# Lugano resistance drop re-visited

Could the drop in apparent resistance of the reactor heater element in the Lugano report be in part due to the core ceramic conducting at high temperature and partial shorting between the 3 phases?

There have been burn outs, core breaches, melted heater wire, hot spots and arcing thermocouples over the months.

Alan Goldwater deduced there was a drop in resistance in the ultra pure Al203 Coorstek cores we were using.

Discussion: http://goo.gl/TxtE0h

Direct Link: https://goo.gl/guRhzl

Additionally, SiO2 based ceramics resulted in plain shorts causing loss of thermocouples.

More recently, Adriano Bassignana made us aware of the Nernst Lamp

https://goo.gl/R89Zbz

Of course, most replicators reactors to date have been single phase. So the differential between two adjacent coil loops where ceramic conductivity has increased would be lower than that between a parallel, helically wound, 3 phase coil.

## Comments

12Quote: was probably intended to be:

Quote: Try imagining this as it was Rossi saying it on the phone to his patent attorney.

(sorry, I didn't know where else to put this)

Closing the loop would do it too. i.e. all control and stimulation power harnessed from output to run reactor for beyond chemical possibility.

This is the kind of Discussion that Bob Higgins and I had.

This and the wrong use of Emissivity means that for me, it all comes down to seeing statistically significant transmutations.

You will not perhaps think this relevant but:

3 phase makes it very easy to get an apparent COP=3 for free by inverting one of the input clamps on the power meter.

Without a proper power meter 3 phase makes input power measurement problematic.

The complexity makes errors in input power measurement, whether unintentional or deliberate, much more possible.

All quantities RMS

For the same 3 resistive elements of equal resistance and 3 phase full waveform of line voltage V (we can revisit that later):

Y - resistor voltage is V, current is V/R, Power is V^2R

Delta -

resistor voltage is V|(1- (-0.5+0.866j))| = sqrt(3)V,

resistor current is sqrt(3)V/R,

power is I^2R = 3V^2R

The key thing is that delta increases element voltage by sqrt(3) over line voltage. compared with Y.

The Y equivalent resistance is reduced by the square of this (a factor of 3).

see wikipedia for details.

So for this to work we need:

Dummy run: Wye

Active run: Delta

Motivation? For limited range of V in from control box delta with give higher power than Y.

There are other darker possible motivations.

But motivation is all speculation.

What is fact is that the numbers given make sense for this change in setup, and given that the reactor must have been disconnected and reconnected such a change is possible.

The testers should have recorded this change, but we know that their work had issues in other areas - most notably a singular lack of controls of any kind (no independent temperature measurement, no dummy reactor at same power) and an error in how they calculated temperatures.

Nick Oseyko said

"I had tried heating mullite pipe with torch back in May. Becomes very conductive before melting. 1 cm gap was about 100 ohms."

Given that there was barely more than a few millimetres between phases in the Lugano reactor, and a large potential difference between them, this effect cannot be dismissed easily.

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