So I Wuz Thinkin' About Bore Sighting...

[gitano]

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). 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, 0.01" 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:

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

Caliber is in inches
Range is in feet
bullseye diameter is in inches

Given that equation, let's first plug in the numbers for a .30 caliber rifle at 100 feet (33.3 yd).

0.300 * 0.33 * 100 = 9.9 inches (It works!)

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. 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. Algebraically manipulating the equation we get:

Bullseye Diameter / (Caliber * Constant)= Range

Substituting numbers in:

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

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 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.

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.

Paul

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

Paul