- #1
aseylys
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So I was sitting around pondering about forces, and I found myself thinking about shock-waves. Whether they're caused by supersonic travel, or explosions, the pressure at the front of the wave-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=Kp*KR*√[pv*pg](M2-1)1/8he-3/4L3/4*Ks
Kp: Pressure Amplification Factor
KR: Reflection Factor (assumed 2.0)
pv: Atmospheric pressure of aircraft (Pa)
pg: Atmospheric pressure of ground (101325 Pa)
M: Mach number
he: Altitude
L: Length of object
Ks: Object shape factor
To find Ks, we have this equation:
Ks=.685*√[Ae]/(L¾*Le¼)
Ae: Effective area
L: Length
Le: Effective length
Because it's a cube, L=Le
So we get
Ks=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 Kp and he. So without doing those we get
Δp=111244*Kp/he¾
Now, Kp 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.
he 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 Kp, 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/m2. 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
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=Kp*KR*√[pv*pg](M2-1)1/8he-3/4L3/4*Ks
Kp: Pressure Amplification Factor
KR: Reflection Factor (assumed 2.0)
pv: Atmospheric pressure of aircraft (Pa)
pg: Atmospheric pressure of ground (101325 Pa)
M: Mach number
he: Altitude
L: Length of object
Ks: Object shape factor
To find Ks, we have this equation:
Ks=.685*√[Ae]/(L¾*Le¼)
Ae: Effective area
L: Length
Le: Effective length
Because it's a cube, L=Le
So we get
Ks=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 Kp and he. So without doing those we get
Δp=111244*Kp/he¾
Now, Kp 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.
he 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 Kp, 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/m2. 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