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Well... sure enough Centripetal force is a force but what we are looking =
for
is expended or generated energy. It will take X amount of energy to=20
rotate this mass to a certain velocity, then without friction.. no =
energy=20
should be needed to keep it spinning.... all without friction. However =
that
is not the case as there always is friction and actually is a velocity =
cubed
function.. the faster the speed the higher the friction. This very true =
with
automobiles 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.=20
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=20
arm. In a Clem conical drum which is composed of many Phi ratio =
spiral tubes=20
there is considerable reactive thrust from the nozzles which drive =
the rotation of=20
the unit while the centrifugal force is drawing the water into the =
hub. =20
This thrust was taken into account.. the mass flow rate along with the =
number of=20
exit nozzles.
Your lawn=20
sprinkler depends on the city water pressure of 60 PSI or more to =
provide the=20
driving force to turn the head array. In an artificial gravity set =
up such as the=20
Clem engine the radius of the unit is critical as is RPM to achieve =
the necessary=20
pressures needed to drive the assembly beyond the point of =
frictional losses=20
in the distribution tubes. =20
The lawn sprinkler was only an example as NO pressure was specified. It =
could 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. This
also means that efficiency goes down.. as energy is lost. Efficiency =
does not
got up.... and there is no point that I know of.. in physics... where =
one can
overcome this frictional loss.. especially at higher velocities and =
pressures.=20
Please give an example where this is so.. or equations that tell me.. =
you can
overcome frictional losses.
The reason that the distribution tubes are based on a Phi=20
ratio is to provide a practically free fall flow through the tubes =
as envisioned by=20
Schauberger in his device. I believe, as does Jerry that there are =
thermal factors=20
and viscosity variations in the oil that are lurking here also. =
These are the factors=20
that we don't have a firm handle on until a unit is built and run. =
Norm=20
The way I was always taught.. was to first Hyposthesize about your =
theories and
then plan an experiment around it.. test it in a lab environment and =
document
results and revise the theory/hyposthesis more.. etc etc. Fluid =
dynamics is
the key to this all.. not nessarily physics... this is where I have =
learned it.. from=20
college books.. partly mechanical...but graduated BSEE. The professor =
who
taught the class I remember was from NASA... cool class! =20
The oil my guess was to overcome the water boiling problem with =
hypersonic
fluid flow... creating heating effect from the frictional losses... must =
have been
great! hey.. self contained... definately need to get that heat out! =
Which... leads
me to believe.. it was not efficient.
v/r Ken Carrigan=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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