When sighting in a flinter, how do you know how much to move the sites? Like if its shooting left approxamately 5 inches how do you determin how much to move the site to compensate? This is the first time I have tried to sight in a gun, and I am a biginner like my username sugests.
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Sighting in
- Thread starter bejohnson
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First off figure out how many inches it is to the target you were shooting. Lets say it was 25 yards.
There's 3 feet in a yard so that is 3 X 25 or 75 feet.
There's 12 inches in a foot so that is 75 X 12 or 900 inches.
Divide the distance of the group from the target center by 900 inches.
In this case that is 5 inches divided by 900 or .0055
Measure the distance from the rear sight to the front sight. I will guess its 24 inches. Whatever the distance is multiply it by the .0055 (or what ever the real answer is) you calculated.
In my make believe case that is .0055 X 24 which is .132 total movement is required.
For this case the sights will have to be moved .132 total. I say total because it is the total change from one sight to the other. For instance if it was elevation we were adjusting we could move the front sight only or the rear sight only, down or up .132, OR we could move the front sight .066 one way and the rear sight .066 the other way.
All of this is a matter of direct ratios so if the target was at 50 yards, and 50 yards is twice 25 yards if you go thru the numbers you will find the final answer is exactly half of the answer we came up with before so it would be .066 total movement required.
If you don't want to do the math, post the targets range and the distance between your front and rear sight here and someone here will calculate the real answer for you.
There's 3 feet in a yard so that is 3 X 25 or 75 feet.
There's 12 inches in a foot so that is 75 X 12 or 900 inches.
Divide the distance of the group from the target center by 900 inches.
In this case that is 5 inches divided by 900 or .0055
Measure the distance from the rear sight to the front sight. I will guess its 24 inches. Whatever the distance is multiply it by the .0055 (or what ever the real answer is) you calculated.
In my make believe case that is .0055 X 24 which is .132 total movement is required.
For this case the sights will have to be moved .132 total. I say total because it is the total change from one sight to the other. For instance if it was elevation we were adjusting we could move the front sight only or the rear sight only, down or up .132, OR we could move the front sight .066 one way and the rear sight .066 the other way.
All of this is a matter of direct ratios so if the target was at 50 yards, and 50 yards is twice 25 yards if you go thru the numbers you will find the final answer is exactly half of the answer we came up with before so it would be .066 total movement required.
If you don't want to do the math, post the targets range and the distance between your front and rear sight here and someone here will calculate the real answer for you.
Well, you could determine it exactly using that bad-old high school triginometry, or even algebra. You didn't say at what range you are 5" left or the distance between your front & rear sights - which makes an actual measurement answer tough. So, 100 yds = 3,600 inches. 5"/3,600" = X/sight radius, where "X" is the amount you'll have to drift the rear sight.
Or, you could do like I really do and just start whacking the sight until it moves a tiny amount and firing three more shots to see what happened.
Or, you could do like I really do and just start whacking the sight until it moves a tiny amount and firing three more shots to see what happened.
Stumpkiller's "Whacking" method is a bunch more fun and probably quicker. It surely is easier on the brain pan than doing all that math and trying to figure out how to measure the actual location of that calculated sight move. When you're finished with all the tech stuff you STILL must shoot to check shot location. Whack the sight and get on with it.
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I'm'a "whack'er",.... 'cause I ain't smart 'nuff to be a "figger'er"!! :: :: ::
