MOA definition

[mete]

It's called dispersion, measured in minutes of angle ,conveniently - 1 MOA @ 100 yds is almost exactly 1".

 

[ltc444]

Except for top tier marksman, most firearms shoot better than the operator.

A number of writers and testers have reported rifles and loads that shoot smaller spreads at longer distances than at shorter ranges.

Example writer ___ reports that factory Xs ammo shot 1.5 inch group at 100 yds, 1.75" at 200 yds and then opened up to 3" at 300 yards.

Is this rifle a sub minute rifle, and over minute rifle, or a minute rifle?

In the end do you hit what you are shooting at and does it put the target on the ground.

 

[maestro pistolero]

One Minute of Angle = 10.47" at 1000 yards

Just under 1/2 deviation even at 1000 yards! This illustrates why using 1 inch @ 100 yards is close enough for just about any civilian work. Still, my first serious scope is in mils. No deviation whatsoever. Great thread.

 

[Flopsweat]

... The easiest way to get a center to center measurement, which is correct Art, is to measure the group at it greatest dimension and subtract from that measure one bullet diameter. ...

But of course some of us can't leave it that simple. Lately I've been subtracting the average size of the two holes I'm measuring from, since the holes are rarely the exact diameter of the bullets. They can vary from one type of paper to another and even from one hole to another on some paper. Now that I've finally gotten under .5 MOA it can make a noticeable difference. And I use calipers now too, so I can get 100ths of an inch, or 1000ths if I ever feel the need. Probably won't need the 1000ths this year or next. But digital calipers sure make short work of it.

Flops

 

[brickeyee]

The easiest way to get a center to center measurement, which is correct Art, is to measure the group at it greatest dimension and subtract from that measure one bullet diameter.

Unless you are shooting wadcutters the bullet holes are not all that likely to be one caliber to 0.001 inch tolerance.

tan^-1(group size ÷ range) x 60 = MOA...

tan^-1 (1.5÷3600) x60=1.4 MOA (yah, yah, 1.4324 etc. but sig figs!)

More than significant figures, an approximation.

You can get the correct answer by dividing the group size by 2, computing the angle,and then doubling it.
MOA is not a perpendicular concept, but a circular measurement.

 

[Buzzcook]

I don't think I've seen an older thread resurrected.
Still pretty useful info.

 

[(*_*)]

Use the name that all the other industry's use and info is easy to find...
http://en.wikipedia.org/wiki/Minute_of_arc

 

[Chuck Dye]

You can get the correct answer by dividing the group size by 2, computing the angle,and then doubling it.
MOA is not a perpendicular concept, but a circular measurement.

Using my ancient H-P 15C, that 1.5 inch group yields 1.432394405, twice a .75 incher produces 1.432394467, a 4.3284x10^-06% difference. I can live with the error.

 

[brickeyee]

Using my ancient H-P 15C, that 1.5 inch group yields 1.432394405, twice a .75 incher produces 1.432394467, a 4.3284x10-06% difference. I can live with the error

.

Fine for you, but it is still wrong.
How about telling folks the correct way and letting them decide on the error they are wiling to put up with?