Help with some MOA math

[SG4247]

So, by my calculation, if you position the rifle bore off of dead center in a typical fclass rifle, by more the about .010” you are at risk of shooting a 9, in perfect conditions, on a 1 MOA 10 ring.

I do it all the time.

 

[OLeveraction]

a 28 inch barrel .008145 would equal 1 MOA

It the distance between the front sight and the rear sight that should be used to figure that.

 

[SG4247]

In my example, I was thinking the length from the rear bag resting point, to the muzzle for an fclass set up. Not much room for bad manners.

 

[creedfan]

It the distance between the front sight and the rear sight that should be used to figure that.

the OP didn’t say anything about having sights he didn’t even say if it was a rifle

 

[AccelR8]

I do it all the time.

Hah! Gotcha beat...my last fclass target has THREE...count 'em, three nines!

 

[Ned Ludd]

What’s the equation for how much the scope reticle moves for 1 MOA? Another way to ask is how much does the angle of the barrel move for 1 degree in inches or thousands?
I’m trying to understand how little or how much the barrel moves to change the POI down range. Is it 1/60?

My advice is to stick with the angular measurements for most things in shooting because they never change. The answer to the question highlighted in red is 1 degree. A one degree angle is one degree at all distances. One MOA is one MOA at all distances. However, I understand the measurement you're trying to determine.

What you are referring to is the relationship between the arc length that is subtended by a certain angle for a circle with a given radius, not the angle itself. When we refer to group spread using a linear measurement, we are using an approximation of arc length. What we actually measure is the distance between the bullet hole centers on a flat surface. The arc is actually not flat, but at the distances we shoot, linear measurements taken from a flat surface are a reasonable approximation of arc length. What you want to know is how arc length at some distance (i.e. radius) is affected by slight changes in position of the rifle/barrel or turret adjustment on the scope. To do this, you need to convert that movement into an angle in degrees, then use the following formula:



The angle (theta) will need to be estimated first. There are various ways to do this, but one would be to use the above formula for some movement at the muzzle, which is located at some known distance (r) from the shooters eye, and back-calculate to solve for theta. Once you have theta, you can plug in any radius length you wish to determine the effect on arc length at some distance for that change in angle (theta) at the rifle.

For example, if the muzzle of a rifle is 20 inches away from the shooters eye and the muzzle travels one tenth of an inch laterally or vertically between two successive shots, how far apart will the shots land at 100 yards distance, assuming no other outside interference such as wind?


To start, estimate theta:

r = 20", arc length = 0.1", plug and play in the above formula and solve for theta:

(2)(3.1416)(20")(theta/360 degrees) = 0.1"

theta = (0.1")(360 degrees)/(2)(3.1416)(20") = ~0.29 degrees

Note: even only one tenth inch movement at the muzzle is a huge change in theta, approximately 17.4 MOA (0.29 degrees)



At 100 yd, a muzzle deflection of 0.1" (i.e. ~0.29 degrees) can then be estimated to have the following effect between the impact of two shots at 100 yd distance, again using the above formula:

Arc length = (2)(pi)(r)(theta/360)

Arc length = (2)(3.1416)(3600")(0.29 deg/360 deg)

Arc length = ~18.2"

In other words, a muzzle deflection of only one tenth of an inch for a rifle with barrel length consistent to put the muzzle 20" away from the eye would result in a shot deflection of slightly over 18" at 100 yds distance. One could more easily arrive at the same value if they know that 1 MOA angular dispersion is approximately equal to arc length of 1.0472" at 100 yd distance. The bottom line is that it doesn't take much. Very small movement of the rifle or minute adjustment of a scope turret can have a profound effect on shot placement as the distance increases.

To make a similar estimate for a scope adjustment (i.e. reticle movement), simply use the distance from the eye to the reticle as "r", and the reticle travel or movement as the arc length to solve for theta.

 

[Zero333]

1 MOA = 0.016666654061044 degrees

 

[Fairchaser]

My advice is to stick with the angular measurements for most things in shooting because they never change. The answer to the question highlighted in red is 1 degree. A one degree angle is one degree at all distances. One MOA is one MOA at all distances. However, I understand the measurement you're trying to determine.

What you are referring to is the relationship between the arc length that is subtended by a certain angle for a circle with a given radius, not the angle itself. When we refer to group spread using a linear measurement, we are using an approximation of arc length. What we actually measure is the distance between the bullet hole centers on a flat surface. The arc is actually not flat, but at the distances we shoot, linear measurements taken from a flat surface are a reasonable approximation of arc length. What you want to know is how arc length at some distance (i.e. radius) is affected by slight changes in position of the rifle/barrel or turret adjustment on the scope. To do this, you need to convert that movement into an angle in degrees, then use the following formula:

View attachment 1469146

The angle (theta) will need to be estimated first. There are various ways to do this, but one would be to use the above formula for some movement at the muzzle, which is located at some known distance (r) from the shooters eye, and back-calculate to solve for theta. Once you have theta, you can plug in any radius length you wish to determine the effect on arc length at some distance for that change in angle (theta) at the rifle.

For example, if the muzzle of a rifle is 20 inches away from the shooters eye and the muzzle travels one tenth of an inch laterally or vertically between two successive shots, how far apart will the shots land at 100 yards distance, assuming no other outside interference such as wind?


To start, estimate theta:

r = 20", arc length = 0.1", plug and play in the above formula and solve for theta:

(2)(3.1416)(20")(theta/360 degrees) = 0.1"

theta = (0.1")(360 degrees)/(2)(3.1416)(20") = ~0.29 degrees

Note: even only one tenth inch movement at the muzzle is a huge change in theta, approximately 17.4 MOA (0.29 degrees)



At 100 yd, a muzzle deflection of 0.1" (i.e. ~0.29 degrees) can then be estimated to have the following effect between the impact of two shots at 100 yd distance, again using the above formula:

Arc length = (2)(pi)(r)(theta/360)

Arc length = (2)(3.1416)(3600")(0.29 deg/360 deg)

Arc length = ~18.2"

In other words, a muzzle deflection of only one tenth of an inch for a rifle with barrel length consistent to put the muzzle 20" away from the eye would result in a shot deflection of slightly over 18" at 100 yds distance. One could more easily arrive at the same value if they know that 1 MOA angular dispersion is approximately equal to arc length of 1.0472" at 100 yd distance. The bottom line is that it doesn't take much. Very small movement of the rifle or minute adjustment of a scope turret can have a profound effect on shot placement as the distance increases.

To make a similar estimate for a scope adjustment (i.e. reticle movement), simply use the distance from the eye to the reticle as "r", and the reticle travel or movement as the arc length to solve for theta.

Excellent comment and answers my original question as poorly worded as it was. Thank you!

 

[Coonman300]

Threads like this remind of 2 things.
1. There are a lot of smart people on this site regarding calculus or algebra.
2. I ain’t one of them.