射擊精度的單位---MOA(組圖)

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射擊精度的單位　　

MOA-Minute of Angle，翻譯成中文就是分角，即是指360度內的1度其中之60份之1

　　假設一個圓的半徑的長度是50米，那麽它的直徑就是100米 ( 50 X 2 )，計算MOA夾角大小的公式如下:

　　( 100 X 3.14 ) X ( 1 MOA / 21600 )

　　= 314 X 0.000046

　　= 0.014444 (米)

　　= 1.44 cm

　　即是在50米的距離下1 MOA的夾角相當於1.44 cm ，亦即是說一把1 MOA精度的狙擊槍歸零後，射擊距離50米的目標下會有 1.44 cm 的隨機誤差。靶子中心點畫一個直徑1.44 cm的圓，接著射擊的時候彈著點可能會偏上、偏左、偏右或偏下，但隻要不超出這個直徑 1.44 cm的圓就可以了。一些槍械製造商會在300米的距離下射擊50次來作準 (每一槍也是在冷槍管的情況下射擊)，誤差不可超過8.7cm ( 600 X 3.14 ) X ( 1 MOA / 21600)







What is a Minute of Angle (MOA)?by Robert W. Current ([e-mail]) and Trevor M. Riedemann ([e-mail]) .

It is a measurement of angle, a subdivision of a degree. There are 360 degrees in a full circle

4 quadrants = 1 circle
90 degrees = 1 quadrant
60 minutes = 1 degree
60 seconds = 1 minute
21,600 minutes = 1 circle

So they are all just measurements of an angle, should be listed in any fact finder or such.

The reason it is so widely used in the shooting sports is because if you measure an angle of one minute the lines almost run parallel at short distances, but at long distances you can see them separate. And quite conveniently they separate to 1 inch apart when you get 100 yards from the point they cross. That works out great for measuring rifle accuracy (in an idealized situation, no wind, ...):

Shots spread at an angle, increasing group size as distance from the target increases.

Diagram 1

 

"M" is the muzzle and "d" is the distance to the target, now it is easy to see the closer you are to the Target, the more likely you are to hit it (we all know that, at least intuitively) and it can be seen it is some function of an angle. That is why accuracy is ideally measured in angles not just shot spread.

So, if you have a 1 Minute of Angle (1 MOA) rifle, you can expect about one inch groups at 100 yards, 2 inch groups at 200 yards, 3 inch groups at 300 yards.

To explain this if full detail requires the use of trigonometry/geometry:

Diagram 2

 

At a distance of 100 yards from the target, length "d" on the diagram, you shoot a group that measures one inch, segment "C" on the diagram.

Then:
tan(q/2) = C/2d
(look at the addendum below to see how this is found)

Solving for theta
q = 2 tan^-1(C/2d)

For C = 1 inch
d = 100 yards = 3600 inches
q = 0.0159155 degrees = 0.9549 arcmin

Alternatively this can be made by an approximation:
S = Rq
where q is in radians solve for theta
q = S/R
S is approximately equal to C (1.0000004" ~ 1")

R is approximately equal to d (100.000001 yards ~ 100 yards)
q = C/d = 1 inch/3600 inch = 0.00027777 radians = 0.9549 arcmin

So, your one inch group at 100 yards is really 0.95 MOA group!

Want to go into more detail? Well a degree is symbolized by a little o above and to the right of the number, a minute is symbolized by a single apostrophe (') and a second is abbreviated with a double apostrophe ("). You know what that means? I could say I shot a 1' group (a "one minute" group), and it could mean I shot a group that was 0.95 inches at 100 yards, or even a 1.8 inch group at 200 yards. This use of the symbols is really one of my pet peeves, that I never really rant about, I just let people sound silly and laugh to myself later. I shot a 1" group. Wow, did I read that article right, he said 1" that is one second right? It has to be, he didn't say 1 inch at a 100 yard distance, he didn't even mention distance, so he must be reporting accuracy in MOA, and that would work out to be 1/60 (one sixtieth) of an inch at 100 yards. Man those guys are so good, I must really suck, I will never be able to shoot that well.

But, even this all seems to be an ideal way of measuring things for shooting, you have to keep in mind that the real world is rarely ideal. You really have to take a few more things into account, at short range they matter less but get out to 1000 yards, and well, wind, luck, drag coefficients, and all kinds of junk I admittedly don't even fully understand will start to come into play. For very long range projectiles the rotation of the earth will alter the trajectory significantly (Coriolis force)!

---

ADDENDUM

For inquiring minds, this addendum demonstrates the derivation of the equation [tan(q/2) = C/2d] used in the above analysis. Redrawing Diagram 1:

Diagram 3

 

There are two right triangles. Focus on the right triangle on top defined by the angle "a" and the line segments of length "x", "y", and "R."

From Diagrams 1 and 2
q = 2a (or a = q/2)
C = 2y
or
y = (C/2)
x = d
whre d is distance to the target

The basic trigonometric function "tangent" is defined as
tan a = y/x

Replace a, x and y the expressions above
tan q/2 = (C/2)/d
which is also
tan q/2 = C/2d

Solving for q/2 gives the expression
q/2 = tan^-1(C/2d) or
q = 2 tan^-1(C/2d)

Another approximation (similar to the one presented above) for the small angle involved is
a = tan a
where "a" is in radians, then
a = y/x
again, replace a, x and y the expressions above
q/2 = (C/2)/d
q/2 = C/2d
multiply both sides by 2, and
q = C/d

Which is the same as our previous approximation. This is a great "real world" application problem for kids who are doing trig and geometry - and enjoy shooting.