need help from the math gurus

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[ARCHIVED THREAD] - need help from the math gurus

Posted: 6/5/2008 8:50:39 AM EDT

I may have this all screwed up, higher math never was my thing, but i am trying to figure out how many MOA are subtended at 100 yds by the width of a standard A2 front sight post .070"

HELP my brain is exploding

Posted: 6/5/2008 9:22:16 AM EDT

[#1]

How far is your eye from the FSP?

Posted: 6/5/2008 9:23:28 AM EDT

[#2]

make it easy say 14"

Posted: 6/5/2008 9:39:15 AM EDT

[#3]

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 back-to-back.

In this case the right triangle has an opposite side of .035 and an adjacent side of 14.

Need more?

(edited for typos)

Posted: 6/5/2008 10:01:35 AM EDT

[#4]

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)

~~(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

Posted: 6/5/2008 10:08:23 AM EDT

[#5]

The Front Sight of my 16" Mid-Length 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)

~~(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.

Posted: 6/5/2008 10:19:37 AM EDT

[#6]

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

Posted: 6/5/2008 10:41:37 AM EDT

[#7]

ok so can someone figure the front sight coverage on a standard A2 rifle and a A2 carbine length?

okay thanks I believe i have it by jove!

Posted: 6/5/2008 10:47:35 AM EDT

[#8]

Boy Golly!!! Do I appreciate the ancient Greeks, or what?

Everyone should study and be able to apply geometry.

Posted: 6/5/2008 10:48:13 AM EDT

[#9]

Quoted:
ok so can someone figure the front sight coverage on a standard A2 rifle and a A2 carbine length?

You do it! I bet you can, if you wanted to!

Posted: 6/5/2008 10:49:01 AM EDT

[#10]

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...

Posted: 6/5/2008 10:49:40 AM EDT

[#11]

Quoted:
ok so can someone figure the front sight coverage on a standard A2 rifle and a A2 carbine length?

A M-4 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!

Posted: 6/5/2008 10:50:18 AM EDT

[#12]

Quoted:
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...

Ooooh! I love a cook out!

Posted: 6/5/2008 10:58:00 AM EDT

[#13]

okay okay, I have proverbially learned to fish. Thanks for the help. I bet i am not the only one who benefits from my ignorance.

Posted: 6/5/2008 11:06:17 AM EDT

[#14]

Quoted:
how many MOA are subtended at 100 yds by the width of a standard A2 front sight post .070

BTW the wise-guy 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:

inv tan( opposite side / adjacent side ) * 60 * 2

Where opposite side = (width of the sight / 2) and adjacent side equals the distance from your eye to the sight.

Posted: 6/5/2008 11:11:51 AM EDT

[#15]

Quoted:

For future reference, if you are playing fast and loose, the MOA coverage at 100 yards for a sight is approx:

inv tan( opposite side / adjacent side ) * 60 * 2

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:

inv tan ( width of FSP / distance from eye to FSP) * 60

Posted: 6/5/2008 11:58:32 AM EDT

[#16]

[ARCHIVED THREAD] - need help from the math gurus

AR-15 » Optics, Mounts, and Sights

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