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CALCULATIONS FOR FRONT OFFENSIVE PLANNING
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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.
The following examples illustrate this.
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The alternative will be to have the forces in contact (if any) to hold
part of the front line in the front's sector in order to enable the
front to launch the attack by four armies with the following
variations. Diagram of front operation with two armies.
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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.
-----F=640 ÷ (1.25 x 14 x 70)=0.56
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:
1. Main sector Vs 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 Vs 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.
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