Question for the MOA aficionados

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tpcollins

New member
I realize this isn’t a firearm example but the answer should work the same way.

I have a trajectory that has a 10.17” drop at 50 yards with a 30 yard zero.

If I divide 10.17 by .5 (50 yards / 100 yards) = 20.3 x 1.047 = 21.3 MOA drop . . . yeah or nay?


I assume holdover equals drop but “clicks up” is different. Thanks.




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tpcollins

New member
Actually I think I should divide by 1.047, not multiply . . .


= 20.3 x 1.047 = 21.3 MOA drop versus 20.3 / 1.047 = 19.39 MOA
 
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HiBC

New member
It may involve more variables than you are considering. If you are involved in shooting a firearm,you must consider the line of your sights is offset from the line of the bore.

If you are just working out a geometry problem,make a reasonably nice drawing, Use a straight edge. X-Y grid paper may help. Identify the triangles
and team up with Pythagoras.Its mostly basic trig.

If you go to the Hornady Bullet Website, you can access a free ballistic software to play with. Its an easy to use,good tool
 
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stinkeypete

New member
I will chime in that the science idea of "Unit Cancellation" will help in almost all science problems. Let's pick Inches as our "convenient" unit of measure. (People that live in countries with the metric system are going to laugh in disbelief at what we have to do here.)

50 yards x (3 feet/yard) x (12 inches/foot) = 1800 inches "adjacent side" for the base.
10.17 inches is the "opposite side" of the triangle (You can use -10.17 if you like, too)

Tan(A) = Opposite/Adjacent
Inverse Tangent( Tan(A)) = Inverse Tangent (Opp/Adj)
A = .3237177 DEGREES. (Note: find the inverse tangent of 1. That should = 45 degrees or your calculator is in radian unit mode.)

0.3237177 Degrees x (60 minutes/1 degree) = 19.42306 MINUTES of angle.
The deviation from the point of impact from the laser-sight boreline of the barrel is 19.42306 minutes of angle.

There are 60 minutes of angle per degree, and 60 seconds of angle per angular minute. These wacky units are convenient when doing land surveying and considering longitude and latitude back in the days when one had to do all divisions by hand, and before the decimal number system was widely understood. If you want to divide (30 degrees 40 minutes 20 seconds) in half, it's easy to see it's 15 degrees 20 minutes 10 seconds.

With modern electronics, it's simply obsolete. We forget that modern electronics have only been around for about 50 years. In our hobby, we use other inconvenient units like grains of powder, drams (shotgun shells), ounces (shotgun pellets), minutes, yards, miles.. and I am sure more. History going back a couple of hundred years.

To be exact, a Minute of Angle is close to 1" at 100 yards but it's really 1.047 inches per 100 yards but 8.37" at 800 yards. Except we use fractions of inches (just to drive people crazy) and I'd say ".37 is about .375" which everyone knows is 3/8 of an inch. So at 800 yards the difference is 3/8". Maybe not enough for you to care about, but since we pulled out the calculator anyhow, why not do it right?



This way of doing things doesn't use rough approximations, other than rounding off the degrees to 7 decimal places. the inverse tangent function is various places on various scientific calculators or calculator apps.
 
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stinkeypete

New member
Well, my Honors Geometry students would all be able to do that calculation. It’s basic unit conversion and triangle trig, freshman/sophomore level math.
 
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HiBC

New member
In shop math language, you have to convert all the numbers to the same units of measure (inches or feet or cubits,doesn't matter. But you may need your range to the target in inches if you measure your sight radius in inches,etc.)
 
T

TXAZ

New member
tpcollins said:
I realize this isn’t a firearm example but the answer should work the same way.

I have a trajectory that has a 10.17” drop at 50 yards with a 30 yard zero.

If I divide 10.17 by .5 (50 yards / 100 yards) = 20.3 x 1.047 = 21.3 MOA drop . . . yeah or nay?


I assume holdover equals drop but “clicks up” is different. Thanks.




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Click to expand...
Sounds like a high end slingshot at 660 ft/ sec…
 
S

std7mag

New member
Just shoot the freak'n thing already!!!

Took longer to ask the question, and get a response, than to shoot it (whatever it is) and get meaningful field data. Otherwise known as dope.

I never got past first year algebra.
 
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JohnKSa

Administrator
The first thing you need to know is what 1 MOA is at the distance in question.

Here's a really simple formula for calculating the value of 1 MOA in inches at any distance given in yards.

Distance (yards) / 95.5 = MOA (inches)

That will give you an answer with an error of less than 0.01%

50 / 95.5 is about 0.524 which means that the value of 1 MOA in inches, at 50 yards is about 0.524"

You want to know what 10.17" at 50 yards works out to in MOA. 10.17" /0.524" is about 19.4 MOA.
 
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JohnKSa

Administrator
In this case, let's say we approximate 1MOA as 0.5" at 50 yards, instead of using its actual value of 0.524"

Then we would calculate that 10.17" was 20.3 MOA instead of the correct value of 19.4 MOA.

If we dialed the 20.3 MOA value into the scope, we would be off by 4 clicks in a scope with 1/4 MOA clicks--half an inch at 50 yards and an inch at 100 yards.

Not a huge error, but definitely a lot more than 0.047".
 
B

Bart B.

New member
https://www.nssf.org/shooting/minute-angle-moa/

Most exact number I've seen is this one to a hundred places...

1.04719753642832854694747069666400334739860873986429830552235157457471965151538005004775737357536725837...

.... inches per hundred yards.

However, most bullseye targets in the USA have rings in inch spaced sizes. They are set at ranges measured in yards. Therefore, using MOA values in inches simplifies many issues.

Aperture rear sights made in the USA normally have 40 tpi adjustment screws. With 12 clicks per turn, the standard 30 inch sight radius from the front sight enables 4 clicks to move impact exactly 1 inch per hundred yards.
 
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JohnKSa

Administrator
Therefore, using MOA values in inches simplifies many issues.
Click to expand...
I don't believe anyone is advocating anything else.

For those who are, 1 MOA at 100m is about 2.909cm--or one can round that to 3cm. :D

Interestingly enough, approximating 1MOA as 3cm at 100m is only in error by about 3% as opposed to the roughly 5% error generated by approximating 1MOA as 1" at 100yds.
 
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