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Mechanical free energy devices => mechanic => Topic started by: TommeyLReed on April 06, 2019, 07:10:54 PM

Hello all,
I want to show my new build of a 4ft diameter rotor design for a big Clem engine build.

You should resize the images before you post them.
I have to scroll 4 feet to see the whole picture.
Vidar

Odd, why the to photos are big?
Wont let me change the settings.

Really nice build Tommey. :thumbsup:

This is a small video with some specs.
https://www.youtube.com/watch?v=U101rHzM_Hg&feature=youtu.be

Hello all,
My theories on the Clem engine is questionable, but my theories is something different.
I believe I came up with this design with many prototypes that did not work as plan, but each design had some small parts to this theory.
Just what is this unit?
This is based on a tornado water spout.
Tornado are heat engines as a vortex is created, temperature is lower and could reach 273k.As the temperature drops heat is pulled into the vortex to increase rotation size.Pressure becomes negative causing a implosion effect or vacuum.
This unit is no different.
As the rotation starts, fluid is pumped upward into the drum causing it to fill due to small rim jets at the end of each tube.
Heat will start to build due to fluid friction in the pump/motor.
As the rpm's increase, centrifugal forces are generated causing pump load to decrease in pressure.
At a given speed (not known) fluid is being forced outward more then the pump can handle, causing the motor to accelerate.
In other word if the pump is moving 20gpm at 1000 rpm's and the rotating mass is pulling 30gpm, the motor will increase causing a chain reaction in the acceleration until the system blows itself apart.Control valves are need to control flow.Temperature will effect velocity of oil flow, increasing flow rate.
A vacuum is needed to prevent cavitation and air drag, this will also prevent oxygen in a flammable fuel due to temperature.
Tom.

Hello all,
This is a big build and here are more pictures.

Hi All,
Lets look at some real mathematics when things rotated.
Centrifugal force is calculated by a formula:
Force= M*V^2/R
In other words:
http://www.calctool.org/CALC/phys/newtonian/centrifugal
At a mass of water 8.33lb of fluid rotating at 2000rpm and a diameter of 4ft = 27713.6 pound of fluid force is created.
10 gallons of fluid is inside the rotating mass while it rotates at all times.
This will produce over 100,000 pounds of fluid force rotating 2000+rpm's. This mean also that the drive pump/motor will have to accelerate to keep up with all the fluid force ejecting from the rotating pistondrum.
16 pipe at 2.25IDx14in holds about 5gal.
20" drum x 4.5 hold about 5gal231cubic inch =1gal.

New video
https://www.youtube.com/watch?v=Y1Adgc4DAfA

Hi Tom,
Nice build so far,I assume your using oil for the fluid?
Also that the fluid is being recycled?
With each cycle it will gain temperature, what prevents ignition?
artv

I plan to have a complete vacuum 2030" inch of mercury while running. This will keep any type of explosion from happening. No oxygen, no fire.
Oil will be cooking oil due to high flash point and in a vacuum while heated.

I plan to have a complete vacuum 2030" inch of mercury while running. This will keep any type of explosion from happening
Details on this would be interesting.
Thanks artv

Hi All,
This is a simple animation of how this implosion would work.
https://www.youtube.com/watch?v=6UtzXx48k3E

Hi All,
This is past information to show static head pressure vs rotation pressure difference.
This show when you rotate at a high velocity it creates forces that could be use to generate mechanical work.
https://www.youtube.com/watch?v=m1YPZe9RdwM

https://www.youtube.com/watch?v=xYFyNQm7H2U&t=9s

Hi all,
It’s getting big.

Hello All,
I have been very busy building this new version of the Clem engine I call a hydraulic implosion motor.
I have a few short videos to show off now of the build.
https://www.youtube.com/watch?v=6ITWZ20Gj6w (https://www.youtube.com/watch?v=6ITWZ20Gj6w)
I will add that it's not completed yet because of testing the speed of rotation.
I will try to answer your question and some specs information.
Tom

I have a lot of short videos of the build.
https://www.youtube.com/watch?v=4Q7vfucSq_8 (https://www.youtube.com/watch?v=4Q7vfucSq_8)
Tom.

Hi Tommey,
Can you tell me why you are building this build so large ??
Just curious.
Graeme

Did you ever finish this build? Any more videos you can post directly here, all your YouTube videos are down.

https://www.youtube.com/watch?v=Ji0GAtq2EM (https://www.youtube.com/watch?v=Ji0GAtq2EM)
Your going to get a kick out of this.

When your ready with the external combustion engine I want to work with you on it. I have a way of increasing it's efficiency with magnetic repulsion. No bologna.

The experiment for R&D on James Roney's Contour Magnetic Runway Plate
Calculations
Joules of work = force x distance
Gravity 9.8m/sec ^2
1 netwon / 1 meter = .7376 ft/lbs per second is wattage
 unit of power is Watt, defined as joule per second. or .7376 ft/lbs
For example we have a 6.2 gram magnet and a distance of 10 centimeters
10 centimeters is .1 meter to find netwon we multiply magnet weight by gravity
distance second squared.
.0062 x 9.8 = 0.0608 newton
now multiply the newton by meter to get Joules
0.0608 x .1 meter = .00608 Joules
Work is measured in Joules but lets get the foot pounds
.00608 x .7376 = 0.04484608 ft/lbs
lets now do some more calculations of kinetic Wattages and real Work wattage using the
example.
0.0608 newton x .1m = 0.0608 Joules
formula "Joules to Watts"
the power P in watts (W) is equal to the energy E in Joules (J), divided by the time period T in seconds
(S)... so the problem is E(J)/T(S) = P(W)
Magnet rolling down
.00608 Joules / .47 seconds = 0.01293617 watts(Kinetic)

Magnet rolling up
.00608 Joules / .52 seconds = 0.011692308 watts(Work)
so if you take the difference of the watts you see gravity adds
0.001243862
We can see here magnet rolling up is basically free energy since it is
going against gravity with magnetic attraction/repulsion and only a slight loss in watts
Now if we apply this to a real world working engine, you can calculate the
boost by the speed / distance the rotor travels and have a fixed magnet.
or have a magnet rotating and a fixed metal plate. either way you can
calculate the boost.
Now the above calculations might not seem like much until you realize a 6 gram magnet is very small. If you use a 100 gram magnet now your getting 1 joule of power or 1 watt
and 300 gram would be 3 watt etc.. and we start getting into the 10s and 100s and 1000s of watts the heavier the magnet gets with gravity and velocity.
I am including a picture of a 100 gram magnet rolling up, 1 watt of "free energy"