Range Report Minute of angle question

[Unknown]

When figuring out how many minutes of angle a target's surface area represents at a given distance, are the minutes of angle figured in square inches, or is it based on a round measurement?

For example, a 12 inch round target has roughly 113.14 square inches of surface area, while a 12 inch square target has 144 square inches of surface area. This is a considerable difference in surface area, and so far, Wikepedia, and my google fu has been unable to answer this question.

Said another way, is target representing roughly 1 MOA at 600 yards, a 6 inch circle, or a 6 inch square?

 

[CEL317]

You're over thinking it.

MOA is an arc measurement of distance not surface area. In your example, 1 MOA at 600 yards equalling 6" (actually 6.28314") means the size of the target is 6" on a given line. Or the distance from one edge to the opposite edge is 6" (1 MOA). For a circle it would be the diameter and for a square the side.

Does that make more sense?

 

[Absit]

To get super technical if you wanted ONLY a 1 minute target at 600 yards it'd be a 6.28314" diameter circle. The square would technically be larger than 6 minutes corner to corner if it measured 6 minutes side to side, and less than 6 side to side if it measured 6 corner to corner. Ignore the surface area, it just can't be longer than the inch conversion at any point..if you want to be super technical with it.

 

[kansas]

It's easy to forget that a MOA is a measurement of part of a circle. There are 360 degrees in a circle, 60 minutes in a degree and 60 seconds in a minute. As Absit said a MOA is actually just a tad bit more than an inch at 100 yard.

 

[Greg Langelius *]

MOA is an angular measurement, which spans a linear distance across the target face proportional to the distance. Think of it as a drawing/drafting divider.

In English units, one MOA will span 1.047" at 100yds, 2.094" at 200yd, 3.141" at 300yd, etc., etc., and 10.47" at 1000yd.

It's safe enough to call it 1" per 100yd, as the imprecision, even at 1000yd, only amounts to just under 1/2", which is way too thin a slice for me to correct my aim at 300yd, let alone 1000yd distance.

Greg

 

[kmose73]

Said another way, is target representing roughly 1 MOA at 600 yards, a 6 inch circle, or a 6 inch square?[/QUOTE]

Both. The target would be 1 moa wide by 1 moa tall. The vertical part matters for drop and the horizontal for windage. Don't think in terms of inches. It's just moa.

 

[gstaylorg]

Here's a picture that may help clarify what everyone has been describing:

If you take two radii lines (mrad or moa, doesn't matter) and rotate them 360 degrees around a center axis extending from the center point to the edge of the big circle, you will get a cone. Picture in your mind that the two radii axis as shown above are a slice of pie. Rotate the pie around the axis extending from the point of the pie through the center of the crust. The edges of the crust will describe a circle as you rotate the pie. That [flat] circle represented by the open end of the cone on the target (or the rotated pie crust) is the boundary for the area described by an angular measurement at some distance. Note that this circle is perpendicular to the one actually shown in the picture. It will always be a circle, not a square, so the difference in area will be [pi*r1e2] - [pi*r2e2] where "r1" and "r2" are the radii of the larger and smaller circles, respectively.

Unless you're doing something like figuring out the difference between the area of the X-ring on an F-Class target and Standard High Power target, I'm not really sure why you are going down this road. If you simply want to understand the math, that's ok and I get it, but you really don't want to be thinking in terms of "inches" or "square inches" when you're shooting at a target. And you definitely don't want to be converting angular measurements into linear measurements while you're shooting. It's not necessary and will likely mess up your shot while you're sitting there trying to calculate in your head. Let the reticle do the work for you and simply focus on angular measurements.

Whether you use mrad or moa, an angle is an angle, and it doesn't change regardless of how far out you go. All that changes is the size of the circle at the open end of the cone I described above. The arc subtended by 1 moa (ie. the circle at the end of the cone) at 100 yd is roughly 1.0472"; at 1000 yd, it will be 10.472"; and at 500 yd it will be half that. The angle doesn't change no matter how far you extend the axis. This is really most of what you need to know to use your reticle for aiming and/or correction purposes.

 

[Unknown]

Gstaylorg,
That is what I was looking for! Thanks to everyone.

I was looking for this info because I run local rifle matches, and was trying to think up stages where the MOA was the same for various targets at various distances. I realize that one MOA is not truly one inch, but I have read it called a "shooter's MOA", or "SMOA", and it is represented as one inch per hundred yards. I wanted to be able to calculate this correctly, so I needed to know whether I should one shape VS another to represent one MOA.

As I currently understand it, I would use the standard formula for figuring the square inches for a given circle, lets say 6 inch round at 300 yards would equal roughly 2 MOA. Then, if I placed a 12 inch circle at 600 yards, this would also represent roughly 2 MOA (or SMOA if you prefer).

I wanted to find out whether the correct formula for figuring out MOA would be based on a circle, (Pi, R squared) or square (height X width). I now know it is a circle. I suspected as much because the whole MOA system is based off portions of angles of a circle.

I run local rifle matches and wanted to run some stages where targets and point value is based on the MOA each target represents at each distance. A target that is 2 MOA would be worth twice as many points as a 4 MOA target at the same distance. I realize I could have simply used square inches, but I have had lots of inquiries about how many MOA our "normal" targets represent. So I wanted to be able to properly figure out ROUGHLY how many MOA a square or rectangular target would represent. I realize that unusual target shapes can't actually be converted from square inches to MOA, but it will help me approximate what the various shapes represent if I figure square inches for round targets, and then use the same square inches for oddly shaped targets. (Realizing of course that MOA is round to account for both elevation and windage in accuracy)

Thanks to everyone. Now back to your regularly scheduled programming.

 

[gstaylorg]

Sounds like you've got it squared away. It can make a difference for what you're doing because with a round target, if your elevation is slightly low or high of center but still within the circle, your shot can fall outside if you're too far off center to one side or another. Thus a circular target is slightly more challenging to hit than a square of dimensions equal to the diameter of the circle. For your targets of unusual dimensions, it sounds like total area would be the way to go, but you could also score them based simply on the moa of the widest dimension. Good luck with your matches.

 

[elfster1234]

http://m.youtube.com/watch?v=VA2PZBD5Tjg

 

[Greg Langelius *]

OK, here's another monkey wrench for ya...

Rifle performance (on average) is not a straight proportional progression based on distance alone. If it were, the cone of diversion would be a pure cone with arrow-straight sides.

But that's not how rifles shoot at distances varying from up-close to waywayout there. That dispersion cone flairs outward as it gets longer, like the bell of a herald's trumpet.

As a far greater shooter than I once put it, "a good 1/2MOA rifle and shooter at 100yd becomes a great 2MOA rifle and shooter at 1000yd.", or somesuch...

Greg

 

[RTK]

As a far greater shooter than I once put it, "a good 1/2MOA rifle and shooter at 100yd becomes a great 2MOA rifle and shooter at 1000yd.", or somesuch...

Greg

Seems to always work that way for me too

 

[Deleted member 10043]

The most logical answer is in angle measurements. But you can measure a plane in both distance and mils or moa as part of navigation. Navigation?

For mils=(inches/3.6)/(yards/100). Not as complicated as changing course in degrees on a ship to reach a target but in the same ballpark. You try the formula yourself if your target is too far for your reticle to measure. Example, group @650 yards is 8" above bullseye. That will be an adjustment of .342 mils. Easy enough. Go back to the firing line and adjust 3 clicks down and try it again. I use it for trajectory validation for both the load and the scope as a part of zeroing the rifle. It kills two birds with one stone.

Now, I wouldn't explain that to the rancher sitting next to me. He would think I'm full of shit. He knows to lower his shot. Anyway, his scope has a mil reticle and moa turret. he doesn't know the difference and can care less. He is shooting as good or better than anybody else. It just depends at what level do you want to study this sort of thing.