I've heard that also the humming bird and bumble bee exhibit the same
'impossible' flight characteristics you mention in the beetle....
Norman Wootan wrote:
>
> 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...
> >
> >
> >
>
> ---------------------------------------------------------------
> 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...
>
>
>
>