Okay you brilliant people..... how does one solve for this question?
Say the rifle is 36" long, and pivots from the butt end. How far (in thousandths of an inch) must the muzzle deflect to change the point of impact 1moa at 100 yards?
Yes, I know it is simple math. No, I am not having the knowleging of the mathiness ways.
Lessthandependent

You know that things like plastic bullets, beanbag rounds, rubber buckshot,
etc. are called “Lessthanlethal”, right? It used to be they were called
“non...
16 minutes ago
13 comments:
Fortunately this is a pretty simple arrangement, making the assumption we're in PhysicsLand where everything is straight lines and sunshine.
1 minute of angle is equal to one inch per one hundred yards  so the muzzle needs to deflect one minute of angle.
Since we're at ONE yard from the pivot, we need to move 1/100 of the correction needed at the target ... or 1/100".
I think. If my math is right. (And this is ignoring the few hundredths of an inch per hundred yards that is an exact MOA...)
Ooohhh...looks like your Shoot Boss needs to start the whippings now if you're going to teach IMC! :o)
As you already know, angular deflection is the same no matter what the distance...in other words, 1 MOA (minute of angle) is the same at 100 yards as it is at 200 yards as it is at 25 yards as it is at 1 yard (36"). The LINEAR distance that the 1 MOA subtends is what varies.
This isn't exact but for field/Appleseed shooting purposes, it is as much as the practical rifleman needs to know. 1 MOA = 1 inch per 100 yards. (it's actually 1.047"...but who's counting).
1 MOA = 1" at 100 yards
1 MOA = 2" at 200 yards
1 MOA = 143.7" at 14,370 yards
1 MOA = .25" at 25 yards
1 MOA = .01" at 1 yard
The muzzle would need to deflect approximately 1/100th of an inch at the muzzle to deflect the bullet impact 1" at 100 yards.
That's why those steady hold factors, six steps and natural point of aim stuff we teach is so important!!
Kris (SteelThunder  RWVA Shoot Boss / Instructor)
Throwing out the vagaries of external ballistics, your answer is 10(0.010).
Carteach, all the math you need to know is to count to 20. "THE RULE OF 20" should be burned into your neurons. "THE RULE OF 20" helps with all your shooting needs. Whether prepping for hunting, or strict marksmanship and competition...
What exactly is "THE RULE OF 20" you say?
Most ammo comes packaged in a box of 20 rounds. Keep shooting until you get it right. LOL
If your gun of choice should be a .22LR then you can apply "THE RULE OF 50".
Indeed the good and gentle men commenting before me are correct. A small error may arise in that the butt of the rifle is actually 101 yards from the target. I think we may ignore this without fear of great harm.
We can use your 36" dimension as the radius of a circle. We multiply the radius 2 times to get the diameter of 72".
Next,we multiply this diameter by pi to get the circumference:
72 x 3.1416= 226.1952
We next divide the circumference by 360 to get the arc length for one degree:
226.1952/360= .6283
We know there are 60 minutes of angle in one degree, so:
.6283/60 = .01047
Or, in round numbers,1/100th of an inch arc length for one minute of angle at a 36" radius.
Late to the post, the others are correct... :)
I'll accept .010" as the (rounded) answer, but I'm afraid that's merely an approximation, because it assumes trajectory is not a variable. As departure angle increases, flight time increases, lengthening the time gravity has to act on the projectile, not to mention weather conditions.
For a 100 yard problem, especially involving moderate to high velocity small arms, it would not be unreasonable to ignore the trajectory variable.
At target distances measured in miles, however, I'd suggest some calculus, a good calculator, and reliable temperature and humidity data, or a prepared table from the Dept. of Ordnance.
A) You folks is Schmartz
B) Sheesh, that was simple. My thinker must be busted. I had visions of some complicated degrees of angle problem with conversions and moon phases and such.
So.... that means, discarding all other factors for the moment, that someone shooting a half inch group off the bench at 100 yards managed to fire each shot with the rifle less than .0025" off a centerline, or .005" total movement.
About the thickness of a sheet of paper.
It is not really the buttplate to muzzle you need to consider. It it the rear sight to front sight. Consider a line from the center of the rear sight passing over the top of the front sight and passing through the center of the target.
The distance between the rear and front sight is known. Let's say it's 16 inches, but you could measure your particular rifle. 100yards = 3600inches, so imagine a very narrow triangle 3600 inches long. If the hypotenuse is 1 inch and both sides are 3600 inches, the angle of movement to move the point of impact one inch is .0159155 degrees. At 16 inches, this means you move the front sight .00445 inches to move the point of impact 1 inch at 100 yards.
I used a calculator.
The easiest way to approximate the math is divide the sight radius by 3600 (number of minutes in a circle).
36"/3600= 0.01" per minute of angle.
100 yards = 300 feet = 3600 inches.
However, since the sight radius of a rifle is rarely 3 feet long, you'll generally get smaller numbers than that.
The actual math would be, a circle with a radius 100 yards has a circumference of 200yards (2r) times pi = 22619.448 inches. Divide that number by 369 degrees gives you 62.8318 inches, divide that by 60 minutes gives you 1.0471966 inches = 1 MOA at 100 yards.
Doing the same math at 36" radius (pivot point at the butt) gives you 72x3.15159=226.19448 divide that by 360 degrees gives you 0.628318" and divide that by 60 minutes gives you 0.010771966".
But for simple math, just dividing the sight radius (or pivot radius) by 3600 will get you a good enough answer.
The math is different almost every shot. In a rest would be the the most consistant.
On the part of the shooter consider resistance applied to the rifle by the hands, shoulder, and of course gravity.
On the part of the rifle there are many variables. Assuming it shoots .10 moa at 100 yards you still have barrels thickness and temp that effect the numbers.
All that being said. The straight lines math makes it a quick and acceptable assumption because the work to get real numbers would make your head explode. Lol
Yup, JoeD is right on.
This also means you can work out corrections for smaller sight radii and shorter distances. A CETME with an 18" sight radius would need .005 of an inch correction at 100 yards for one inch, or .02 of an inch at 25 yards for one inch of correction.
Post a Comment