Norman Wootan wrote:
>
> Hi! Ken: Let's try an Arthur C. Clark's approach as in 2001 Space
> Oddesy to our
> little puzzle. Let's assume that we are on a hugh space station that
> has been spun
> up to an RPM of rotation that provides exactly one G at the outer rim of
> the station.
> Let us have an artificial environment with a river of water around the
> station. We know
> that it only takes an initial amount of thrust to spin the station up to
> the RPM needed
> for the G effect we need. Now we have a small meteorite strike the
> station at a
> tangential angle and penetrates the hull into our artificial river
> around the outer edge
> of the hull. Water is now jetting out into space at a tangent to the
> hull and providing
> 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
> increasing. Now this effect is going to go exponential because it is
> directly related
> to the velocity squared. It will only require a short period of time
> before the space
> station will suffer a catastrophic centrifugal structural failure. We
> are all pinned to
> the floor of the station by the multiplying G forces and are helpless to
> prevent the
> destruction. Now try your math on this situation. Norm
>
> Ken Carrigan wrote:
>
> > 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 = -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 = moment) or M*w = P for Power! For all who
> > would like the in to one equation .. here...P = (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...
>
> ---------------------------------------------------------------
> Hi! Ken: Let's try an Arthur C. Clark's approach as in 2001 Space
> Oddesy to our
> little puzzle. Let's assume that we are on a hugh space station that
> has been spun
> up to an RPM of rotation that provides exactly one G at the outer rim
> of the station.
> Let us have an artificial environment with a river of water around the
> station. We know
> that it only takes an initial amount of thrust to spin the station up
> to the RPM needed
> for the G effect we need. Now we have a small meteorite strike the
> station at a
> tangential angle and penetrates the hull into our artificial river
> around the outer edge
> of the hull. Water is now jetting out into space at a tangent to the
> hull and providing
> 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
> increasing. Now this effect is going to go exponential because it is
> directly related
> to the velocity squared. It will only require a short period of time
> before the space
> station will suffer a catastrophic centrifugal structural failure. We
> are all pinned to
> the floor of the station by the multiplying G forces and are helpless
> to prevent the
> destruction. Now try your math on this situation. Norm
>
> Ken Carrigan wrote:
>
> 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 = -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 = moment) or M*w = P for
> Power! For all who would like the in to one equation ..
> here...P = (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...
>
>