.50 / 20mm as AT

[Mr. Johnson--]

Interesting Jason, though I don't know If I agree with you percentages for KOs.

To find out the proper numbers, we need to build some ww2 tanks, guns, and ammo. And proceed to test out every combination extensivly. Which is not going to happen.

Don't know if you guys have tried making some scenarios with lots of german 50mmAT guns supplied with like 50 T rounds and no AP vs lots of allied tanks. You see lots of non-KO penetrations. Very diffrent then normal CMBO battles. Totally gives you a feel for the early war years when Tungesten was more abundant.

[Jarmo]

Hit points for the tanks!

[Puff the Magic Dragon]

<blockquote>quote:</font><hr>Originally posted by Mr. Johnson-<THC>-:

To find out the proper numbers, we need to build some ww2 tanks, guns, and ammo. And proceed to test out every combination extensivly. Which is not going to happen.

<hr></blockquote>

Why this? This has been done already - both on shooting ranges and from 1939 - 1945 on the battlefield. I really can't believe that there are no stats about it!? Thousands of eyewitness have reported their experiences. Where is all this material?

[Redwolf]

Hm, Jason, don't you think the energy values must be taken from *after* the penetration? After passing a nearly-stopping amount of steel, a big round is obviously in the same class as a smaller round after less resistance.

I also think this cannot go without possible exits of the round. A 88 L/71 round hitting a Greyhound doesn't leave much energy in the vehicle, but goes through with most of it.

[JasonC]

There are two cases - barely penetrated and overpenetrated.

You might think the first would mean little chance of damage, but actually that is not the case. The energy is conserved. If the armor lets the round stop in the tank, all the energy is going to still be there. The round may be slowed, but only by accelerating pieces of the armor. This is different from a successful richocet, where the surviving round carries way most of the energy, and the remainder is spread over the whole weight of the tank. So even marginal penetrations, if achieved by high enough energy rounds, are going to do substantial damage.

In fact, the heaviest guns (e.g. the Russian 122mm, at 7.5 million Joules, or the German 128mm at 10 million) could probably wreck most tanks even without a penetration. You are getting up near energies high enough to lift the turret off. They just usually penetrate, too.

Then there is the case of overpenetration. It is quite true that an M3 halftrack is not going to absorb 5 mJ when an 88L71 passes clean through it. Most of the energy goes out the other side, as the remaining kinetic energy of the departing shell. As a result, the total energy delivered to the M3, is going to plateau, and a short 75mm AP (1.3 mJ) or an 88L71 (5 mJ) are both going to do comparable things.

However, those comparable things are still going to be knock-out nasty. A 3-3.5 inch hole in the side of the vehicle, with all the armor that use to be there transformed into secondary projectile, in quite small splinters going very fast. Plus killing anyone and breaking anything along the actual flight path.

There will always be some chance of the part of the vehicle that gets hit, being less than critical. That is why even the heaviest guns still have chances of only M-killing the vehicle, for example. And if you plot my KO chance figures against the absolute muzzle energy, I think you will see there is already significant diminishing returns. Thus, 1.3 mJ Short 75 e.g.) already has .75 probability of a KO, and increasing that to 3 mJ - a factor of 2.3 times - only increases the KO chance to .85, or only 1.13 times. Double that again, and the knock out chance only increases marginally.

It is easy to see that this is the only reasonable way to model the fall off in hit effects do to "wasted energy" from overpenetrations. It is not going to be the case that e.g. a (German) 37mm gun has a higher chance of KOing a halftrack than an 88mm. The -portion- of the 37mm's energy delivered is certainly higher. But there is a lot more in the 88mm.

Then look at the "combination" effects of a 90% KO chance for one, and a 40% KO chance for the other. Does it mean the 88mm will -always- do more damage? No. 4% of the time, the 88mm would only damage the vehicle, while the 37mm would KO it (lowest 10% "roll" by the 88, high 40% "roll" by the 37mm). But more than half the time, the 88mm would kill it while the 37mm would only damage it. The 88mm round has 20 times the energy, but only 2.25 times the KO chance.

Otherwise put, 4 rounds of the 37mm, with only 1 mJ all told, will KO the vehicle just as reliably as five times as much energy from 1 round of the 88mm. (.6 ^ 4 = .13, or 87% chance of KO, plus many multiple damage results that will amount to the same). The difference in "delivered energy" needed - better for the 37mm by 5 to 1 - reflects the "wastage" of energy by the 88mm - the fact that 4/5ths of it probably flies out the other side of the halftrack.

That is all captured by the diminishing returns built into the KO chances, therefore. They go up less than linearly as the energy of the round rises. Making the needed "plateau", with random variations from shot to shot.

[Redwolf]

Good points.

I now agree that the energy spent in the actual penetrations is not "lost" for the damage effect. I was assuming that in my last post.

Edited: But whether that applies to HC rounds is a different matter. A nearly-stopped HC round should have lost much of its damage potential, unless I overlook something else.

[ 11-11-2001: Message edited by: redwolf ]</p>