High Velocity AP round hit-zones

[Yoozername]

The following is a example of what high velocity guns hitting percentages are at 'short' battlefield ranges:

Scenario
A Hetzer unit, the first fielded in the East, is in a battle within a large city and must defend against all armor penetrating into the main city area. The main street is 400 meters long and flat. The Hetzer unit have been issued a limited number of APCR rounds to ensure that their combat debut is successful.

Upon arriving in the city, they are warned by the infantry in regards to Soviet Tanks already in the main square up the 400 meter street. The first Hetzer enters the road and turns and faces a Soviet tank several hundred meters up the road directly facing him.

The commander has no time to judge the range and orders "Zone 1, APCR, FIRE!". The gunner immediately sets the elevation to 0.1 degrees, puts his aiming mark at the center of the Soviet tank and fires.

The attached drawing illustrates the following:

Black ellipse is the triple dispersion area (3 x 50% zone of dispersion) for the APCR round if, indeed, the enemy tank is 300 meters away.

Green ellipse shows the triple dispersion area is the enemy tank is 400 meters away. Note it is a larger dispersion and also wider.

The red ellipse is actually the dispersion about the highest point the APCR round would have flown. It is about 150 meters. In other words, any enemy tank between the Hetzer and up to 400 meters has a great probability of being hit.

The triple dispersion of the 50% dispersion gives the area that 99% of the rounds will land.

[Mobius]

Isn't the 50% zone .67448 x 1 std deviation?

[Yoozername]

It's something like this...turned 90 degrees. Edit: It would be 96% not 99%.

[Mobius]

OK. It's 99% fall within 3 std deviations.

[Yoozername]

Another way to look at this data is to apply the 'Standard Range Estimation Percentage Errors'. If the method was to guess the range, then even a 33% range guestimation-error would have little impact. A very high percentage of hits would occur on first round shots. In fact, even if the target tank were 500 meters, then the 'green dispersion ellipse' would sink further down, but it would still overlay the target. A 50% probability of hitting the lower front might be possible.

Range estimation errors are more a function of range. I was in tehe Army and had LAW training and we were taught to use an extended arm with a fist to guage the range to M60 tanks. If a tank was 300 meters away, I could estimate it's range to within 50 meters.

An observation on the dispersion in the drawing...once dispersion is as tight as the black ellipse, then targetting areas of the tank become possible (on a stationary target). Going for a turret shot or the drivers hatch or a track becomes statistically do-able.

[Mobius]

The 'Standard Range Estimation Percentage Error' method used by Bird-Livingston factors in the std deviation of the ranging error.

[Yoozername]

I assume you mean the posts Rexford/Lorrin made at Battlefront? The 'Double-Dispersion' rule of thumb?

[Mobius]

No, the formula in their book. It spells out a theoretical hit percent based on range estimation error and gun dispersion. That is what I primarily use in my ballistic model. (Plus a few other modifications.)

[Mobius]

Here is a diagram of the Bird-Livingston ranging estimation error for 25% ranging error.
(I blurred up the image as it is copyrighted and they might republish their book.)

The lower diagram is my take on the formula. The problem I found is that it pings on the 100% level at around the 350 meter range. Like your diagram with the T-34/85 the dispersion is less than the height or width of the target so theoretically it is impossible to miss even with a ranging error. To make it more realistic for the game I incorporated a human error factor so the accuracy is never 100%.

This extra error could include parallax and what the British call jump and throw-off.

[Yoozername]

The tactic of firing by zones could, of course, be extended to the 'next zone'. That is, 'Zone 2' might extend from 400-750 meters. The zones would naturally be shorter as they are progressed. This is due to the squared effect of gravity. The basic benefit of this system is a better assurance of first round hits in a timely manner. Overlap of the zones, due to higher velocity and flatter trajectory, allows even a panicky crew to have some chance at success. Basically, at short battlefield distances, AT weapons with a high velocity like the German APCR round, the first round hit percentages would be higher than the graphs show as far as 500-750 meters range. A weapon like the Panther gun, which is in this velocity range for its AP39 round, would be a more common example. I have seen the APCR production numbers for both KWK/StuK 40 and PAK 40 weapons. It's actually lower than people think. Mostly because the 'W' (soft) rounds seemed to have been counted in.

I have read of at least two documents that describe this tactic.

[Mobius]

The book Panzer Gunner is from the point of view of a Panzer IVH gunner. It seems to show that the Pz IV H didn't have any APCR when the author was a gunner in the Spring of 1944 and later.

[Yoozername]

I should have used a jagdpanzer/IV L70 firing pangr 39 instead.

I have seen one source that claims APCR for the KWK/SuK 40 guns was nothing more than 30,000 for the tungsten and about twice as many for the 'W' soft rounds. Production halted in August 43 and stocks were very low at the end of the year. Maybe 17,000 combined APCR+W rounds for the L48 guns. Most large caliper weapons like the 88s had the rounds recalled. There were a lot more smaller caliber rounds especially 50mmL42 and 37mm but these really sucked against sloped armor.

The PAK40 had a similar yet smaller number made.

This is supposedly usage numbers but I suspect they are manufacturing data.

1942:
PaK 40 - Total = 118,000
PaK 40 - PzGr 39 = 42,000
PaK 40 - PzGr 40 = 0 (only 7,700 produced)
PaK 40 - Gr.38 Hl/x = 13,000
PaK 40 - SprGr = 63,000

KwK 40 - no data

1943:
PaK 40 - Total = 1,250,000
PaK 40 - PzGr 39 = 359,000
PaK 40 - PzGr 40 = 42,000
PaK 40 - Gr.38 Hl/x = 374,000
PaK 40 - SprGr = 476,000

KwK 40 - Total = 2,167,000
KwK 40 - PzGr 39 = 499,000
KwK 40 - PzGr 40 = 91,000
KwK 40 - Gr.38 Hl/x = 685,000
KwK 40 - SprGr = 891,000

1944:
PaK 40 - Total = 3,218,000
PaK 40 - PzGr 39 = 1,276,000
PaK 40 - PzGr 40 = 0
PaK 40 - Gr.38 Hl/x = 535,000
PaK 40 - SprGr = 1,407,000

KwK 40 - Total = 3,652,000
KwK 40 - PzGr 39 = 1,083,000
KwK 40 - PzGr 40 = 0
KwK 40 - Gr.38 Hl/x = 394,000
KwK 40 - SprGr = 2,175,000

Any idea how old the Yugo test rounds were?

[Mobius]

In the 1950s I assume so about 10 years. There were still APCR and Hl/B and Hl/C still kicking around.

[Mobius]

Going over this dispersion stuff I show what 3 std deviations (99.7%) cover.

[Edit]Edited image

[Yoozername]

It would probably be clearer if you tried to overlap the dark blue ellipse over my green (400m) ellipse. Also, overlap the lt. blue over my black (300m).

[Mobius]

I'm re-reading the Bird-Livingston formula and it says the std dev is from the aim point.
That would make the 50% zone also measured from the aim point. That would put my ellipses at twice the size. I was assuming the 50% zone and std dev to be the size of a box about the aim point. The evidence for this would be the 88mm Flak firing table which where the 50% Dispersion has height and width columns.

[Yoozername]

Not sure if we are on the same page, but tell me if you agree with this drawing.I wanted to do this with 1000m data but used 800m by mistake. In any case, the actual ovals/circles convey what the 50% data and 3X data and also the one std. dev. means to me. I am remembering discussing this type of data with Rexford and the implications of using Double Dispersion.

The scale is off but I am just trying to convey the idea. Mostly, in order to zero a gun, there must be sufficient accuracy. Even with t6he accuracy that is shown here, zeroing might take a dozen or so rounds.

[Mobius]

From what I'm finding now is that the 50% width should be divided by 2 to find the horizontal dispersion from the aim point. Then that number divided by .67448 to get 1 std dev from the aim point. (Similarly the 50% height should be divided by 2 then divided by .67448 to get the vertical dispersion.) 3 x that distance from the aim point would include 99.72% of possible hits. It looks like Bird (Rexford) wrote such in a post in 2000. But in his book the example he gives with the 88mm at 800m (pages 95-96) it seems he left out the divide by 2 part.

Note:
At 800m range, 50% of the rounds fired at a constant aim will be within a box...
88L56 APCBC
800m....0.3m/0.2m

In the book the SD of the vertical dispersion of the 88mm@800m is given as 0.44. But 0.3m/2 = 0.15m, 0.15m/.67448 =0.222m <> 0.44m
Likewise in the book example the sd of the horizontal dispersion is given as 0.3m. But 0.2m/2 = 0.1, 0.1/.67448 = 0.148m. Not 0.3m.

[Yoozername]

My example would be:

75mmL48 at 800 meters APCBC vert. dipersion=0.4m

To convert to normal distribution, see attached graph. convert the one side of bell curve as follows...

Relate half side of 0.4m converts to 0.2m (divide by two). Red line under graph shows line that is related to yellow cumulative percentage on bottom and total length of red line is 0.4m. This is then 25%. Relate 25% to 19.1% (0.5 of Z score). This is 0.1528m.

0.2mx(.191/.25)=0.1528m

This is then multiplied by 6 to get 99% or multiplied by 5 to get 96%. Note this is just one side of distribution. As per the example, it would be 0.9168 below the Bell Curve and 0.9168 above. This would make 99% dispersion 1.833m on a vertical target at 800m. At 1000m, with a vertical dispersion of 50% and 0.6 meters, it really gets larger than a standard 2m by 2.5m target. Zeroing would be questionable.

[Mobius]

All right, that's one way to look at it.