Does air friction increase a gunshot bullet's range?

[Hivoyer]

Would a bullet have longer trajectory,because the air helps it ''glide''?Does this mean that if you fire the same gun and bullet in vacuum and in the air the one fired in the vacuum will fall to the ground a lot more quickly?

 

[LJW]

Do you mean create lift?

In reality, nope. Bullets will (well, any accurate ones) rotate very rapidly and be uniform about the rotational axis. Ergo, no lift produced. A bullet with an aerofoil cross section would be terribly hard to stabilize.

You've asked two questions. Air friction would reduce the range. However, both bullets would remain in the air for very similar amounts of time.

 

[Nessdude14]

Air friction will produce heat (as does any friction) in the air and in the bullet. This heat energy has to come from somewhere according to the conservation of energy. The friction actually takes away from the kinetic energy of the bullet, making it into heat energy. A decrease in kinetic energy means the bullet is being slowed down by this friction.

To answer your question more specifically, the bullet will be pulled to Earth by gravity at the same rate with or without the air friction because, as LJW said, it doesn't create any lift.

 

[K^2]

Actually, the fired bullet does stay in the air slightly longer. Even though it does not glide, the vertical drag is effectively increased. But it's not nearly enough to offset the loss of speed due to friction, so the range is still significantly reduced.

 

[fawk3s]

Might I add that in air the bullet also experiences bouyancy, which reduces the vertical net force on the bullet.

 

[LJW]

Note that I said "very similar", not "the same"

It's true that the bullet in the atmosphere will stay in the air for very slightly longer. For practical purposes, not significant.

 

[K^2]

If by practical purposes you mean something like aiming, that's absolutely correct. The way you usually aim at a distant target is compute the time of flight (usually from table), and use that to get the drop distance (assuming free fall in vacuum) and distance carried by wind (time difference with flight in vacuum multiplied by wind velocity). So the only relevant effect of drag is impact on time of flight.

 

[A.T.]

Actually, the fired bullet does stay in the air slightly longer. Even though it does not glide,

Why can't a bullet glide? At least theoretically. If you fire at a distant target, you don't fire the bullet horizontally, but slightly inclined. If the bullet had enough spin, it should keep that inclined axis, even when its vertical velocity is reduced and negated by gravity. So the bullet axis is not aligned with the velocity anymore, and a lift component is possible.

I guess you would need bullets with more spin, and less drag (pointy back) to make advantage of this. And the Magnus effect would pull the spinning bullet sideways.

 

[K^2]

Because if its center of pressure is even slightly off center of mass, which is almost a guarantee, it will precess.

 

[ben682]

And the Magnus effect would pull the spinning bullet sideways.

i was wondering if anyone knew how to derive the magnus effect equation?

 

[mfb]

Actually, the fired bullet does stay in the air slightly longer. Even though it does not glide, the vertical drag is effectively increased. But it's not nearly enough to offset the loss of speed due to friction, so the range is still significantly reduced.

I can confirm this with a quick simulation, under the assumption of a spherical bullet without rotation - but I do not think those things change the result.

 

[K^2]

i was wondering if anyone knew how to derive the magnus effect equation?

Sort of. At least, in a somewhat idealized setup, which should give you a qualitative answer, at least.

Using Kutta-Joukowski theorem, lift per infinitesimal slice can be computed via circulation.

$$dF_L = \rho \Gamma v dl$$

Where gamma is the circulation. Technically, all three of these should be taken at infinity. However, if you assume a laminar flow, the circulation is going to be the same everywhere, and you know circulation at the boundary of the sphere. Circulation is given by the integral along a closed curve.

$$\Gamma = \int_C v\cdot ds$$

And the layer at the boundary of the sphere moves along with the sphere. That gives Γ=2ωπr², where ω is angular velocity component perpendicular to v. The total lift can then be computed by integrating over all slices of the sphere.

$$F_L = \int_{-R}^{R} 2\pi \rho r^2 (\omega \times v) dl = \frac{8}{3}\pi \rho R^3 \omega \times v$$

Which gives you the correct ωxv dependence of lift. How close the coefficient is going to be, however, I have no idea.

Edit: I just sort of assumed spherical shape in that last integral. If it's a cylinder, or whatever, you'll need to correct for whatever r(l) dependence you actually have.

 

[Emilyjoint]

I can go along with AT because I have seen a video of a shell fired from a tank gun. It definitely spins and stays at an angle that would give lift. I don't know how big the effect would be because that was not the main feature of the video...it was about high speed photography.

 

[ben682]

thankyou k^2

 

[Khashishi]

Even if the time spent in the air is longer, I doubt that the distance traveled is longer, since the velocity is decreased by friction.