QQ

bryceh12321 · March 16, 2009, 01:45 PM

If a 30 MM set of rings comes with .10" inserts, would that be equivelant to making the rings canted to 10 MOA?

Jimro · March 16, 2009, 02:37 PM

Depends on how far apart they are set.

1 MOA approx equals 1" at 100 yards.

that means .010 inches (one one hundredth of an inch) approx equals 1 MOA at 1 yard.

We will assume that your rings are space 6" apart, which is 1/6 of a yard. So a .010" offset will mean a difference of approx 6 MOA. If your rings are 5" apart then the difference is about 7 MOA. The math is 1 MOA divided by the fraction of the yard distance (which is the same as multiplied by the inverse). 1 divided by 1/6 equals 6, (6" equals 1/6 of a yard, 5" approx equals 1/7.2 of a yard)

Simple math, it could be wrong but I hope it helps. If you use two inserts (one in each ring) double the MOA gained. Just remember to put the rear insert thick on the bottom, and the front insert thick on the top.

Jimro

Scorch · March 16, 2009, 06:51 PM

Quote:

If you use two inserts (one in each ring) double the MOA gained.

If you put one under the scope in the rear and one on top of the scope in the front, that woud be the same as just putting one under the scope in the rear. The bottom of the scope in either case would be lifted .010" off the rings in the rear and touch the ring in the front, the shim in the front would do nothing but hold it against the bottom of the front ring, which is exactly what the front ring half is already doing. If you want to double the elevation gained, you would have to put both under the rear of the scope.

Jimro · March 16, 2009, 10:49 PM

Scorch,

The full quote is this:

Quote:

If you use two inserts (one in each ring) double the MOA gained. Just remember to put the rear insert thick on the bottom, and the front insert thick on the top.

Pushing the rear up .010" and pushing the front down .010" gives twice as much offset, the same effect achieved by using a .020" offset insert in the rear and a centering insert up front, which is probably a better solution unless you need to rotate a front insert to correct/center windage adjustment.

Jimro

Jimro · March 17, 2009, 10:15 AM

Bryce,

Checked my math with the guys over at snipersparadise, and the easy math is this.

The TAN of the angle you wish to achieve (20 MOA) multiplied by the distance between the scope mounts, center to center, (3.625) equals the distance the rear mount needs to come up (.021), to check this you can merely change the distance number to a yardage in inches ie 3600 for 100 yds which comes out to 20.94" or approximately 20 MOA.

But you have a Ruger Hawkeye and I don't know the exact center to center distance between your rings.

So what you need to do is measure the distance between the ring centers, and use this formula.

Tan(X)= (shim offset)/(distance between ring centers)

That means InverseTan(shim offset/distance between rings) = X

X is in degrees so multiply by 60 to turn it into minutes.

For example, InverseTan(0.010"/4") = 0.1432. 0.1432x60= 8.6 MOA

You can also play around with the formula to figure out how much shimming you need to get the MOA you want. Say you want 15 MOA, (15 minutes is one quarter of a degree, 15/60=0.25).

Tan(.25) = X/(distance between rings). Lets assume 5 inches this between rings. Tan(.25)x5"= 0.0218" of offset.

I hope this gives you the tools you need to achieve the effect you desire.

Jimro

March 16, 2009, 01:45 PM	#1
bryceh12321 Senior Member Join Date: December 9, 2008 Posts: 136	QQ If a 30 MM set of rings comes with .10" inserts, would that be equivelant to making the rings canted to 10 MOA?

March 16, 2009, 02:37 PM	#2
Jimro Senior Member Join Date: October 18, 2006 Posts: 7,097	Depends on how far apart they are set. 1 MOA approx equals 1" at 100 yards. that means .010 inches (one one hundredth of an inch) approx equals 1 MOA at 1 yard. We will assume that your rings are space 6" apart, which is 1/6 of a yard. So a .010" offset will mean a difference of approx 6 MOA. If your rings are 5" apart then the difference is about 7 MOA. The math is 1 MOA divided by the fraction of the yard distance (which is the same as multiplied by the inverse). 1 divided by 1/6 equals 6, (6" equals 1/6 of a yard, 5" approx equals 1/7.2 of a yard) Simple math, it could be wrong but I hope it helps. If you use two inserts (one in each ring) double the MOA gained. Just remember to put the rear insert thick on the bottom, and the front insert thick on the top. Jimro __________________ Machine guns are awesome until you have to carry one. Last edited by Jimro; March 16, 2009 at 02:43 PM. Reason: unsure of my trig...

March 17, 2009, 10:15 AM	#5
Jimro Senior Member Join Date: October 18, 2006 Posts: 7,097	Bryce, Checked my math with the guys over at snipersparadise, and the easy math is this. The TAN of the angle you wish to achieve (20 MOA) multiplied by the distance between the scope mounts, center to center, (3.625) equals the distance the rear mount needs to come up (.021), to check this you can merely change the distance number to a yardage in inches ie 3600 for 100 yds which comes out to 20.94" or approximately 20 MOA. But you have a Ruger Hawkeye and I don't know the exact center to center distance between your rings. So what you need to do is measure the distance between the ring centers, and use this formula. Tan(X)= (shim offset)/(distance between ring centers) That means InverseTan(shim offset/distance between rings) = X X is in degrees so multiply by 60 to turn it into minutes. For example, InverseTan(0.010"/4") = 0.1432. 0.1432x60= 8.6 MOA You can also play around with the formula to figure out how much shimming you need to get the MOA you want. Say you want 15 MOA, (15 minutes is one quarter of a degree, 15/60=0.25). Tan(.25) = X/(distance between rings). Lets assume 5 inches this between rings. Tan(.25)x5"= 0.0218" of offset. I hope this gives you the tools you need to achieve the effect you desire. Jimro __________________ Machine guns are awesome until you have to carry one.