 This topic has 2 replies, 3 voices, and was last updated 6 days, 14 hours ago by Guy Farrish.

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09/07/2024 at 16:53 #200421John D SaltParticipant
Ivan Sorensen’s thread on how many shots it needs to hit a tank seems to have sunk beneath the tide of new articles. I resurrect the question here for the purposes of scaring people further with some more numbers, or, at least, further torture of existing numbers, in accordance with Guy Farrish’s idea to “make spuriously precise hit tables … to lend an air of gravitas to wobbly home made rule sets”.
We had, you might recall, three data points for WW2 hit probabilities against tanks:
12 rounds at 300 metres (Biryukov and Melnikov, “Antitank Warfare”, 1972)
910 rounds at 1000 metres (ibid)
20 rounds at 1500 metres (TRADOC Bulletin 8, “Modern Weapons on the Modern Battlefield”, 1975)Let us imagine for a moment that we are mad, and want to produce a hitting table for a wargame using 2d6 and no better numerical basis than this.
Since we are mad, we are entitled to make a few simplifying assumptions (“Assume all cows are perfectly spherical”, things like that). One assumption will be that the target we are shooting at is the same size as a NATO tank target, 2.5 by 2.5 metres, giving a target area of 6.25 square metres.
We will also assume that tanks are a little bit like cows. We already know that they look smaller when they are further away, and can use the mil relation to work out how much. Although the NATO tank target is a rectangle, it will be convenient to treat it as a circle of the same area.
As well as being mad, we are lucky, so it happens that we have lying around a spreadsheet written some years ago at the request of a nice man from the Bisley target committee that calculates the CEP that corresponds to a given hit probability against a circle of specified radius at a specified range. CEP, I am sure we all know, is the Circular Error Probable, the radius of a circle within which 50% of shots will fall.
Using this, we can calculate that the CEP you would need (in mils) to get hit probabilities corresponding to our three data points against a circle 6.25 square metres in area would be:
Range (m) P(hit) CEP (mils) 300 0.75 2.103 1000 0.11 2.150 1500 0.05 2.186
The arithmetic mean of the CEPs is 2.146, but as it seems a bit fiddlesome to be specifying things to a precision of 1mm at 1000 metres, we will take 2.15 as a good enough number.
It is now very easy [using the formula P(hit) = 1 – 0.5^((r/CEP)^2), where r is the target radius] to produce a graph of hit probabilities for all ranges up to, ooh, let’s say 3000 metres, and round these off to match the probabilities one gets from rolling 2d6.
Range Roll 150 3 200 4 250 5 300 6 400 7 500 8 700 9 900 10 1500 11 3000 12
If the following graph has come out right, it will show that it’s not a bad approximation to the P(hit) calculation we did.
Since rolling 2d6 to hit reminds me of the hit rolls used in Charles Grant’s wonderful old “Battle!” rules, I include the hit probabilities he suggested, which have lived on in a multitude of “modified Grant” rules such as “Hans und Panzer” ever since.
All the best,
John
09/07/2024 at 19:33 #200435Darkest Star GamesParticipantThank you, Jon. This is helpful indeed, and i appreciate that it can be taken without having to worry about the “what targeting site was used, how cold is the powder inside the shell, was the casing made on a Tuesday after 3pm, was the gunner caffeinated…” caveats that usually come with scientific information and allows the rules writer to create modifiers specific for their game.
"I saw this in a cartoon once, but I'm pretty sure I can do it..."
09/07/2024 at 19:55 #200438Guy FarrishParticipantThanks John, interesting and distressingly useful. I may have to do something with it!

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