Quoted:
nevermind.. 12.665 MOA is what someone posted back in 2009 on here..
if someone has a SIMPLE math way to do this.. chime in..
This question comes up from time to time. It is a simple equation.
The Front Sight of a M-4 Carbine is ~19" from your eye, if you shoot NTCH, roughly.
0.035/19 = 0.001842105
100 yards is 3,600 inches.
Then,
X/3,600 = 0.001842105
(3,600) * X/3,600 = 0.001842105 * (3,600)
(3,600) * X/3,600 = 0.001842105 * (3,600)
X = 0.001842105* (3,600)
X = 6.63158
Since this was an Isosceles Triangle, you need to double this number.
Then means that a 0.070" Front Sight Post, 19" from your eye, covers ~13.26" at 100 yards, or 12.665 MOA
There is another equation to get you where you need to go.
Site width is W (in this case 0.070")
Distance from eye to sight is R (in this case 19")
A circle has 360 degrees, and there are 60 minutes per degree, so a circle has 21,600 minutes
To find the MOA of your sight:
21,600 * (W / (2 * 3.14159 * R))
The circumference of a circle with your eye at the center and the sight at the edge:
(2 * 3.14159 * 19) = 6.28318 * 19 = 119.38042
How much of it the sight covers:
(W/119.38042) = 0.070/119.38042 = 0.0005863608
Multiply that times the minutes in a circle:
0.0005863608 * 21,600 = 12.665 minutes.
In this case (19"), your front sight would cover about 12.7 MOA.