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Author Topic: D.I.Y. Calorimetry  (Read 2952 times)

Offline joegatt

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  • Posts: 47
D.I.Y. Calorimetry
« on: March 21, 2010, 09:51:26 PM »
To account for all the energy transformations, we need to know the resistive loss in the coil.
This would best be determined by encasing the toroid in some kind of calorimeter.
The next best thing is to replicate this resistance outside the equipment.
A resistor connected in series to the toroidal coil can serve this purpose.
The energy dissipation in this external resistor is proportional to the ohmic dissipation inside the toroidal coil.
(If the values of these resistances is the same, the energy dissipation will be identical.)

This external resistor can be conveniently installed with waterproof seals, at the bottom of a polystyrene cup.
The cup is filled with a measured amount of distilled water.
A polystyrene lid is added, with a thermometer poking through ...
This should work nicely.

Regards
Joseph

Free Energy | searching for free energy and discussing free energy

D.I.Y. Calorimetry
« on: March 21, 2010, 09:51:26 PM »

Offline Omega_0

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Re: D.I.Y. Calorimetry
« Reply #1 on: March 22, 2010, 07:29:28 AM »

To account for all the energy transformations, we need to know the resistive loss in the coil.


Resistive loss of the wire is not the only energy expense here. There are other losses in the core material (and may be drop of temperature, as claimed by some). So your measurement will not be accurate enough if you replace the coil/core system with a simple resistor.

Plus, if energy is being transferred to the rotor by some means (this is important to test), it won't reflect in such measurements. In OU there are no shortcuts.

Offline joegatt

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Re: D.I.Y. Calorimetry
« Reply #2 on: March 23, 2010, 12:24:58 AM »
Yes I agree. This method does not account for the hysteresis loss in the core.  However, considering that ferrites are normally designed to have low hysteresis I would expect that loss, to be very small.  It would simply mean, that less power is recovered from the magnetic field by the coil, at the end of each pulse.

Now I would consider:

Power from supply = resistive loss + hysteresis loss + power recovered by flywheel diode + frictional torque x angular velocity

to be a fairly good approximation of the system.  If there was a strong overunity effect, this equation would be thrown off balance.  Certainly, an output that is three times the input, as was claimed by Steorn, would be glaringly obvious.  An overunity effect that is very small, small enough to remain hidden in my estimation errors, is beyond my capabilities.  I would be quite happy to sit back and let some leading-edge laboratory show us the way.
 
Regards
Joseph

Free Energy | searching for free energy and discussing free energy

Re: D.I.Y. Calorimetry
« Reply #2 on: March 23, 2010, 12:24:58 AM »
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