How to relate target accuracy to an angle

I know I can figure out how to compute these things using trigonometry, but thought I'd ask here first.... Is there a website somewhere, that relates:

• distance to target
• distance from center of target to bullet hole

....and computes the angle?


And perhaps the other way as well, where you can select an angle, perhaps one degree, and for any given shooting distance, look up how far the bullet hole mathematically should be from the aiming point on a target.

For any of the above to be most useful, the gun would need to be sighted in quite well, but that's a separate issue.

CWL - no, nothing like that.

MEHavey, no, I was thinking, in an ideal world, with no air resistance, or anything of the kind, if the bullet hole at say, 10 yards, was constantly two inches to the right of the bullseye, how many degrees (theoretically) would the barrel be aimed to the right instead of straight ahead. (....and not specifically any of these numbers, just enough information to calculate this for any situation, again, theoretically.)

BCRider, yes, that is exactly what I mean - although I was wondering if it's all been worked out on a web table where I can instantly look it up. No big deal - what you wrote is a perfect answer - I worked it out in my head, but hadn't gotten to actually do it, which you've done. Thank you! Now I just need to print this out, and keep it somewhere where I can easily find it in the future. Again, thanks, both for the answer, and the explanation!!!

MEHavey said:

Again:

Error (degrees) = 1.6 x Target Offset(inches) ÷ Range(yards)

This is not rocket science...
just 10th grade geometry.

Click to expand...

Hi, and thanks - the earlier answer was the 10th grade geometry answer. Your answer here is even simpler, and also simpler than looking it up in a table, even if there were a table to look it up in.

So, the angle "error" for a bullet hole two inches from the bullseye, at 15 yards, is:

1.6 x 2 ÷ 15 = 0.2 degrees

-----------------------------------------------------------------

As a different, but related question, regarding adjustable sights....

If one were turn the windage or elevation screw one full turn, say, on a gun with 6 inches (or whatever) between front and rear sights, is there a quick way to determine how much of an effect it should have on where the bullet lands, at a specified distance to the target?

Or, for a simpler way of asking this, suppose you somehow knew that your bullet was hitting the target two inches to the right, on a gun with a 6" distance between sights, at a range of 10 yards.

(Change any of the numbers if you wish. They're just examples, to make it easier to explain the question.)



Unless I'm missing something, if you knew this for any given gun, and knew how far your shots were missing the bullseye by, in one adjustment for windage and elevation, you should be able to get your sights almost perfect.

BCRider said:

Mike, there's no way to know that other than for a particular gun. The amount the sight moves is a function of the adjuster's thread pitch. But you can measure it for yourself with a set of calipers or a micrometer or a suitable setup and a dial gauge. And armed with that do the math.

Or you can use ratios. If it's 6 inches between the sights and 12 yards to the target then it's 36feet/0.5feet x rear sight shift= difference at the target.

Click to expand...

Thanks! I didn't know if all guns used the same thread pitch, but you're right, I should work this out for a specific gun.

==================================================

The following comment is only for people who love playing around with math and statistics, and so on, and who consider it "fun", not "work"...... One of the neat things that comes from calculating the CEP of a group of say, 50 shots, is that it will tell you EXACTLY where the center of all 50 of them is, from the bullseye. Theoretically, in an ideal world, using that exact distance, it should be possible to center the sights in a single adjustment, knowing how many "turns" or "clicks" equals a certain distance, at that range. The math is very straight forward.


I'm not in any way suggesting that anyone else actually do this. For me, it's all very enjoyable, and in no way detracts from the enjoyment I get out of going to the range. I usually do the math the same evening.

Others may find it all very useless - but some of those same people shoot three shots in a very tight grouping, and say how great they're doing, and that they better stop before they mess up that target by shooting some more bullets at it. As I see it, for most of them, any of their shots may fall anyplace with a three or four inch circle, and that fact that the three were close together was more due to luck, than skill.

(From watching some of the videos of the skill of snipers with high performance rifles and scopes, I think they know exactly where their bullet is going to go, and know exactly how to use their scopes and adjustments. For them, it's skill + science that makes them so lethal, and if what I've been reading is true, this was true long ago, as well.)

Thank you for posting. This is the first I've read about "sight radius".

For your example, 20 yard distance, and typical adjustable sights on a 1911 pistol, how much of a distance is covered by the adjustments on the rear sight? From one extreme to the other, how much of a change would it make on bullets hitting the target?

Back to your example - just out of curiousity, how easy is it to change a front sight? Can the gun owner do it, or is this a job for a gunsmith?