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Overunity Machines Forum



Thane Heins Perepiteia Replications

Started by hartiberlin, May 28, 2009, 05:54:52 PM

Previous topic - Next topic

0 Members and 15 Guests are viewing this topic.

i_ron

Quote from: wattsup on August 25, 2009, 08:07:25 AM
@TH

Don't know if you ever saw this patent from our dearly departed @JH.
It may help open a whole new avenue and take this to a next level.

http://purco.qc.ca/ftp/Inventors/Jack%20W%20Hildenbrand/patent-7453341/
Click on US7453341.pdf.

You can also go up one level and see some photos.

Basically, he takes either two 20 lbs pull magnets and gets 80 lbs pull force or he does it with one magnet and one coil.

So imagine if the 80 lbs pull force will also give you 80 lb repulsion force if the coil is timed to pulse as it leaves a generator core maximizing repulsion spin.

Also may be useful for your transformer builds.
Anyways just another angle.

wattsup

Good post Wattsup, somewhat similar to the Flynn patent, which also works. You might be interested in my experiments on the Hilden Brand,,,

http://www.youtube.com/watch?v=5prfWVc--dc

http://www.youtube.com/watch?v=kYcskBvSg6Q

Ron

i_ron

Quote from: supermuble on August 22, 2009, 01:21:26 PM
I thought core drag was very minimal, so minimal that it isn't even worth discussing?

If you spin an unloaded generator, there is not enough core drag to really notice right? Doesn't the attraction energy IN almost equal the energy drag OUT?

In my tests all cores have drag. In this case the core drag could be 5 to 15 watts, for each core. This is accumulative, so with a machine of say 10 cores, the drag could be 50 to 150 watts drag penalty. That Thanes machine has drag is obvious in that with two or three cores it wont start without the cores on a movable door which is moved into position after the machine has reached a certain RPM. Same with the California machine with the cores on a rail.You are correct in that attract in force is nearly balanced by retard out force but this does not address Hysteresis loss, which is the major portion thereof.

This is one point that T has never addressed, despite repeated requests.

As I have said so many times, the rotor should be run up with no cores present... then a second reading with the cores in place. Then it is possible to see how much the core drag is reduced with the shorted HV coil. To be OU, as T has stated, the drag must be lower than the first baseline reading. I have never seen this in my experiments.

If anybody has... then please post your results!

Ron




capthook

Quote from: i_ron on August 26, 2009, 06:29:07 PM
As I have said so many times, the rotor should be run up with no cores present... then a second reading with the cores in place. Then it is possible to see how much the core drag is reduced with the shorted HV coil. To be OU, as T has stated, the drag must be lower than the first baseline reading. I have never seen this in my experiments.
Ron

The large core losses are only modestly mitigated by the HV coils.
Running with no cores/coils = X watts input
w/ cores/coils = X+Y (to overcome core losses, and Y, as you mention, is large)
w/ cores and shorted HV coils = X+Y-HV(small)
w/ cores and HV/HC coils = X+Y-HV+HC (call it Z)

X < Z
What is X divided by Z?  Less than .5? (50%)

hoptoad

Quote from: capthook on August 27, 2009, 12:15:32 AM
The large core losses are only modestly mitigated by the HV coils.
Running with no cores/coils = X watts input
w/ cores/coils = X+Y (to overcome core losses, and Y, as you mention, is large)
w/ cores and shorted HV coils = X+Y-HV(small)
w/ cores and HV/HC coils = X+Y-HV+HC (call it Z)

X < Z
What is X divided by Z?  Less than .5? (50%)

@Capthook  .... That seems to sums it all up   ;)  For the moment ... LOL     KneeDeep

Cheers

baroutologos

I am afraid a dead-end has been reached in this direction and a break-through is required.

Regards,
Baroutologos