Another Ballistics Question


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Moondancer
April 23, 2003, 07:05 PM
After reading "Point of Impact" (based on THR recommendations, I might add), I have a question about ballistics as mentioned in the book.

Specifically, it was stated by the author several times that when shooting uphill or downhill you had to hold over due to the effects of gravity.

This goes contrary to some old discussions I remember. It seems to me that the prevailing theories at that time were this:

Hold over when shooting uphill to combat the effects as gravity will pull your bullet down more than you think it will.
Hold under when shooting downhill as gravity will pull your bullet down less than you think it will.

IIRC, the answer to all this was that you had to judge bullet drop by "true" distance or maybe a better way to explain it would be by "time to target". Therefore the answer in both cases, either uphill or downhill, was that gravity's effect on bullet drop would be more than anticipated because your bullet was traveling more than the horizontal distance.

That would agree with the author's statement. Am I right in what I remember????

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Hkmp5sd
April 23, 2003, 07:49 PM
When shooting uphill or downhill, you compensate for both by aiming low.

Bullet drop is how much a bullet falls when fired from a rifle that is horizontal to the ground. When firing up or downhill, bullet drop will cause your bullet to strike higher than your point of aim. To compensate for this, you aim low based on the bullet characteristics, distance and angle up/down.

For example, when shooting a Federal .308 Match 168 grain BTHP at 100 yards with a 60 degree angle, you compensate by firing 1.3" low. For a 500 yard shot, you compensate 41".

md2lgyk
April 24, 2003, 01:54 PM
Hkmp5sd is correct. It's a simple (?) matter of algebra. Assume you are on flat ground, shooting at a target exactly 100 yards away. Gravity acts downward on the bullet for the whole distance. Now assume you are shooting at the same target, still 100 yards away, but uphill (or downhill) a 30-degree angle. The HORIZONTAL distance gravity acts downward on the bullet would be only 86.6 yards so the bullet won't drop as much. Thus, you must aim low in both cases.

Doc
April 24, 2003, 02:31 PM
you guys are GOOD!

However, the effect of angle of hillside is neglible under 300 yds and immerceptible under 200 for high power rifle cartridges.

There are other factors as well including temperature, altitude, crosswind and parallax if I recall correctly. None of which are significant under 200 yds

md2lgyk
April 24, 2003, 03:09 PM
True, but with my trusty .45ACP. . . .

hksw
April 24, 2003, 06:25 PM
md2lgyk,

The horizontal distance between a target 100 yds @ 0° and 100yds @ 30° is different and if taken the POI of the horizontal distance of the 30° bullet it certainly would be above the point of aim as you have state. However, that geometric explanation, IMO, is not quite what is causing the bullet to impact above the POA.

If you look at the bullet path of the projectile shot horizontally, 0°, the path is quite different than that of a bullet shot at 30°. If the geometric theory were applicable, than the POI of a bullet shot straight up 90° at 100 yds would be ~1.75" below the POA as the horizontal distance would be 0 yds. At 0 yds, the bullet would be exiting the barrel and most scopes are ~1.75" above the bore axis.

The flight path of the bullet shot at a particular angle has to be taken into consideration. As you increase the angle (to 90°), the flight path flattens. When it reaches 90°, it will, of course, be flat. The best way I can describe the phenomenon is to think about the path of a bullet shot at 0°. A shallow parabola, the path crosses the LOS once as it flies above the line of sight and crosses the LOS again when it reaches the center of the target, the zero of the scope at 100 yds. To go the the extreme, the flight path of a bullet shot 90° up (or down) will be absolutely flat (not taking into consideration climate, altitude, Earth spin, coriolis force, etc.). The path would cross the LOS once and continue on in a straight line above the LOS (after crossing it at some point closer than when the bullet crossed the LOS when shot horizontally). Between 0° and 90° the POI goes from 0" (zero) to a POI where the flight path of the bullet is laser straight to some point above the point of aim.

telewinz
April 24, 2003, 06:41 PM
So now I have a headache but I least I know I need to replace my 30/30 with a .300 Winchester Magnum and aim dead-on.:(

benewton
April 25, 2003, 02:58 PM
I had to do a ballistics calculator for the 40mm grenade launcher, as in M79/M203, and so got to play with more calculus that I'd done since college, since, while you can ignore 100 yard ranges at 3200 fps, it gets a good deal more fun at 78 mps.

The best text, that I'm aware of, for ballistics in detail is a 1928 (or so, I've got a copy around here somewhere...) text on battleship gunnery. Ah, I'd review my math texts first...

To simplify the mess, you'll shoot over, both up hill and downhill. It's not only the horizontal range factor, but the midrange impact point of the round to the range that you zeroed the weapon to.

If you merely launched the bullet parallel to the ground when you zeroed, and then used Kentucky windage to account for range, you'd hit the target every time.

While I'm sure most everybody picks up a standard ballistics package for their computer, try to lay hands on "Modern Practical Ballistics" for a better explaination of the up/down hill thing.

Then, sit yourself down, and explain to yourself how it has to work!

md2lgyk
April 25, 2003, 03:06 PM
hksw: Yes, you are technically correct. My explanation was intended to be basic - after all, we're not all engineers here.

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