Need some elaboration on sight radius and elevation

E

Ervin

Moderator
Looking through some vintage American Rifleman entries by Col Whelen and I came across an article on the very basics of rifle sights.

He calculates 3600 inches in 100yds and divides that number by 24in (sight radius)
Result = 150.

Seems straight forward, only he doesnt clear up how to use that number (150).


Would I be on the right track if I was to say 1mm in elevation of the rear sight would equal 150mm (about 1/2") on target?

Ive seen some really complex diagrams on this same topic in the past and if I understand this correctly, it's gonna be real helpful.
Its nice to fully understand the geometries of a trajectory and line of sight.
 
I'll have to dig out Whelen book to see it he mentions it, but the correct fomula is:

For 100 yards sight radius divided by 3600 = the amount you need to move the sight for 1 moa change.

100 yards X 36 inche is 3600 inches. (36 inches per yard)

Lets say you are shooting 8 inches high and your sight radius is 24 inches.

24/3600 = .00666, 8 X .00666 = .0533

So if you move your rear sight down .05333 you should be on at 100 yards.

Hope this clears it up a bit.
 
Here's a snippet of the actual article.
Whelen divided range by radius to get 150, not the other way around. Thats whats confusing me.


389334_189186187865512_100003223957936_292046_1873628249_n.jpg
 
It comes out the same. 1/25th of an inch is .04, As he said, that moves the impact 6 inches at 100 yards

The number I gave is .006666 movement per inch.

The article says 1/25 of an inch or .04 will move the impact 6 inches

6 X .006666 = .04

It comes out the same, just which method is easier to figure out.
 
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