A standard Warner rear sight, our reference unit, has 40 pitch lead screws. This equals 0.025 inches of movement of the rear sight for each revolution of either the elevation or windage knob. This 0.025 of movement equates to 3 minutes of the sight aperture displacement, not at the target. By dividing the 0.025 of movement by 3, one minute would be worth 0.0083333333 inches of aperture displacement.

It is now a simple proportional problem.

One Min @ 3600 Inches (100 Yards) = One Min Rear Sight Movement

(1 Min @ 100 Yds) (1 Min @ Rear Sight)

1.047197536 0.008333333

___________ = _______________

3600 Inches X

(100 Yards) (Distance Between Sights)

(0.008333333)(3600)

X = _________________ = 28.64788903 Inches

1.047197536

The 28.64788903 Inches is the theoretical distance required between the front and rear sight aperture planes to obtain One Minute of precise movement at the target.

To calculate a correction factor for a specific set of sights, measure the distance between the two aperture planes. Determine this distance to within 1/8th of an inch. Divide this measurement by the 28.6479. The results will be your correction factor for your rifle.

Example
Should your sight planes be 36.5 inches apart and you want to know how many minutes to move your rear sight for 600 yards to 1000 yards, this is all you need to do.

Divide the 36.5 by 28.64788903, this equals 1.274 (rounded off). This is your correction factor. Normally a 30 caliber 155 grain Sierra Palma bullet traveling at 2950 fps needs 19.5 minutes of elevation to go from 600 yards to 1000 yards. Multiply the 19.5 minutes by the 1.274 correction factor and you get 24.84 (again rounded off). This is the actual number of minutes you need to move your sights for this particular rifle to obtain a 1000 elevation.