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I see what you mean. I went through this by hand and found a discrepancy. The way the formula works is to divide the target height above the aim point by the std. Dev. of the dispersion. 1.22/.74 = 1.648=N Then find the area under the curve you plug that into this: vertical hit%=-.0659 +80.685xN - 2.164xN^2 - 13.276xN^3 + 3.053xN^4 horizontal hit% is done the same way with the ratio of target width to SD of the horizontal dispersion.

horizontal hit% x vertical hit% = final hit %

I found that I had the decimal place off by one in the “-.0659“. Correcting this fixes the dispersion.

Is that error in the game also or just your program?

Can you run a 7,5cmL48 APCBC through 1200m? I want to compare your numbers with the gunner testing that they had to pass. Please run with no error and 20%.

At ~1000m, and no ranging error, you have 62.4%. This would make zeroing the weapon at this range a waste of ammunition. The vert. 50% was 0.6m and the horiz. was 0.4m at 1000m. You can quadruple those numbers and still be at the size of the target.

The 20% range error would almost wash out the majority of the student gunners in the class. That is without motion error either.

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quote:

ORIGINAL: Yoozername At ~1000m, and no ranging error, you have 62.4%. This would make zeroing the weapon at this range a waste of ammunition. The vert. 50% was 0.6m and the horiz. was 0.4m at 1000m. You can quadruple those numbers and still be at the size of the target.

The basic difference comes back to using 1/2 the 50% dispersion to find the std. dev. or the full 50% span. Mathematically it looks like 1/2 should be used. But using that in the formula doesn't produce real results. You'd have to find Lorrin to explain why. Using 1/2 the 50% zone to find s/d gets this:

That's a real result. It looks fairly correct as far as a known range and the dispersion that is listed for the weapon. It may be an ideal condition, but, hopefully, wargames model weapons under a known starting condition.

Not sure why you think I need to find Lorrin. I haven't used his formulas. You have.

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I think I know why there is two times the dispersion in the formula. It is suppose to be calculating the total hit % for a ranging error that overshoots the target and one that under shoots it by a set amount. Instead of doing separate calculations it doubles the hit % from a ranging error of an overshoot. This isn't too much of a problem when the overshoot/undershoot amount is large like more than the height dispersion. But when it is small it must be counting the same error (the dispersion) twice.

That brings up a question in regards to your ranging error math. It seems that you always are modeling some over-range error. It would seem to me, since I actually had military training for this, that the person doing the 'guesstimete' could be under or over or even very close. How do you model this variable?

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quote:

ORIGINAL: Yoozername That brings up a question in regards to your ranging error math. It seems that you always are modeling some over-range error. It would seem to me, since I actually had military training for this, that the person doing the 'guesstimete' could be under or over or even very close. How do you model this variable?

This is a curious part of the formula. Lorrin divides the stated ranging error value by 1.32 to get what he says is the standard deviation of the ranging error. So if the average ranging error is 25% that is divided by 1.32 giving 18.94. The range is then found to 118.94% of the correct range. So a target at 800m the error aim would be at 952m.

I am redoing some parts of the program to try to make it converge at dispersion only hit % when the ranging error goes to 0.

< Message edited by Mobius -- 12/10/2012 8:24:54 PM >

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Continuing the calculations: 1. Find the height of the trajectory as it passes over the aim spot when firing at 952m. Lorrin has some scruffy little formula, I use the naval ballistics program. Let's say it is 1.1m.

2. Find the dispersion standard deviation at the aim point. It is 0 .44m.

3. Find the combined vertical std. dev. Which is the square root of the sum of the squares: sqrt( (step 1)^2 + (step 2)^2) = 1.19m.

4. Now divide half the target (2.5m x 2m) height by step 3. 1.0/1.19=0.84

5. Plug this into the formula that finds the area under the curve of step 4 (S4) std deviations. This is the -.0659 + 80.685*S4 - 2.164*S4^2 - 13.267*S4^3 + 3.053*s4^4. The area under the curve is the hit %. In this case 60%.

6. The horizontal error is just the SD of dispersion at 800m which is .3m. Divide half the target width by this 1.25/.3 which is 4.17 std deviations. The area under the curve is 100%.

7. Multiply the vertical hit % times the horizontal hit% = 60%.

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I'm going to make some changes in the program. The function that figures the standard deviation isn't all too accurate. That needs to be fixed.

Since the program uses a Naval Ballistics program to find the trajectory it fires at sea level elevation. Since tanks don't often fight at sea level I'll raise the elevation to an average of Aberdeen, Bovington and Kummersdorf.

There seems to be some different wording of ranging error magnitudes in reports etc. Some call for 25% average error. Some call for maximum absolute error. Absolute error usually means +/- distance at most. But, 25% average error is not exactly clear.

There is also the issue of ranging to unknown areas, areas with unknown sized targets.

Personally, I think most troops with binoculars ranging on a known height target should get the range +/-20 Maximum. This should include some skill modifier as well as the lucky guess. In other words, the program should use a routine to calculate the error that needs to be applied.

One range estimating skill, useful in the defense, is to use a map. Maps are made very precisely. If you know where you are on the map, and you can sight a known feature (like a T intersection of two roads, then you can easily find a very accurate reading from using the map scale. You can then judge objects as either closer of further than that distance.

< Message edited by Yoozername -- 12/12/2012 5:28:48 AM >

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Yeah, there is no agreement on what a range estimation error is. Is it the maximum percentage, is it the average, is it the standard deviation of all range estimates? The British give this as a percentage sometimes in WO memos and in distance error at specific ranges in others.

This for any accuracy calculation is the biggest fudge factor. If part of a calculation it is at least a consistent FF.

Lorrin in his formula divides the range estimation by 1.32. I have no idea where that came from or why. Except I know what it's affect is. It reduces the number that is combined with his doubling of the dispersion number so the square root of their squares works out to a realistic number. But if dividing the error by 1.32 gives the s.d. then the range error is about at the 81% level of possible error.

What I do is use the range estimation factor as a place where a combination of standard crew training and optics quality show up.

Seems gun accuracy is always a topic on forums of tactical military games.

< Message edited by Mobius -- 12/13/2012 3:22:08 PM >

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I've been working to improve the formula. It doesn't double the dispersion factor like in the B-L equations. What happens now is that the ranging error miss distance becomes relatively more important to accuracy. For regular ammo the change is only about 2-3% difference. But high velocity, low over-aim point of the target flight does see a definite improvement. Here the old formula is compared to the new. The APCR now shows that it is more accurate than the APHE ammo even though it has a higher dispersion.

There are also other benefits with higher velocity ammunition. Shooting at moving vehicles is improved due to decreased flight time as well as flatter trajectory. Shooting, and observation of shot, is also improved due to decreased flight time.

SOP may have been observed as far as range limitation, i.e. 'Not to be used over 800 meters, but in certain circumstances, it may have been used even if APCBC would have penetrated. By this I mean a circumstance where enemy vehicles are advancing within a dangerous range to the shooter. The shooter would not care in this circumstance if the targets were T34/76 or JS-1. They would have used the APCR to quickly get more assurance on hits and destruction since time, even mere seconds, is of the essence. Granted, I have no proof other than common sense.

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Surprisingly all the APCR, even for Soviets does as well in accuracy or better than the standard rounds now. At least for under 2000m. As for what ranging error means, I'm going to have to go with the standard deviation of the range error.

They are taking questions a WoT blog on what to ask Doyle and some Russian expert on German tanks and stuff. I posed the 75mm/L46 muzzle velocity quandary to someone gathering questions for them.

In the book by the Panzer IV gunner. In 1944 he only had a choice of 3 types of rounds for the 75mm/L48. APHE, HE and smoke.

< Message edited by Mobius -- 12/20/2012 8:49:27 PM >

In late 43 and 1944, it was certainly pulled for certain calibers. Especially 75mm/88mm. I really wonder about the Yugo tests and how they got any.

But there is some very late war evidence that it may have been issued if it was still around. The Eisenhower report has claims by US tankers of Germans using "souped-up" ammunition. But who knows what they meant. The German 75mmL70 and 88mmL71 were souped up already.

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We of course can speculate. Preparing for invasion when fresh units came to occupy Normandy and other occupied Western European territory they would of brought along their weapons and the ammo. If they hadn't used the APCR already they would of just stored it. Also, if ammo was shipped in 1943 when it was being produced some would have gone to Western Europe. There was no need to use any of it until June 1944.

Given the accuracy of the laser range-finders in modern weapons systems, and the removal of the range error this gives...the over-accuracy of even a 25mm weapon system like this is highly effective.

quote:

Findings: For a dispersion level (0.8 mils) that was near the maximum allowed valuefor training ammunition, HPs were (a) 90% or greater for an 8-foot square zeroing'target for ranges of 800 meters or less and (b) less than 50% for a fully-exposed frontal view of a BMP at ranges as short as 1600 meters. For the maximum allowed dispersion level (0.5 mils) for armor-piercing ammunition that is fired from a AFV, analysis indicated (a) a 90% or greater HP for an 8-foot square zeroing target at ranges of 1200 meters and less, (b) a 68% HP for a fully-exposed frontal view of a BMP near the tracer-burnout range, and (c) a 90% dispersion zone that was about 2 mils in diameter.

A pretty good site. I wish he finds a more secure means of linking pics. I remember a long thread at BF where I argued about SF being used for range finding. It seems that some versions could do so!

I was also re-reading parts of 'Panzer Gunner'. He clearly uses the 'map-method' to estimate a range of 1500 meters during an ambush on Soviet Armor and resupply trucks. The Germans also had a 'pen-sized' measuring device that could be used with a map to measure distance. It basically had a small wheel and mechanism inside. It could be used to measure a road march or strait distances.

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After getting a few little bugs out of my Ballistics program I am now able to recreate German accuracy tests to a high degree in most cases. It is also useful in reverse engineer and find the dispersion values the Germans used for the tests. Someone posted the 88mm/L71 firing tables and with the proper drag formula the program generates it.

The first column of Hit% is for dispersion only. The second Hit% column is for 2 x dispersion.

I have seen the first column data on WIKI and FPRADO, etc. But the footnote for 1) seems to say "Double the 50% Accuracy (dispersion)"?

In any case, the other columns are interesting in that they seem to support my first post in regards to 'range' settings for 'hit-zones'. Similar to 'Battlefield' sight settings, setting a range is a quick way to assure a high percetage hit for each zone. The target height, the critical data point, is only 2 meters in these tables. The other column seems to give maximum height and also the range that achieves maximum height.

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They must of calculated by hand using the lateral and vertical dispersions listed of each row range and not used a single number multiplied by range like I do. Since they round to a single decimal place I was surprised that my numbers came out so close.

The third/fourth columns on the right page may be doing something like your first post was saying.

Also, knowing that their 'accuracy' is based on pure dispersion and 'firing' on 2 x dispersion you can work back to what they were using for dispersion on guns. Even if you cannot find the original firing tables of the weapons, which is often the case.

What sticks out to me is the crossover point at 1000 m for the 'known range' dispersion-only accuracy. It is 100% at 1000 meters and then decreases with further range. Knowing that, and knowing the target height of 2 meters, the 0.5 m 'Hohe' (height) dispersion value of 50% is easily applied. Basically, the full dispersion of 100% is 2 meters!(see drawing and previous posts regarding dispersion curves)

If we look at the 2100 m range values, we see that the 50% height dispersion has doubled to 1 meter. This results in a 82% chance of a hit. This doubling of the 50% dispersion can be thought of as the rectangles in the drawing increasing in height by double. This would move the 2% and 7% rectangles off the target. 2+7+2+7=18. 100-18=82%. If we go back to the 1000 m values, we see roughly 82% for the 'double-dispersion' column.

As far as I am concerned, this agrees with what I have been saying about 'double-dispersion'. That is my story and I'm sticking to it.

I suppose one way to look at this is...A crew, with a EM34 (WWII German Rangefinder) firing at a stationary enemy T34 (76) at 2100 meters, has about the same chance as a crew firing at 1000 meters under battle conditions and using range estimation (stationary target). Sort of a rule of thumb I suppose.

In actuality, the dispersion is rather loose at 1000 meters. Considering that is the range that the gun would use as a zeroing point. In engineering terms, there is no band-gap assurance. In other words, even out-liers are on the bleeding edge.

I really wonder what percentage of wargamers have a clue about statistics. They seem to 'understand' data but have no idea how the engineering world uses it. Hit or miss, I suppose.

< Message edited by Yoozername -- 7/27/2013 5:00:54 AM >