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Impact of Front Sight Height
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Does anyone have a source of reference or a link of how I could solve for the implications or impact of how the height of a front sight affects trajectory?

Example: Win 94 30-30 has to have front sight replaced. Factory orginal size is 0.360". Available replacements are either 0.343" or 0.375" respectively. Difference on the +/-Y-axis is relatively identical but I have no reference material here to actually determine the impact.

Any help would be very much appericated, and thanks in advance.

[ 07-18-2002, 06:17: Message edited by: Alex Szabo ]
 
Posts: 902 | Location: USA | Registered: 01 March 2001Reply With Quote
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If I understand your question correctly, there is a fairly simple solution, using the concept of similar triangles, even though the trajectory of the bullet is not perfectly flat.

The increase in the height of the point of impact is about the distance to the target times the increase in front sight height, divided by the sight radius.

In my 29" Swede, .001" at the front sight works out to about 1/8" at 100 yards.
 
Posts: 2281 | Location: Layton, UT USA | Registered: 09 February 2001Reply With Quote
<Don G>
posted
What is the distance between the front and rear sights? Call that SightRadius in inches, then the change in POI at 100 yards is:

Change in POI = (.360-.343)/SightRadius*3600

For a twenty inch SightRadius, this works out to 3.06 inches higher.

If the SightRadius is twenty inches the higher (.375) sight will drop the point of impact 2.7 inches at 100 yards.

Don
 
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quote:
Originally posted by denton:
If I understand your question correctly, ....

Hi Denton,

Thanks for the rapid response. Yes, you are on target. Before this, I drew a few lines on a piece of paper, and determined that I could calculate [trigonometrically] the plus or minus theta angle(s) if I had two points of reference. Because the difference is only a linear measurement, I do not know how to convert the linear unit to the angle that it represents.

If I understand your approach and methodology however, you are simply expressing this same problem with a horizontal Z-axis representing trajectory (or perhaps the bore axis) and then using your coefficient multiplier example over the sight radius.

What do you mean by sight radius? Is is the metric radius of the round portion of the front sight or the radius of the target in its actual linear units?

Thanks again!
 
Posts: 902 | Location: USA | Registered: 01 March 2001Reply With Quote
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quote:
Originally posted by Don G:
What is the distance between the front and rear sights? Call that SightRadius in inches, then the change in POI at 100 yards is:

Change in POI = (.360-.343)/SightRadius*3600

For a twenty inch SightRadius, this works out to 3.06 inches higher.

If the SightRadius is twenty inches the higher (.375) sight will drop the point of impact 2.7 inches at 100 yards.

Don

Thanks Don!!!! I like simpler anyway than the path I was attempting to go down.

Where did you and Denton pick up this knowledge? I would like to read this information to internalize it myself.

Thanks in advance...
 
Posts: 902 | Location: USA | Registered: 01 March 2001Reply With Quote
<Don G>
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Alex,

I think I learned about similar triangles and ratios when I was twelve years old. Since I'm an engineer, the concept and it's uses have been driven home countless times since.

Don
 
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Don,

The answer to my question comes in the similar angles solutions that both you & Denton asserted.

Because you have simply scaled back the starting metric to the firearm itself, the right angle begins at the bore axis (representing side b) and then is raised or lowered based upon the sight (representing side a) one can now determine the theta on the tan or cot relationship. (pound my forehead...).

Thanks again.
 
Posts: 902 | Location: USA | Registered: 01 March 2001Reply With Quote
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Sight radius is just the distance from the rear sight to the front sight.

The formula comes from solving the proportion X4/X3=X1/X2, where

X3 is the distance to the target
X4 is the amount POI is raised or lowered
X1 is the amount the front sight is raised or lowered
X2 is the distance from the front sight to the rear sight.

Solving for X4 gives the equation I posted.

Anyway, the long sight radius of the Swede gives a decent basis for comparison.... shorter sight radius will give more shift at target per change in sight height... so between 1/8" per thousandth for a really long barrel, and maybe 3/16" per thou for a short rifle is in the ballpark.

Good luck!
 
Posts: 2281 | Location: Layton, UT USA | Registered: 09 February 2001Reply With Quote
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Thanks for elaborating Denton! The information you two have provided, enabled me to precisely articulate to my customer the buying decision criterion.

I finally asserted to him that it is based upon his personal technique and style in order to make the correct decision for him.

Naturally, because he was an ole ground pounding jarhead, he knew which decision to make immediately.

Thanks again!
 
Posts: 902 | Location: USA | Registered: 01 March 2001Reply With Quote
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Alex,
Order a Brownells catalog and the complete mathamatical formula for iron sights with any lenth barrel is there for your information, its in the iron sight section.
 
Posts: 42314 | Location: Twin Falls, Idaho | Registered: 04 June 2000Reply With Quote
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quote:
Originally posted by Atkinson:
Alex,
Order a Brownells catalog ...

Thanks Ray! Yep, its right here on page 265, never bothered to look...

[ 07-18-2002, 22:08: Message edited by: Alex Szabo ]
 
Posts: 902 | Location: USA | Registered: 01 March 2001Reply With Quote
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