Front sight thickness

[bobanderson]

Does anyone know the formula or calculation to make my front sight appear the same width as the target at a fixed distance?

For instance, I have an 18" sight radius on my Sharps and want the front sight to be the right thickness to appear the same width as a clay bird (is that 4"?) at 50 yards.

I was told that this information might be available in back Skirmish Line issues, but I took the lazy way out and asked my question here.

[George Gompf]

Kind of makes you wish you had paid more attention to the trigonometry lessons way back when doesn’t it?

You have a basic trig problem. You are missing one point though, that is the distance your eye is from the front sight, not the sight radius Set up a trig problem with your eye at zero. You can turn the problem into a right angle issue by having the base leg at 50 yards as 2” on a 90 degree angle; if you are calculating a 4” diameter target at 50 yards. Figure your angles and hypotenuse then shorten it up to the length the front sight is from your eye.

Using this method, you can double the width of the base leg at the front sight and find out what the front blade width should be. You can then use the same calculation layout to see how wide the front sight will appear at 100 yards.

You are going to need a scientific calculator and possibly an engineer with a PhD to operate it. I have told you how to do it but absolutely not answered your question. Complete this homework assignment and I will give you an “A” for the work and thanks for a practical application I can use in my technical math class for the students to calculate. I have been using an artillery problem to locate the elevation and range of a gun position on a hill an unknown distance away. This will be a good addition to the homework or classroom problems.

Thanks
George

[rachbobo]

Does anyone know the formula or calculation to make my front sight appear the same width as the target at a fixed distance?

For instance, I have an 18" sight radius on my Sharps and want the front sight to be the right thickness to appear the same width as a clay bird (is that 4"?) at 50 yards.

I was told that this information might be available in back Skirmish Line issues, but I took the lazy way out and asked my question here.

Clamp the weapon in a gun vise aimed at the target at the known distance.
Look through the sights as a second person holds a set of calipers in front of the front sight blade. Have the second person open and close the calipers till you find the right width.
Done deal.

Bill Cheek
Cockade Rifles

[Ron/The Old Reb]

George's way sounds like more fun.

[Greg Ogdan 110th OVI]

It's a ratio problem. If the black is 4" at 50yds, how wide would it be at the distance from your eye to the rear of the front sight. Sort of kind of like figuring sight height adjustment. Yes, you could do it George's way, but why so complicated?!

[bobanderson]

George,
I was distracted by Mary Lou Whatshername during class. There were much more interesting angles to consider. I realized when I sent the note originally that sight radius wasn't necessary. The distance from my eye to the face of the front sight on my Pedersoli 1859 Sharps is 27".

Bill,
Thanks. I've done the multiple pairs of hands and the dial caliper trick, but I figured there was a more elegant solution I could use in the shop.

And finally to Dave of the Chartiers Valley Guards, who sent me the note below...


Dear Mr. Anderson,
This is a simple geometry problem. Have a friend measure the eye-to-front-sight distance in inches, as suggested by Mr. Gomph. Send me that number at my E-Mail address. I will calculate the sight width for you ... and provide the basic trigonometry formula that applies to any weapon.
My best/Good shooting,
Dave M. Chartiers Valley Guards



I told him in my email reply that I would post it here as well so others could use the formula.

Thanks to all that responded.

[George Gompf]

The Mary Lou’s of the world have distracted many a lesson. I had one young lady come in class and announce that she was divorcing her husband, was working as a stripper and invited all the guys to go to the club in Memphis to watch her dance. I never could get the attention of any of the guys in class the rest of the evening. Her dancing career apparently worked out as she did not return to class.

Using a ratio of 4 inches to 1800 inches, I got a sight width of .060 based on the eye to sight length of 27”. I dont know why everyone insists on doing things the simple way. A long drawn out trig problem gives hours of lasting entertainment.

George

[DaveCVG]

Dear N-SSA Shooters,
Just got back from the Family Reunion and sat down to do the calculation ... I agree with Mr. Gompf that the answer for Mr. Anderson is sight width of 0.060".
This is basically a "tanget" problem, where in trigonometry (or in the N-SSA ... trigger-nometry) the "tangent" of an angle in a right triangle is defined as "tangent = opposite length divided by adjacent length". The "adjacent length" is the distance from Mr. Anderson's eye to the center on the bird, or 50 yards equals 1800 inches. The opposite length of the right triangle is half the diameter of the bird, or 2 inches.
Tangent problems like this are ratio problems ... 1800 inches from eye to bird, 27 inches from eye to front sight, half bird diameter of 2 inches, unknown value we will call "x" ...
2/1800 = x/27 ...
we solve for "x" and get 0.030 inches (1800x = 54, and x = 54/1800 = 0.030 inches).
The "x" of 0.030 inches covers half the bird as stated above, so we multiply the x of 0.030 inches by 2 and get 0.060" for sight width to cover the bird.
If you are working the same question for a rifle or musket, just replace the "27" with your new "eye-to-sight" distance and repeat the arithmetic to solve for "x" and multiply "x" by 2.
For me, the tough part starts now ... align sights, place focused front sight on fuzzy target, and pull trigger.
Best/Good shooting,
Dave M. Chartiers Valley Guards