The Panther's H.Dv.119/325 "Firing Table"

[critical mass]

Am still on research leave and away from my books. I apologize if I didn´t made it clear enough:

(Pzgr39 striking at <650m/s (?) with 704m/s muzzle velocity)

that less than 650m/s was my estimate for 600m distance, which I cannot verify here but Mobius seems to have been able to find the reference velocity (i.e. 638) in the meantime. By no way I intent to attribute this exact velocity to the -IVl at this range.
The difference in penetration between both data is subtle, and the higher velocity data was also choosen to guard against my own error.
Therefore, the refernce penetration at 638 is even lower, yet the projectile defeated the frontal aspect in two out of two cases (I presume under ideal conditions in regard to target angle), where it was at least 120mm thick -in possession of sufficient section thickness to theoretically completely keep out the 75mm Pzgr39, had the cast armor been of the same quality as german RHA.
OKH mentioned that the armor quality of the IS2 (presumably the early, stepped front type) cast was too low, allowing successful penetrations by PAK 40 and KWK40 from usual "battle range". "Quality" however, can have multiple charakters. Speculating here, but I use to think partly because a later remedy was attempted by substituting softer plates in IS2 production, that overly hard armor in original production (good against soviet AP projectiles but a poor choice vs non-deformable (under those conditions) Pzgr39) was part of the problem. Of course, it could also have included one or more of the following issues, which are commonly encountered in thick cast armor:

inferior non-metallic inclusions,
inferior quantity of gas bubbles in the cast matrix,
improper heat treatment, or
britellness caused by differential cooling rates between surface and centre of thick sections subjected to air cooling

[Yoozername]

I wonder if anyone has the exact methodology the Germans used to create the FT height and width? How many rounds were fired at each range? I imagine it would be something like a dozen rounds for each range? To get a good statistical dispersion data set, 32 would be a good number, but that would be quite a number of rounds. I suppose the data is taken from a new weapon that has been 'broken in'?

The 'rectangle' (or square) describes an area where 50% of the rounds dispersion will land. Each shot is therefore measured for a distance in the height and width away from the aim point. This data is then combined for all the rounds to find the 50% median values. I always thought that a better way would be 68% or a Probability density function. 50% being +/- 0.6745 sigma.

[critical mass]

The following pages detail how the FT dispersion was taken, what defines an outlier data (wild shot), how large the sample size was. The actual source is a ww2 KM gunnery naval document, though the therein described principles are relevant also for the Army to the best of my knowledge.

[Yoozername]

double

[Yoozername]

Wish I could read it!

I think I have posted this link before in other threads...pg. 120-123 are Panther

http://wwii.germandocsinrussia.org/ru/n ... ect/zoom/5

[critical mass]

A short summary in engl.

[+] while the dispersion was theoretically endless whenever the number of test rounds is endless, in practical terms the total dispersion (sans "wild shots") was to be found within an area, four times as large as the 50% dispersion zone

[+] The exact half of the 50% dispersion zone (either horizontally or vertically, or in range), was called the "probable deviation "w" ("wahrscheinliche Abweichung"), and contained approx. 25% of all falls of shots. The total dispersion area is roughly eight times as large in size as "w" (graph).

[+] The area of the probable deviation "w" was calculated as follows, each seperately for deflection ("Seite"), height ("Höhe") and distance ("Länge"):
[1] the arithmetic mean of all shots is calculated and thus provides the mean point of impact (notice: mean point of impact is not the target centre)
[2] now the individual deviation for each shot is calculated in reference to the mean point of impact
[3] the mean of all deviation distances is calculated and called "mean deviation" ("durchschnittliche Abweichung")
[4] according to practical experiences and records of the prooving ground, as well as supported by the probability theory, there is always a relationship between the "mean deviation" and the "probable deviation "w"". The "probable deviation "w" is always 0.8453 times the "mean deviation".
[5] according to this relationship, the "probable deviation" and the "50% dispersion" (beeing two times as large as the "probable deviation") can be readily calculated with the mean deviation. This 50% dispersion is then to be entered in the "Schußtafeln" under the columns "50% Abweichung in Höhe, Seite und Länge".
[6] Before final entry, a check has to be made against wild shots. Wild shots ("Ausreißer") have to be excluded from the calculation.
A wild shot is considered a fall of shot which exceeds the 50% dispersion by a ratio defined in dependency to the number of test shots (9 to 80, table).
[7] if a wild shot has been positively identified, the steps [1] to [7] have to be repeated

[Yoozername]

Thanks! I appreciate the efforts you put in this. But, just to be clear, they would fire 9-80 rounds at a range? Surely, 9 for close ranges and 80 for long ranges??? Technically, they are not shooting for accuracy, but rather for dispersion. Or, another way to say, they are looking for the characteristic. They are not trying to zero the weapon IOW. It is a bit 'retentive' and many would be satisfied with 95% 'assurance' and leave the outliers out of it. 3 times the 50% would give 95%?

How would this compare to other nationalities?

[critical mass]

The concept of creating the "mean deviation" ("durchschnittliche Abweichung") in german procedure is identic to what the US Army and US Navy called "true mean dispersion". Similarely, the 50% zone of dispersion as well as the probable deviation are known, too, under the terms "50% zone" and "probable error".

US: True Mean Dispersion x 0.846 = probable error
german: "mean deviation" ("durchschnittliche Abweichung") x 0.8453 = "probable deviation "w" ("wahrscheinliche Abweichung")

US True Mean Dispersion x 1.692 = 50% zone
german: "mean deviation" ("durchschnittliche Abweichung") x 0.8453 x 2.0 = "50% Streuung"

There are differences in what constitutes an outlier as well as how many shots are to be fired.

Soviet practice is not too different, though they only refer to mean deviation (true mean dispersion in US terminology). That is, the average deviation distance from the mean point of impact (50% radius from centre, unfortuna). Presuming a normal distribution pattern, and in order to produce a 50% zone instead, one would have to check and multiply their data by 1.692.

A more elegant approach would be to normalize ALL data to standart deviation figures. It follows according to the probability theory that a standart deviation equals:

sigma = 1.2545 times the true mean dispersion / mean deviation
sigma = 0.742 times the 50% dispersion zone
sigma = 1.484 times the probable error

[Mobius]

I'll post this again.

Rheinmettal handbook formula.jpg (17.94 KiB) Viewed 1140 times

I believe the Russians used the term срединныe отклонение 'median deviation' not 'mean deviation'. You were correct over at the Tank Archives were you said their average is 1/2 the 50% zone. I.e multiply by 2 to get the 50% zone.

[Yoozername]

From D 2003 dated July 1, 1943

[critical mass]

I believe the Russians used the term срединныe отклонение 'median deviation' not 'mean deviation'. You were correct over at the Tank Archives were you said their average is 1/2 the 50% zone. I.e multiply by 2 to get the 50% zone.

I hope they don´t use median. Median is not suited to metric scale levels. Instead it is best applied to ordinal scale levels. The factual use of median instead of mean would drag the consequence behind that the data are not comparable at all with other data, which use arithmetic means.

example for mean and median:
five shots
10
10
11
23
27

mean = 16.2
median: 11

The median only parts a distribution into two similarely large samples while ignoring the individual data (unlike the mean, it´s only a seperation of datapoints, not a seperation of the data themselves, in the example above ´11´lies in the centre of the group with two datapoints in front each and behind). It completely escapes me why one would use statistics for ordinal scale levels while the data at hand are metric scale levels...

There are cases, where mean and median would fall together (a perfectly symmetrical distribution) but because the scale level is different, comparison should not be attempted.

[Thoddy]

[quote="Yoozername"] It is a bit 'retentive' and many would be satisfied with 95% 'assurance' and leave the outliers out of it. 3 times the 50% would give 95%?
[quote]
three times the 50% gives the 96% Zone

visualisation of the 50% zone within the area of events (the 100% area is not exactly 100% but slightly less)

[Mobius]

I hope they don´t use median. Median is not suited to metric scale levels. Instead it is best applied to ordinal scale levels. The factual use of median instead of mean would drag the consequence behind that the data are not comparable at all with other data, which use arithmetic means.

Remember this is theoretically a Guassian distribution - your example is not.
This is how the Russians are using it. (Interesting they use shell ballistics in an ordinary online statistics site.)
Beside it is a portion of a Russian firing table.

[critical mass]

Thanks Mobius.

Normal distributions will be encountered but not necessarely in symmetrical condition. Once the pattern has a skewness towards either centre or peripherie (or both in a bimodal distribution pattern), the median will make for incorrect data. But whatever that will be, Your graph shows clearly that the soviet methodology is approximate similar to that used elsewhere, except for that they do use in their FT tables what the germans call "wahrscheinliche Abweichung" and what the US and british call "probable error", instead of the more commonly encountered 50% dispersion zone.
This is also substantiated by a close comparison with Thoddy´s graph above, also showing the pattern sizes. Both graphs show EXACTLY the same ratios:

2%-7%-16%-25%-25%-16%-7%-2%
(bold: 50% zone composed of two brackets with 25% each. Soviet dispersion is reported in one bracket, german dispersion are given in two brackets)

i.e. the soviet Wa (deviation in height) and Wb(deviation in deflection) represents exactly half the 50% dispersion zone (50% Streuung in Höhe / Breite, respectively). Therefore, Your position is right and mine has been wrong.

.

[seppw]

Panther gun next pages:

Hello Miles,
thank you for your contribution. Do you have the Kommandotafel for the PzGr39(p.10,11,12?), too? I'd appreciate it a lot, because I'm currently working on a spreadsheet visualizing the hit probability(parameters: target size, range est. error) for as many WWII guns(+ammo) as possible.
I need the shell travel times in order to calculate the vertical displacement between the center of the target and the actual aim point.
If you were so kind to upload the Kommandotafel for the PzGr39, you'd save me a 12h trip to the library and back.

Regards Sepp