Author Topic: Target Size and Sight Picture  (Read 417 times)

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Offline Jim B.

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Target Size and Sight Picture
« on: June 25, 2004, 04:28:47 AM »
Thanks very much to everyone who responded to my questions about target size and preferred sight picture.  My objective in starting this discussion was to find out how to design a better target for my 209x50 for use in sighting in with an aperture sight and standard diameter front bead.  

After some discussion with a colleague we concluded that it is a simple case of similar triangles.  The solution we came up with is as follows (for my rifle):

Given:
- The apex of a triangle is at the aperture of the receiver sight
- The width of the front sight forms the base of a triangle with the apex
- The width of the target forms the base of similar triangle with the apex

Therefore:
The ratio of the distance from the apex (receiver sight) to the front sight divided by the width of the front sight should be same as the ratio of the distance to the target from the apex divided by the diameter of a circle that will be exactly subtended by the front sight.

The equation looks like: a/b = x/y
Or, rearranged to solve for "y": y=bx/a

Where:
- "a" is the distance from the aperture (apex) to the face of the front sight
- "b" is the width of the front sight
- "x" is the distance from the apex to the target
- "y" is the diameter of a target that will be exactly subtended (covered) by the front sight at distance "x"
- Units must be consistent!

In practice for my rifle, the solution looks like this:
- For simplicity, I use the form of the equation: y = (b/a)x
- The front sight is 0.095" wide (b)
- The face of the front sight is 25.06" from the face of the aperture sight (a)(apex)
- Those two measurements give me a ratio (b/a) of 0.00379 for my rifle
- At a distance of 50 yards (1800") from the muzzle (distance from muzzle to aperture could be added in but should not be significant) the solution is y = 0.00379 x 1800 => y = 6.8"
- At a distance of 100 yards (3600") from the muzzle the solution is y = 0.00379 x 3600 => y = 13.6"

What this tells me is that at 50 yards my front sight will just cover a circle about 7" in diameter.  If I make a target designed like a ring with an inner diameter of 8 or 9 inches I should be able to center my front sight in the ring with visible white space around the front bead - kind of like a globe sight in reverse.  Alternatively, I could make a solid color target disc about 8 or 9 inches in diameter and hold the front sight so there would be an even color rim around the bead.

Please keep in mind that this is just for targets for sighting-in and load development.  It is also only relevant to those who subtend the target with their front sight.  I would greatly appreciate any comments, corrections or suggestions.

Jim

Offline Roger_Dailey

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Re: Target Size and Sight Picture
« Reply #1 on: June 25, 2004, 10:00:53 AM »
Quote from: Jim B.

Please keep in mind that this is just for targets for sighting-in and load development.  It is also only relevant to those who subtend the target with their front sight.  I would greatly appreciate any comments, corrections or suggestions.

Jim


   That should work quite well for your purpose.  If you need to figure things with a little more precision, use the distance from your eye to the bead instead of the distance from the rear sight to the bead.  The distance between the sights would be the proper number to calculate sight adjustment.

Offline Jim B.

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Target Size and Sight Picture
« Reply #2 on: June 25, 2004, 10:04:05 AM »
Roger,

You are correct - the eye is the apex of the triangle, not the aperture.  Thanks for correcting my error.  The distance from the eye to the front sight is the correct distance "a".

Jim