Panther turret weak point

[Guest Big Time Software]

Lewis,

The shot angle calculation was an approximate one, but here's how I did it. Figure the time of flight for the shell (say 1.3 seconds at 1000 meters for the 76mm). Then, knowing that gravity accelerates at 9.8 m/s, we know that the vertical velocity (V, in m/s) of the shell at time T (in seconds) is:

V = S - 9.8*T (where S is the starting vertical velocity)

Integrating, we know that the vertical position (P) of the shell is:

P = ST - 4.9*T*T

Assuming that firer and target are at vertical position P=0, we can solve this equation:

0 = S(1.3) - 4.9(1.3)(1.3)

or

S = 6.37

In other words, the gun had to be angled upward enough to impart a vertical velocity of 6.37 m/s for the shell to arc up, then down, to the same height as the firer after traveling 1000m.

Now we solve to get that angle. Let's break the velocity of the shell down into vertical and horizontal components. This gives us a right triangle. The hypoteneuse is 790 (m/s) - the gun's muzzle velocity - and is the (initial) path of the shell. The vertical component - the side opposite the angle at which we're "firing upward" (A) - is 6.37 as we calculated above.

The law of sines tells us that

sin (90deg) / 790 = sin (A) / 6.37

or

sin (A) = 0.00806

or

A = 0.46 degrees

If we assume that the shell arcs downward at the same angle at which it arced upward, then A is also the angle at which the shell strikes the target. (Note: this is not quite correct, as Lewis pointed out, but it's close enough for our purposes).

So that's where I got the half-degree result. Note that I have not taken into consideration things like dynamic deceleration of the shell due to air resistance, etc. So this is just an approximate calculation, but it gives a pretty good ballpark result. You mentioned that the real striking angle is higher than the firing angle, and that makes sense due to the decelerative effects of air resistance. But even if that doubled the angle (and I doubt it would) we'd still only have one degree of change here, which isn't enough to make much difference.

<BLOCKQUOTE>quote:</font><HR>Are hit area calculations on a percentage distribution? That is, once you get a hit on a tank, is the exact area like the lower hull a percentage basis on its square area?<HR></BLOCKQUOTE>

Basically yes, but we reduced the chance of a lower hull hit further, figuring that often times at least a small part of the lower hull would be obscured by intervening terrain (not enough for hull-down, but enough to cover some of the tank).

- - -

Paul,

You're saying that we should lower the Panther's armor quality further? Are you sure that wouldn't unfairly make the Panther an easy target? I want to make sure I understand you correctly.

Please note that CM accounts for BHN separately from the armor quality rating. So a lower BHN is already modeled without having to lower the quality rating as an approximation.

Charles

[Paul Lakowski]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Big Time Software:

Paul,

You're saying that we should lower the Panther's armor quality further? Are you sure that wouldn't unfairly make the Panther an easy target? I want to make sure I understand you correctly.

<HR></BLOCKQUOTE>

Yes and No. The armor in the vicinity of these areas should result in weakened zones but often the armor is thicker to ofset this effect,some times this works like in the Tigers front mantle 100mm cast mantle plus 100mm turret plate = ~ 140mm effective resistance similare to the mantle thickness in the middle.

The problem is when you read of projectile x penetrating armor Y is it due to hitting a vunerable area....to that end any 'poor quality of german armor by wars end is going to be expressed as localized weakened areas or plates and not nessesarly a across the board loss for all tanks of that type/year.

We just got a report showing that Russian T-34s in an effort to speed up production suffered from poor or incomplete welding ...the result was visible gaps throught the joints of some of the models exhaimed..this doesn't apply to all, more of a quality control problem.

[Guest Big Time Software]

Paul,

Very interesting. If you're right, then that would suggest that perhaps a better way to simulate, say, 85% quality would be to randomize it and on any given shell hit the percentage could be anywhere from 70% to 100%. So the average would be 85%, but you could get higher or lower on any given hit.

The only problem with this is that I wonder if it might allow too many "fluke" kills on a Panther.

Charles

[Lee]

Hmm, you could lessen that "fluke" kill problem by narrowing the range a bit.

Say random 80-95% armor resistance. The extra 5% on the high side

would acount for the chance of hitting heavier resistance as in

the case with the Tiger's mantle. If anything, that would short

change the Panther a bit, as much heavier resistance could be met if

a hit took place where there was some thick armor overlap. But, for now

at least, I think this would be a reasonable solution. Perhaps later

a more thorough examination could be made of extra armor resistance in

the Panther at key points and some more precise numbers offered for

a more accurate simulation of this.

Charles, Paul, thoughts?

[This message has been edited by Lee (edited 11-30-2000).]

[Paul Lakowski]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Big Time Software:

Paul,

Very interesting. If you're right, then that would suggest that perhaps a better way to simulate, say, 85% quality would be to randomize it and on any given shell hit the percentage could be anywhere from 70% to 100%. So the average would be 85%, but you could get higher or lower on any given hit.

The only problem with this is that I wonder if it might allow too many "fluke" kills on a Panther.

Charles<HR></BLOCKQUOTE>

You've just described a "Ballistic limit S curve", yes this does happen not only with AP but APFSDS and HEAT, but hold of for now I've just put the question 'Why" to Robert Livingston and he had several interesting things to say [ as always]

< i >

Paul, the data is based on several test shots against Panther glacis' in NW Europe, using standard issue US ammo, as well as data from multi-shot test programs using US ammo against US plate over a range of hardness, with measured amounts of flaws from "none" to "a lot"

Paul,

The reason that armor has less ballistic resistance is generally that the armor does not cohere under impact. As you are well aware, this can be expressed as hardness/softness or ductile/brittle continuums. Both hardness and ductility are essential. However, the steel must also be homogeneous, and if a gradation of hardness is deemed desirable, as in KC ship armor, or Face Hardened tank armor, the transition from hard to less-hard areas must be gradual and even. Any unevenness in the interior molecular structure of the steel creates stress-points which accumulate stress and may initiate interior cracking under load.

These stress points may be caused by laminations (layering in the steel) or by inclusions (specks of carbon, dirt, phosphorous, sulfur). Stress points are also created at air bubbles and hot tears (found in T-34 bow, and KV turret castings, respectively). The Panther glacis I mentioned in the other message (tested by the Allies) had a zone of brittle steel in the center section, like the center layer of a three-ply piece of plywood. Hardness was about 262 BHN throughout the thickness. The interior layering would weaken the armor, even though thoe outer layers were OK for hardness and brittleness.

I recall that the lower bow plate (45mm) of a T-34 tested at Aberdeen also showed this layering. Another source of stress-point formation is incorrect heat treatment. The Germans used a multi-step "timed quench" process which involved hoisting steel in and out of a bath several times; times were given to the second, but given the size of the pieces, different areas would get a different time of immersion.

Imagine this with a multi-ton Panther glacis. Stress points can be created by this kind of uneven quenching. The US used water jets within the quenching bath to even-out the hardening and tempering of large pieces like Sherman turrets and hulls. Their problem was they were simply too thin, not that the armor was of poor quality (except the pre-'44 made Shermans, which had a nasty tendency to be too thin AND flawed.) -- Robert

</ i >

Heres a quote from Jonas Zukas [ one of the world leaders in ballistics research] talking about the problems of quality control applied to modern ballistics tests….

< i >

"For example, rolled homogeneous armor (RHA) steel is used extensively in military construction. It can safely be said that RHA is rolled. It is also used as armor. However, the military speci"cations that govern the production of RHA have wide tolerances so that it is anything but homogeneous. Material properties (primarily hardness) are known to vary by as much as 10% within a lot of RHA and up to 30% from lot to lot. This makes single tests (the famous ` one-shot statistics a ) useless and correlation between numerical results and experiments unlikely unless a statistically meaningful number of tests have been done. Simple go/no}go ballistic tests can cost upward of $2000 each. Instrumented "eld tests can run from $10,000 to $100,000 each. As a rule, then, a statistically signi"cant data set is almost never available."

</i >

OK I'll ask him about the ballistic limit S curve and see what he says...

[Lee]

I want to clarify something here. The extra 5% of possible quality

would only apply to front turret hits, since I assume that is where the extra

armor would be encountered. Perhaps leave the hull at 85% and just make

the front turret 85-100% quality? That would seem to be a pretty good fix

until some more exact thickness numbers can be had for this armor overlap.

[:USERNAME:]

<BLOCKQUOTE>quote:</font><HR>Originally posted by Big Time Software:

Lewis,

The shot angle calculation was an approximate one, but here's how I did it. Figure the time of flight for the shell (say 1.3 seconds at 1000 meters for the 76mm). Then, knowing that gravity accelerates at 9.8 m/s, we know that the vertical velocity (V, in m/s) of the shell at time T (in seconds) is:

V = S - 9.8*T (where S is the starting vertical velocity)

Integrating, we know that the vertical position (P) of the shell is:

P = ST - 4.9*T*T

Assuming that firer and target are at vertical position P=0, we can solve this equation:

0 = S(1.3) - 4.9(1.3)(1.3)

or

S = 6.37

In other words, the gun had to be angled upward enough to impart a vertical velocity of 6.37 m/s for the shell to arc up, then down, to the same height as the firer after traveling 1000m.

Now we solve to get that angle. Let's break the velocity of the shell down into vertical and horizontal components. This gives us a right triangle. The hypoteneuse is 790 (m/s) - the gun's muzzle velocity - and is the (initial) path of the shell. The vertical component - the side opposite the angle at which we're "firing upward" (A) - is 6.37 as we calculated above.

The law of sines tells us that

sin (90deg) / 790 = sin (A) / 6.37

or

sin (A) = 0.00806

or

A = 0.46 degrees

If we assume that the shell arcs downward at the same angle at which it arced upward, then A is also the angle at which the shell strikes the target. (Note: this is not quite correct, as Lewis pointed out, but it's close enough for our purposes).

So that's where I got the half-degree result. Note that I have not taken into consideration things like dynamic deceleration of the shell due to air resistance, etc. So this is just an approximate calculation, but it gives a pretty good ballpark result. You mentioned that the real striking angle is higher than the firing angle, and that makes sense due to the decelerative effects of air resistance. But even if that doubled the angle (and I doubt it would) we'd still only have one degree of change here, which isn't enough to make much difference.

Basically yes, but we reduced the chance of a lower hull hit further, figuring that often times at least a small part of the lower hull would be obscured by intervening terrain (not enough for hull-down, but enough to cover some of the tank).

- - -

Charles<HR></BLOCKQUOTE>

WOW! A detailed answer to one of my questions from Charles. I am honored.

While I agree with you that the angle is not great for a high velocity round; there are other things going on here.

First off, you solved for a constant velocity or "rocket" application on a planet with no atmosphere. That is; you are assuming that the velocity of the projectile is constant (more about this later) and the atmosphere isnt there. But you are "correct" in that the angle for this particular application does not get much larger than what you have calculated.

While you roughly used some physics and threw in a bit of calculus; Let me build on it.

Your missing bit of info for this boundary value problem is the final velocity. Its the value that should be used in the law of sines equation. Again it is not that great a difference , I figured from using 1/2MV^2 and the differences in penetrations at 100 and 1000 meters that its probably around 725 M/s at 1000 meters or so and AGAIN you are right; it does not make that big a difference.

The height difference between the gun on a sherman and the lower hull armor on a panther should be used. But it also would only add a tenth of a degree or so. Not so big also.

BUT. I believe something else is going on here that has escaped most people.

Gravity. It occurs to me that the vector force of gravity is in fact aiding the downward sloped armor!

While "sloped-back" armor (like the upper hull on the panther) is TRYING to bounce the penetrator off itself but is being HINDERED by gravity, the lower hull is actually being helped by gravity. That and the angle that the shell really is descending at made this area less vulnerable than it seems like ingames like PE and CM. There should be a different formula for downward sloped armor!

I cant prove it but it seems logical. Anyway, the germans DID lower the value of armor in this area from 60mm to 50mm. They also did not match the lower hull armor on the king tiger with the upper hull armor. It was 100mm vs 150 mm for the lower and upper hull respectively. Its actually pretty close to the ratio of the panther; 50 to 80mm. I also think its the reason that the panther had such a problem with the turret front armor deflecting shot downward. It really was a design flaw and exploited by allied armor that got close enough to aim for it.

Lewis

[This message has been edited by :USERNAME: (edited 12-04-2000).]

[This message has been edited by :USERNAME: (edited 12-04-2000).]

[:USERNAME:]

It also occurs to me that close range plays a big part here also.

A gun can only depress so much (so in cases like the soviet tanks with 2-3 degrees of depression, this area could not be hit at certain minimum ranges) and the downward angle is definitely going to help the lower hull area.