Long Range Shooting Question

mram10

Please read, before you say, "This has been touched on a million times", please.

I was speaking with a physics friend (BS) and he told me if I fired a bullet at a target 1000 yds away, while simultaneously dropping one the same distance as the computed drop(255in), they would strike the ground at the same time. I understand in a vacuum how that could work, however with the kinetic energy behind the fired bullet, along with the shape of the projectile and the viscosity of air, this seems like a different animal entirely.

Just to clarify the test:
- bullet fired from high powered rifle at a target 1000 yds away.
- barrel in parallel to the ground (not arcing path)
- both bullets are 300gr and alike

Thanks guys.

tiny-tim

hi mram10! welcome to pf!

let's see …

we're only interested in the vertical ("y") component of motion

let the velocity be the vector v = (v_x,v_y) = (dx/dt,dy/dt)

i] if we assume that air resistance is proportional (and opposite) to velocity, A = -kv,

then the component of air resistance in the vertical direction is kv_y, = kdy/dt

and so the vertical equation of motion is md²y/dt² = mg - kdy/dt, and this is irrespective of whether there is any horizontal motion

ii] however, if air resistance is proportional to speed-squared, A = -k|v|v,

then the component of air resistance in the vertical direction is k|v|v_y, = k|v|dy/dt = k√((dx/dt)² + (dy/dt)²)dy/dt

which is obviously smallest if dx/dt = 0

in other words: if air resistance is proportional to the square of the speed, then it has a greater effect on reducing vertical motion if the bullet is shot than if it is dropped

(but if air resistance is proportional to the speed itself, then there's no difference)

PaulS1950

The vertical velocity is due to the acceleration of gravity. If the rifle barrel is level with the ground (90 degrees to gravity) then both bullets will hit the ground at the same time.
The one fired from the rifle will trave a lot farther but it falls at the same speed. The bullet does not produce lift.

rcgldr

There could be a difference if the vertical component of drag was different between a dropped bullet versus one where the spin keeps it horizontal. If the bullet's center of mass versus the center of drag allowed the dropped bullet to rotate towards a vertical positiion during the drop, then there could be less vertical drag on the dropped bullet, but I don't think this is being considered in these type of thought problems. You could change the comparason so that the dropped bullet had the same spin as the fired bullet to eliminate any pitch effects of the bullet. Other issues like curvature of the earth, or spinning bullet, cross wind, and magnus effect (could result in lift or sink) are usually not considered either.

AlephZero

Tiny Tim's analysis is a reasonable starting point if you ignore the shape of the bullet (or assume it is a non-spinning sphere). For any realistic speed the drag force will be roughly proportional to the square of the speed, not the speed.

Any additional forces caused by the bullet "flying" will tend to make it fall slower than a sphere, so all the effects "add up" rather than "cancel out".

A practical way to investigate this is look at the adjustments to the rifle sights, and figure out the how the aiming position changes for different ranges. That will tell you how far the bullet actually "falls" below a straight line of flight.

mram10

Wow, most of this is over my head, but I am slowly understanding. Thank you for your time guys. I will run some ballistic tests to see the true drop from a 90deg to gravity barrel so I can compare the times. I know it drops 255" and takes 1.34sec to reach 1000yds, however, the barrel is aiming slightly upward (arc).

rorix_bw

Good luck! This is the correct approach - testing by experiment.

As others have said,

The *theory* is that they will strike the ground at the same time as the gun only provides horizontal and not vertical movemet to the bullet. All of the vertical movement comes from gravity, which is the same on both bullets.

However, in practice many things might affect this. Don't expect the results to be *exactly* identical. Remember your friend is talking about a theoretical system devoid of practical effects.

BTW if you were in the army artillery they would have taught you what your friend said about the time being equal (at least in the UK) as they teach ballistics to gunners (or at least they did, when my friend was in).

mram10

Not an artillary guy. I just build and shoot long range rifles. My 338 edge is my long range elk rifle and does a great job. I digress.. :)

According to my ballistic software it takes 1.34 seconds to travel 1000 yds and drop 255 from a parallel barrel. Now I just need to drop a 300gr bullet from 255 inches and see how long it takes. Anyone have an easy formula that can find the time to fall?

Nugatory

Sure...
General formula for something starting at rest and accelerating at a constant rate is [tex]s=\frac{1}{2}at^{2}[/tex] where a is the acceleration, t is the time, and s is the distance. Turn that around because we know s and want to know t, and we have [itex]t=\sqrt{\frac{2s}{a}}[/itex]. Here s is 255 inches, a is 32 ft/sec² or 384 in/sec²... So t is theoretically 1.15 seconds, a bit longer allowing for air resistance.

That's more different from 1.34 seconds than I expected to see when I started the calculation... Maybe your ballistic software knows about some interaction between a fast-moving bullet and the air that tends to keep the bullet up a bit longer? No way the numbers should be different in a vacuum.

rorix_bw

Earlier you had the same numbers with a slightly elevated barrel.

Ballistics software I have seen takes altitude, air temperature and wind speed and direction into account. What parameters are you using?

A ball round will rotate in flight and could generate lift from that - not sure how that works.

mram10

Parameters:
3000ft
40deg F
300gr
.758 Ballistic Coefficient
2810 feet per second

If the program takes into account a non-level barrel, then the time would be slightly less for a level barrel, correct(due to less effect from gravity)?

Nugatory

Upwards-pointing barrel keeps the bullet in the air a bit longer than a level barrel, downwards-pointing barrel keeps the bullet in the air for a bit less time. Level barrel is the one that should be compared with the dropped bullet.

mikeph

I can't see how there can be any lift, or in fact any perpendicular forces due to the air resistance on a spherically symmetric spinning bullet.

rorix_bw

It's possible the ballistic software is measuring the drop relative to the where the sights are aimed, and not where the barrel is pointed - these won't be in line, as the shooter "zeroes in" the gun (aligns the sights with the impact point) at a chosen range.

I found this, which describes the lift generated by a spinning ball.

http://www.grc.nasa.gov/WWW/k-12/airplane/beach.html

mfb

This requires an axis of rotation which is not aligned with the direction of movement - something you will never see in bullets. In table tennis, it is common to use this effect.

Is it possible to "zero in" the gun for a distance of 0? This should allow horizontal shots.

mram10

I tried to zero it at 0 and the program wouldn't let me. I can try to do it for 1 yd and have it shoot 2" low since the scope sits 2" higher than the barrel center.

rorix_bw

Except in a crosswind ... and some weird situations I don't remember.