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Posted: 6/5/2008 7:50:39 AM EST



How far is your eye from the FSP?



make it easy say 14"



The triangle formed by your eye and the sides of the FSP is an Isosceles triangle. I will be using the following page for for an example:
mathworld.wolfram.com/IsoscelesTriangle.html As you can see from the diagram, this is equivalent to two right triangles backtoback. In this case the right triangle has an opposite side of .035 and an adjacent side of 14. Need more? (edited for typos) 


So, 0.035/14 = 0.0025
100 yards is 3,600 inches. Then, X/3,600 = 0.0025 (3,600) * X/3,600 = 0.0025 * (3,600) X = 0.0025 * (3,600) X = 9 Since this was an Isosceles Triangle, you need to double this number. Then means that a 0.070" Front Sight Post, 14" from your eye, covers 18" at 100 yards. One MOA = 1.04719755" 18/1.04719755 = 17.1887 MOA 


The Front Sight of my 16" MidLength is 21" from my eye.
0.035/21 = 0.00166 100 yards is 3,600 inches. Then, X/3,600 = 0.00166 (3,600) * X/3,600 = 0.00166 * (3,600) X = 0.00166 * (3,600) X = 6 Since this was an Isosceles Triangle, you need to double this number. Then means that a 0.070" Front Sight Post, 21" from my eye, covers 12" at 100 yards. 


There's another way that's easier, IMO.
Site width is W Distance from eye to sight is R A circle has 360 degrees * 60 minutes/degree = 21,600 minutes To find the MOA of your sight: 21,600 * (W / (2 * 3.14 * R)) The circumference of a circle with your eye at the center and the sight at the edge: 2 * 3.14 * R = 6.18 * 14 = 87.92" How much of it the sight covers: W/87.92 = 0.07/87.92 = 0.000796 Multiply that times the minutes in a circle: 0.000796 * 21,600 = 17.19 minutes. In this case (14"), your front sight would be 17.2 MOA. For the 21" distance: 21,600 * (0.07 / (2 * 3.14 * 21)) = 11.46 MOA 



Boy Golly!!! Do I appreciate the ancient Greeks, or what?
Everyone should study and be able to apply geometry. 

Don't be so openminded that your brains fall out.
General education should not be mere training of the hands to work, but training of the mind to properly reason. http://www.welltrainedmind.com/classed.php 
You do it! I bet you can, if you wanted to! 

Don't be so openminded that your brains fall out.
General education should not be mere training of the hands to work, but training of the mind to properly reason. http://www.welltrainedmind.com/classed.php 
If you build a man a fire he is warm for a day. If you set a man on fire he is warm for the rest of his life...



A M4 carbine length should have an eye to front sight distance of ~19", depending upon where you place your eye. An A1/A2 Rifle should have an eye to front sight distance of ~24", depending upon where you place your eye. You now have two different formula's to do this. Go forth and share what you have learned here! 


Ooooh! I love a cook out! 



BTW the wiseguy answer is "the same as at 1000 yards" as MOA is an angular measurement. For future reference, if you are playing fast and loose, the MOA coverage at 100 yards for a sight is approx:
Where opposite side = (width of the sight / 2) and adjacent side equals the distance from your eye to the sight. 


Thinking about this for a moment, if you aren't working for NASA, just do:




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