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Old October 20, 2012, 07:13 PM   #34
JohnKSa
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Join Date: February 12, 2001
Location: DFW Area
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Quote:
And the biggest group will be 1 + 2 = 3. 3 MOA. This happens when all the errors or variables add up directly.
Correct.

Assuming that the 1 and 2 MOA figures are hard limits--no shots ever go larger than those limits for the shooter and rifle respectively. (This isn't really a good assumption. In reality the figures aren't hard limits, they are a very simplified representation of the probability distribution of the shots on the target. That's why the MOA figures for 10 shots groups is larger than that obtained shooting 3 shot groups. The more shots you get, the better the representation of the probability distribution you get and the farther the extreme shots are likely to be from each other. But the math gets complicated and I don't want to mess with it so I'm going to assume these are hard limits. It will serve to illustrate the probabilities that explain why it's problematic to think about it in terms of straight addition of accuracy figures.)

AND

Assuming the shot distribution over the magnitude limits are uniform. (This isn't really a good assumption either but it will suffice for illustration. The shot distribution would probably be better modeled with a normal (Gaussian) distrubition but I don't want to fiddle with the math.)

AND

Assuming that we accept magnitudes that are 95% or greater as being sufficiently large.

Then you'd need a group containing about 400 shots in order to get JUST the magnitudes to line up so you'd have a really good chance of getting a group that was 2.9MOA or greater.

That completely ignores the error angles which would also have to line up just right to get a group that large. So...

Making the same general assumptions as above and assuming that the error angles have to line up within 18 degrees (5%) to make everything work, then you'd end up again needing a group with maybe 400 shots before you'd have a really good chance of having JUST the angles lined up just right.

If we assume that the magnitude and angle errors are independent of each other and independent from shot to shot.

Then, to get the magnitudes AND angles to line up just right all at the same time--based on the above assumptions, we would need a group with something like 160,000 shots in it to have a really good chance of getting everything to line up just right.

That's why it's much better to talk about averages when we combine accuracy figures than to try to define maximums. Averages provide useful information. If you talk about maximum group sizes instead of averages, the resultant figures aren't very representative of what is likely to be encountered in the real world.
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Last edited by JohnKSa; October 21, 2012 at 11:59 AM. Reason: Corrected exponent on probability calculations.
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