Up/Downhill corrections

[Buffalobob]

I think. If you took this to an extreme, a bullet shot straight up and then falling back to earth would be going at the same speed when it got back to earth and hit you in the head. They don't.

But we agree gravity is a downward force and it only affects things in one direction - downward.

So to reword the example I use to support my hypothesis ( I might not actually be correct).

If you go 2000 yards up in the sky and shoot a bullet straight down then the acceleration from gravity will add about 197 fps to the velocity you observe as it strikes the ground.

If you fire a bullet straight up then deceleration by gravity will reduce the velocity observed at 2000 feet by about 197 fps.

There will be a total difference in velocity of 394 fps approximately (the effects of air drag will cause the actually added or subtracted velocity to be somewhat less than calculated).

This is true irregardless of whether you shoot in a vacuum or not.

[phorwath]

Originally Posted by Buffalobob

But we agree gravity is a downward force and it only affects things in one direction - downward.

(the effects of air drag will cause the actually added or subtracted velocity to be somewhat less than calculated).

This is true irregardless of whether you shoot in a vacuum or not.

I calculated lots of these types of examples in college 30-35 years ago in an engineering curriculum, but we always assumed a vacuum because that simplifying factor was required to allow the straightforward determination of a "correct" answer. I agree gravity is an ever present downward source of acceleration. It's ultimately the primary force that keeps a bullet that's been fired straight up from leaving earth's atmosphere. And it keeps a bullet moving downward until it's stopped by impact with the earth (or when we do everything correctly - a game animal).

But the atmosphere is present and it's a huge (overwhelming) factor controlling bullet velocity (deceleration) and ballistics (predictive modelling of the bullet in flight), making your estimate on the relative addition to, or subtraction from, velocity in the example a complete guesstimate. Gravity's affect on the horizontal velocity component (horizontal deceleration) of a bullet fired on a horizontal plane is truly negligible during exact horizontal fight, because the bullet is travelling perpendicular to the forces of gravity. And for angled shots somewhat above or below true horizontal to the face of the earth, the effect on the horizontal component of a bullet's velocity still borders on the negligible at high velocity in comparison to atmospheric friction, which is why it's not a necessary input factor in the computer models like Exbal.

I'm not a ballistician. Otherwise I'd be able to explain this more clearly. And even they won't try to explain the theory to an average layman. Do I think gravity plays any role in bullet flight in upward versus downward angled shots with high velocity bullets? Yes. Do I believe that role is significant enough to be worthy of consideration and incorporation in a predictive ballistics software program. No, because its affect borders on the negligible. It might serve a purpose if modelling the fight of an arrow or BB in predicting the long range angled shot ballistics of those slow speed objects.

[Jon A]

Originally Posted by Buffalobob

The acceleration of gravity is 32.2 ft per second squared.

The equations for calculating the velocity effect is 1/2 GxTxT. So for a time of travel of one second which would be about a 1,000 yard shot, the acceleration would generate a velocity of = 1/2x32.2x1x1 = 16.1 fps.

This is incorrect. The above equation gives you the distance traveled, assuming an initial velocity of zero. 16.1 feet. It is provable by the following: If accelerating from zero at 32.2 f/s/s for 1 second obviously the terminal velocity is a * t = 32.2 * 1 = 32.2 ft/s. Since you started at zero and your acceleration was constant your average speed is (0+32.2)/2 = 16.1 ft/s. Average that velocity for 1 second and you travel 16.1 feet.

So the added or subtracted velocity = 1/2x 32.2x3.5x3.5 = 197 fps.

So the terminal velocity difference using a * t = 32.2*3.5=112.7 ft/s.

[Buffalobob]

"This is incorrect."

Well, it is good one of us can remember simple physics.