SOVIET VOROSHILOV ACADEMY LECTURES
DIVISION OPERATIONS

Exercise 1

On the basis of Army's initial instruction (attached) determine the time required for one regiment with 2 artillery battalions to move to the border area and deploy there when:
- depth of regiment column - 30 km
- depth of the area of deployment - 5 km
- average speed of march - 20 km
- distance to the border area - 120 km

Answer Use the formula:

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

Exercise 2

On the basis of the same instructions determine 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

Answer

To solve the problem a combination of calculations and application of norms can be used. Calculations are done both on 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=5 + 1 + 45÷15 + 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)

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 same formula)

T=170÷25 + 1 + (30 - 10) ÷ (0.6 x 25) + 8 ......(2)

T=6:48 + 1 + 1:20 + 8=17:08 hrs

3. For the rear service

(use the same formula)

T=195÷25 + 1 + (10 - 5) ÷ (0.6 x 25) + 8=17:08 hrs

Notes

(1) Movement distance for first echelon regiments is assumed to be 150 km on the basis that the regiments should move another 25 km from the division deployment line to reach their 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).

Exercise 3

As the commander of 16 MRD read the instructions of the army commander (attached), clarify your mission and calculate the following.
- 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

Answer
1. The depth and width of the division mission is measured on the map in the following manner

a. Depth of the immediate mission 17 km with a width of 8 km
b. Depth of long range mission is 21 km and a width of 12 km
c. 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. There 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 "recognaisrovka".

Exercise 4

Given the available time to prepare and plan the divisions offensive battle (from _____________ to _____________) schedule the measures in a table to include time to organize, prepare and plan the battle.

Answer

(Issue prepared table)

Exercise 5

On the basis of the mission assigned to the division by the army and the deductions of the clarification of the mission prepare initial instructions to regiments and a separate combat instructions to BMP regiment to cover the border.

Answer

(Issue prepared texts)

Exercise 6

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.

Answer

To solve the problem use the rate of advance nomogram. To use the nomogram first you have to 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

Exercise 7

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;

Answer

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

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)=8 ÷ 2.3=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 instance use 1.3:1 for secondary sector, then:

W_m=8 (3 - 1.3) ÷ (4.3 - 1.3)=8 x 1.7 ÷ 3=4.5 km

Exercise 8

Prepare the table of correlation of forces and means and summarize deductions on the basis of the correlation of forces and means. For calculation purposes the enemy forces and means are filled in the table.

Answer

Issue prepared table. The following issues are to be clarified.
- does overall correlation match the requirement of the rate of advance?
- can the required correlation of forces and means be established in the main direction?
- based on the enemy's commitment of his second echelon, determine when second echelon of our troops should be committed into battle, (the soonest and the latest time of the commitment).

Exercise 9

Calculate the time for 16 MRD to advance from the assembly area (departure area) and deploy for shift in to 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:

Answer

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 the 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=(D_a x 60) ÷ V_a=(1 x 60) ÷ 8=7.5 [ H - 7.5 min or H - 00:07]

t_r=t_a + (D_r x 90) ÷ V=7.5 + (4 x 90) ÷ 24=7.5 + 15=H - 22.5 min

t_b=t_r + (D_b x 60) ÷ V=22.5 + (12 x 60) ÷24=22.5 + 30=52.5 min

t_b=H - 00:53

t_rr=t_b + (D_rr x 60) V=52.5 + (20 x 60) ÷ 24=52.5 + 50

t_rr=H - 01:43

t_i=t_rr + (D_i x 60) ÷ V=01:43 + (40 x 60) ÷ 24=01:43 + 01:40

t_i=H - 03:23 hrs

t_vit=t_i + (D x 90) ÷ V=03:23 + (5 x 90) ÷24=03:23 + 00:19

t_vit=H - 03:42

t_i'=t_i + (G_K

b. Using the tables:

The same calculations can be done by using the tables 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.



Exercise 10

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.

Answer - Use the following formula:

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

Solution:
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.



Exercise 11
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:

Answer

To determine the number of anti-tank weapons you can use the formulas or the nomogram.

1. Formulas:

M_n=degree of destruction of artillery tanks (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.

2. The nomogram can be used for up to 40 tanks. However for over 40 tanks calculation can be made on the basis of what can be calculated by the nomograms: 60=40 + 20, or 77=40 + 37 etc.

In our example the following calculation can be done on the nomogram:

draw a perpendicular line from 0.5 make 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.



Exercise 12

The division commander decided to use the divisional AT reserve at the line of repulsion of the enemy's counter-attack, which per exercise 10 will be 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.


Answer Use the following formula:

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/

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.

Exercise 13

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.

Answer

To determine the duration of artillery strike we have to find out 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)

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;

Solution:

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.


Exercise 14

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.

Answer

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

1. First determine how long does it take the enemy to cross 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;


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 the following calculation can be done, 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 interception of the enemy 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



Exercise 15

Issue combat instructions to first echelon requirements and AT reserve on repulsion of enemy's counter-attack.

Answer

Issue prepared text


Exercise 16

Prepare combat instructions to the second echelon regiment on its commitment into combat.

Answer

Issue prepared text.


Exercise 17

Prepare combat situation report to the Army staff on repulsion of enemy's counter-attack and commitment of the second echelon.

Answer

Issue prepared text.



Exercise 18

Prepare combat instructions to the forward detachment to move to the river, intercept the retreating enemy and to establish bridgehead on the river.