Centrifuge at hypersonic speeds?

I'm interesting in testing some aerospace propulsion ideas at actual
hypersonic velocities, perhaps up to even orbital velocity. Of course
hypersonic wind tunnels are quite expensive and none even go up to
orbital
velocity.
What I wanted to do was have a rotor move at rim velocity at this
speed
range. For current materials this would be far beyond their tensile
strenth if
you are using a uniform rotor. However, I was wondering if tapering
might make
it possible using materials in common use.
Here's a post to sci.astro presenting the calculation for rotating
spheres made
of diamond:
Newsgroups: sci.physics
From: snipped-for-privacy@cars3.uchicago.edu
Date: Fri, 09 Jul 2004 19:56:04 GMT
Local: Fri, Jul 9 2004 3:56 pm
Subject: Re: What is the Fastest spinning man made Object?
formatting link

The example given there for a rotating sphere resulting in:
v_lim = sqrt(2*S/rho), where S is the yield strength and rho the
density,
suggests that it might work, since this is larger than for a uniform
rotating
ring in which v_lim is:
v_lim = sqrt(S/rho).
So I'm thinking a sufficiently tapered rotor might be able be able to
get
orbital velocity rim speed using current materials (not diamond).
I know that the highest speed flywheels are able to get a rim
velocity in the
range of 1,100 m/s using carbon fiber composites. But this is an
anisomorphic
material and might be difficult and expensive to produce in the right
shape I
need. So I'm thinking of using high strength aluminum, titanium, or
steel
alloys.
I know that the theory for tethers proposed for the "space elevator"
use
exponential tethering. Then this tapering should also work if you were
using a
rotating rod-like structure. However, for my usage I need a disk-like
or
torus-like rotor.
What's puzzling me is that with these kinds of shapes you need to
worry about
tangential stresses, not just radial ones. So my question is if you
used the
right tapering to take care of the radial stresses, would that be
sufficient
to take care of the tangential stresses as well?
Bob Clark
Reply to
Robert Clark
Loading thread data ...
[snip rest of crap]
Have you ever been correct about anything?
idiot
The limiting equatorial velocity, for a rotating sphere ( velocity is independent of the size of the object) is given by
v_lim = sqrt(2*S/rho)
where S is the yield strength and rho the density. Now, 10 tonnes/mm^2 translates to about 10^11 Pa (100 gigapascals). Diamond density is about 3500 kg/m^3. Throwing this into the formula obtain
v_lim = 7600 m/s (approximately)
i.e. a tad less than 5 miles/sec.
It arises from forces acting on a surface element with area dA of the rotating body. For an infinitsimally thin surface element the only stresses acting are tangential since radial stresses go to zero on the surface. Said tangential stresses still yield a radial force component for a curved surface. It is a general result from the theory of elasticity (first by Laplace, that given a surface element with area dA and thickness dt, with tangential stresses present in the surface, the normal (to the surface) force acting on this element is
dF_s = (S1/R1 + S2/R2)*dAdt
where R1, R2 are the main radii of curvature of the surface (at the point of evaluation) and S1, S2 are the stresses along the directions of the corresponding axes of curvature).
Take the ball rotating with an angular velocity w, and evaluate the forces acting on an equatorial surface of element. Do the evaluation in the rotating reference frame. The surface element is acted upon by two forces. First, there is the elastic force, described above, that simplifies to
dF_s = (S1 + S2)*
dAdt/R
since R1 = R2 = R, the radius of the sphere. This force is pulling the surface element inwards, towards the rotation axis. The second force is the centripetal one
dF_c = R*w^2*dm
where dm is the mass of the suface element, given by
dm = dAdt*
rho
where rho is the density. So, we get
dF_c = R*w^2*rho*dAdt
This force acts ouwards, away from the rotation axis. Since the surface element remains stationary in the rotating frame (until the sphere pops), the two forces dF_s and dF_c must be equal. So, we've
R*w^2*rho*dAdt = (S1 + S2)*dAdt/R
Cancelling the common factors and reorganizing we get
(R*w)^2 = (S1 + S2)/rho
R*w is simply v, the velocity of the equatorial point. As for S1 and S2, neither of them can be larger than the tensile strength S. So, we get
v^2
Reply to
Uncle Al
[snip rest of crap]
Go ahead, build it. We'll wait. Empirical physical reality is Uncle Al's ally - and a powerful ally it is.
Reply to
Uncle Al
Ya know, Uncle Al, I bet he could learn more about wind tunnels. But I bet you can stop being such an asshole.
Anyone wanna bet?
Jonathan
s
/lajos.htm#a2
Reply to
Jonathan
Think shock and expansion tubes. The later can go to super-orbital velocities, and are being used for re-entry tests of planetary bodies other than earth. See the NASA HyPulse facility at ATK-GASL, CALSPAN, T5 at GALCIT, the newer tube in Germany and the X2, X3 facilities in Australia. Shock and expansion tubes are routinely used for scramjet experiments. In fact NASA showed quite good compaisons between the Hyper-X Mach 10 flight test and ground test data obtained in HyPulse.
Reply to
me

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