I found some figures using google, but they didn't seem right. It calculated my sight to have a width of 20.05 MOA.  However when I am at the range I know my post just covers an 8" target. I guess I could just go with that, but hoped for a bit more accurate number. Thanks for any help.

Measure the actual width of the post, then the distance from the rear sight to the front post, figure MOA from there.

Assuming your 8" target is covered at 100 yards, your front post if 8 MOA wide.

http://nssf.org/video/facts/MOA.cfm

Just covers an 8" target at what range? The distance from your eye of both the post and the target matters.

The sight post is 0.070" and my sight radius is 12". And that makes 20.05 MOA according to the math equation I found.  But I assume this is wrong and it is actually an 8 like I see at the range. I have never checked if my range is exactly 100 yards or not and was curious as to whether there was a more accurate equation.

My eye is about 16.5" from the front sight post.

I don't think there is a more accurate equation, unless you want one which accounts for differences in resolving power from eye to eye. The equation is just right angle trig.

L = Distance from observer to post
W = Post Width
Theta = Arc whose arclength is post at the sight radius in radians
MOA = Arc whose arclength is post at the sight radius in arcminutes

Theta = 2*atan(W/2L)

MOA = (3438) Theta

If we use sight radius that gives 20 MOA like you said. If we try 16.5 instead of 12 that gives you 14.8 MOA, which is (only a bit) closer to what you observed.

I think you should use distance between eye and FSP to figure FSP arcwidth. The rear sight is only there to ensure alignment. The calculation using sight radius is made under the assumption that you are shooting "nose to charging handle," that is, your eye is right behind the rear sight.

Hmmm, that seems high to me.  When shooting at 4moa targets, my fsp covers them with about 2moa left on each side.  I would've guessed about 8moa as well, I wonder why there's such a big discrepancy between our experience and the math?

Carbine Gas System
12.7 MOA

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.

Thanks for the math help guys! I'm gonna stop out to the range and double check the distance-to-target as it is my only variable that could be throwing my measurement off by 6" at 100 yards.

In case you were curious why my equation differed from RedFalconBill's, mine is the "exact" expression. His expression is valid when the argument of the arctangent W/2L is a small number ( << 1), so that atan(W/2L) is approximately W/2L. This is usually always the case when talking about shooting, so the difference between the two equations is small enough to ignore.

A standard front sight on an AR is 0.070" in width.

With an A2/A4 and fixed stock the front sight would cover 10 MOA (+/-), depending upon the distance of your eye to the front sight.

Just an update.  Went to my local range and it measured 298 ft to the target. Not too bad. I also double checked my target coverage and it appears to cover just a bit more than the 8" target. So everything seems to match with a 10-12" coverage at 100 yards. Thanks again for the math help.