Moa question

C

codyb1991

New member
I'm kinda new to shooting so bare with me. I shot a 4 shot group at 100 yards today and the group is 1" wide and 1.5" tall. Is it a 1" group or a 1.5" or a 1.25" group?
 
My understanding of MOA

I may be all wet and someone will no doubt correct me if necessary but, based on your measurements, this would be considered a 1.5 in. group. An MOA group is 1 in. between the farthest 2 bullet holes.

I don't know what you are shooting for but a 1.5 inch group (at 100 yds) will definitely make a dead deer:)
 
The group size is measured as the largest dimension minus the caliber (diameter) of the bullet.

A 1.5" wide group (outer edge to outer edge) fired with a .308 would be correctly reported as 1.192 inches.
The same group fired with a .223 would be reported as 1.277 inches.

The idea is that you're reporting the largest distance from the center of the holes of the two farthest apart shots. Since centers are hard to measure, outer edge to outer edge minus one bullet width gives you the same number.
 
Remington 700 ADL in .270 Win. I'm happy with the results, although I've only shot a total of 60 rounds through it. I'm thinking that my barrel isn't "broken in" yet. And I need to replace the crappy Remington scope mounted at the factory.
 
Peetzakilla, so its the outer edge of the two most distant bullet holes? And is measured horizontally and not vertically? Then I have 1"group?
 
No, it is the longest distance period.

It should truly be measured center to center, but the easiest way to do it is measure outside to outside, then subtract the diameter of the bullet.

It it was a perfect 1.5 inches, outside edge of the farthest 2 holes, minus .277 (the diameter of a .270 Winchester bullet) you get 1.233.
 
If you're going to measure outside edge to outside edge and then subtract the nominal caliber from the measurement then be sure of two things, especially with small groups.

1. Make sure that the nominal caliber is the actual diameter of the bullet.
2. Make sure that the holes in the paper are actually the size of the bullet diameter. I have seen low-velocity FMJ/roundnose bullets poke holes in paper that are considerably smaller than full caliber. On a light background target you can often see "wipe" marks outside the actual hole, but on a dark background those marks are sometimes not visible.
 
Convention has it--for well over a century--that a group's size is measured from center to center of the farthest bullet holes. Same as max distance minus one bullet's diameter.
 
Just find the 2 holes the greatest distance apart. Measure from the outside edge of one, to the inside edge of the other. This automatically gives you the exact center to center measurement without any math. If all the shots are in a group small enough that there is literally only 1 hole in the target and you cannot determine where the inside edge is then you have to measure the extreme outside spread then subtract bullet diameter.
 
well..

if you are measuring the outside dimensions of a group shot in the shape of a square or rectangle then we have some math to do.

(a^2) + (b^2) = C^2

1^2 + 1.5^2 =C^2 so... 1+2.25=c^2 so the squuare root of 3.25 is 1.8027....

so 1.803 - the caliber of bullet (.270) is 1.537 that is your group size.
 
quepasakimosabe - I was going to let this go, but your equation only fits if the shots were a perfect square or rectangle. The shots would have to line up perfectly. Measuring shot group size is not the same as measuring the area of the group. Group size is measured as jmr40 or Art explains it.

Now let me break down why your a2+b2=c2 does not work.

Say you have a group that is 2" wide by 3" high. Measure from farthest point of width and height from points not of equal distance to another. Conventional method says your group size is 3" minus caliber.Your equation would equal to 3.6 minus caliber. That's a difference of .6".

As the size goes up, 3" high by 4" wide, the difference is even bigger, 4" minus caliber versus 5" minus caliber. That's a whole inch difference.

I can see what you are trying to say but it's not the way its done.
 
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I do it like jmr40.

There are a lot of different figures of merit for accuracy and "group diameter" taken as center to center of the widest shots is only one of them.
 
Wow I never thought measuring the distance between the two furthest holes could be so complicated.....:confused: Could you imagine a conference call between a few aeronautical engineers.....:eek:
 
I do agree it is the distance between the furthest shots and would have included a diagram of lines with O's in it to explain what I was saying. The poster lead me to believe that he had two w\shots in the opposite corner of the square/rectangle which is the only reason I wold even measure a group in a square. I should have taken into accout that maybe that is not the case, in which you would meausre the furthest two shots, subtract the caliber of round and bingo.
 
Make is easy on yourself and download the On Target tool demo copy.
http://www.ontargetshooting.com/download.html

Scan in your target,
select two points that you know the separation for (like the edges of a 3 inch target ring) and enter that data,
select the caliber and place the cursor over the bullet holes and tell the tool where you aimed.

The tool will calculate the separation and the MOA automatically and show it on your target.

Here is a sample.

Savage10FPH4895418gr168grSMK2805112610websmall.jpg


If you like the tool, buy the newest version that has all the bells and whistles. I have used this tool for about 2 years for over 2000 group measurements. I swear by it.
 
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