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- Thread starter Hivoyer
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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.

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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.

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K^2

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

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K^2

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A.T.

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Actually, the fired bullet does stay in the airslightlylonger. 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.

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K^2

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And the Magnus effect would pull the spinning bullet sideways.

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

- #11

mfb

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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.slightlylonger. 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.

- #12

K^2

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Sort of. At least, in a somewhat idealized setup, which should give you a qualitative answer, at least.i was wondering if anyone knew how to derive the magnus effect equation?

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

[tex]dF_L = \rho \Gamma v dl[/tex]

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.

[tex]\Gamma = \int_C v\cdot ds[/tex]

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.

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

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.

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thankyou k^2

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Khashishi

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