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MichelanJell-O t1_jah78bk wrote

Reply to comment by interstellargator in Does the spinning of a bullet have any effect on its drag? by Eastern-Ability8411

Excellent point. The latter. If you divide the motion of the bullet into a linear component and a rotational component, the additional friction drag only acts against the rotation. So the spinning bullet will slow its rotation over time, but the two bullets should have almost exactly the same speed.

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phdoofus t1_jagtybh wrote

Reply to Does the spinning of a bullet have any effect on its drag? by Eastern-Ability8411

If a bullet could maintain straight line flight, with the bullet axis coincident with the flight axis, then the effect should theoretically be zero.

However, a variety of factors can cause the two axes to NOT be coincident. In this case, imagine you're driving down the road with the window down and you put your hand out. If you angle your hand relative to the flow of air past it, you can easily feel the increased drag on your hand.

The whole point of spinning the bullet (gyroscopic stabilization) is to keep the two axes coincident through the expected/desired flight path of the bullet.

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The_Illist_Physicist t1_jagtuez wrote

Reply to Does the spinning of a bullet have any effect on its drag? by Eastern-Ability8411

Seeing as though nobody is really answering your question I'll take a swing at things. As a disclaimer, while I am a student of physics, my specialty is optics and not fluid dynamics.

From the Wikipedia page on drag coefficients, it looks like the total drag coefficient (C_d) is defined as the sum of pressure drag (C_p) and friction drag (C_f). These can be thought of respectively as:

The component of drag that comes from the traditional resistive force we think of due to the fluid exerting a pressure on the face of the projectile, and a "negative pressure" behind the projectile. If you look at the expression to compute this, there's a dot product between the free stream flow unit vector and the negative surface element normal unit vector (inside a surface integral). What this means is for a cube in a non-turbulent fluid, only the front and back faces of the cube contribute to this C_p term. By thinking about the symmetry, a spinning bullet should have no effect here.

The component of drag that comes from a shear stress due to fluid friction along the sides of the projectile. Again in this surface integral there's a dot product, this time between the shear stress unit vector and and the free stream flow unit vector. This is then scaled by multiplying it by the total shear stress at the differential surface element. For a spinning projectile, this total shear stress value is larger than if it was not spinning, so by this reasoning a spinning bullet should experience more drag than a stable, non rotating bullet.

Now I suppose this is all for a very idealized scenario. However a quick search for literature on friction drag simulations turned up this paper https://www.hindawi.com/journals/ijae/2020/6043721/ which seems to suggest that the spinning effect does indeed have some small contribution to the total drag, although it gets becomes less important at higher speeds (below Mach 1).

TL;DR: Yes it appears a bullet does incur a little bit of extra drag from it spinning, however it's necessary to stabilize the bullet and is fairly negligible when the projectile speed is high.

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