Pz IV tactics?

[Pyewacket]

Just one question bewteen your discussion:

After reading this thread I made some tests/training with tanks vs. tanks (1943) . Pz IV and Pz.III vs T34. So here is my question, why the Pz.III can eat lots of Russian shells before being KO while the Pz. IV blew up mostly after the first hit? Pz.IV armor is 80 and Pz.III 50+20. So who knows the reason?

[Bone_Vulture]

Probably just tough luck. I've noticed that tanks can go on an incredible streak of resilience, especially against small caliber gun fire. My KV-1 '40 took some five upper hull / turret front penetrations from a German 50mm AT gun, and survived without a single crew casualty. The crew obviously panicked, but otherwise the tank was like brand new.

[Joachim]

Originally posted by Pyewacket:

Just one question bewteen your discussion:

After reading this thread I made some tests/training with tanks vs. tanks (1943) . Pz IV and Pz.III vs T34. So here is my question, why the Pz.III can eat lots of Russian shells before being KO while the Pz. IV blew up mostly after the first hit? Pz.IV armor is 80 and Pz.III 50+20. So who knows the reason?

Jens,

the PzIV turret is 50mm even for the late models. This is the weak area. Especially when hull down most of the hits hit the turret.

Turret hits and the poor ole Pz-IV

Gruß

Joachim

[Keke]

Originally posted by Pyewacket:

Just one question bewteen your discussion:

After reading this thread I made some tests/training with tanks vs. tanks (1943) . Pz IV and Pz.III vs T34. So here is my question, why the Pz.III can eat lots of Russian shells before being KO while the Pz. IV blew up mostly after the first hit? Pz.IV armor is 80 and Pz.III 50+20. So who knows the reason?

In reality Pz-III's spaced armor (50+20) was execptionally resistant against APHE-ammo, even that of Soviet 122mm. I don't know how well this fact is modelled though, but I do know that late Pz-IIIs have better front turret armor than Pz-IVs, and in this game, it means a lot.

[Pyewacket]

Ah, so it's the space and the turret difference!! Thanks, good explanation!

[coe]

question on the effects of hull down.

Let's say you you have an AT gun shooting at

a PZ IV and the AT gun can kill it.

Statistically will the PZIV last longer in

hull down mode or non hull down mode.

I know that in hull down mode theoretically the

AT gun must aim for the turret but since that is

a smaller object wouldn't it be harder to hit?

Likewise in a non hull down PZ IV would the AT gun just aim for the center of the mass?

So in summary, does the supposed harder to hit probability of a PZIV hull down compensate for the fact that a hit (in the turret) is more likely to kill?

[Sardaukar]

Originally posted by Pyewacket:

Just one question bewteen your discussion:

After reading this thread I made some tests/training with tanks vs. tanks (1943) . Pz IV and Pz.III vs T34. So here is my question, why the Pz.III can eat lots of Russian shells before being KO while the Pz. IV blew up mostly after the first hit? Pz.IV armor is 80 and Pz.III 50+20. So who knows the reason?

Spaced armour in Pz III causes the HE-burster in APHE shells to detonate before 50 mm plate. Those shells can be seen in Soviet unit menu having "large HE burster". Thus, 50+20 is lot more than plain 80 mm. Big difference between big shell exploding outside 50 mm armour and big shell exploding inside 80 mm armour .

Cheers,

M.S.

[Dschugaschwili]

Originally posted by coe:

question on the effects of hull down.

Let's say you you have an AT gun shooting at

a PZ IV and the AT gun can kill it.

Statistically will the PZIV last longer in

hull down mode or non hull down mode.

If the turret is the only vulnerable part of the PzIV, it will last longer hull up. That's because while the size of the vulnerable area of the PzIV stays the same, the aim point will be closer to this area (assuming center-of-mass aim) if the tank is hull down, making hits against the vulnerable area more likely.

Dschugaschwili

[Bone_Vulture]

Wasn't there a consensus some time ago that there's always a base chance of hitting the turret, even when the tank is fully exposed? Meaníng that the after the hit probability you see, there's a fixed probability that the damage is dealt to the turret, even when the hull would be an easier target?

Basically, this would mean that it is just as likely that the any scored hit will land on the turret, whether the tank is hull down or up.

[Joachim]

Originally posted by Bone_Vulture:

Wasn't there a consensus some time ago that there's always a base chance of hitting the turret, even when the tank is fully exposed? Meaníng that the after the hit probability you see, there's a fixed probability that the damage is dealt to the turret, even when the hull would be an easier target?

Basically, this would mean that it is just as likely that the any scored hit will land on the turret, whether the tank is hull down or up.

Tests regarding this in the thread I mentioned above suggest that hull down tanks receive more turret hits. Data posted on the hit probabilities support that claim.

Under certain conditions a PzIV lasts longer when hull up.

Gruß

Joachim

[Dschugaschwili]

Originally posted by Bone_Vulture:

Wasn't there a consensus some time ago that there's always a base chance of hitting the turret, even when the tank is fully exposed? Meaníng that the after the hit probability you see, there's a fixed probability that the damage is dealt to the turret, even when the hull would be an easier target?

IIRC the hit distribution for a hull-up tank was:

1/6 lower hull

1/2 upper hull

1/3 turret

For a hull-down tank:

1/4 upper hull

3/4 turret

My above statement about turret hits being more likely against hull-down tanks still stands. I could bore you with a mathematical model that shows this, but I will try to make it easier to understand.

Common sense says that the chance to hit a certain target rises the closer to it you place your aim point (assuming that your sights are properly aligned). If we now define "target" as "PzIV turret" we immediately see that the chance of hitting it increases if you aim at a higher point of the tank (but no higher than the turret center), which is precisely what you do when the tank goes hull down and you're aiming at the center of the visible mass.

Dschugaschwili

[George MC]

I've been following this discussion with interest. I just received a copy of the new Nafziger publication "The German Tank Platoon in WWII: It's Training and Employment in Battle". It has some scenarios which were used in training tank crews/platoon leaders in combat tactics. Anyways the discussion thread and one of the scenarios seem to go hand in hand. So I've built a wee scenario for budding PzIVH company commanders to play with . I've just posted it on the scenario depot so give it a few days for Keith to set it up. Its on the CMBB Battle section, titled "TBP No1 Attack against PAK". Not the most original title in the world. I'd appreciate any feedback on it if you decide to play it.

[Bone_Vulture]

Originally posted by Dschugaschwili:

I could bore you with a mathematical model that shows this, but I will try to make it easier to understand.

Please do bore me, I just finished my homework on statistical math.

[Dschugaschwili]

I feared that somebody would ask me to come up with the mathematical model, so here we go:

Assumptions:

1. The shot distribution is a normal distribution around the aim point (in other words, shaped like a bell curve).

2. I'll ignore shot spread left-to-right and only look at the spread in up-down direction. (This shouldn't be a problem since the different plates are located above each other)

There's a certain problem with the analysis because there seems to be no closed formula for the integral of exp(-x^2), which would be needed to calculate the exact values.

But no matter. What we want is the probability of hitting a plate with a certain size (height) that is located at a certain distance (center) from the aim point (zero). This would translate into the above integral from (center - 0.5*height) to (center + 0.5*height). As mentioned above, getting the exact values isn't really possible, but we can use a table with percentiles for the normal distribution that can be found in every statistics book to approximate. We just have to take the difference between the percentiles for two values that are (height) apart from each other to get the chance of hitting a spot between both edges. That's the hit chance against a plate with size (height).

You can do a couple of example tries, and you'll see that the hit chance against a certain plate decreases the farther the center of the plate is from the mean value of the distribution (the aim point).

When you look at a bell curve, you can see the reason for it (remember, the area under a part of the curve is your hit chance), but I'll leave this as an exercise for the reader.

In gunnery terms, it's exactly what I wrote in my last post: the closer you place the aim point to the center of the target (or the closer you place the target to the aim point), the higher your chance of hitting it.

Dschugaschwili

PS: I'm quite busy right now, so replies from me can take some time.

[Joachim]

Originally posted by Dschugaschwili:

I feared that somebody would ask me to come up with the mathematical model, so here we go:

Assumptions:

1. The shot distribution is a normal distribution around the aim point (in other words, shaped like a bell curve).

2. I'll ignore shot spread left-to-right and only look at the spread in up-down direction. (This shouldn't be a problem since the different plates are located above each other)

There's a certain problem with the analysis because there seems to be no closed formula for the integral of exp(-x^2), which would be needed to calculate the exact values.

But no matter. What we want is the probability of hitting a plate with a certain size (height) that is located at a certain distance (center) from the aim point (zero). This would translate into the above integral from (center - 0.5*height) to (center + 0.5*height). As mentioned above, getting the exact values isn't really possible, but we can use a table with percentiles for the normal distribution that can be found in every statistics book to approximate. We just have to take the difference between the percentiles for two values that are (height) apart from each other to get the chance of hitting a spot between both edges. That's the hit chance against a plate with size (height).

You can do a couple of example tries, and you'll see that the hit chance against a certain plate decreases the farther the center of the plate is from the mean value of the distribution (the aim point).

When you look at a bell curve, you can see the reason for it (remember, the area under a part of the curve is your hit chance), but I'll leave this as an exercise for the reader.

In gunnery terms, it's exactly what I wrote in my last post: the closer you place the aim point to the center of the target (or the closer you place the target to the aim point), the higher your chance of hitting it.

Dschugaschwili

PS: I'm quite busy right now, so replies from me can take some time.

Not to mention that this simple model has the drawback of being one-dimensional... but it delivers the message.

Project the visible parts of a tank on a (huge ) piece of paper. Form a bell from the function of a two-dimensional normal distribution modelling the hit porb. Yes, this looks like a real bell put on the paper

Now the likelihood to hit the tank is the volume of that bell that is above the outlines of the tank in relation to the volume that is outside the boundaries of the tank. The likelihood to hit a certain part is the relation of the volume above that part in relation to the rest. The center of the bell is the aim point.

Gruß

Joachim

[Pyewacket]

IIRC the hit distribution for a hull-up tank was:

1/6 lower hull

1/2 upper hull

1/3 turret

For a hull-down tank:

1/4 upper hull

3/4 turret

mmmh, what I'd like to know is simple the ratio of hitting the tank and missing it.

e.g. let's say in the first test the PzIV was hit with every shot. and in the second every 3rd shot hit, then we would have:

1/6 lower hull * 1,0 = 1/6

1/2 upper hull * 1,0 = 1/2

1/3 turret * 1,0 = 1/3

For a hull-down tank:

1/4 upper hull * 1/3 = 1/12

3/4 turret * 1/3 = 1/4

This would mean that in hull down you get less hits in the turret........

[Dschugaschwili]

Originally posted by Joachim:

Not to mention that this simple model has the drawback of being one-dimensional... but it delivers the message.

In 2d the mathematical model is much more complicated and much harder to understand, mostly because the standard deviations are usually different for both dimensions and the shape of the target area is a problem, too.

But the hit chance still goes down when you move the target away from the aim point in a straight line, so the model is still valid for our purposes.

Dschugaschwili

[Bone_Vulture]

Originally posted by Pyewacket:

What you say?!

Damn, you people are turning this complicated. I think the easiest way to present the formula would be the following:

Cover factor * tank segment factor = hit probability

When the tank is fully exposed, the cover factor is 1. I don't know if tanks can be more or less "hull down", but whatever the case, the formula would go like this:

cover factor (0...1) * tank segment factor (eg. turret, 3/4) = hit probability

More simple... Or not?

[Joachim]

Originally posted by Dschugaschwili:

</font><blockquote>quote:</font><hr />Originally posted by Joachim:

Not to mention that this simple model has the drawback of being one-dimensional... but it delivers the message.

In 2d the mathematical model is much more complicated and much harder to understand, mostly because the standard deviations are usually different for both dimensions and the shape of the target area is a problem, too.

But the hit chance still goes down when you move the target away from the aim point in a straight line, so the model is still valid for our purposes.

Dschugaschwili </font>

[Dschugaschwili]

Originally posted by Pyewacket:

mmmh, what I'd like to know is simple the ratio of hitting the tank and missing it.

The ratio between the hit probabilities against a hull-up and a hull-down tank is highly dependant on the gun, range and crew experience. I suspect that the silhouette rating of a hull-down tank is reduced, so the chance to hit it is calculated against a smaller target. How much it is reduced is BFC's secret.

But since the entire turret and a part of the upper hull remains visible, the probability of hitting a hull-down tank anywhere must be at least as high as the probability of hitting the turret of a hull-up tank. In fact, according to the model I outlined above, the chance to hit only the turret of a hull-down tank must already be higher than the chance to hit the turret of a hull-up tank.

Dschugaschwili

[Dschugaschwili]

Originally posted by Joachim:

But if the turret is not as broad as the hull, the model is no longer a good approximation. The hit chance goes down much faster if you move the aim point towards the turret than towards the ground when you have a small turret like the PzIV. This even increases the effect you want to show.

But I guess most posters here don't like to hear statistics on that level .... sniff

Gruß

Joachim

I was aware that a smaller turret will increase the effect of my model, but I didn't want to scare away potential readers by making it more complicated than necessary.

Dschugaschwili

[Joachim]

The test for CMAK are here

[Pyewacket]

What you say?!

Damn, you people are turning this complicated. I think the easiest way to present the formula would be the following:

LOL, but you're right!

I just wanted to know how many hits the PzIV gets in his turret in hull down and how many when fully visible

More simple... Or not?

no!

[Bone_Vulture]

Originally posted by Pyewacket:

no!

Goddamnit, DIE! :mad:

Or not.

[ February 19, 2004, 01:58 PM: Message edited by: Bone_Vulture ]

[Yggdrasill]

Jeez, and yet people wonder why statistics is confusing. I'm going to try to clarify this, if only to get it straight in my own head. In simple English the question is, for a tank such as the PzIV, that has weaker turret armor than upper or lower hull (frontal), is it safer to be in a hull-down or exposed position?

we assume (common sense) -

1. the probability to hit a hull-down tank (anywhere) is less than the probability to hit a fully exposed tank (anywhere). Joachim gives the statistical rationale:

"Project the visible parts of a tank on a (huge ) piece of paper. Form a bell from the function of a two-dimensional normal distribution modelling the hit prob. Yes, this looks like a real bell put on the paper

Now the likelihood to hit the tank is the volume of that bell that is above the outlines of the tank in relation to the volume that is outside the boundaries of the tank. The likelihood to hit a certain part is the relation of the volume above that part in relation to the rest. The center of the bell is the aim point."

2. given a hit, the probability that it will be on the turret is higher on a hull-down tank than on a fully-exposed tank. This makes sense because the turret on a hull-down tank constitutes a greater portion of the target area than it does on an exposed tank. In Dschugaschwili's model, which looks at the vertical axis only, the turret is closer to the center of the bell curve (fewer standard deviations away) in hull-down than in fully exposed position.

To answer the original question, it depends on:

1. how much reduced the hit probability is overall for an hull-down tank; and

2. what the turret hit distribution is for both hull-down and exposed.

Now, Dschugaschwili argues:

"But since the entire turret and a part of the upper hull remains visible, the probability of hitting a hull-down tank anywhere must be at least as high as the probability of hitting the turret of a hull-up tank. In fact, according to the model I outlined above, the chance to hit only the turret of a hull-down tank must already be higher than the chance to hit the turret of a hull-up tank."

The second sentence is true if we're assuming a hit. But is it true for any shot before it hits? D claims that it is, but the second sentence doesn't necessarily follow when you introduce the second (lateral) dimension into his model (as Joachim pointed out). Pyewacket presented a simple mathematical example of a case where a turret hit is less likely in hull down position. It all depends on the overall to hit probabilities and the hit distribution probabilities for exposed and hull-down. Since we don't know either of these, isn't the question moot?

Here's my question: what game test can we devise that will provide useful data for this question?

thor