M14 Forum

stevenmac2

Hi Everyone,

I see there were some changes to categories so I hope that I can post this question here. I have a few questions that I would like to ask:

First question - I am curious if there is a reference guide, like a Combat Geometry 101 or 102 guide/reference that is typically taught to snipers that I could read through. I would like to have a deeper understanding to some of the calculations most commonly they use.

Second question is this as I was trying to learn about the method of using Triangulation if your Ranger Finder isn't working. There is where I may need further assistance. I saw an example from a show, but I can't get the math to work out right. The shot was close up to a mile away using .50 cal.

My Super Sniper 10x scope specs 1 mil @ 3.44 inches, but to make the math easier (since I stink at it), lets say 1 mil = 3.6 inches.

Now if we take the mil-dots presented, does each mil-dot from cross-hair represent a degree of angle? Is this the correct degree numbers no matter what scope you would use? I broke out the 0.5.

For example:

Mil (degrees)

..x..........x..........x...........x........|
0.20----0.15----0.10---0.0573---0.0


Now, the formula I saw applied was
x(ft) + tan (deg) = Distance.
but the math doesn't seem to work out for me.

Their example was they had to take out a mortar spotting crew. This crew was standing in a window. They estimated the window to be about 3 ft wide.
In the program, they showed 3ft + tan (0.33) = 5280 feet for their distance. Then dialed in. I couldn't derive to the same distance applying the addition - I have to be missing something.


So I tried what I am familiar with from the little bit of knowledge I have with a small example of trying to derive to distance by measuring the length.

Target is 3.6 inches (L) x 3.6 inches (H) - 1 mil target for our purpose. We know that 3.6 inches = 1 mil at 100 yards or 36 inches at 1000 yards. So lets check this.


1st Example - target size 3.6 inches through 10x power scope.
x = 3.6 inches length
1 mil = 0.573 degrees

x * tan (theta) = Distance
tan (0.573) = 0.01
3.6 * 0.01 = 0.036
0.036 * (10 x scope power) * (100) = 36 inches
36/36 = 1
1 mil / 1 = 1 * 100 = 100 yards


2nd Example - target size 9 inches through 4x power scope.
x = 9 inches length
1 mil = 0.573 degrees (or 9 inches at 100 yards)

x * tan (theta) = Distance
tan (0.573) = 0.01
9 * 0.01 = 0.09
0.09 * (4 x scope power) * (100) = 36 inches
36/36 = 1
1 mil / 1 = 1 * 100 = 100 yards

Look right or is there an easier way to do this?


Thanks... Steven

jmoore

1 mil

1 inch at 1000 inches (or one yard at 1000yards- which was probably the original "artey" use)

But NOT .573 degrees!

~3.6 MOA to one mil

60 MOA to one degree

So one mil would be 3.6/60= 0.06 degrees. I reckon you lost a zero there...

45.308

Basic ranging math:
target size in inches x 27.77 (constant for yards) divided by mil of the target = yards

Acquire reference target = window
standard window size = 36"

mil window at .5 mils (.575 mils would get range to around 5280 feet)

36 x 27.77 = 999.72

999.72 divided .5 = 1999 yards

jmoore

1yd/1000yd=1mil

So 1yd (36" window width)at 1760 yds (target distance in yards) would be...0 .568 mils, yes? Right convenient of them to be exactly a mile away!

Whatsinaname181

45 has it right.

we use math for a few different things.

range estimation formula is (and this is for "baseball" mils and theres a few variations of it) is- height in target in inchesx 25.4/number of mils high. USMC uses "football" milaradians and a different set of formulas. The Tactical Mildot Reticle is down available is makes things a lot easier now too. Mils and half-mils in hash marks. makes reading a lot quicker too.

this formula also works for width. the bigger the target, the more accurate the formula.

We do a lot of KISS and SWAGing in the sniper world (scientific wild ass guess). some guys like to get super duper high speed gee wizz mathmatical about things but honestly it doesent help any and just slows you down. first of all, you cant shoot that good, second of all, neither can your weapon or ammo. a sniper has about 3,782 other things to think about than the rotation of the earth and oprea winfreys new diet plan while hes in the hide, like "after i off these guys, I have to worry about getting the hell out of here" or "how many of this Aholes buddies are in the area, ready to respond when he drops?"

We also use wind and lead formulas. we use formulas for temp change (if you zero at 80 deg and shoot when its 100 there will be a change of POI).

changes in elevation (1 MOA for every 1500 feet in elevation)

We use the pathagorean theorum for high angle shooting. This is in conjuction with cosign formula (basically what youre doing is trying to find "true" distance. thats why when youre shooting up or down hill you always hit high to one degree or another.

We call 1 mil 3.5 MOA. its easier to devide by and keeps things simple. Your BDC on your M3A or your MKIV has 1 MOA adjustments on it.

theres quite a bit of math involved in the sniper world BUT, the best thing to always do is Keep It Simple Stupid. We can get CSI and super sniper Horatio Cane all day long about this stuff, but when youre "working", things are different. most of the time, we dont use that fancy stuff.

wbm

+1

"Those things which can be accomplished with little are done in vain with more."

RAMMAC

This might help you out a little.

http://www.ultimatesniper.com/Docs/46.PDF

stevenmac2

Where I got the .573 for the 1 mil was from a PDF from over at Mil-Dot.com called "MILS and MOA - A Guide to understanding what they are and How to derive the Range Estimation Equations" By Robert J. Simeone
http://www.mil-dot.com/media/1027/th..._equations.pdf

If you goto his page 4, it shows:

(360 degrees in a circle) / (6283.2 mils in a circle) = 0.573 deg/mil

On Page 6 is where I found a better reference to the calculation above using the tan (theta) and he goes on to say:

We know the length of one side of the triangle, and the angle. But to use the tangent function, we need to convert the angle that is expressed in “mils” into an angle that is expressed in “degrees”. Remember from (C) on page 4, 1 mil = .0573 degrees. Now we
can solve for “x”.

Fig. 5
1) Tan θ = opposite / adjacent
2) Tan .0573° = x / adjacent 3,600 in.
3) 3,600 (Tan .0573) = x
(NOTE: using your calculator, the tangent of .0573 is .001)
4) 3,600 (.001) = x
5) x = 3.6 inches
So 1 mil at 100 yards equals 3.6 inches.


(my comment) I typically use a height reference for standard range estimation formulas, but I never tried using a formula to estimate the range using width through the tan function, and just was very interested in that after I heard it used. I found the reference from the program. It is from a program that aired on the History Channel called Sniper: Bulletproof and was from where two U.S. Army snipers were on a mission to take the targets out. The show said the range finder was no good, so they had to do a quick calculation. There are some downloads in RAR format from this link: (it was earlier into the program)

Here is the video you can watch immediately online:
Video from History Channel showing U.S. Army Snipers in 2003 using this calculation. Where the calculation is shown is at 7:10 minutes into the video ===> watch here
http://www.videobb.com/watch_video.php?v=IVid5wu80IHZ


Downloadable RARs.
http://avaxhome.ws/video/History_Cha...lletproof.html
http://www.wupload.com/folder/841861

**UPDATE**
When I watched the program, I thought it was a x + tan() = D. That was wrong. I watched the video link above again, and it was x (divided by) tan() = D. They showed 3 ft / tan (0.33) = 5208ft in the program.

So does it now work out to what they are saying???
x = 3ft window
deg = they sighted slightly below one mil (0.5) to (0.33)

3 ft width / tan (0.33) = D
3 ft / 0.00575965 = D
D = 520.86
I suppose if you then multiply by 10 x scope power you get
D = 520.86 to be 5208 ft
D = 5208 ft /3 yards = 1736 yards

So lets go back to try the 1 mil @ 3.6 inches and knowing that 1 mil equates also to 0.573 deg.

3.6 inches width / tan (0.573) = D
3.6 inches / 12 inches per feet = 0.3 ft
0.3 ft / tan (0.573) = D
0.3 ft / 0.01 = D
D = 30 * 10x power scope = 300 ft
D = 300 ft / 3 yards = 100 yards - look right?

Target 36 inches @ 1000 yards = 1 mil with 10x power scope.

1 mil = 0.573 deg
36 inches width / tan (0.573) = D
36 inches /12 inches per feet = 3 ft
3 ft / tan (0.573) = D
3 ft / 0.01 = D
D = 300 * 10 x power scope = 3000 ft
D = 3000 ft / 3 yards = 100 yards



Now with my 4x power scope with target 9x9 inches.

9 inch width / tan (0.573) = D
9 inches / 12 inches per feet = 0.75 ft
0.75 ft / tan (0.573) = D
0.75 ft / 0.01 = D
D = 75 * 4x power scope = 300 ft
D = 300 ft / 3 yards = 100 yards

Does this look right to everyone?

Thanks... Steven Mac

RAMMAC

You're making it more complicated than necessary.

That document that you reference is a mathematical proof, it explains why the the concept works. The end result are the formulas on page 8 section G and page 11 section J, they are both the same formula we use in the field but they are written slightly differently.

Swamp Rat

If the math has you boggled, try this:

http://www.mildot.com/

jmoore

360/6283.2 = 0.0572956... deg/mil

So HE lost a zero! It's still missing, regardless.

RAMMAC

The math is wrong. The decimal point is in the wrong place, probably due to entering 0.33 degrees in to the calculator instead of 0.033 degrees.

I get -
tan (0.33) = 0.0005759587168
3ft/0.0005759587168 = 5208.71 feet

In the video they used 0.0331 degrees rather than 0.033 degrees. That would result in a slightly shorter range.

tan (0.331) = 0.0005777040467
3ft/0.0005777040467 = 5192.97 feet

High Hat

Boggled me! All I can say is do what?

stevenmac2

Okay coolness! Glad you guys took a look at this! So do you really think he did the calculation in the field or had some look-up information with him?

RAMMAC

If you notice in the video they showed that the window was less than a MIL across, he would have had to resolve his MILs to at least the nearest 0.025 MIL in order to get the accuracy required to determine an angle of 0.331 degrees; that's pretty tough to do. I'm not saying he couldn't do it because snipers practice calling MILs for that reason but the guy must have used a little bit of SWAG (scientific wild a$$ guess). That's probably why his POI was 3 MILs off with the first shot.

Could he do the math without a calculator? I doubt it, he probably has a range card with the tangent values on it.

• He MILs something that he's pretty sure of it's size that is close to the target.
• Then he determines the angle based on how many MILs cover that object.
• Then he divides the size of that object, in feet, by the tangent of that angle.

That gives him an answer in feet and he can divide again by 3 and get an answer in yards.