Calculating MOA group size at odd ranges?

Bigeclipse

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I apologize for sounding dumb but I am having a hard time wrapping my head around calculating MOA group size at odd ranges. For example, I understand that 1MOA at 100yards is approximately 1inch. So a 1 MOA group at 300 yards would be 3 inches. Well let's say you set a target up at 123 yards and you shot a .9inch group. What math do I do to convert that to MOA group size?
 

Bigeclipse

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yards x 0.01047 will give you 1 MOA at that Yardage.
123 x 0.01047 = 1.287 " = 1 MOA
Divide Group inch size into 1.287
0.9 / 1.287 = .699 moa group
Thank you. That is exactly what I needed. I was having such a hard time wrapping my head around it and knew it was probably something easy.
 

ShtrRdy

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It flows better, and you don't have to stop and write anything down.
I was looking at it a little more and you could do the following:

Take the group size and shift the decimal point right two places.
Then divide by 1.047, and then divide by the true distance in yards.

0.9" right shift decimal two places ==> 90
Divide 90 by 1.047, and then divide by 123
You get 0.699 MOA.
 

Calamity

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I apologize for sounding dumb but I am having a hard time wrapping my head around calculating MOA group size at odd ranges. For example, I understand that 1MOA at 100yards is approximately 1inch. So a 1 MOA group at 300 yards would be 3 inches. Well let's say you set a target up at 123 yards and you shot a .9inch group. What math do I do to convert that to MOA group size?
Divide the distance in yards by 100, and multiply by the group size in inches. Example: 123 yds = 1.23. (1.23 x 0.9) = 1.107 MOA. That is not exact because a minute of angle is not exactly one inch at 100 yds, but it will do for anyone but the most exacting.
 

Calamity

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yards x 0.01047 will give you 1 MOA at that Yardage.
123 x 0.01047 = 1.287 " = 1 MOA
Divide Group inch size into 1.287
0.9 / 1.287 = .699 moa group
JLM6401; I think your math is off my friend. Common slip. If you multiply 1.267 x 0.9= 1.158 MOA it will be closer to the truth.
 

Calamity

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JLM8401: I owe you and Bigeclipse an apology. I am the one who had it bass ackwards and should have divided instead of multiplying. Extracted head from ***--better now.
 

DMP25-06

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JLM6401; I think your math is off my friend. Common slip. If you multiply 1.267 x 0.9= 1.158 MOA it will be closer to the truth.
Calamity ,

Pardon me for saying , but YOUR MATH is wrong .
JLM6401 is CORRECT .
Just think about the fact that if YOU shoot a group size of 0.9" @ 100yards , you are approximately .9 MOA , because you are smaller than 1" @ 100 yards .
Now , if you shoot that 0.9" group at a longer range , HOW CAN IT BE A LARGER MOA THAN 0.9 MOA ?

No disrespect to you .

DMP25-06
 

Weaselthis

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OK, I'll risk showing my ignorance.
If you shoot a .9" group at 123 yards and subtract the bullet diameter for a center to center MOA figure, (for eg. say a 7mm bullet .284), would the Center to center group MOA then be:
.9-.284= .616 x 1.23= .757 MOA at 123yds.?
Thanks for the question OP.
A chance to learn from this talent pool!
 

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