Angle Fire Primer

[echo3mike]

In an ongoing quest for a better understanding of LR shooting, the subject of sloped / angled shooting caught my attention. While many shooters are fortunate enough to shoot on level ground, that prize trip to hunt in the Rockies can present some challenges in this area.

An angled shot can be either up or down hill from the shooter. Regardless, the shot will go high if the angle of fire is left uncorrected. This is caused, according to the Sierra Rifle Reloading Manual, 4th ed, by a discrepencies in the line of sight, the line of departure, and the bullets trajectory. It's a little difficult to explain properly without the graphic figures, but

"...the line of sight is still located...in a perpendicular direction away from the line of of departure. Because of the firing elevation [or depression] angle, the bullet trajectory no longer intersects the line of sight at the [target's] range. ...In other words, the bullet shoots high from the shooter's viewpoint as he aims the rifle..." p622

The easiest way to mitigate the effects of this descrepency is to alter the "percieved" range adjustments to more accurately reflect the actual range to the target. In order to accomodate for the slope angle and ranging targets, some math is in order. Sorry.

(1)

First, a right triangle. Side c is the hypotnuse, and is the considered the "slope range" to the target. Side a is the opposite side from angle A, and is the elevation of the shooter above the target in a downhill shot. Side b is the adjacent side, which represent the "horizontal range" to the target. Problems arise when the shooter uses the "slope range" when adjusting his scope for the shot. What's needed in this case is the use of the "horizontal range" for both the elevation and windage adjustments. The question is then, how is the "horizontal range" acquired? Several methods exist, depending on what data the shooter has available to him.

The most direct way is the Pythagorian theorum ( a squared + b squared = c squared), which needs the lengths of two of the sides. If the "slope range" (c) is 169yds, and the elevation of the shooter (a) is 120yds, then you'll need the square root of (c squared minus a squared) = the square root of 14161= 119 yds for the " horizontal range".

If the shooter has:

-one side + the shot angle*:
if a = 15yds, A = 41 degrees, then
c= a/sin A = 15/sin41 = 22.86yds
b= a/tan A = 15/tan41 = 17.26yds

Getting in a little deeper, some necessary functions can be found through Sines, Cosines and Tangents. The neumonic SOHCAHTOA may come to mind; to obtain
Sines: sin A = a/c; sin B = b/c
Cosines: cos A = b/c; cos B = a/c
Tangents: tan A = a/b; tan B = b/a

In order to obtain the shot angle, I need the use of the ARCTAN function on a calcualtor;
arctan (a/b) = A
arctan (b/a) = B



*Of course, there's an easier way... and it's really the only thing a shooter needs. Since the shooter can get the slope range through whatever means he has available and can either acquire or SWAG the shot angle, he can obtain the horizontal range through:

cos A = (horizontal range) / (slope range)

which can be manipulated to

Horizontal range = cos A X slope range
or
Slope range = (horizontal range) / cos A
(2)

And there's an easier way still. My beloved Marine Corps has an Angle Firing Chart, which has the cosines of slope angles laid out for easy useage. It works with all calibers and velocities. It can be found at this web page, if you scroll down. From there, it's just a simple matter of having the form in your log / data book, finding the slope range and slope angle and then plug and chug. If you're in the situation where you need to take a sloped shot, it's a heck of alot easier than :banghead: with a bunch of trig!

Good Luck!
S.

1) image from http://www.oakroadsystems.com
2) Lau, M. 1998. The Military and Police Sniper . Precision Shooting. Manchester, CT. p182

 

[benewton]

echo:

The true horizontal range is the major part of the accurate shot, but there others, including altitude, air pressure and temperature, which effect the bullet velocity out of the barrel, and the zero method used to set weapon sights.

Pick up "Modern Pracitical Ballistics" by Arthur Pejsa for a fun discussion of the zero problem: it's cheap enough, and an interesting read, assuming math doesn't bother you much.
But, to get you thinking, when you zero the weapon, you compensate for the gravitational attraction of the earth on the bullet over a known distance. But gravitational force is a vector quantity, and, since we normally zero over a flat "surface", it operates at 90 degrees to the line of flight, and hence has the maximum effect on the bullet's trajectory.
When you fire on a slope, the gravitational force isn't at 90 degrees, but at some lesser angle. Thus you'll fire over the intended point of impact every time if you your sights normally.