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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.

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.


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.

-----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: {short description of image}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

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

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

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.