Ry pointed out this thread to me on ARFCOM. Probably many people will want to stop at the picture and move on after that but the more interesting part to me is solving the sighting problem.

Here is my thought process on the problem:

The drop is the same regardless of the gun orientation. Keep in mind that drop is independent of point of impact (POI) relative to point of aim (POA).

To solve this problem in general look up the drop for this range on the ballistics table for your ammo.

With the gun zeroed for this range the barrel is angled up such that it compensates for both the drop and the height of the sight (Sight Height or SH) above the bore.

Suppose the drop is 2 inches and the sight height is 1.5 inches. Hence the angle of the barrel is such that the bullet rises, relative to the muzzle, 3.5 inches between the muzzle and the target.

When you invert the gun you have the angle of the barrel giving 3.5 inches additional “drop” to the gravity induced drop for a total of 5.5 inches.

But you have the sight below the barrel which means you “get back” twice the sight height of the total. So the gun will be shooting -5.5 + (2 x 1.5) or 2.5” low.

Hence, the general solution for a gun zeroed at a given range when you turn it upside-down it will have a POI of:

POI = POA + SH – 2 x Drop

Or probably more useful is the POA relative to the POI:

POA = POI + (2 x Drop) – SH

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