Hi! Ken: While is was attending Embry-Riddle Aeronautical University at
Daytona
Beach FL. we had professors from the Cape show us mathematically that
the
Allomyrhino Dichotoma, Giant Japanese Rhino Beetle absolutely could not
possibly
fly for it's body mass vs wing area was so far out of proportion it was
ridiculous.
Someone forgot to tell the beetle that flight was futile. So it goes
with math when
trying to explain anomalies in nature. Since you love to apply math to
our conver-
sations, solve this for Garry and me. Given: a drum (right
cylindrical section) 60
inches in diameter, 6 inches tall with a coaxial shaft, spinning in a
horizontal plane
of rotation at 1800 RPM. Inside this spinning drum we have a vertical
wall of
water 6 inches thick being maintained in this configuration by
centrifugal force
(artificial gravity of this system). Now please calculate the water
pressure in
PSI against the outer wall of this spinning drum. Thanks in advance for
your
help with this design. We all should have fun while trying to validate
the Clem.
Norm PS: I too completed my BS but have opened my mind to
other
possibilities and explanations, the reason for looking into this
free-energy field.
Ken Carrigan wrote:
> Well... sure enough Centripetal force is a force but what we are
> looking foris expended or generated energy. It will take X amount of
> energy torotate this mass to a certain velocity, then without
> friction.. no energyshould be needed to keep it spinning.... all
> without friction. However thatis not the case as there always is
> friction and actually is a velocity cubedfunction.. the faster the
> speed the higher the friction. This very true withautomobiles
> friction to air flow.. when speeding down a road -- velocity cubed.
>
>
>
> Hi! Ken: Maybe I have a different physics book than the one
> you are using.
> Centripetal force = mass x velocity squared / radius
> F=mv2/r Now go back and calculate the water pressure at the
> exit nozzle based on the volume of water in the
> arm. In a Clem conical drum which is composed of many Phi
> ratio spiral tubes
> there is considerable reactive thrust from the nozzles which
> drive the rotation of
> the unit while the centrifugal force is drawing the water
> into the hub.
>
> This thrust was taken into account.. the mass flow rate along with the
> number ofexit nozzles.
>
> Your lawn
> sprinkler depends on the city water pressure of 60 PSI or
> more to provide the
> driving force to turn the head array. In an artificial
> gravity set up such as the
> Clem engine the radius of the unit is critical as is RPM to
> achieve the necessary
> pressures needed to drive the assembly beyond the point of
> frictional losses
> in the distribution tubes.
>
> The lawn sprinkler was only an example as NO pressure was specified.
> Itcould easily take in 2000 psi or whatever pressures you so desire.
> However,as the pressures go up.. so does the frictional losses and
> heating effects. Thisalso means that efficiency goes down.. as energy
> is lost. Efficiency does notgot up.... and there is no point that I
> know of.. in physics... where one canovercome this frictional loss..
> especially at higher velocities and pressures.Please give an example
> where this is so.. or equations that tell me.. you canovercome
> frictional losses.
>
> The reason that the distribution tubes are based on a Phi
> ratio is to provide a practically free fall flow through the
> tubes as envisioned by
> Schauberger in his device. I believe, as does Jerry that
> there are thermal factors
> and viscosity variations in the oil that are lurking here
> also. These are the factors
> that we don't have a firm handle on until a unit is built
> and run. Norm
>
> The way I was always taught.. was to first Hyposthesize about your
> theories andthen plan an experiment around it.. test it in a lab
> environment and documentresults and revise the theory/hyposthesis
> more.. etc etc. Fluid dynamics isthe key to this all.. not nessarily
> physics... this is where I have learned it.. fromcollege books..
> partly mechanical...but graduated BSEE. The professor whotaught the
> class I remember was from NASA... cool class! The oil my guess was to
> overcome the water boiling problem with hypersonicfluid flow...
> creating heating effect from the frictional losses... must have
> beengreat! hey.. self contained... definately need to get that heat
> out! Which... leadsme to believe.. it was not efficient. v/r Ken
> Carrigan
>
> > 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...
>
>
>
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Hi! Ken: While is was attending Embry-Riddle Aeronautical Universityat Daytona
Ken Carrigan wrote:
Well... sureenough Centripetal force is a force but what we are looking forisexpended or generated energy. It will take X amount of energy torotatethis mass to a certain velocity, then without friction.. no energyshouldbe needed to keep it spinning.... all without friction. However thatisnot the case as there always is friction and actually is a velocity cubedfunction..the faster the speed the higher the friction. This very true withautomobilesfriction to air flow.. when speeding down a road -- velocity cubed.
Hi!Ken: Maybe I have a different physics book than the one you are using.This thrust was taken into account.. the mass flowrate along with the number ofexit nozzles.
Centripetal force = mass x velocity squared / radius F=mv2/r Now go back and calculate the water pressure at the exit nozzle based onthe volume of water in the
arm. In a Clem conical drum which is composed of many Phi ratiospiral tubes
there is considerable reactive thrust from the nozzles which drivethe rotation of
the unit while the centrifugal force is drawing the water into thehub.YourlawnThe lawn sprinkler was only an example as NO pressurewas specified. Itcould easily take in 2000 psi or whateverpressures you so desire. However,as the pressures go up.. so doesthe frictional losses and heating effects. Thisalso means that efficiencygoes down.. as energy is lost. Efficiency does notgot up.... andthere is no point that I know of.. in physics... where one canovercomethis frictional loss.. especially at higher velocities and pressures.Pleasegive an example where this is so.. or equations that tell me.. you canovercomefrictional losses.
sprinkler depends on the city water pressure of 60 PSI or more to providethe
driving force to turn the head array. In an artificial gravityset up such as the
Clem engine the radius of the unit is critical as is RPM to achievethe necessary
pressures needed to drive the assembly beyond the point of frictionallosses
in the distribution tubes.Thereason that the distribution tubes are based on a PhiThe way I was always taught.. was to first Hyposthesize about your theoriesandthen plan an experiment around it.. test it in a lab environment anddocumentresults and revise the theory/hyposthesis more.. etc etc. Fluid dynamics isthe key to this all.. not nessarily physics... this iswhere I have learned it.. fromcollege books.. partly mechanical...but graduatedBSEE. The professor whotaught the class I remember was from NASA...cool class! The oil my guess was to overcome the water boiling problemwith hypersonicfluid flow... creating heating effect from the frictionallosses... must have beengreat! hey.. self contained... definatelyneed to get that heat out! Which... leadsme to believe.. it was notefficient. v/r Ken Carrigan
ratio is to provide a practically free fall flow through the tubesas envisioned by
Schauberger in his device. I believe, as does Jerry that there arethermal factors
and viscosity variations in the oil that are lurking here also. Theseare the factors
that we don't have a firm handle on until a unit is built and run. NormJusthad to do this.. it's been bugging me as to how this exit forcefrom a centrfugalnozzle head could produce overunity. So here issome math (sorry Engineerhere) that will explain what a Normal"sprinker head" would achievein power. Forgot you said this.. butthe raduis squared does comeinto play.. however.. still do notsee where the "kick over" occurs forover 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 nozzlesare "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 conciderthat laminerflow is occuring and no resistance is taken in to account (frictionorheating). 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*Ais called the mass efflus of the nozzle.. or lets say "m"for now. The moment going into the hub.. enterance of water... iszero cause theradius "r" is zero. Thus the equation reduces to onlyone sum.. or(r x V)*m. Now we have to take into account for therotational velocityand 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 thesystem by the generator wouldbe -r*(Ve - W*r)*m. BUT the momentapplied to the generator is thenegative of the moment of the generatoron the arms. So the equationbecomes positive... no biggy!The power now is simple...one we have the moment.. It isthe 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 propermks units are used for example ... RPM hasto be changed to RPS.. etc etc..and don;t forget 2*pi for rotationalvelocity! v/rKen Carrigan PS.. NOW... WHERE IS THE OVERUNITY!!!Helpme understand...
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