"What force would I need...." Topic

My game table is 6x9. My rules give 1" = 100 yards. So my table is 4 miles wide, 6 deep. What's the biggest Soviet unit that would attack along that front?

I'm thinking that;s probably frontage for 2, maybe 3 battalions?

So a regiment is about right, no?

Probably a Regiment with follow on forces. Figure 5km is a NATO Battalion front or Soviet Regiment

Take a look here:

link

Thanks

Extra Crispy:

That frontage is a bit big for a Soviet-style reinforced regimental attack unless it's part of a manoeuvre battle. If it's a breakthrough battle that frontage of 6.5 -7.0 Km frontage (4 miles) would be closer to a divisional frontage. See the section after about p.50 on divisional and lower level attacks in the following document for references.

PDF link

Cheers and good gaming!
Rod Robertson.

As above, a Soviet regimental frontage in the attack was around 6km. In an environment of nuclear threat, there would be gaps between the regiments.

A NATO battlegroup typically held a zone around 4km Square, again with gaps.

In the covering battle units would be more dispersed, and obviously every Soviet regiment would have another one behind it..

Does this help?

Hello, Extra Crispy!
Standard Soviet motorized infantry company comes on a front of 500 m.
For mechanised infantry battalion, reinforced by a tank company, the front is up to 1000 m (taking into account that the enemy has about 80 tanks, AIFWs with ATGM, ATGM TOW and Dragon, as well as taking into account that the enemy is exposed to artillery, missile and other effects before the attack).
So in this area (if I counted, about 6500 m), you can safely attack by regiment.

6 miles = 9.7 kilometers

Standard Soviet attack width against a defending enemy is approximately 4KM per regiment (two battalions up, and the third/forth battalion following 15-30 KM behind).

so your width would support a division-level attack against a defending enemy, with a two regiment front (4 battalions total).

Keep in mind the following battalions of those first two regiments would be at LEAST 9.5 miles behind the front. (Or the equivalent of 1.5 to 3 game table widths. And the division follow on regiment would be MORE than 3 game table widths behind the front. (I capitalized the words because it most wargames the regiment or divisional follow on forces are on the table way too early.)

For example, assuming the minimum distance the 2d echelon battalion would be following behind (9.5 miles or 1.5 board widths), and assuming that battalion moves 12 inches per turn (kinda' generous and to keep it simple, we'll say 9 feet in the rear), it would take the first follow on battalion at least 9 game turns before it even reached the board.

The Division's follow on regiment would be at least 45KM to the rear, or 30 miles from the front. Using your game table width of 6 miles, that would but the follow on regiment FIVE game tables to the rear or 30 feet -- it would take 30 turns for the follow on divisional regiment to get to the table.

(I think I have the math right. But you get the point that, given the constraints, a Soviet division in the attack in a normal wargame would only be able to put 4 battalions, on line, on the table during the whole game. Unless you played a long game or eliminate the time/distance factor.)

My source for all this is the classic US FM 100-2-1, The Soviet Army: Operations and Tactics.

Thank you that is very helpful/

In general, this is the formula for which is necessary to calculate the forces necessary for the attack:
You need to know the number of anti-tank systems (including AIFWs) and tanks in the defending force (denoted for X, in pieces) and the length of the offensive front (denoted for to Y, in km).
The amount available to the advancing units of tanks and anti-tank systems (including BMP) is denoted by Z, in pieces.

The length of the offensive front (for this unit, in km) = 5 (or 6) * ((0,4 * 0,7 * X) / Y) / (Z * 0,2)

0.4 – the number of remaining anti-tank and tank after shelling and processing of aviation
0.7 – the number of vehicles in the first line
5 (or 6) – a fivefold or sixfold superiority upcoming
0.2 – allowable losses for advanced forces

Source: Тактика (рота, батальон). /под редакцией генарл-полковника Е.И. Крылова. Часть 2. М: Военное издательство, 1991. Pages 199-204, chapter "Advance".

So, my 4x8 table playing the short/wide way should probably have 2 full battalions of Red Army/WARPAC troops, representing the leading elements of an entire regiment?

Ouch. Will have to tell the guys that at the game session today!

Yea, looking this up, I'm planning on using 20" to a kilometer, and a 6' wide table, and running full battalions attacks down the length of it.

If the first one gets chewed up, the second and third waves are there to reinforce the attack, and to break through the NATO defenders.

You can use the same ground scale for 6mm, 12mm, or 15mm vehicles and troops, depending upon what you have available, and what level of game scale compression you can live with.

UsmanK,

Can you check the formua?

"You need to know the number of anti-tank systems (including AIFWs) and tanks in the defending force (denoted for X, in pieces) and the length of the offensive front (denoted for to Y, in km)."

This makes the formula:-

Y = 5 (or 6) * ((0,4 * 0,7 * X) / Y) / (Z * 0,2)

This is unusual as it has 2 Y's AS THE DEFINITION APPEARS TO BE THE SAME. This is OK but unusual. OR has something been lost in translation.

Whichever, it interesting. The US manuals may have a diffrerent opininion on the fact that the shelling reduced their forces to 40% of the original. They claim proivided everybody is in there positions the calualties would be minimal. ;-)

Thanks for a very interesting view.

rayle3,
15 bounds doews seem a long time. I recon if called the next battalion being say 20 km behind would take about an 1 1/2 hrs to get up to get up. Thios is roughly 18km at 25 km/k in colum about 45 mins, deploy to line and move at 8 km/ht about 15 min. Total about an hour plus messing about another 1/2 hour. In our game that would be about 9 bounds. Proably a significant difference to 15 bounds. We use 10 min per bound.

Hello, UshCha!
Yes, I've been lost in translation (more precisely, in the explanation). In general, Y – is a defensive front for the X pieces of defender's tanks and ATGM. Y and X are used to calculate the approximate number of defender's anti-tank systems and tanks for 1 km.
Briefly, the method is – calculated number of enemy vehicles per 1 km, created five-six-fold superiority for the attackers; then simply divided by available in the company / battalion / regiment number of anti-tank systems and tanks. We get the front of the offensive for a company / battalion / regiment.
Honestly, I think it looks a bit schizophrenic, but it is written in the textbook ;)

In this book, there is still a formula for calculating the front attack by counting the average density of small arms fire on a kilometer.

And about the loss – I also think all this is too optimistic. Just do in peacetime military theorists tend to be optimistic ;)

UshCha……great discussion by all. Obviously the time/distance factor we use is the driver for how many turns it takes. One thing I like is that all this points out that in too many fictional WW3 battles the follow on battalions/regiments arrive way too soon. It makes a significant difference.