another small breakthrough on our NERD technology.

poynt99 · March 13, 2012, 12:38:22 AM

Quote from: Rosemary Ainslie on March 13, 2012, 12:30:30 AM
Poynty Point. You say OF COURSE? That means that your accede that the the current is flowing in both directions through the that rail of LED's notwithstanding their polarity? Because what I'm telling you is that NOT ONLY does only that one rail light up. It also STAYS permanently lit. Even TK's video sample showed evidence of this. Not sure why you should all find this of so little interest? But there you go. One can take the horse to water - but that's about it.

Kindest regards as it's much needed
Rosie Pose.

You don't understand.

No that's not what I said. Read it again:

Quote from: poynt99 on March 12, 2012, 11:50:04 PM
Of course only one LED stays lit, the one pointing to the circuit.

The one (or many) pointing to the circuit is the one that lights when current is being DELIVERED by the battery.

Rosemary Ainslie · March 13, 2012, 12:41:50 AM

Quote from: poynt99 on March 13, 2012, 12:38:22 AM
You don't understand.

No that's not what I said. Read it again:

The one (or many) pointing to the circuit is the one that lights when current is being DELIVERED by the battery.

I DO understand. That same one STAYS LIT when the current is NOT being delivered by the battery - is my point. Poynty Point. Pay attention.

Rosie Pose,

poynt99 · March 13, 2012, 12:46:05 AM

If there is no current being delivered by the battery, and the diode is pointing towards the circuit, then the LED would not be lit.

Since it is always lit and it is directly in series with the battery, current is always being delivered by the battery.

Rosemary Ainslie · March 13, 2012, 01:05:17 AM

Quote from: poynt99 on March 13, 2012, 12:46:05 AM
If there is no current being delivered by the battery, and the diode is pointing towards the circuit, then the LED would not be lit.

Since it is always lit and it is directly in series with the battery, current is always being delivered by the battery.

NO Poynty. Where's all that intellect I've accused you of? We have two rails of LED's. The one takes current from the battery supply. The other takes current from CEMF. TWO OPTIONAL PATHS. Now. We've got an oscillation. Voltage moves equally - above and below zero. Therefore, correspondingly, the current flows above and below zero in each oscillation. Therefore - one would expect the current to move through either one or other of those rails - depending on that polarity. NOW. NOTA BENE. There is NO CORRUPTION OF THAT OSCILLATING WAVEFORM - ON EITHER SIDE OF THOSE LED's. YET. Only ONE RAIL STAYS LIT. And the other doesn't even turn on. Not EXACTLY what our standard model would predict. I would have thought?

Rosie Pose

Rosemary Ainslie · March 13, 2012, 01:07:58 AM

Quote from: Rosemary Ainslie on March 13, 2012, 01:05:17 AM

NO Poynty. Where's all that intellect I've accused you of? We have two rails of LED's. The one takes current from the battery supply. The other takes current from CEMF. TWO OPTIONAL PATHS. Now. We've got an oscillation. Voltage moves equally - above and below zero. Therefore, correspondingly, the current flows above and below zero in each oscillation. Therefore - one would expect the current to move through either one or other of those rails - depending on that polarity. NOW. NOTA BENE. There is NO CORRUPTION OF THAT OSCILLATING WAVEFORM - ON EITHER SIDE OF THOSE LED's. YET. Only ONE RAIL STAYS LIT. And the other doesn't even turn on. Not EXACTLY what our standard model would predict. I would have thought?

Rosie Pose

And when you've digested this much - then refer to our second paper. There's a full on proposed explanation. It's what this claim is all about. LOL.

Again, and ever,
Rosie posie pose.