
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:
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):
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 roughlevel
.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 preprepared 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 counterattacking 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:
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 counterattack,
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 counterattack 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
counterattack. This calculation can also be conducted by filling in the
preprepared form.
Example calculation: On the basis of the assumptions which were mentioned in
exercise 10 determine the number of antitank 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 antitank
weapons use the formulas or the nomogram.
Formulas;
M_{n}=degree of destruction
of tanks by artillery (in percentage);
P_{1}=probability of destruction by one weapon in one shot;
N=required number of antitank 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 counterattack, 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 counterattack 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:
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 counterattack 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.
\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 secondechelon, 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 secondechelon 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)
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 secondechelons 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 ÷ 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 firstechelon 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 (90110). 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:
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
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 tacticaltechnical 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 operationaltactical 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. Operationaltactical
traffic carrying capability of a route of march is determined according to the
following formula:
where:
P=the operationaltactical 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
operationaltactical 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  P_{1}=25.5 ÷ 200=0.6; for route 2 
P_{2}=19.5 ÷ 200=0.5;
for route 3  P_{3}=28.5 ÷ 200=0.7; for route 4 
P_{4}=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 OT, 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 a_{1}, 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 1015 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
crosscountry routes. And, conversely, if troops are displaced over
considerable distances, then to reduce march time it will probably be more
advantageous to designate fewer roads providing they permit a higher speed of
movement.



Table for value of coefficient a:
Value of Coefficient
a_{1} 
Given value: a=S_{av} ÷
S_{fm} 
Value of coefficient a_{1} 
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:
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



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) + 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):
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 ÷ KTV_{max};
D=distance (depth) of the mission
K=terrain coefficient:
1.25 level
1.00 roughlevel
.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 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:
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



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:
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 preprepared 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
counterattacking 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:
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 counterattack,
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 counterattack 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
counterattack. This calculation can also be conducted by filling in the
preprepared form.



EXERCISE ELEVEN
On the basis of the assumptions which were mentioned in exercise 10
determine the number of antitank 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 antitank weapons use the formulas or
the nomogram.
Formulas;
M_{n}=degree of destruction
of tanks by artillery (in percentage);
P_{1}=probability of destruction by one weapon in one shot;
N=required number of antitank 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 counterattack, 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 counterattack 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:
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 counterattack 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 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
secondechelon, 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 secondechelon 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=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;
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



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 + (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



EXERCISE FIFTEEN
Issue combat instructions to the first echelon regiments and the antitank
reserve on repulsion of enemy counterattack.



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 counterattack 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.

