So I Wuz Thinkin' About Bore Sighting...

Started by gitano, October 14, 2015, 08:36:39 PM

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I've been doing a lot of bore sighting lately, and trying to bore-sight .17 caliber bores (the Riedl in .17 Remington and the .17 Predator), with 64-year-old eyes is a bit challenging even in the best of circumstances. Things work out mostly because I refuse to give up, but while the results are "satisfactory", (meaning get on paper without firing a shot), they could be less challenging and more 'satisfactory' (meaning closer to "on" bullseye instead of just on paper). The biggest challenge is centering the target on the bore. Most of the targets I use have relatively small bullseyes - ~ one inch or so. That's small at 100 yd. Even at 35 or 40 yards it's small relative to the diameter of the field of view looking through a bore. "There must be a better way..." I think there is, and I have the simple solution.

Normally, in preparation for sighting a rifle in I consult a trajectory table for a given load and note where the bullet path FIRST crosses the Line of Sight. For most scope-sighted, center-fire rifles, this occurs between 25 and 40 yds. That's where I set my target to bore sight. However, since I am USUALLY working on a load, the exact trajectory (and therefore crossing point) is not really known. However, we do know something that is independent of our load! We know what a "minute of an angle" is. Rounding, a minute of an angle "subtends one inch at 100 yards" (or 300 feet). Therefore, 0.001" at the rifle equals ~1" at 100 yd. (Actually 1.047 inches, but I'm going to ignore the .047 for the sake of simplicity AND it really isn't necessary or meaningful at short ranges.)

Given the above geometric relationships, we can therefore say that 0.300" at the rifle equals 30 inches at 100 yd. That means that when one looks at a target at 100 yd through the 0.300" bore of a .308 caliber rifle, the bore 'outlines' a circle 30 inches in diameter. (For those that regularly bore-sight, that should sound 'about right'.) SO... if the target at 100 yd had a black 30" circle, or some other form of bore-sized circle, one could "perfectly" center the rifle's bore on the target by 'filling' the bore with the bullseye.

A 30" circle at 100 yd is not easy to come by. However, because of the geometry, that circle gets smaller as we shorten the distance to the target. Because the relationship is linear, at 50 yd the size of the circle is 15" (half of 100 yd = half of 30 inches). Coming closer, say within that "first crossing" window of 25 to 40 yd, we can get that circle down to say 10 inches at 33 yd, or 7.5 inches at 25 yd. Those sized 'bullseyes' are much easier and more convenient to come by.

Using a .30 caliber is probably close enough for most folks, but "close enough" isn't good enough for me when 'right on' is just a matter of 'doing the math'. So... I derived a formula to determine the exact size of the bullseye needed for a given caliber at a given range. "You" only need to remember - or write down - one number: 0.33, (out to as many "3s" as you care to add). The equation is:

Bore * 0.33 * Range (in FEET - very important - NOT yards) = bullseye diameter

Bore is in inches
Range is in feet
bullseye diameter is in inches

Given that equation, let's first plug in the numbers for a .308 caliber rifle at 100 feet (33.3 yd).

0.300 (the bore) * 0.33 * 100 = 9.9 inches (It works!) (0.300 is the bore diameter, not the caliber.)

Now let me give an example using a .17 caliber bore, one considerably removed from .30 caliber.

The actual bore of a .17 caliber centerfire rifle is 0.168", therefore the equation is:

0.168 * 0.33 * 100 = 5.5"

How about for a .45-70 Gov't at 25 yd?

.45 * 0.33 * 75 = 11.1

If you want to do some algebra, the ratio of 5.5 to 9.9 is - within rounding error - equal to the ratio of 0.168 to 0.300. Also, by doing some algebra, we can determine the range at which we would have to put a given bullseye for a given caliber (bore). This would come up when you have a target in hand and can't make a 'custom' one for your specific 'zero-crossing' point. For example, let's say we have a target with a 10-inch bullseye and we are sighting in a .45-70 Gov't rifle. Algebraically manipulating the equation we get:

Bullseye Diameter / (Bore * Constant)= Range

Substituting numbers in:

10" / (.45" * 0.33) = 67.34 feet or ~22 yd. (More precisely, 22.4 yd).

So if you have a target with a 10" bull and you want to bore sight your .45-70 (or any .458 caliber rifle) using this method, you would set the target at 22 yards and the bull would fill the bore when you looked through it.

I intend to fabricate some targets with either circles or black "bullseyes" to facilitate more precise bore sighting of my .17 caliber rifles. With those, I can simply set the target at 33 yd and look down the bore until the bullseye fills the bore. The bore of the rifle will then be CENTERED on the bulls eye and I can adjust the scope as necessary. That also facilitates returning to the exact previous "center" when you bump your rifle as you are moving from looking through the bore to looking through the scope. (Something I seem to do with regularity.)

There are NRA "standard" targets with 10" bullseyes. I don't recall what their specific use is, but I have several of them. I THINK they are 50-yd .22 RF targets. They would be 'perfect' for bore sighting my .308 chambered rifles at 100'/33 yd.


PS - Here's how to "say" the equation in words:
Divide the caliber(bore) by 3 and multiply the result by the range. The result is the target diameter.
One third of the caliber(bore) times the range equals the target diameter.

For finding the range instead of the target diameter:
Divide the target diameter by one-third of the caliber(bore). The result is the range.


PS - If you live in one of those countries that equates the caliber of a firearm to the bore diameter instead of the groove diameter, you can interchange the words (and dimensions) "bore" and "caliber" in the above equations. In the US, USE THE BORE DIAMETER, NOT THE GROOVE DIAMETER (AKA CALIBER).

Be nicer than necessary.


I found the targets I was thinking of:

The bullseye (black, "7-ring"), is exactly 5.50" in diameter. So for my .17 caliber rifles, I would put the target at:

5.5 / (.168/3) = 98.2 feet or 33 yd.

33 yd is very close to the first 'zero-crossing' point. Therefore, if I put the bullseye in the center of my bore and the crosshairs in the center of the bullseye, I should be very close to right "on" when I move the target out to 100 yd.


Be nicer than necessary.



All of the 'detail' was to provide the foundation for the simple equation. There are just two operations:  One division, (bore divided by 3), and one multiplication (result of division times the range). "You" - whomever "you" are - may not  find it particularly useful, but there really isn't much simpler math in all of ballistics/reloading than "X divided by Y times Z = R".

Be nicer than necessary.


Well I does lead to I have a x bore diameter and a y diameter target what distance do I set it up for. A neat way to toe. A person can use a  five gallon bucket ice cream pail margarine tub etc what ever is round and laying around.


I was going to say that "most" people know the bore diameter of their rifles, but quickly caught myself. That probably isn't true. In the UK, there are a lot of people (but not "most" I think), that not only don't know the weight of the bullets they shoot, they think it's 'crass' to talk about what bullet one uses, its weight, the type, charge, distance to the target, etc., and think American's "obsession" with that is "boorish". I'm certain that there are MANY American gunners that haven't got a clue what the diameter of the bore of their rifles are. However, here at THL, I'd be pretty surprised to find someone that didn't.

Here's a table of common bore diameters and that value divided by 3.

Be nicer than necessary.


I have a Bushnell bore sighted than is so so at best. This is much likely a better way


So... I was trying to figure out ways to make it even simpler. I can. How about just ONE arithmetic operation - multiplication.

Looking at the constant - 0.33 - and the fact that I was using FEET instead of yards, (because I wanted precision), but then going back to yards, I realized that one can eliminate the constant completely. In other words, you can drop the 0.33 constant, and just multiply the bore diameter by the range IN YARDS. For example:

.308 Caliber bore (.300), times the range IN YARDS (33) = bullseye diameter in inches

.300 x 33 = 9.9

Which is identical to the number we got in the original post. Using yards instead of feet causes a reduction in precision that is insignificant.  I don't think I can make it simpler than that.


PS - This of course also simplifies the equation for determining the range at which to set a target with a fixed bullseye size. Divide the bullseye size (in inches) by the bore diameter. The result is the range at which that bullseye will "fill" the bore. For example:

5.5" / 0.168" = 32.7

5.5 equals the bullseye diameter of the target in post #2,
0.168 is the bore diameter of the .17 caliber, and
33 (rounded) is the range in yards to set the target to 'fill' the bore.

Be nicer than necessary.