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OK.. I think what you are saying is there is a spacecraft spinning with =
a constant
angular velocity, w, about the z-axis at the same time its mass center, =
O, is=20
traveling with a velocity,Vo, in the y direction. (Rotating around the =
earth sort of).
Instead of using something hitting this craft.. we'll use a Force, F, =
tangential
that uses hydrogen - perroxide jet. When this is fired...we want to =
know the
absolute acceleration of the point where it fired on the rim of the =
spacecraft.
The radius of gyration of the craft well call, k and it's mass, m.
=20
Answer in short... accelleration (a) =3D -Fr^2/(m*k^2)i - (F/m - r*w^2)j
=20
v/r Ken Carrigan
=20
PS.. It is a real number and not infinite..
=20
Hi! Ken: Let's try an Arthur C. Clark's approach as in 2001 Space =
Oddesy to our=20
little puzzle. Let's assume that we are on a hugh space station =
that has been spun=20
up to an RPM of rotation that provides exactly one G at the outer =
rim of the station.=20
Let us have an artificial environment with a river of water around =
the station. We know=20
that it only takes an initial amount of thrust to spin the station =
up to the RPM needed=20
for the G effect we need. Now we have a small meteorite strike the =
station at a=20
tangential angle and penetrates the hull into our artificial river =
around the outer edge=20
of the hull. Water is now jetting out into space at a tangent to the =
hull and providing=20
thrust which in turn is causing an acceleration of our spin rate. =
The G forces are multiplying rapidly now and the pressures at the =
tangential hull penetration are=20
increasing. Now this effect is going to go exponential because it is =
directly related=20
to the velocity squared. It will only require a short period of time =
before the space=20
station will suffer a catastrophic centrifugal structural failure. =
We are all pinned to=20
the floor of the station by the multiplying G forces and are =
helpless to prevent the=20
destruction. Now try your math on this situation. Norm=20
Ken Carrigan wrote:=20
Just had to do this.. it's been bugging me as to how this exit =
forcefrom a centrfugal nozzle head could produce overunity. So here =
issome math (sorry Engineer here) that will explain what a =
Normal"sprinker head" would achieve in power. Forgot you said this.. =
butthe raduis squared does come into play.. however.. still do notsee =
where the "kick over" occurs for over unity. OK.. make believe we have a =
sprinkler (twisted L shaped on eachside) which is feed water in the =
center of it. THis sprinker is turningat a velocity of "w" and the =
length of the arms (in the L forms) are"L". The exit velocity realitve =
to the nozzles are "Ve" for exit velocity,and the area of the exit =
nozzle is "A". I think that now covers it.Assumptions have to be made =
now.. and we concider that laminerflow is occuring and no resistance is =
taken in to account (friction orheating). What we want to find is the =
power produced from thissprinkler. So..the moment around the z-axis.. =
will be the summationsof all (r xV)*p*V*A . Opps now "r" is radius, and =
"p" is density ofwater. p*V*A is called the mass efflus of the nozzle.. =
or lets say "m"for now. The moment going into the hub.. enterance of =
water... iszero cause the radius "r" is zero. Thus the equation reduces =
to onlyone sum.. or (r x V)*m. Now we have to take into account for =
therotational velocity and the velocity of the zozzle exit jet... in to =
thenormal atmosphere. Now we can subsitute r X V =3D -r*(Ve - w*r).Thus =
the moment applied to the system by the generator wouldbe -r*(Ve - =
W*r)*m. BUT the moment applied to the generator is thenegative of the =
moment of the generator on the arms. So the equationbecomes positive... =
no biggy! The power now is simple...one we have the moment.. It is the =
momenttime the rotational velocity.. (M =3D moment) or M*w =3D P for =
Power! For all who would like the in to one equation .. here...P =3D =
(w*r*Ve - w*w*r*r)*m ENSURE.. that proper mks units are used for example =
.... RPM hasto be changed to RPS.. etc etc.. and don;t forget 2*pi for =
rotationalvelocity! v/r Ken Carrigan PS.. NOW... WHERE IS THE =
OVERUNITY!!!Help me understand...=20
=20
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Ken Carrigan wrote:=20