The_Illist_Physicist

The_Illist_Physicist t1_jd2bhym wrote

** better in a lot of ways.

I only say this because moissanite is a 9 on the Mohs hardness scale. Other than that though it definitely creams diamonds.

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The_Illist_Physicist t1_jagtuez wrote

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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The_Illist_Physicist t1_j23drw8 wrote

It's a fairly common laser amplification technique, the lab I work in uses Chirped Pulse Amplification for a lot of our pulsed lasers.

In order to get an ultrashort pulse of light (short in the time domain) it must be made of a large spectrum of frequencies/colors (long in the spectral domain). This is essentially the Fourier transform relationship, similar to the uncertainty principle.

The problem with high energy, ultrashort pulses is that they like to burn optics. You'll toast a lens if you send a beam with enough power through it. So what you do is separate the pulse into its many colors, amplify them individually, and then recombine. It's simple yet brilliant at the same time.

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