V =3D 2 pi r /p then a =3D 4pi2xr/p2
Given drum shaped satelite 60 inches (5 ft) diameter spinning
1800 RPM containing a wall of water 6 inches thick on the inner wall
we have:
a =3D 4 x pi squared x 2.5 ft. divided by .033 sec squared or
98.70 divided by .001 =3D 98,702 for an acceleration force (a).
Now according to the little handbook of hydraulics, head can be figured
as height of water colume multiplied by .434 with the results in PSI at
1 g. Our colume of water is .5 ft x .434 =3D.217 PSI Now we have an
acceleration or g force of 98,702 x .217 =3D 21,418 PSI. If we have 24
nozzles that have a total cross section area of 1 square inch then you
can see how much HP can be generated. Ken, please verify these figures
for I only had 2 Phd=92s here and I am only a BS. Ha! Ha! Norm
PS: I hope you followed my p calculation which is 1800 RPM =3D 30 turns /
sec which takes .033 sec/turn. OK!
Subject: Re: Clem... design and equations
Date: Mon, 14 Sep 98 10:19:23 PDT
From: "Garry Whitman" <whitman@kskc.net>
To: "Norman Wootan" <normw@fastlane.net>
Hi Norm
Ran the numbers you gave Ken
force on the water =3D 544410 lb.
divided by 1131 square in. of drum wall =3D 481 PSI
481 times 24 nozzles with an area of 1 sq. in. =3D 11544 pounds
11544 times 2.5 ft radius =3D28860 ft. lbs. torque
(28860 times 1800 RPM) divided by 63025 =3D 825 HP
Now I know why you like the idea.
Garry
----------
> 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 =3D mass x velocity squared / radius
> > F=3Dmv2/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 =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...
> >
> >
> >
>