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OPERATIONAL CALCULATIONS

 

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:

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Use the formula:

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

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

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

 
 

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

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


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

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

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

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

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where:
Wm=width of the main sector
Wo=width of the overall area
Co=overall correlation of force
Cm=required correlation of force
Cs=correlation of force below which one can not drop in the rest of the action area.
Wm=8 (3 - 2) ÷ (4.3 - 2)=3.5 km


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

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

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

a. Using formulas:

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ta=time for crossing final deployment into line of attack minus (H) in minutes:
Da=distance of line for going into attack formation from the forward edge of the enemy position in km:
Va=rate of movement in attack formation in km/h:
tr=time for crossing the line of deployment into company column (H -minutes):
Dr=distance of line of deployment into company columns from the line of deployment into attack formation in km:
v=average rate of movement of units in mounted formation during march:
tb=time for crossing line of deployment into battalion column (H -minutes):
Db=distance of line of deployment into battalion columns from line of deployment into company columns in km:
trr=time for crossing the last regulation line (LRL) prior to deployment into battalion columns (BC):
Drr=distance of LRL to BC:
ti=time for passing the start line (SL):
Di=distance of SL from LRL:
tvit=time to begin movement from assembly area
D=distance of SL from assembly area:
ti'=time for crossing the SL by second echelon:
Gk=depth of the mounted column of first echelon in km:
Dk=distance between the tail of the first echelon and the head of the second echelon in km:
60 and 90=coefficient for conversion of time in minutes for the average speed during the deployment on each line:
ta=(1 x 60 ÷ 8=7.5 ( H - 7.5 min)
tr=7.5 + (4 x 90) ÷ 24=7.5 + 15 min=(H - 22.5 min)
|tb=22.5 + (12 x 60 ÷ 24=22.5 + 30 min=(H - 53 min)
trr=52.5 + (20 x 60) ÷ 24=52.5 + 50=(H - 1hr,43 min)
ti=01:43 + (40 x 60) ÷ 24=01:43 + 01:40=(H - 03:23)
tvit=03:23 + (5 x 90) ÷ 24=03:23 + 00:19=(H - 03:42)
ti'=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:

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

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

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

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

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

Example calculation: On the basis of the assumptions which were mentioned in exercise 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;

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

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

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

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

Example calculation: The division commander decided to use the divisional AT reserve at the line for repulsing the enemy's counter-attack, which previous exercise set at 7 km up from the current forward line of the first echelon troops. The AT reserve is now 8 km from the forward line. The enemy is 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:

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

1. First determine the tn:


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tn=time for AT reserve to move and prepare for action;
Vn=speed of movement of AT reserve;
tr=time to prepare for action;
tn=(8 + 7) ÷ 12 + 0.5=1.75.

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


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

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

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

Use equations:

t=ta + tr ...........(1)



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

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

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

Example calculation: The division has accomplished its immediate mission and continues the attack in depth to complete the destruction of the enemy forces in its tactical zone and accomplish the long range mission by the end of the day. The first echelon regiments are fighting with the enemy forces, which conduct delaying action and cover the withdrawal of its main forces across the Schmalin River. Two enemy battalion size columns 15 km from the river are withdrawing to the river apparently to establish defense on the river. The division commander has decided to assign a forward detachment to prevent the arrival of these enemy battalions on the river. The distance to the enemy columns from the head of the assigned forward detachment is 15 km. A 20 minute delay is expected in the movement of enemy columns due to planned friendly air strikes. The speed of movement of enemy's columns is 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 + (tp x Vp) ÷ Vp
t=time it takes the enemy to reach the river;
D=distance of the enemy to the river;
tp=expected delay in enemy's movement;
Vp=speed of enemy's movement;

Results:
t=[15 + (0.5 x 15)] ÷ 15
t=(15 + 7.5) ÷ 15=22.5 ÷ 15=1.5
t=1 hr and 30 minutes
Therefore the enemy's columns must be overtaken within less than 1.5 hours. \

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

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

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

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

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

 
 

Calculations for Front Offensive Planning

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

Example calculation: Presume the initial instructions received by the front commander establish the following: the front is assigned a mission for offensive operation with the scope:
overall depth: 640 km
depth of immediate mission: 280 km
depth of long range mission: 360 km
width of the frontage: 340 km
duration of operations 14 days
The front is composed of four combined arms armies on D day, a tank army will join the front on D + 2. During the clarification of the mission the commander determines the following:
number of attack directions and breakthrough areas;
number of armies in the first echelon;
rate of advance, and required correlation of forces for such rate of advance.

Answer:
Since the direction of the main attack is determined by the superior commander, the front will facilitate the establishment of the appropriate grouping of forces and means and support of the axis. In this case the front also has a choice to determine the number of supporting attacks and forces allocated to them.
1. On the basis of initial data the front has one direction of main attack. According to theory a major part of the forces should be allocated to this direction: two armies with a frontage of not more than 60 km in a European type of terrain (60 x 2=120 km of front). Now what is left is 340 - 120 + 220 km. Therefore this 220 km is to be covered by two more armies.
If the front launches a supporting attack with one army, it cannot give the army a sector of more than 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"):
f (factor)=D ÷ KTVmax;
f=correlation factor;
k=terrain factor (1.25 for open terrain);
Vmax=theoretical speed in km/day.

On the nomogram read that a correlation of 4.6 to 1 is required to achieve the accomplishment of the mission in 14 days. This is a rough calculation and the enemy's detailed capabilities are not taken into consideration. This correlation mostly applies to the axes of main attack, while on axes with holding attacks a lower correlation can be accepted.
Assume that the four divisions attacking in the right flank army and the three divisions attacking in the army adjacent to it to the south are the main direction.
On the supporting direction three divisions of the left flank army attack in the first-echelon with one division of the adjacent army to the north participating in the breakthrough. Determine the overall breakthrough area and number of artillery pieces required.

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

Exercise calculation: Determine the correlation of forces and means in the holding area when overall correlation of forces and means is 3:1 and as discussed in the above exercise, the width of the main sector is 120 km, of supporting attack sector 80 km, and overall width of front's operation 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:

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1. Main sector versus the entire front:

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

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


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

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

3. Suppose that the army which will be assigned in this 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 - Cs) ÷ (3 - Cs)
60 - 20 Cs=119 - 140 Cs
120 Cs=59; Cs=59 ÷ 120
Cs=0.5

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


 
 

Calculating Operational Scale Rates of Movement

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

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:



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

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

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

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where:

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

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

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

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

On each of them moves a column of troops consisting of five troops consisting of five troop units with a given composition and an overall depth of march formation of (De) equaling 200 km. The traffic carrying capacity of each route is determined according to formula (2) and the following results are obtained:
for route 1 -- P1=25.5 ÷ 200=0.6; for route 2 -- P2=19.5 ÷ 200=0.5;
for route 3 -- P3=28.5 ÷ 200=0.7; for route 4 -- P4=20.5 ÷ 200=0.5.

Operational tactical traffic carrying capability for the entire zone is determined according to formula (3) -- Pz=0.6 + 0.5 + 0.7 + 0.5=2.3. This means that over the given zone in the period of one hour there can pass through two troop units with a typical or characteristic depth of march formation for the determined conditions as provided in the given instance.

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

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

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

{short description of image}

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

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

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

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

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

The nomogram provided is constructed on the assumption that:
a=Sav ÷ Sfm=2.

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

In formula (4) and in the nomogram the same speed of movement is used over the entire route of march. In practice, even in preliminary evaluation of march routes, the speed can vary. In case the difference in speed of the march does not increase more than 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 De is determined. After this the remaining quantity of troops is distributed over the routes of march which permit slower speeds of movement and calculations are made according to the same formula.

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

 
 

Table for value of coefficient a:

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

EXERCISE ONE

On the basis of army's initial instruction the division staff calculates the time its lead regiment requires to move to the border area and deploy there when:

depth of regiment column - 30 km
depth of the area of deployment - 6 km
average speed of march - 20 km
distance to the border area - 120 km


Solution: Use the formula:

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

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



 
 

EXERCISE TWO


On the basis of the same instructions the division determines the time for the division to deploy and occupy the designated departure area when:

depth of division columns on 3 routes - 80 km
depth of the area of deployment - 35 km
average speed of march - 25 km
distance to departure area from start line - 125 km
time to complete engineer work - 8 hours

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

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

B. Calculations in specific terms

1. For the first echelon regiments (depth of column 30 km, depth of area of deployment 10 km):

{short description of image}

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

2. For second echelon regiments use the same formula:
T=170 ÷ 25 + 1 + (30 - 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)



 
 

EXERCISE THREE


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.

 
 

ANSWER THREE


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".


 
 

EXERCISE FOUR

Given the available time to prepare and plan the division's offensive battle (from _______ to _____), schedule the measures in a table to include time to organize, prepare, and plan the battle.


 
 

EXERCISE FIVE

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 separagte combat instructions to BMP regiment to cover the border.


 
 

EXERCISE SIX

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


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

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



 
 

EXERCISE SEVEN

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 SEVEN

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

{short description of image}

Wm=width of the main sector
Wo=width of the overall area
Co=overall correlation of force
Cm=required correlation of force
Cs=correlation of force below which one can not drop in the rest of the action area.
Wm=8 (3 - 2) ÷ (4.3 - 2)=3.5 km

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



 
 

EXERCISE EIGHT

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


 
 

ANSWER EIGHT

The following issues are to be clarified.

does overall correlation match the requirements 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 the division second echelon should be committed into battle, (the soonest and th4e latges time of commitment).


 
 

EXERCISE NINE

Calculate the time for the 16thMRD 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:


 
 

ANSWER NINE

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:

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ta=time for crossing final deployment into line of attack minus (H) in minutes:
Da=distance of line for going into attack formation from the forward edge of the enemy position in km:
Va=rate of movement in attack formation in km/h:
tr=time for crossing the line of deployment into company column (H -minutes):
Dr=distance of line of deployment into company columns from the line of deployment into attack formation in km:
v=average rate of movement of units in mounted formation during march:
tb=time for crossing line of deployment into battalion column (H -minutes):
Db=distance of line of deployment into battalion columns from line of deployment into company columns in km:
trr=time for crossing the last regulation line (LRL) prior to deployment into battalion columns (BC):
Drr=distance of LRL to BC:
ti=time for passing the start line (SL):
Di=distance of SL from LRL:
tvit=time to begin movement from assembly area
D=distance of SL from assembly area:
ti'=time for crossing the SL by second echelon:
Gk=depth of the mounted column of first echelon in km:
Dk=distance between the tail of the first echelon and the head of the second echelon in km:
60 and 90=coefficient for conversion of time in minutes for the average speed during the deployment on each line:
ta=(1 x 60 ÷ 8=7.5 ( H - 7.5 min)
tr=7.5 + (4 x 90) ÷ 24=7.5 + 15 min=(H - 22.5 min)
tb=22.5 + (12 x 60 ÷ 24=22.5 + 30 min=(H - 53 min)
trr=52.5 + (20 x 60) ÷ 24=52.5 + 50=(H - 1hr,43 min)
ti=01:43 + (40 x 60) ÷ 24=01:43 + 01:40=(H - 03:23)
tvit=03:23 + (5 x 90) ÷ 24=03:23 + 00:19=(H - 03:42)
ti'=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.

 
 

EXERCISE TEN

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 TEN

Use the following formula:

{short description of image}

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

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

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


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

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

 
 

EXERCISE ELEVEN

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 ELEVEN

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

Formulas;


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

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


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

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

 
 

EXERCISE TWELVE

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.

 
 

ANSWER TWELVE

Use the following formula:

{short description of image}

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

1. First determine the tn:

{short description of image}

tn=time for AT reserve to move and prepare for action;
Vn=speed of movement of AT reserve;
tr=time to prepare for action;
tn=(8 + 7) ÷ 12 + 0.5=1.75.
2. Now determine the t:
t={[D + (tp x Vp)] - d} ÷ Vp - tn
t={[21 + (1.15 x 15)] 2} ÷ 15 - 1.75
t=[(21 + 17.5) - 2] ÷ 15 - 1.75
t=(38 - 2) ÷ 15 - 1.75
t=36 ÷ 15 - 1.75
t=2.25 - 1.75=0.5
t=30 minutes.

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


 
 

EXERCISE THIRTEEN

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 THIRTEEN

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=ta + tr ...........(1)




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t=time of artillery strike;





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

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


 
 

EXERCISE FOURTEEN

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 FOURTEEN

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

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

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

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

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

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

3. In order to find the required speed of movement of the forward detachment to overtake the enemy in one hour perform the following calculation, using the formulas:
Vn={[D - tp x Vp)] + (to x Vp)} ÷ to
Vn=[15 - (0.5 x 15)] + [(1 x 15)] ÷ 1
Vn=(7.5 + 15) ÷ 1=22.5 ÷ 1=22.5
Vn=22.5 km/hr.

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

 
 

EXERCISE FIFTEEN

Issue combat instructions to the first echelon regiments and the anti-tank reserve on repulsion of enemy counter-attack.


 
 

EXERCISE SIXTEEN

Prepare combat instructions to the second echelon regiment on its committment into combat


 
 

EXERCISE SEVENTEEN

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


 
 

EXERCISE EIGHTEEN

Prepare combat instructions to the forward detachement to move to the river, intercept the retreating enemy and establish a bridghead on the river.