First Round Accuracy of German Tungsten Core Ammo

[rexford]

The general approach to calculating hit % against rectangle targets follows this long drawn out process:

1. take the 50% dimensions and convert to a 68.26% zone, which is one standard deviation.

Say 50% zones of 1m vertical and 0.6m lateral, which become 68.26% zones of 1.5m and 0.9m.

2. Compute vertical and lateral hit probability.

Assume 2.0m vertical height. Divide 2m by 68.26% vertical zone to obtain 1.33 standard deviations.

Assume 2.5m lateral width. Divide 2.5m by 68.26% lateral zone to obtain 2.78.

Using a standard deviation vs area table for normal distributions:

1.33 standard deviations relates to a probability of 82% that the vertical scatter will fall within the top and bottom heights of the target.

2.78 standard deviations result in a 99.5% chance that the lateral scatter will end up with the left and right limits of the target width.

3. find overall hit %

Multiply the vertical probability by the lateral probability to obtain the overall hit %, or 82% x 99.5% = 81.6%.

There are many ways to do the calculations but the above method is "easiest" for rectangles.

It took me several months to get this straight about 20 years ago.

So a 1m vertical 50% zone has an 82% chance of landing within the height limits on a 2m high target if we disregard lateral scatter.

If we double the test dispersion the 50% vertical zone becomes 2m, the 68.26% vertical zone becomes 3m and the vertical hit % decreases to 50% (2m/3m = 0.667 standard deviations).

In this case, doubling the vertical 50% zone decreases the vertical hit probability from 82% to 50%.

And doubling the lateral 50% zone from 0.6m to 1.2m lowers the lateral hit % from 99.5% to 84% (2.5m/1.8m standard deviations).

So doubling the lateral and vertical 50% zones lowers the overall hit % from 82% to 42%.

Note that MANY writers indicate that the double dispersion hit percentage is what a gunner could do under a whole lot of different situations. I don't know if the double dispersion is anything but a guess as to how wide rounds with constant aim might scatter during the stress of combat.

[rexford]

Originally posted by Mr. Tittles:

These are numbers collected by British Army Operational Research Sections during WWII (summarized in WO 291/180) Ranges are in yards, report indicates that the target is assumed to be a approximately the size of a Tiger Ie. Hit probability also assumes no crew error in line or range estimation.

Versus a Hull-Up static Target

6 pdr @ 500yrds…..100 percent chance of a First Round Hit (FRH: first round hit)

6 pdr @ 1000yrds…100 percent FRH

6 pdr @ 1500yrds…96 percent FRH

6 pdr @ 2000yrds…87 percent FRH

17 pdr @ 500yrds……100 percent FRH

17 pdr @ 1000yrds….100 percent FRH

17 pdr @ 1500yrds….100 percent FRH

17 pdr @ 2000yrds….98 percent FRH

17 pdr @ 2500yrds….93 percent FRH

Probability of a hit on first round, hull down static Tiger Ie sized target, assumed no error in line or range by crew.

Versus a Hull-down static Target

6 pdr @ 500yrds…..85 percent FRH

6 pdr @ 1000yrds…43 percent FRH

6 pdr @ 1500yrds…22 percent FRH

6 pdr @ 2000yrds…14 percent FRH

Versus a Hull-down static Target

17 pdr @ 500yrds…..88 percent FRH

17 pdr @ 1000yrds…51 percent FRH

17 pdr @ 1500yrds…29 percent FRH

17 pdr @ 2000yrds…18 percent FRH

17 pdr @ 2500yrds…12 percent FRH

If a Tiger is assumed to have a frontal area of about 3.2m width and 2.6m height, taking into account ground clearance area and a turret that is narrower than the hull and tracks, the 1500m single vertical dispersion hit %'s would be (calculated from German 50% zone data an disregarding lateral line scatter and range estimation errors):

The 50L60 APC would score 100%.

The 88L56 APCBC would score 100%.

The 88L71 APCBC would score 99%.

The 75L48 APCBC would score 92%.

So the British 6 pdr gun and ammo would be slightly higher than the German 75L48 APCBC, and the 17 pdr would be superior. I suspect that the British were shooting uncapped AP but who knows for sure.

Against a hulldown Tiger at 1500m (0.6m turret height), the 75L48 APCBC hit % would be 31% based on the 50% zone height of 1m. This is higher than the Brits got for the 6 and 17 pdr guns and ammo, which suggests that the British results were more spread out than the German (German results bunched around middle like bell shaped curve, British results may not be as close to bell shape curve due to limited number of rounds).

[ August 23, 2004, 05:30 PM: Message edited by: rexford ]

[Mr. Tittles]

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Originally posted by Mr. Tittles:

Thats what I thought it was. You seem to have been saying that the zone lengths were both multiplied by the factor. That would spread out the rounds and make zeroing a nightmare.

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REXFORD: They are both multiplied by the conversion factor.

If the 50% zone is 1m high and 0.5m wide, the 90% zone is 2.44m high and 1.22m wide (1.22m left and right of center and 0.61m left and right of center).

And the doubled dispersion for the 90% zone would be 4.88m high and 2.44m wide.

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Rexford: Plot the data on a piece of graph paper.

For a total single dispersion distance of 1.0m (lateral and vertically), the spread of shots is a smooth curve with 50% within 0.5m of the center. And 68% within 0.75m of the center. And 95% within 1.0m.

Mr. Tittles: You seem to be saying two different things. We are just talking single dispersion here. My origional question still stands; if 50% of the rounds land in 1sq m, where do the rest land?

Could you address this? It is very contradictory.

[rexford]

"For a total single dispersion distance of 1.0m (lateral and vertically), the spread of shots is a smooth curve with 50% within 0.5m of the center. And 68% within 0.75m of the center. And 95% within 1.0m."

Oops.

If 50% vertical zone is 1m long (half the shots land within a 1m vertical distance, half above mid-point, half below):

50% will be within 0.50m of center of box

68.26% within 0.74m of center (used 1.48 multiplier)

80% within 0.925m of center

95% within 1.45m of center

95.5% within 1.48m of center

[Mr. Tittles]

75mmL48 APCBC 1000m

If 50% vertical zone is 0.6m long (half the shots land within a 0.6m vertical distance, half above mid-point, half below):

50% will be within 0.30m of center of box

68.26% within 0.444m of center (used 1.48 multiplier)

95.5% within 0.888m of center

So nearly every round would fall in a 1.8m height difference at 1000m. This is so close to a 2m target height that zeroing at 1000m would be tricky and certainly would need some ammo expenditure.

I would like to reiterate a point I tried to convey in another thread. That is, in the field, the Germans would not be able to zero their tank guns so that 1000m would land at 1000m. This is in regards to teh Fibels calling for extra range because the guns are zeroed to a range. Nonsense. They are zeroed to a target at a range. Ie they already have height built into the zeroing.

[Mr. Tittles]

75L48 APCBC

100m....0.1m/0.0m

300m....0.2m/0.2m

500m....0.3m/0.2m

800m....0.4m/0.4m

1000m..0.6m/0.5m

1300m..0.8m/0.7m

1500m..1.0m/0.9m

2000m..1.6m/1.3m

2500m..2.4m/1.8m

3000m..3.3m/2.3m

So for a 75mmL48 APCBC firing at 1500m, the vertical spread could be up to 3 m.

This is so sloppy that walking rounds in would be nearly impossible. The main technical difficulty is that the overall area spread of the rounds is greater than the target area size. And this is with a known range mind you. Trying to adjust after the first round is a waste of time. You would have to fire two or three rounds without adjusting just to get some idea where your 'firing-box' is centered.

The fact that 50% of the rounds WOULD have the needed precision makes me question still the methodology used here.

If 50% of the rounds land in 1 sq m and the other 50% land in the surrounding 8 sq m; is this a bunching up bell function?

[ August 24, 2004, 09:08 AM: Message edited by: Mr. Tittles ]

[rexford]

A German speaking friend of mine at work, whose father flew German aircraft at a production center during the war, translated the ballistic table column headings for me. Here are some of the more interesting ones:

a. Hit Percentage for a target area of 2.5m x 2m

two percentages are given, second one is in brackets

The footnote associated with hit percentage in brackets translates as:

" Parenthetical values valid for battle condition dispersion (double 50% dispersion)"

b. Area for a target height of 2m

This column gives the range where the trajectory height is 2m or less when the gun and the aim point are at the same elevation, which roughly corresponds to a battlesight type aim (aim at target bottom).

The double dispersion is nothing more than a guess as to how well the aim could be placed on the target during hectic battle conditions, and doubling the dispersion has an accuracy impact equal to adding a random dispersion of 1.73 times the base dispersion.

If the base scatter 50% zone is 1m, and a 50% zone dispersion of 1.73m is added, the statistical sum is 2m.

If the 68.26% zone is 1.48m (1m x 1.48) and another 68.26% zone of 2.56m (1.73 x 1.48) is added, the statistical sum is 2.96m (2m x 1.48).

In other words, if a bell shaped curve with a 50% zone of 1.0m is combined with another bell shaped curve with the same 50% zone, the 50% zone that results from the two curves has a length of 1.41m. To obtain a 50% zone length of 2.0m one must combine 50% zones of 1.0m and 1.73m.

Assuming that battle introduces a random scatter zone equal to 1.73 times the base value is rather severe, and the doubling may be a rough estimate without much real validity. We have not been able to find anything more on the justification for double dispersion.

[ August 24, 2004, 04:55 PM: Message edited by: rexford ]

[rexford]

Here are some additional dispersion figures for the 50% zone based on German wartime tests (muzzle velocity in brackets):

German 50mm Pak 38 HE (550 m/s)

1000m: 0.4m vert and 0.3m lat

1500m: 0.8m vert and lat

German 76.2mm Pak 36 HE (550 m/s)

1000m: 0.7m vert and 0.6m lat

1500m: 1.5m vert and 1.1m lat

2000m: 2.2m vert and 1.5m lat

German 88L56 HE (810 m/s)

2000m: 2m vert

German 88L56 HEAT (600 m/s)

1000m: 0.7m vert and 0.4m lat

1500m: 1.2m vert and 0.6m lat

2000m: 1.8m vert and 0.8m lat

German 76.2mm Pak 36 HEAT (450 m/s)

1000m: 1m vert and lat

1500m: 2m vert and 1m lat

2000m: 3m vert and 1m lat

German 75mm JG 37 u.42 (Patr 38 H1/A)(355 m/s)

HEAT

1000m: 0.84m vert and 0.70m lat

1500m: 2.13m vert and 1.15m lat

Russian 122mm HE (800 m/s)

1050m: 0.65m vert and 0.3m lat

2050m: 1.40m vert and 0.7m lat

Russian 76.2mm field gun BR354A APBC (662 m/s)

1050m: 0.55m vert and 0.55m lat

2050m: 1.30m vert and 1.20m lat

[rexford]

Originally posted by Mr. Tittles:

75L48 APCBC

100m....0.1m/0.0m

300m....0.2m/0.2m

500m....0.3m/0.2m

800m....0.4m/0.4m

1000m..0.6m/0.5m

1300m..0.8m/0.7m

1500m..1.0m/0.9m

2000m..1.6m/1.3m

2500m..2.4m/1.8m

3000m..3.3m/2.3m

So for a 75mmL48 APCBC firing at 1500m, the vertical spread could be up to 3 m.

This is so sloppy that walking rounds in would be nearly impossible. The main technical difficulty is that the overall area spread of the rounds is greater than the target area size. And this is with a known range mind you. Trying to adjust after the first round is a waste of time. You would have to fire two or three rounds without adjusting just to get some idea where your 'firing-box' is centered.

The fact that 50% of the rounds WOULD have the needed precision makes me question still the methodology used here.

If 50% of the rounds land in 1 sq m and the other 50% land in the surrounding 8 sq m; is this a bunching up bell function?

First off, they did not fire at a 2m by 2.5m target, which is a point I have made several times. The actual target was much bigger.

It would help if you would stop thinking about the area within that box.

Secondly, you have not looked at how the rounds would be distributed through the entire area with a 1m vertical and 0.9m lateral 50% zone.

Those 50% zones (1.0m/0.9m) are associated with 68.26% zones of 1.44m/1.30m vertical/lateral.

50.00% are outside 1.00m/0.90m.

31.74% are outside 1.44m/1.30m.

25.00% are outside 1.66m/1.49m

20.00% are outside 1.85m/1.67m

15.00% are outside 2.07m/1.87m

10.00% are outside 2.37m/2.14m

05.00% are outside 2.82m/2.55m

01.00% are outside 3.72m/3.36m

00.50% are outside 4.03m/3.64m

2.07m vertical and 1.87m lateral captures 85% of the random scatter, and the size has to been approximately doubled to capture another 14.5% (4.03m/3.64m).

My point has been made.

[Mr. Tittles]

50.00% are outside 1.00m/0.90m.

31.74% are outside 1.44m/1.30m.

25.00% are outside 1.66m/1.49m

20.00% are outside 1.85m/1.67m

15.00% are outside 2.07m/1.87m

10.00% are outside 2.37m/2.14m

05.00% are outside 2.82m/2.55m

01.00% are outside 3.72m/3.36m

00.50% are outside 4.03m/3.64m

You truly think outside the box.

Half are in 0.9 sq m. The other half are spread out in a little over 8 sq m. If we double the 0.9 sq m to 1.87 sq m we only get 68%. If we triple it, we get about 3/4.

It just seems that there is a concentration and too much non-concentrated.

[ August 24, 2004, 05:01 PM: Message edited by: Mr. Tittles ]

[rexford]

British report WO 291/762 discusses a firing test with five 6 pdr armed Churchill IV's against a target representing a Panther turret (2' high and 5' wide).

The difference in accuracy between the best three and worst two is eye opening:

AVERAGE HIT PERCENTAGE WITH KNOWN RANGE

FOR APCBC ROUNDS

Churchill.500.800.1000.1500 yards

BEST 3......89%.84%..81%..62%

WORST 2...52%.52%..34%..12%

AVG FOR 5.74%.73%..62%..42%

---------------------------------------

75L48........90%.71%..57%..30%

75L48 hit % based on single dispersion.

The German dispersion figures probably ran in a similar fashion to the British, although the small sample of five Churchills may overstate the typical differences one might observe.

At any rate, it is likely that ALL tanks and guns from WW II followed a bell shaped distribution curve, with extremely accurate guns that could place shots in a close ring, and very bad guns that put shots "all over the place".

[rexford]

Originally posted by Mr. Tittles:

50.00% are outside 1.00m/0.90m.

31.74% are outside 1.44m/1.30m.

25.00% are outside 1.66m/1.49m

20.00% are outside 1.85m/1.67m

15.00% are outside 2.07m/1.87m

10.00% are outside 2.37m/2.14m

05.00% are outside 2.82m/2.55m

01.00% are outside 3.72m/3.36m

00.50% are outside 4.03m/3.64m

Thats a 158.24%

You truly think outside the box.

That's not what it means.

If one looks at the distribution of vertical dispersions within the curve, 50% are greater than 1.0m and 50% are below 1.0m.

5% are greater than 2.82m and 95% are below 2.82m.

Take a few days and think about it.

[Mr. Tittles]

A German speaking friend of mine at work, whose father flew German aircraft at a production center during the war, translated the ballistic table column headings for me. Here are some of the more interesting ones:

a. Hit Percentage for a target area of 2.5m x 2m

two percentages are given, second one is in brackets

The footnote associated with hit percentage in brackets translates as:

" Parenthetical values valid for battle condition dispersion (double 50% dispersion)"

HIT PERCENTAGE AT A KNOWN RANGE

TUNGSTEN CORE AMMUNITION

DOUBLED DISPERSION

RANGE...50L60...75L48...88L56...88L71

100m..........100......100........100.......100

300m..........100......100........100.......100

500m............98........99........100.......100

700m............84........88..........95.........98

900m........................69..........87.........93

1100m......................52..........74.........85

1300m......................37..........63.........76

1500m......................25..........52.........66

1700m......................19.......................58

1900m......................14.......................50

Just to be clear, did the German report specify that its for a known range? Shouldnt this be titled 'First Round Hit Percentage...'?

Could you post both columns? I take it this data is second column? I would still like to see 75mmL48 HE and HEAT data (both columns)

Thanks

PS I edited my post above. Almost within a minute of your post.

[Mr. Tittles]

Ive written many test procedures and the manufacture of tank guns and antitank guns would have certainly had an acceptance criteria.

More than likely it would be something like:

1. mount test sight (this would be the same sight used for all tests, the sights would also have an individual acceptance test but the gun is proofed with a 'known-good' sight).

2. Boresight weapon to 25m to get sight and barrel aligned. This could be a large piece of paper with two crosses on it. One for the barrel and one for the sight. This will rough in the weapon.

3. Fire 10 rounds at 100m. 5 must be in a rectangle no larger than x/y. 8 must be within X/Y. Adjust shot group to center of target to align sight. Refire test.

4. Fire 10 rounds at 1000m. 5 rounds must be in x1/y1. 8 must be within X1/Y1. Adjust as in the first test. If more than one round is outside X2/Y2, then repeat test.

The x's and y's are obviously derived from the test data. The 50% zone is certainly known.

The gun is therefore calibrated for repeatabilty and accuracy. To me, this would be a minimal test procedure. The gun may actually need to be fired several times before testing just to break it in.

[rexford]

Originally posted by Mr. Tittles:

A German speaking friend of mine at work, whose father flew German aircraft at a production center during the war, translated the ballistic table column headings for me. Here are some of the more interesting ones:

a. Hit Percentage for a target area of 2.5m x 2m

two percentages are given, second one is in brackets

The footnote associated with hit percentage in brackets translates as:

" Parenthetical values valid for battle condition dispersion (double 50% dispersion)"

HIT PERCENTAGE AT A KNOWN RANGE

TUNGSTEN CORE AMMUNITION

DOUBLED DISPERSION

RANGE...50L60...75L48...88L56...88L71

100m..........100......100........100.......100

300m..........100......100........100.......100

500m............98........99........100.......100

700m............84........88..........95.........98

900m........................69..........87.........93

1100m......................52..........74.........85

1300m......................37..........63.........76

1500m......................25..........52.........66

1700m......................19.......................58

1900m......................14.......................50

Just to be clear, did the German report specify that its for a known range? Shouldnt this be titled 'First Round Hit Percentage...'?

Could you post both columns? I take it this data is second column? I would still like to see 75mmL48 HE and HEAT data (both columns)

Thanks

PS I edited my post above. Almost within a minute of your post.

They didn't call it first round hit percentage because it would be for any shot where the range was precisely known, which almost never occurs on the battlefield.

In the next few days I'll look into how dispersion influences second and follow-up round corrections after a miss.

I looked in the Sherman gunnery manual for sight adjusting of the gun, which included a procedure with a distant aiming point. To adjust for a shot at 1000 yards, it did not involve any actual shots. One checked to see that the gun and sight were laid on precisely the same point.

The next section addressed ADJUSTMENT OF SIGHTS USING TESTING TARGET, which was used when distant aiming points were not available and was a second choice to distant aiming.

"Place the testing target from 80 to 120 feet in front of the tank, at approximately the same height from the ground as the gun." One then "adjust the sight to put the zero range line and the vertical lind of the sight on the mark for the sight."

The next procedure is ADJUSTMENT OF SIGHTS FOR 1,000 INCH FIRING. With 1,000 inch firing trials (about 83 feet or 25.4m) dispersion would not be an issue.

The Americans checked the sights without having to use 1000 yard shots, and thus avoided having to deal with random dispersion in the vertical and lateral directions.

[Mr. Tittles]

The next procedure is ADJUSTMENT OF SIGHTS FOR 1,000 INCH FIRING. With 1,000 inch firing trials (about 83 feet or 25.4m) dispersion would not be an issue.

The Americans checked the sights without having to use 1000 yard shots, and thus avoided having to deal with random dispersion in the vertical and lateral directions.

This is laughable.

[rexford]

Originally posted by Mr. Tittles:

The next procedure is ADJUSTMENT OF SIGHTS FOR 1,000 INCH FIRING. With 1,000 inch firing trials (about 83 feet or 25.4m) dispersion would not be an issue.

The Americans checked the sights without having to use 1000 yard shots, and thus avoided having to deal with random dispersion in the vertical and lateral directions.

This is laughable.

Please explain why it is so funny, and also detail how the Germans boresighted their guns.

[Mr. Tittles]

At 25m it would not be apparent that a gun had unacceptable dispersion. The rounds would make holes that overlap or very nearly overlap. Even a 75mm Sherman could do this. The repeatability would allow the gun to be zeroed for center fire and adjustment to the crosshairs. But unless the gun is fired at some range (hopefully 500m+), the real accuracy of the gun is not apparent.

Its basically an open ended rough-in at best. By zeroing at a close range and then testing at a real battle range, you have a test grounded at two points and a basis for accuracy. The real nature of the dispersion must be taken into account if it follows a distribution as you say.

Look at all the data for the German guns at 100m. In some cases its 0 vertical and horizontal. Can you really say what the dispersion is by looking at the 100m data? What about 25m data? Do you think the dispersion goes away by ignoring it?

Ana analogy is as follows:

You put a 4 in by 4 in pole in the ground. Its 4 feet high. You climb to the top and nail another 4 by 4 at the top and its nailed at its very end. Its hanging off into space 4 feet.

You claim that you used the overlapping 4 inch area to insure its square and therefore you know that at 4 feet away from the pole, you can predict where the other end is.

[ August 28, 2004, 08:56 AM: Message edited by: Mr. Tittles ]

[Mr. Tittles]

http://tiger1.info/EN/Target.html

Heres the German gun calibration target for roughing in the gun at 50m. I have seen pics of Germans putting fine wire across the end of the gun tube for just this purpose. The next procedure would be to fire at a known range.

Recently I read of a Tiger I that could not adjust on Target. Every correction brought the rounds either too close or too far. The range was around 1000m and the crew would have nailed the enemy long before the third round. A technician arrives and finds the sight was loose! Every firing moved the sight a little so that corrections were meaningless. He tightens sight, roughs it in and fires at 1000m to zero the gun.

[Mr. Tittles]

In Tiger Tanks-Green on pg 46, there is a report from British testing of a Tiger I gun. Its stated that a 5 shot group of 16in by 18in was obtained. Note that it is not saying that is the 50% zone, its saying that all 5 rounds landed in that area at 1200 yards (1100m~). 3 out of 5 rounds fired at a moving target (15mph) were also hits. I would assume they used a known range method.

[Mr. Tittles]

Krupp’s Main Range, Meppen

For guns required to be very accurate (eg. tank and anti-tank guns), about 95 per cent of rounds fell within a rectangle 60 cm. high and 30 cm. wide on a vertical target at 1,000 metres. It was stated that the dispersion for line is less than for height. The ratio of lateral to vertical dispersion remains substantially constant at all "flat-trajectory" ranges for any gun. The angular dispersion also remains nearly constant with range in the flat-trajectory region.

Herr Hengefeld, interviewed at his house in Meppen, gave the following figures for bore tolerances, which were required throughout the length of the bore 7.5 cm gun : High 7.51 cm. Low 7.50 cm.

8.8 cm. gun : High 8.81 cm. Low 8.80 cm.

The shoulders of all shot were finished by centerless grinding.

[Mr. Tittles]

60cm*30cm=278 sq in

16in * 18in = 288 sq in!

I would like to see any dispersion data besides 50% zone data. I would actually like to see 95% data and compare it to 50% data.

[rexford]

Originally posted by Mr. Tittles:

At 25m it would not be apparent that a gun had unacceptable dispersion. The rounds would make holes that overlap or very nearly overlap. Even a 75mm Sherman could do this. The repeatability would allow the gun to be zeroed for center fire and adjustment to the crosshairs. But unless the gun is fired at some range (hopefully 500m+), the real accuracy of the gun is not apparent.

Its basically an open ended rough-in at best. By zeroing at a close range and then testing at a real battle range, you have a test grounded at two points and a basis for accuracy. The real nature of the dispersion must be taken into account if it follows a distribution as you say.

Look at all the data for the German guns at 100m. In some cases its 0 vertical and horizontal. Can you really say what the dispersion is by looking at the 100m data? What about 25m data? Do you think the dispersion goes away by ignoring it?

Ana analogy is as follows:

You put a 4 in by 4 in pole in the ground. Its 4 feet high. You climb to the top and nail another 4 by 4 at the top and its nailed at its very end. Its hanging off into space 4 feet.

You claim that you used the overlapping 4 inch area to insure its square and therefore you know that at 4 feet away from the pole, you can predict where the other end is.

They were not testing for dispersion, they were checking the mean point of impact and a short range was effective because dispersion would be so small as to be a non-factor.

[rexford]

Originally posted by Mr. Tittles:

60cm*30cm=278 sq in

16in * 18in = 288 sq in!

I would like to see any dispersion data besides 50% zone data. I would actually like to see 95% data and compare it to 50% data.

British 90% zone for 17 pdr APCBC is 4.1 minutes vertical and 3.46 minutes lateral at 1000m.

Compare that to 88L56 APCBC 50% zone at 1000m and see which is better.

[rexford]

“In Tiger Tanks-Green on pg 46, there is a report from British testing of a Tiger I gun. Its stated that a 5 shot group of 16in by 18in was obtained. Note that it is not saying that is the 50% zone, its saying that all 5 rounds landed in that area at 1200 yards (1100m).”

Range is close to 1100m.

All 5 test rounds land within a rectangle of 0.41m lateral by 0.46m vertical. German dispersion data has a 50% zone at 1100m of 0.44m vertical by 0.26m lateral.

So it would seem, at first glance, that the Tiger test mentioned above was an above-average weapon or the ammunition it fired was better than average, interms of vertical dispersion. The lateral dispersion isn't out of line with a 50% zone of 0.26m at 1100m.

5 rounds does not prove anything because even a poor gun with large dispersion could conceivably place 5 rounds in the small area.

It also has to be remembered, as shown in the test with five Churchills, that wide variations could occur from one tank to another and that the average dispersion would not apply to most tanks.

“Krupp’s Main Range, Meppen

For guns required to be very accurate (eg. tank and anti-tank guns), about 95 per cent of rounds fell within a rectangle 60 cm. high and 30 cm. wide on a vertical target at 1,000 metres. It was stated that the dispersion for line is less than for height. The ratio of lateral to vertical dispersion remains substantially constant at all "flat-trajectory" ranges for any gun. The angular dispersion also remains nearly constant with range in the flat-trajectory region.”

95% of rounds expected to land within a 0.60m vertical x 0.30m lateral box at 1000m. Tiger dispersion data at 1000m has 50% zone of 0.40m vertical and 0.23m lateral, which relates to a 68.26% zone of 0.60m vertical and 0.345m lateral.

95% zone based on Tiger dispersion data for 50% zone works out to 0.60m x 1.96 or 1.18m vertical, and 0.345m x 1.96 or 0.68m lateral.

Looking at the 95% zone from the German 50% zone data and the statement in quotations, it seems that the author of the above statement forgot to double the 0.60m x 0.30m figure, which probably represented the bottom or top half of the target rectangle instead of the entire upper and lower halves.

I would also guess that his statement was an approximation. 0.6m x 0.3m. It is obvious that the 75L48 APCBC could not come close to the 0.6m x 0.3m 95% zone.

British 17 pdr APCBC has a 90% zone of 4.1 minutes vertical and 3.46 minutes lateral, which is much greater than 0.6m x 0.3m at 1000m.

I question the above statement and feel it is the half size box.