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Old October 11, 2012, 10:32 PM   #26
JohnKSa
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Join Date: February 12, 2001
Location: DFW Area
Posts: 17,959
Quote:
It is exactly a simple addition of shooter and bullet error on target.
That is correct if we consider only a single shot.

However, it's important to understand that both the rifle and shooter errors are random so they don't always end up pointing in the same direction. The rifle's error for a particular shot might be in the 5 O'Clock direction while the shooter's error for that same shot might be in the 11 O'Clock direction (opposite the rifle's error) from the aiming point. The effect on the target in that case would actually be a shorter distance from the center than either of the errors ALONE would have caused.

In fact, the odds that both errors will end up pointing in exactly the same direction (or in exactly opposite directions) are vanishingly small.

So when you try to calculate the group sizes generated by a 1MOA shooter and a 2MOA rifle you have to consider that both the magnitude of the error (distance from the aiming point) and the angle of the error (direction of the error from the aiming point--say 10 O'Clock or 6 O'Clock) are random.

For the expected group size to be generated by a simple addition of 1MOA and 2MOA, the angular error for the rifle and the shooter would have to line up for each shot which is impossible in any practical sense. To calculate the actual expected group sizes of a 1MOA shooter using a 2MOA rifle, one must use techniques for adding random variables.
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