes and artillery's fire). The speed of advance of the enemy is 15 km/hr the effective range of AT weapons is 3 km. The AT reserve will need 30 minutes to deploy on the fire line and prepare for action. It's speed of movement 12 km/hr.

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


Solution:

Use the following formula:

where:
t=time available for the commander and his staff to assign mission;
D=distance of the enemy to the line of contact;
t_p=total delay time for enemy forces;
V_p=speed of movement of enemy forces;
d=effective range of AT weapons;
t_n=total time required for AT reserve to move to the line of the repulsion of the enemy's counter-attack and time to prepare for action.

1. First determine the t_n:

where:
t_n=time for AT reserve to move and prepare for action;
V_n=speed of movement of AT reserve;
t_r=time to prepare for action;
t_n=(8 + 7) ÷ 12 + 0.5=1.75.

2. Now determine the t:
t={[D + (t_p x V_p)] - d} ÷ V_p - t_n
t={[21 + (1.15 x 15)] 2} ÷ 15 - 1.75
t=[(21 + 17.5) - 2] ÷ 15 - 1.75
t=(38 - 2) ÷ 15 - 1.75
t=36 ÷ 15 - 1.75
t=2.25 - 1.75=0.5
t=30 minutes.


This means that the division commander and staff should assign mission to the AT reserve within 30 minutes (not later) so that the AT reserve will arrive and get prepared on the line of repulsion before the enemy tanks reach the effective range of AT weapons at the line.

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

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

Use equations:

t=t_a + t_r ...........(1)

where:
t=time of artillery strike;
t_a=time for crossing the attack line up to the fire safety line;
t_r=time spent from deployment in the company columns up to the line of attack;
D_a=distance of attack line from enemy position;
d=distance of fire safety line to the enemy position;
V_a=speed of movement in attack;
D_r=distance of deployment into company columns;
V=speed of movement of troops;

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

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


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

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

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

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

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

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


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

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


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

V_n={[D - t_p x V_p)] + (t_o x V_p)} ÷ t_o
V_n=[15 - (0.5 x 15)] + [(1 x 15)] ÷ 1
V_n=(7.5 + 15) ÷ 1=22.5 ÷ 1=22.5
V_n=22.5 km/hr.

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

Calculations for Front Offensive Planning

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

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

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

Answer:

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

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

If the front launches a supporting attack with one army, it cannot give the army a sector of more than 80 km.
(therefore: 220 - 80=140 km).

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

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

The determination of the number and overall width of the breakthrough areas depends on the number of armies and divisions in the first echelon. This is again a tentative and rough assessment, while the more detailed calculation can be conducted later, on the basis of artillery capability and number and type of enemy targets.
Penetration areas:
----- main direction: 2 armies each 12 km=24 km
----- supporting direction: 1 army 10 km - 10=total of 34 km


2. Rate of advance:
----- a. 640 ÷ 14=45 km/day
This is the average rate of daily advance which should be maintained. In order to see if this rate of advance can be maintained, make an overall and summary correlation of forces and means. A 3:1 correlation supports an average rate of advance of 40 to 60 km/day.
----- b. To find out the required correlation of forces and means (basically on the main direction) use the formula/nomogram of ("Correlation of forces needed for rates of advance") That is Figure 103 above:
----- f (factor)=D ÷ KTV_max;
----- f=correlation factor;
----- k=terrain factor (1.25 for open terrain);
----- V_max=theoretical speed in km/day.

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

Assume that the four divisions attacking in the right flank army and the three divisions attacking in the army adjacent to it to the south are the main direction.

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


Answer:

1. Main direction: 4 divisions + 3 divisions=7 divisions: the norm for the width of the breakthrough area per division is 4 km. Therefore 7 x 4=28 km. General norm for number of artillery pieces required per km of breakthrough area is 100 (90-110). Therefore 28 x 100=2800 artillery pieces.
2. Supporting direction: 3 divisions + 1 division=4 divisions: 4 x 4=16 km (width of breakthrough area): 16 x 100=1600 artillery pieces.


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

Answer: Use the following formula:

1. Main sector versus the entire front:

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

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


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

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


3. Suppose that the army which will be assigned in this 140 km sector between the main sector (120 km) and supporting attack sector (80 km) decides to launch attack by one division in a 20 km sector with a 3:1 correlation to support the flank of the main or supporting attack sectors. In this case the overall correlation of forces and means in this sector will further drop from its original 0.85:1. Here is how it calculates using the same formula above.
20=140 (0.85 - C_s) ÷ (3 - C_s)
60 - 20 C_s=119 - 140 C_s
120 C_s=59; C_s=59 ÷ 120
C_s=0.5


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

Calculating Operational Scale Rates of Movement⁽⁴⁾

The necessity for evaluating terrain passability in a zone of prospective operations arises equally with the study and analysis of its protective and maskirovka qualities every time an offensive operation is planned and conducted or when troops are regrouping or maneuvering. This is natural since passability of terrain is an important operational factor for successful troop movements and actions in an operation.⁽⁵⁾

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

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

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

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

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

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

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

This formula reflects the tactical and technical sides of the problem. However, for operational calculations it cannot be used in this form for two reasons. First, this concept of traffic carrying capability does not take into account the peculiarity of troop movement in march formation, when a definite distance is established between separate columns. Second, according to the computed value for traffic carrying capability it is difficult to draw the correct conclusions required for solution of operational problems. Obviously, for operational calculations it is more expedient to introduce the concept of operational-tactical traffic carrying capability, which can be determined by the quantity of formations and units passing through in a prescribed time period over a given route of march. Operational-tactical traffic carrying capability of a route of march is determined according to the following formula:

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

As a result the traffic carrying capability as well as the speed of movement computed according to the given formula can describe the passability of a route of march in the form of a numerical indicator. When planning a major maneuver of troops in any particular zone it is important to determine not only the passability of individual routes of march, but also the overall passability of terrain in the zone. The characteristic passability of a zone can be its traffic carrying capability, which will depend on the quantity of routes of march, the speeds of movement over them by large columns, and on the organization of the march formation. Traffic carrying capability in a particular zone (P_z) can be defined as the sum of operational-tactical traffic carrying capabilities available on its routes of march (P), i.e.:
P_z=/P (3)

Thus, the criterion for traffic carrying capability of a zone can be taken as the quantity of large units and units, with a typical length of march formation for the given conditions, passing through the entire zone in the given period of time.
For example: In the zone there are four routes of march. On the basis of operational and tactical requirements and the evaluation of conditions on the routes of march the average rates of speed were established as follows: on route 1 -- 25 km/hr; on route 2 -- 19 km/hr; on route 3 -- 28 km/hr and on route 4 -- 20 km/hr.
On each of them moves a column of troops consisting of five troops consisting of five troop units with a given composition and an overall depth of march formation of (D_e) equaling 200 km. The traffic carrying capacity of each route is determined according to formula (2) and the following results are obtained:
-----for route 1 -- P₁=25.5 ÷ 200=0.6; for route 2 -- P₂=19.5 ÷ 200=0.5;
-----for route 3 -- P₃=28.5 ÷ 200=0.7; for route 4 -- P₄=20.5 ÷ 200=0.5.

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

This means that over the given zone in the period of one hour there can pass through two troop units with a typical or characteristic depth of march formation for the determined conditions as provided in the given instance.
The operational/tactical traffic carrying capacity of march routes, as seen from formula (2), depends on the speed of movement, the overall depth of the march formation, and the quantity of large units and units traveling over the route of march.

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

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

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

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

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

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

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

The nomogram provided is constructed on the assumption that:
----- a=S_av ÷ S_fm=2.

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

In formula (4) and in the nomogram the same speed of movement is used over the entire route of march. In practice, even in preliminary evaluation of march routes, the speed can vary. In case the difference in speed of the march does not increase more than 10-15 percent, then during operational calculations (by formula or on the Nomogram) the average speed of movement over all selected routes is used. To assure the simultaneous appearance of troops in the designated area it is advisable to have the troop columns proportional to the possible speeds of movement. If the difference in speeds of movement turns out to be substantial, then initially a determination is made on troop groupings which can complete the march over the best roads in the allotted period of time. Mathematically this means that according to formula (4) with a given value K the depth of the march formation D_e is determined. After this the remaining quantity of troops is distributed over the routes of march which permit slower speeds of movement and calculations are made according to the same formula.
With the aid of formula (4) and the nomogram, analysis of interrelated parameters for passability of terrain leads to certain conclusions. First, establishing the degree of terrain passability makes it possible to determine the optimal quantity of march routes in a given operational tactical situation. Second, with an increase in the number of march routes in the zone of troop operations during constant speed of troop movement over the roads, the periods of time for regrouping are decreased as a result of the reduction in time for deployment of forward movement of columns consisting of shorter march formations on each route of march. Third, the possibility is provided for the rapid and correct determination of the quantity of routes of march for troop movements in anticipation of a meeting engagement or an attack from the march, and also when completing a march in compressed periods of time over short distances. Fourth, the main portion of time for troop movements over large distances is lost on their deployment. Therefore in given cases it is advisable to increase the number of march routes by a maximum use of dirt roads and cross-country routes. And, conversely, if troops are displaced over considerable distances, then to reduce march time it will probably be more advantageous to designate fewer roads providing they permit a higher speed of movement.

Figure 120 Table for value of coefficient a:

Figure 121 nomogram to determine4 quantity of march routes

Methods for Calculating Marches in Complex Situations

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

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

The success of a march also depends on how quickly the requisite calculations are performed and decisions made by the commander pertaining to continuation of the march under conditions of abrupt situation changes or when it is necessary to deploy for combat directly from the march columns. The principal problems which must be solved in deployment of troops from march formation include determination of the distance at which columns following one another could line up with column heads abreast, and the time required for this -- depending on speed, length of columns and column lead. (Shown in Figure 122).

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

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

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

The time (t) it takes for the heads of columns A and B to stand abreast is determined with formulas
t=l ÷ V₁ (3) or t=l + S ÷V (4)

If the distance l was not determined (or does not need to be determined), and it is necessary to determine only the time required for the columns to stand abreast, it may be obtained with formula:
l=S ÷ (V₁ (K - l)). (5)

Example problem: at what distance (l) and how much time will it take for the head of column B to reach a point abreast of the head of column A if the first column (A) proceeds at a speed of V₁=16 km/h, while the second column (B) can proceed at a speed of V₂=24 km/h and if the lead (S) between column B is 8 km (Fig 123).

Figure 122 Conditions of formation shift (a) and deployment (b) of columns on a march

Figure 123 Formation change of columns in movement

Figure 124 Relationship between speed, time, and distance rqeuired for columns to move from sequence head to tail into line abreast

Figure 125 Values of coefficient K

Substituting specific data in formula (1):

we find that:
l=(16 x 8) ÷ (21 - 16)=16 km

According to formula (3): t=l ÷ V₁
we find that t=16 ÷ 16=1 hr

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

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

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

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

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

Time (t) is found on the same graph by drawing a perpendicular line from point 1=60 km to the point of intersection with the horizontal line designating speed V₁ (in this case equal to 20 km/h). The point of intersection of the perpendicular line and this line will show the time it will take for column B to reach a position abreast of column a with the given ratio and specific speeds; the time is found on diagonal line t at which the point of intersection occurs. In our example t=3 hrs.

We can also use the graph top solve inverse problems such as determination of the ratio of speeds V₁ and V₂ at which the columns can reach a position abreast (or deploy for combat) in a give time or at a given distance 1. For example, assume that we must determine at what ratio of speeds (V₁ and V₂) two columns can be brought to a position abreast if this must be accomplished in a distance 1=20 km, and the column lead is S=10 km. In order to solve this problem draw a horizontal line from point "10 km" on line S and a vertical line from point "20 km" on line 1 to their point of intersection. Find in the region of this point of intersection the closest diagonal line

K=V₂ ÷ V₁ .

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

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

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

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

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

K=V₂ ÷ V¹.

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

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

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

where S_tot is the total length of the route or individual (of significant length) route section, while t_tot is the time distance on that section, in calculating particular values of V average one must proceed from road conditions:

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

Footnotes

1. This discussion is taken from a recent Soviet article "Modeling Battle" by Colonel P. Ulugbenikov

2. This very interesting calculation is the only major new formula found in the new book Informatics in military affairs by A. Vainer, published by DOSAAF in 1989. The publication of the book itself, devoted to the use of computers and other methods for the practical application of mathematical analysis by commanders to combat as a text for high school students is an indication of the unrelenting Soviet effort to educate a new generation in the scientific study of leading forces in combat.

3. Among recent discussions of this topic is the article "Toward the question of the creation of strike groupings in offensive operations" by General Major A. E. Tatarchenko in Military Thought May, 1982. In addition to interesting theoretical issues the article contains several useful formulas, one for determining the ratio of widths of strike sector and total width of front in relation to the ratio of correlation of forces in the strike sector and across the total front (see this section here); the second for determining the relationship between initial and final correlations and the losses of two sides; and this formula for the relation of correlation of forces and advance rates.

4. This section is from an article "Method for Evaluating Terrain Passibility", by Colonels V. Korneychuk, K. Lapshin, and Ye. Galitskiy, which appeared in Voyennaya Mysl' in April 1967.

5. This section is taken from an article by Colonels Korneychuk, Lapshin, and Galitskiy which appeared in Military Thought in April 1967.