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**bow shock pressure**is a force.

So I started thinking of a way to generate them without having to go Mach 1 or detonate something. And here is what I came up with, and please be as brutal as you want with the criticism. This is all hypothetical and done out of curiosity.

So imagine, if you will, a block .1 m x .1 m x .1 m of some super dense metal and this block weighs about .5 lbs (.2267 kg). Now imagine this block is at the end of a .1 m rod and this whole contraption rotates around the other end of the rod. (Hoping that you're still with me) This rod will need to rotate at least at 33422 RPM to go 350 m/s which is a little more than Mach 1 (v=r*RPM*.10472). Which isn't hard for today's electric motors, so let's bump it up to Mach 1.1 (so 17905 RPM).

That's the setup. Now for some more math. So going off some math off a great technical paper (linked at the bottom), I was able to calculate the bow shock pressure with some generous assumptions.

Bow Shock Pressure:

Δp=K

_{p}*K

_{R}*√[p

_{v}*p

_{g}](M

^{2}-1)

^{1/8}h

_{e}

^{-3/4}L

^{3/4}*K

_{s}

K

_{p}: Pressure Amplification Factor

K

_{R}: Reflection Factor (assumed 2.0)

p

_{v}: Atmospheric pressure of aircraft (Pa)

p

_{g}: Atmospheric pressure of ground (101325 Pa)

M: Mach number

h

_{e}: Altitude

L: Length of object

K

_{s}: Object shape factor

To find K

_{s}, we have this equation:

K

_{s}=.685*√[A

_{e}]/(L

^{¾}*L

_{e}

^{¼})

A

_{e}: Effective area

L: Length

L

_{e}: Effective length

Because it's a cube, L=L

_{e}

So we get

K

_{s}=2.166 (In that ballpark)

Now onto our assumptions, as crazy as they may be. I'm assuming that the whole system is sealed and the environment is pressurized to 3atm and we have some technology (magic?) to transfer the pressure generated in the system to the outside to be measured.

I think I've assigned all the variables except K

_{p}and h

_{e}. So without doing those we get

Δp=111244*K

_{p}/h

_{e}

^{¾}

Now, K

_{p}is usually in the ballpark of .8-2.1 for supersonic aircraft, so that factor won't have much affect on the order of magnitude of the pressure.

h

_{e}affects how the bow shock travels through the air s it's really not relevant to my system because it's closed. Unless someone can give me an idea of what to assign it, I'll just ignore it, because like K

_{p}, it won't affect the order of magnitude much.

Wrapping this up, we have somewhere in the vicinity of 110000 Pa, which is 110000 N/m

^{2}. Supersonic aircraft are usually within 50-100 Pa, and I was trying to get as large of a pressure as possible.

Can someone comment on this, or just tell me that this idea is completely ludicrous and shut me up?

http://www.pdas.com/refs/tp1122.pdf