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October 12, 2012, 09:30 PM   #30
JohnKSa
Staff

Join Date: February 12, 2001
Location: DFW Area
Posts: 21,028
Quote:
 My point is, the accuracy of a rifle and its ammo is the largest group they shoot. Half its size is the furthest a bullet will strike from where the rifle's aimed. If the rifle's aimed somewhere inside a 1 MOA circle 'cause that's the area it moves around in when held by a shooter, the furthest a shot will land from the edge of that circle is half the size of the rifle and ammo's accuracy. So, a shooter aiming at a point on the target holds within 1/2 MOA of that point; his area where the sights align is 1 MOA. His desired impact point's in the middle of it. His rifle and ammo's largest groups are 2 MOA. The furthest a bullet will strike from where it's aimed is half that amount; 1 MOA. Therefore, the resultant group on target should the shooter fire many shots will be 1 MOA of holding plus 2 MOA of accuracy and that's 3 MOA. The shooter's shot will land somewhere between his desired impact point and 1-1/2 MOA away from it. The fewest number of shots will be at the outside edge, but they're still gonna be there. They have to be counted in the measurement of the group he shoots. Most of the shots will be inside about 2 MOA, but not all of them. About one third will be in the 2 and 3 MOA range.
Again, the only way that they will add up to 3 MOA is in the extremely improbable case where the vector angle of the errors lines up perfectly on two shots that go in opposite directions on the target and the magnitudes of all the errors are at their maximums.

Each shot has a random error associated with it. That error has both a magnitude and an angle. If you consider that the shooter contributes some error and the rifle contributes some error, now each shot is the vector sum of two random errors. That is the combination of two random magnitudes and two random angles.

To get a group of 3MOA, you'd first need a shot where the random error magnitudes of both shooter and rifle are essentially at their maximums and the random error angles of both the shooter and the rifle are lined up in the same direction. Then you need a SECOND shot where both random magnitudes are again essentially at their maximums and the two random angles are lined up with respect to each other but in the OPPOSITE direction of the two random angles of the other shot where everything lined up.

So you need two shots in your group where all of the following are true.
• All four random magnitudes are at or very near their maximums.
• One of the shots must have two random angles line up very closely.
• A second shot must have two random angles that line up very closely and that are opposite or very nearly opposite the other two random angles from the other shot.
The odds of having that happen in any reasonable number of shots (let alone a group shot with 3, 5, or even 10 shots) are astronomical. Essentially impossible from a practical standpoint.

The bottom line is that a X MOA shooter with a Y MOA rifle will shoot groups that are larger than the larger of the two accuracy figures but smaller than X+Y MOA.
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