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Front sight height

#1 User is offline   wolfie 

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Posted 18 November 2009 - 08:05 AM

Someone help me out. I forgot what the formula is for how much your POI changes at 25 yards for every 1 1000ths of an inch in front sight elevation. Trying to get the right front sight for my rear fixed Heinie on a Glock 34. Heinie web page says that G34 often shoot 2.5 to 3" high at 25 yrds and offer a higher front sight should this happen.

#2 User is offline   bulamike 

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Posted 24 November 2009 - 09:57 AM

ooh not sure of the formula, but I went through one hell of a time trying to work adjustable sights on my G34, enough to where I sold it out of frustration.
I recall having to go to a .215 tall sight mated with my adjustable rear, the best setup I had were the sevigny combo, fixed sights, still had a fiber optic front .215 and rear Warren Tactical Rear.

#3 User is offline   kevin c 

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Posted 24 November 2009 - 10:19 AM

I do believe the formula is:

Change in POI equals change in FS height times the distance from the rear sight to the target divided by the distance from the rear sight to the front sight (all units of distance being the same for each measurement - inches, feet, meters or whatever).
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#4 User is offline   Va_Gunner 

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Posted 24 November 2009 - 05:35 PM

I just took my Spartan to Dave Dawson and he is hooking me up for the same issue. I'll let you know where i end up.
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#5 User is offline   Flexmoney 

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Posted 24 November 2009 - 09:50 PM

There is a thread with the proper formula in the FAQ's on the 1911forum

Or, you can do a search here. I've probably posted it a half dozen times in the past.
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#6 User is offline   Guy Neill 

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Posted 24 November 2009 - 10:38 PM

Here's an excerpt from a piece I wrote a while back that may help. Values from the formula I derived match the values listed in the table on Brownell's website.

δF/SR = y/(SR+R) with δF = change in front sight height
SR = sight radius
y = distance between aiming point and impact point
R = range

or δF = (SR ● y) / (SR + R) all values are in inches (● signifies multiplication)

For our example, y = 3”
SR = 7”
R = 25 yards = 900 inches

δF = (7 ● 3) / (7+900) = 21 / 907 = 0.023” (actual number is 0.02315”)


Guy

#7 User is offline   Xfactor 

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Posted 30 November 2009 - 06:00 AM

I worked out a formula a while back and posted it with some simple instructions over at M&P Forum a while back... here's the link

I subsequently found out that this is the same formula the AMU uses, so it works. :cheers:
- Aaron

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