Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[neptune]

@teslaalset .Great idea on the mug heaters .They are also available in 12 and 24 volt varieties with resistance values of 1.2 ohms and unspecified resistance/wattage at 24volt .These may be better as they are nearer in resistance to Roses experiments . They cost about£5 each on Ebay . Several could be used in series if required . @Pirate , do we know the application and resistance of the element you show ?

[poynt99]

Poynty

I should be able to get that oscillation with just one battery.  The scope wires would be connected directly to the battery.  No other wires in the setup. 

Yes Rose, that would be excellent if you could do this.

Will there now be no evidence of that wild voltage oscillation at the battery?  Or will it simply stay level? With the occassional ripple?

Correct. Assuming you can achieve the self-oscillation as before with a single 12V battery, the battery voltage measurement taken directly across its terminals will show a 12VDC value, with an estimated 350mVpp of oscillation ripple riding on top of that.

Will the antiphase condition between the voltage and the shunt now change?  Will this be out of phase and therefore 'no advantage'?

The small ripple voltage seen riding on the 12VDC may still be in anti-phase with the shunt oscillation, but the fundamental battery v(t) x i(t) product is going to look and compute quite differently.

If I apply the math trace - a product of the battery and shunt voltages - will they now show 'positive' as opposed to negative?

You will ostensibly have a steady 12V x  your same oscillating Vcsr voltage. Your battery voltage trace during the oscillation phase may vary from between 11.8V to 12.2V as opposed to what you have now, where Vbat varies between 0V and +250V.

Will the mean average across the shunt change to default always to positive?

I can not reliably predict what the shunt voltage wave form will look like at 12VDC supply (assuming you can make it work at 12VDC), vs. the operation at the 60VDC supply, but the fact that you are measuring the battery voltage at a different point will not affect the shunt voltage measurement, as you will measure that at the same point as before.

Let me know what you expect to see Poynty.  Because this time I want the argument 'up front' if possible.

I believe I have done so.

.99

[poynt99]

yes - agreed

Rosemary

Right, Part 2 then.

What will happen to our PVbat calculation if rather than using this agreed upon equation:

PVbat = V(P1 - P4) x V(P3 - P4)/0.25

we make a small change and use this instead:

PVbat = V(P2 - P4) x V(P3 -P4)/0.25

What happens to the PVbat calculation?

.99

[Rosemary Ainslie]

Ok.  Now.  2 things.  Give me the balance of your argument. And are you able to apply any kind of  moderation on your forum.  Call off your dogs POYNTY.  Or is there a 'free for all' when it comes to trashing my character?  IN which case why am I speaking to you?

You really need to apply some constraint there.  Your forum is losing credibility.  Just look at the readership levels. As for the latest incursion by Fuzzytomcat.  Why do you allow it?  Is it because it satisfies your argument somehow?   Not good Poynty.  Not at all.  I may be an idiot - I may even be a moron.  I don't know.  But I sure as hell am NOT a liar.

Rosemary

PS our posts crossed and I've got a visitor.  I'll get back here tomorrow.

R

[hartiberlin]

Right, Part 2 then.

What will happen to our PVbat calculation if rather than using this agreed upon equation:

PVbat = V(P1 - P4) x V(P3 - P4)/0.25

Okay. 49.99 milliwatts flowing out of the battery.

we make a small change and use this instead:

PVbat = V(P2 - P4) x V(P3 -P4)/0.25

What happens to the PVbat calculation?

.99

Hmm,
what should this be now ?
Only half the power..