# 1 MOA means a target circle how big at 100 yrds?

Forseti

April 14, 2003, 12:26 AM

Forgive me, I can't remember...1 MOA "cut" at a plane of 100 yards means we are talking about a circular area of just what diameter? At a 200 yard plane, what size is a 1 MOA "circle" target area?

If anyone has a good link on this with the math, I'd love to get it so I can bookmark it...

If you enjoyed reading about "1 MOA means a target circle how big at 100 yrds?" here in TheHighRoad.org archive, you'll LOVE our community. Come join

TheHighRoad.org today for the full version!

444

April 14, 2003, 12:33 AM

1 MOA is generally considered to be 1" for each hundred yards. 1" at 100 yards, 2" at 200 yards etc. I belive that 1 MOA is actually slightly larger than that but I don't know the exact measurement.

When measureing a group, the idea isn't to keep all the shots in a circle of that diameter, but rather the group is measured center to center of the two widest shots. I am sure others will disagree with this, so you will get some more input.

Forseti

April 14, 2003, 12:39 AM

Hmmm...according to that, everyone who claims 1 or 2 MOA accuracy at 100 yards on this board is putting all their bullets into a 1 or 2 inch hole....

I've got to get better at shooting. I thought putting everything into a 3.25" hole at 100 yards was pretty good.

444

April 14, 2003, 12:47 AM

That depends on what you are shooting the group with and from what position.

TechBrute

April 14, 2003, 12:53 AM

I thought putting everything into a 3.25" hole at 100 yards was pretty good. Shooting prone off a bipod using a Robar SR-90, 3.25" is likely to get you laughed off the range. Shooting that offhand with a military rifle is going to get you applause.

JimC

April 14, 2003, 08:17 AM

Minute-of-angle or "MOA" is actually 1.047" per 100 yards, sometimes rounded off to 1.05" per 100 or 1/60 th of a degree.

TarpleyG

April 14, 2003, 09:10 AM

This might help. They say you can round it off to 1.00" for easier calculation at shorter ranges.

http://www.imagestation.com/picture/sraid59/p5ee0d7e428ee8805e256f4ad2fcadf8a/fc57354b.jpg

GT

No4Mk1*

April 14, 2003, 10:37 AM

The math:

1 MOA is 1/60 of one degree.

That is 0.0166666666666666 degrees

The tangent of an angle in a right triangle is equal to the lenth of the opposite side divided by the adjacent side...

The angle is 0.01666666 and the tangent is 0.00029088821687.

Since there are 3600 inches in 100 yards...

Opposite side = 0.000290888 * 3600

Thus the opposite side of the angle is 1.04719758 inches.

So the size of one MOA at any range can be calculated by 0.01047198 (inches/yard) * RANGE (yards)

hksw

April 14, 2003, 01:15 PM

But, it's generally taken as 1"/100yds, even though it technically is not exactly 1".

cratz2

April 14, 2003, 04:07 PM

There are a lot of variables. For one thing, shooting off bags on a bench is going to be more steady for most folks than a bipod or prone. Shooting with an informal rest should allow for more accuracy and precision than offhand. And of course, different loads will provide different levels of accuracy in most rifles.

Not to get into a contest of which rifles are more accurate, but in general, a quality bolt action rifle with a heavy barrel is going to be more accurate than a Remington 7400.

3.5" off hand with a 7400 is very decent shooting. A 3.5" group from a Sako TRG42 with good handloads built to the rifle might mean the shooter is doing something wrong. Of course, it might mean one screw isn't tightened properly too.

Forseti

April 14, 2003, 11:40 PM

I was using an AR-15 kit gun with a 24" barrel and a 5x scope. Ammo was Georgia arms 55 grain .223 reloads. Not match ammo, but goes bang everytime.

I have been wanting to try it out with heavier bullets and see how that affects things, since the barrel is pretty long. I think it has a 1/8 twist.

The scope is such that it has a sight picture like this (but without the deer)

http://www.impactguns.com/store/media/atn_reticle.gif

The dot almost covers a 1.5" circle at 100yds. The sight picture above is cruder than the etched glass actual sight.

So, clearly I have room for much improvement.

cratz2

April 15, 2003, 12:59 AM

I have an ATN 5x33 scope and it was not as accurate for me as most scopes are. I had it on my AR for a while to try out then I put it on my Savage 12FVSS in 308 which is my scope evaluation rifle. I just didn't like it too much. I think it would be fine for a shorter tactical rifle but the dot isn't big enough for super quick target acquisition and is too big for very precise work. I guess it serves it's purpose, I just don't have a need for that purpose, I guess.

http://photos.imageevent.com/cratz2/guns//DCP_1515a.jpg

FNFiveSeven

April 15, 2003, 04:52 PM

Then consider this: 1 MOA may be = 1.04719758 inches @ 100 yds, but 2 MOA is not exactly twice that number, it's a little more... think about it.

and something else to keep you thinking in your spare time:

Let's say you are shooting at a range where you need to shoot 24 inches over a target to hit it... there are two different ways to do this. Method 1: You can stand up off the bench and just "shoot straight," or Method 2: You can tilt the gun up and remain on the bench. The question is, do both methods actually accomplish the same thing, or will the bullet impact depend on the method used? :evil:

Art Eatman

April 15, 2003, 07:04 PM

Oooohhh, blackrazor! One MOA is one MOA, period! Whatever length is subtended at whatever range, it will be twice that length at double the range. That's an "It's gotta be!" deal. This ain't self-esteem math we're talkin', here.

As for the second question, what you've got is two feet of drop. You don't say how the sights are set, but it doesn't really matter. You either hold over or you adjust your sights accordingly--that is, you tilt the rifle. But standing up or sitting down doesn't have anything to do with anything.

For instance, I sight my '06 for a 200-yard zero. To hit a target at 400 yards, I hold the horizontal crosshair about 22" to 24" above the desired point of impact--sittin', standin', or missionary position.

:), Art

Groucho

April 15, 2003, 07:24 PM

Art Eatman said, "I hold the horizontal crosshair about 22" to 24" above the desired point of impact--sittin', standin', or missionary position.

:what: I've got to go back to the basics. I don't remember the "missionary position" in shooting rifles:scrutiny: I'm glad I found this place. I'm learning so much:D

Groucho

Just when I think I know something, I realize I don't know nuttin'

FNFiveSeven

April 15, 2003, 07:25 PM

Art... ahhh the questions I pose are in the details. Yes, you are right, if you double the distance you will double the size of the circle at any given angle measurement. BUT... that's not exactly what I said. What I said was that if you double the ANGLE at a FIXED distance, you more than double the size of a circle. This concept becomes more obvious when you look at larger angles. Clearly 90 degrees is an infinitely large "circle" , not just 5400 inches @ 100 yards.

As to the second question, I believe it does make a difference "how" you aim at a target. For example, consider a target 1000 yards away... if you're shooting with a 308 that requires about 400 inches elevation (approx 33 feet) to hit. You can either tilt the rifle upwards at approx 1 degree, or you can go to the 3rd floor of a building. But clearly one situation is different from another, as in this extreme case you can see that by not tilting the rifle but by raising it you are effectively shooting from an elevated position.

Oh well, just some brain candy to think about...

No4Mk1*

April 15, 2003, 07:57 PM

Blackrazor is correct. 2 MOA is 0.00000017723 inches larger than expected because by the time the bullet that is 2 minutes off target reaches paper at 100 yards, the bullet has actually flown 100.000016923 yards. So at this range, 2MOA is slightly larger than you would expect if you assumed the bullet had only flown 100 yards. At such small angles of these this math is only for fun of course.

The other one:

Changing your elevation in relation to the target will change the angle needed to fire on and hit the target. So speaking of aiming the barrel, if you are at the same elevation as the bullseye you must aim (the barrel) above the bull to hit it. If you elevate the firing position there will be some point where the barrel is aimed perfectly horizontally to hit the target. Further elevation of the shooter requires the barrel to be pointed down. Net result is the target is hit every time.

It is easier to just sit there and aim 24 inches high though or adjust the sights.

Art Eatman

April 15, 2003, 08:20 PM

"Clearly 90 degrees is an infinitely large "circle", not just 5400 inches @ 100 yards."

It's one thing to worry about the fifth digit to the right of a decimal point, but I fail to see how 1/4 of a circle is infinitely large. Infinite quadrants?

:), Art

Jim K

April 15, 2003, 09:46 PM

A circle of 5400 inches in diameter would hardly be "infinite". The confusion arises because using minutes of angle assumes that the "target" is an arc of a circle or, to be more exact, a section of a sphere, like a satellite dish, with each part of the target at the same distance from the rifle muzzle.

If it is, then the ratio of minutes of angle to distance is correct for any point on the target. But, if the target is flat, and the center is at the given distance (e.g., 100 yards) then any shot not at the exact center will travel farther than 100 yards. This is insignificant in dealing with a few minutes of angle, but if the best our rifle can do is to keep its shots in a 90 degree angle, the distance to different points on a flat target does become significant.

Of course, maybe a glass bedding job, and check the throat erosion....

Jim

FNFiveSeven

April 16, 2003, 02:00 AM

Mk1,

That's an interesting take, but I was just doing the trig and the numbers fall out. Here's how it goes, at least as far as I figure it:

1 MOA @ 100 yds = tan (1/60)*3600 = 1.04719758 inches

2 MOA @ 100 yds = tan (2/60)*3600 = 2.09439534 inches

Note that we are already seeing how 2 MOA is actually a little more than exactly twice 1 MOA at 100 yds, in this case it's 2.00000017 times bigger.

Now, I know what you all are thinking, who cares!? What difference does a factor of 0.00000017 make?!!?

Just hold on and see where this goes at larger angles...

100 MOA @ 100 yds = tan (10/60)*3600 = 104.74930166 inches

This value is not 100x bigger, but is actually 100.0282x bigger. Error's getting more significant now!

Now just for fun, let's look at 5399 MOA. What's that you say? It's still going to be somewhere around 5399 inches at 100 yards? Let's see...

5399 MOA @ 100 yds = tan (5399/60)*3600 = 12375888 inches o

or ~ 195 miles!!! 5399 MOA isn't even *close* to 5399 inches at 100 yards!

Ha ha!

:scrutiny: :scrutiny: :scrutiny: :scrutiny: :scrutiny: :what:

P.S.

You can see where this is going, 5399 MOA is almost 90 degrees, hence my earlier statement.

Art Eatman

April 16, 2003, 11:36 AM

blackrazor, did you know that way back when, high school math included both plane and spherical trig? My old trig book copyright dates range from 1918 to 1946. :)

Anyhow, I imagine you'll have a long and illustrious career at one of the spice companies, picking fly poop out of pepper. (Sorry; lost control, there. It's the occasional outbreak of a character defect.)

:D, Art

Mal H

April 16, 2003, 02:18 PM

blackrazor - where your math goes wrong (and it does) is in using the tangent of the angle. You should use the sine. In a right triangle, the tangent is the ratio of the opposite side to the adjacent side, whereas the sine is the ratio of the opposite side to the hypotenuse. As the angle increases the adjacent side will approach zero. At 90°, the tan is infinity, which you have shown. But, at 90° the sine is unity, so switching sin for tan in your formula you would end up with 5399 inches as expected (the opposite side is now very close to the length of the hyp). At very small angles (i.e., a few minutes) there is practically no difference between the tangent and the sine of the angle, also as you have shown.

In reality, using a trig function to determine the arc length of a circle isn't valid when the angle is greater than a few minutes of arc. The error in the true length is made even greater for smaller circles. That's why using a trig function for a large circle (like our 100 yd example) is ok for anything we want to do with a gun.

FNFiveSeven

April 16, 2003, 03:03 PM

Mal H,

Not to beat a dead horse here or anything, but I don't think there's anything wrong with my math.

I believe that the use of the tangent is correct, and here's why:

the distance from the shooter to the target is a fixed distance, in this case, defined to be 100 yards. This distance cannot be changed because the shooter and the target are not moving.

Now, you stated that one of the reasons that we should not use the tangent is that the length of the adjacent approaches zero as the angle approaches 90 degrees (aka 5400 MOA), but this is not true. This is the case in the unit circle where the *hypotenuse* is fixed at a distance of "1" but not in our case, where we have fixed the adjacent at a distance of 100 yards, and allow the hypotenuese and the opposite to vary. The opposite is the distance we are missing by, holding over, etc. and the hypotenuse is the distance the bullet travels. Therefore is is appropriate to define angle to the target as the ratio of the opposite (holdover distance) over the adjacent (distance to target). We are not dealing with the unit circle here.

Intuitively, it should also make sense that if you miss the target by 5400 MOA you are missing it by 90 degrees, and clearly missing my 90 degrees means that you've missed it by an infinite amount.

You can try this at home, take your rifle, point at the 10 ring, and then rotate the rifle by 5400 MOA (90 degrees). No matter how big the target is, it's obvious you won't hit the paper.

On second thought, don't try this as home... :what:

Mal H

April 16, 2003, 03:19 PM

Well , ok. I tried.

Now, you stated that one of the reasons that we should not use the tangent is that the length of the adjacent approaches zero as the angle approaches 90 degrees (aka 5400 MOA), but this is not true. Pythagoras and Mrs. O'Brien might disagree.

... clearly missing [b]y 90 degrees means that you've missed it by an infinite amount. Here, Pythagoras, Mrs. O'Brien, Einstein, Planck, Bohr, Galileo, Aristotle, and I all disagree. Schrödinger would disagree with you about 50% of the time. Heisenberg might jump to a completely different conclusion, but you can never be certain what he's thinking about anyway.

FNFiveSeven

April 16, 2003, 04:07 PM

Mal,

How can the adjacent approach zero?!? By definition, we've FIXED the adjacent to be 100.00000000000 yards, the straight line distance from the shooter to the target! It doesn't approach anything as the angle is changed, not at 1 MOA, not at 45 degrees! You don't have to use the unit circle every time you deal with trigonometry.

Also, why do you disagree with my statement regarding shooting 90 degrees off target leads to an "infinite miss?" If you point at target, and then rotate 90 degrees and pull the trigger, you can't seriously be implying that you'll miss the target by only 5400 inches! Right?!

How do you post pictures on this thing?

No4Mk1*

April 16, 2003, 07:52 PM

This is a fun discussion.

I have to agree with Blackrazor on this one. I think the confusing point here is that there are three ways to quantify how far off a shot is. The first is to describe the angle the bullet traveled compared to the intended travel vector. This is Measured in minutes or degrees if you are a lousy shot. Another way is to measure the arc length of a circle at a given distance. The third and most widely used is to measure the distance of deviation of the shot in the plane which the flat 2 dimensional target defines. We are defining the distance of a miss as the distance from the bullseye to where the bullet reaches a plane extending from the 2 dimensional target hanging at 100 yards. This is the common "group" measured in inches. For all practical purposes, arc lenght and group size are interchangable for small angles (few degrees) and group size and arc lenght may be converted into MOA by a simple multiplication factor. Error becomes significant when you are many degrees off target.

The reason I like blackrazor's point of view is because in the real world, we don't directly measure angle or arc length, we measure group size in the plane of the target. Therefore it is appropriate to use a fixed adjacent side of 100 yards and a measured opposite side (shot deviation from bullseye). These are the only values we know. from there we must calculate the angle. The arc lenght can also be calculated from this.

As far as the infinite miss, he is not saying that 90 degrees is an infinite quantity of course. He is saying that a bullet aimed at the center of a plane which is fired 90 degrees off target will never stike the plane. So in the 90 degree miss, the angle is 90 degrees, the arc length is 157.08 yards. Since the bullet will never strike the plane of the target, the measured deviation in that plane is infinite.

Soap

April 16, 2003, 08:09 PM

Anyhow, I imagine you'll have a long and illustrious career at one of the spice companies, picking fly poop out of pepper.

I almost fell off my chair there! :D

FNFiveSeven

April 17, 2003, 02:55 PM

Ditto on that Daniel.

Art... funny, sometimes I *do* feel like that is my career. I guess you had me pegged :o

thanks for backing me up, No4Mk1. Looks like we're on the same page.

dongun

April 17, 2003, 03:14 PM

Ya'll got it all wrong! Einstein said that relative velocity distorts both time and space. Therefore, the bullet actually never reaches the target....:evil: :evil:

FNFiveSeven

April 17, 2003, 03:56 PM

Well, the bullet does reach the target, but it will be a bit "younger" than it would have been if you walked it there. heh heh

Dave P

April 17, 2003, 04:18 PM

Let me wade in on one point: 90 deg being an infinite circle at 100 yrds.

I think when you are looking at these large angles, think of them as +- 45 degrees, not 0 to 90. If I aim 45 degr off to the left of the target, I will hit at 100 yrds on the side of the target (think of a 45 degree triangle). If I now aim 45 to the right of the target, and again hit 100 yrds to the side.

The two holes in the target are now 200 yrds apart (-45 and +45 degr from my POV), which is in the neighborhood of 7200 MOA!

Everybody got it?

FNFiveSeven

April 17, 2003, 04:31 PM

Dave,

yes, I see what you are saying. When I said 90 degrees off, I meant in one direction, or +/- 90 degrees. You can also just think of it as 180 degrees (by your definition), which of course still implies "infinite" missing of the target.

But, from looking at your example of +/- 45 degrees, the error in the statement "each MOA is 1 inch at 100 yards" really becomes obvious. As you said, the two holes in the target are separated by 7200 *inches* (i.e. 200 yards), even though there are actually fired 5400 *MOA* apart! So clearly at increasingly large angles, each additional minute of angle corresponds to much more than 1 inch at 100 yards.

whew.

P.S. I guess I'm a freak...but y'all are too if you've read this far! :neener:

Ebbtide

April 17, 2003, 04:44 PM

Huh?

This thread gave me a headache:)

Mal H

April 17, 2003, 05:37 PM

*sigh*

Peetmoss

April 17, 2003, 08:13 PM

OMG my brain has melted.:uhoh:

FNFiveSeven

April 17, 2003, 08:23 PM

Nothing wrong with a little mental exercise, I think this stuff is fun. Keeps the mind sharp....

:cool:

Shootin' Buddy

April 18, 2003, 04:46 AM

Well, I might as well step in and add my .00000002 cents.

1 MOA @ 100 yds = tan (1/60)*3600 is not quite the right formula in this context.

It is a matter of definition in target group sizes calculations that the center point (what we might for convenience refer to as 0 MOA) is a calculated average of all distances measured center to center between each bullet hole and every other bullet hole considered. Thus, if I had exactly two holes in a target the center point (0 degrees) would be exactly half way between them. If then these two holes measured exactly 1 MOA from 100 yards distant, the deviation from 0 MOA would then actually be 1/2 MOA and so the correct formula for determining the precise distance in inches between the two points is 2*(tan(1/120)*3600).

2*(tan(1/120)*3600) <> tan (1/60)*3600

and that's my .00000002 cents :)

tstr

FNFiveSeven

April 18, 2003, 05:05 AM

Good insight there, shootin' buddy. You are certainly right, because if someone says their rifle shoots 1 MOA groups at 100 yards, they usually mean that they can keep their shots within a 0.5 MOA radius circle around the center of the target. And, if there's one thing we've shown in this thread, it's that at any given distance, 2 x (0.5 MOA) does *not* equal 1 MOA. Your 20 nanocents are appreciated! :D

No4Mk1*

April 18, 2003, 04:30 PM

Good point, that was worth more than .00000002 cents.

pax

April 18, 2003, 04:45 PM

Yeah, closer to .0000000221524 or thereabouts. ;)

pax

There are three kinds of researchers: Those who can do math and those who cannot. -- Tom Rusk Vickery

Steve Smith

April 18, 2003, 04:51 PM

I'm gonna go hide under a rock.

Mal H

April 18, 2003, 05:29 PM

Hey! Steve! Get outa here! This is my rock. Go find your own!

:D

(And it was a good thing I saw who it was, 'cause I had my fingers in my ears and didn't hear you coming.)

If you enjoyed reading about "1 MOA means a target circle how big at 100 yrds?" here in TheHighRoad.org archive, you'll LOVE our community. Come join

TheHighRoad.org today for the full version!

vBulletin® v3.8.6, Copyright ©2000-2015, Jelsoft Enterprises Ltd.