What is "distance" in minute of angle formula?

[Tom]

I am reading an article on determining the minute of angle. The formula is Minute of Angle = Width of Sight in inches / (Distance [D] x Distance to target in inches).

Can someone tell me what "Distance [D]" is?

Thanks.


********

I am editing after my first post. Thanks for the explanations. I'm tinkering with an idea and want to see "mathematically" how much of my front sight covers a 4 inch target at 50 yards and an 8 inch target at 100. That is why I am trying to find out what is Distance [D].

[Greg Ogdan 110th OVI]

Tom, for our purposes, it doesn't really matter. 1 MOA is approximately 1 inch at 100 yards. If you're shooting longer yardages ( past 300 or so) then you may want to be more precise. Hope this answers your question.

Regards

[jonk]

Sight radius has nothing to do with it, other than impacting how you shoot, and with what ease. Minute of angle is just what the name implies.

Picture a circle with you, the shooter, at the center. A radius is drawn to the target. It's length is whatever distance you are shooting at. The circle is tangential to the target, with you in the center.

A circle has 360 degrees. A degree has 60 minutes, and a minute has 60 seconds. It's the same system that a clock is based on.

Now if you consider shooting at 100 yards, you have a radius of just that; 100 yards. A little geometry and algebra (Diameter = Pi (radius) squared) gives a diameter of this circle of being about 628 yards, or 1884 feet. With me so far?

So, one degree of this circle is 1884/360= 5.23 feet. Or, at this point, we might want to consider inches; so 62.8 inches. Remember, 60 minutes per degree, right? 62.8/60=1.047. So one minute is 1.047" at 100 yards. Ergo, shooting a 1" group at that range is referred to as one minute of angle. Roughly 2.1" at 200 yards, .52" at 50 yards, etc. It just depends on the diameter of that imaginary circle that has you at the center and the target at the end of a radius drawn between you and it.

[Kevin Tinny]

Great explanation, Jonk:

Guess this helps explain you selection for Sherman's Bodyguard.

Admiringly yours, (to a point. Haha)
Kevin Tinny

[WBR10654]

I have not "really" ever shot with a scope and was trying to learn a few things and watched this on YouTube it was helpful.

I played with a scope a few times but actually shot the same rifles better without the scope so I figure I was doing something wrong...



https://www.youtube.com/watch?v=VA2PZBD5Tjg



I am reading an article on determining the minute of angle. The formula is Minute of Angle = Width of Sight in inches / (Distance [D] x Distance to target in inches).

Can someone tell me what "Distance [D]" is?

Thanks.


********

I am editing after my first post. Thanks for the explanations. I'm tinkering with an idea and want to see "mathematically" how much of my front sight covers a 4 inch target at 50 yards and an 8 inch target at 100. That is why I am trying to find out what is Distance [D].

[Rob FreemanWBR]

Yes Dave, that is a good video. And the earlier explanation in this thread is also VERY helpful!

As for your scope - you may want to double check to ensure it is properly mounted/affixed to your piece. Once that's verified it's up to you and your understanding of your piece, load and esp. increments that your scope is using when making adjustments. Windage can/will be dependent on mother nature, but drop should be fairly consistent at known distances - using the same ammo/lot.

I prefer to zero mine at 100 yards. Beyond that its just a matter of grouping and from there making the needed/proper adjustments. Just be sure to write down your "dope" (Data of Previous Engagement) so you can go back and reset to zero!!!

[R. McAuley 3014V]

This may be the formula you are looking for:


https://www.youtube.com/watch?v=S5AGsHSIsVo


Basically the distance (to the target) = Height (of the object) x 1000 / number of mildots

Example: a Yankee officer sitting on a horse 16 hands tall x 27.7778 / 3 mildots = 740 yards

[jonk]

Jumping back into this, I guess I'm stumped too as to what the second distance question is; because given that a front sight might be anywhere from 1/64" thick (on my Krag) to 1/4" thick (on my Berthier) if you divide that by ANYTHING other than a stupidly small number... you get a stupidly huge number.

To illustrate:

Minute of Angle = Width of Sight in inches / (Distance [D] x Distance to target in inches)
MOA= .1 (as an example) /? X 3600 (for 100 yards)

Now we already know that one minute of angle is equal to just over an inch, as per my previous post. 1.047". To fill in a value for the question mark that comes close to this, we would have to have something on the order of:
1.047=.1/(x)3600

Solve for X. Dividing by .1 is the same as multiplying by 10. So we can say:
1.047=10(3600)(x)
1.047=36,000x
x=1.047/36000
x=0.000029". Which is to say, 29 millionths of an inch.

Now nothing I can think of on a gun is that fine. Algebra was never a strong suit, but in a simple "solve for X" that is the result I come up with.

I'm curious, where did you find this? Some context might help us solve it. Though the videos may point in the right direction.

[Hal]

I may have misread, but I "assumed" the equation was for determining how many minutes of angle the front sight appeared to be as you sighted down the barrel. Maybe Distance (D) is distance from your eye to the front sight?

Say your front sight was .060" wide, distance from your eye to the front sight was 30" and you were shooting at 50 yards (1800"):

(.060"/30") X 1800" = 3.6"

I could believe a 1/16" wide sight on a carbine would appear to be a hair less than the width of a 4" bull at 50 yards.


Now......actual "Minute of angle"? No, that ain't "Minute of angle"..........

[R. McAuley 3014V]

I suppose the real question is, what is it you want to do? It is that you want to have some idea of how far to the left or right you could aim for a wind correction or changing the impact of the bullet? If you want to correct for wind, then you will need to know the wind speed and its direction to calculate a value, then based on the direction that the wind may affect the bullet, whether to take full value, half value, or no even value. I can teach you how to read for wind but if you don’t have a scope to see the mirage, or its a cloudy day and the mirage is not visible, it would be just a wild guess.
If you want to know how wide the top of your front sight blade appears in relation to the width of the target center bull’s-eye for a given distance, you can calculate this, but while a National Match front sight for the M14 Rifle is the same width of the 200 yard target center, that does you little good for ranges beyond 200 yards. You can certainly determine the proper front sight width for a set distance or range, say for a 4” wide bull’s-eye at 50 yards, but at 100 yards the 8” diameter target center may only be half that width? So, is this what you want to determine?

If I might suggest rather than doing your calculations in inches, try using mils (or 1/1000th of a degree), as there are 6400 mils to a 360 degree circle. Some useful conversions include: one degree is equal to 17.778 mils; while one mil is 0.057296 of a degree, or one mil is equal to 100 mm (3.6 inches) at 100 yards or 10 cm at 100 meters.

[jonk]

I may have misread, but I "assumed" the equation was for determining how many minutes of angle the front sight appeared to be as you sighted down the barrel. Maybe Distance (D) is distance from your eye to the front sight?

Say your front sight was .060" wide, distance from your eye to the front sight was 30" and you were shooting at 50 yards (1800"):

(.060"/30") X 1800" = 3.6"

I could believe a 1/16" wide sight on a carbine would appear to be a hair less than the width of a 4" bull at 50 yards.


Now......actual "Minute of angle"? No, that ain't "Minute of angle"..........
That might be it, and my gut feeling was that the unknown D referred to sight radius, as per the original post. I just couldn't figure out how to square it mathematically.

How much of the front sight covers the target doesn't entirely matter though, if that's it. Whether shooting a peep or open sight, the eye naturally centers the three points of articulation (rear, front, target) so that the front sight is centered in the rear sight, and the front sight is in the middle of the target, whether you take a center hold or 6 o clock hold.

If that's what you're looking to figure out, there is a simple practical way to determine it. Sketch out a target on a piece of paper. Look through your sights. Observe what the front sight is covering/straddling. Note that on the paper. Now measure the reality on the target. That's the "MOA" your front sight is covering.