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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 741309 times)

neptune

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #360 on: March 26, 2011, 07:30:52 PM »
@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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #361 on: March 26, 2011, 07:48:19 PM »
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.

Quote
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.

Quote
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.

Quote
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.

Quote
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.

Quote
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #362 on: March 26, 2011, 07:57:26 PM »
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #363 on: March 26, 2011, 08:00:01 PM »
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

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #364 on: March 26, 2011, 08:07:24 PM »
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.


Quote
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..

Pirate88179

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #365 on: March 26, 2011, 08:10:43 PM »

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.
.99

.99:

Am I missing something here?  If Rose is feeding energy back to the single 12 volt bat. in the form of amps and volts, would this not show up on the bat. terminals?  What I mean is, checking only across the battery terminals will not show which way the energy is flowing, only the energy available at the terminals correct?

Example: 

12 volts at the terminals,  circuit off.  Circuit ON and feeding back 5 volts (just a number for this example) to the battery from the running circuit, would you not measure 17 volts at the terminals with a DMM?

Bill

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #366 on: March 26, 2011, 08:27:37 PM »
.99:

Am I missing something here?  If Rose is feeding energy back to the single 12 volt bat. in the form of amps and volts, would this not show up on the bat. terminals?  What I mean is, checking only across the battery terminals will not show which way the energy is flowing, only the energy available at the terminals correct?

Example: 

12 volts at the terminals,  circuit off.  Circuit ON and feeding back 5 volts (just a number for this example) to the battery from the running circuit, would you not measure 17 volts at the terminals with a DMM?

Bill

Bill,

In order to send power back into the battery, the voltage and current have to be in anti-phase, i.e. a positive voltage higher than the terminal voltage, AND a current going in the direction of the battery (a negative current), both at the same "time".

A DMM will measure the average voltage on the battery terminals. So, if there is a consistent higher voltage and negative current, the DMM will measure that increased voltage.

The goal of this exercise however, is to establish if the present battery voltage measurement is valid. If there is a significant difference between the measurement made directly on the battery terminals vs. on the other end of 22 feet of wire, then there is an obvious problem that must be addressed, and this puts the claims and measurements into question. Do you agree?

.99

hartiberlin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #367 on: March 26, 2011, 08:38:49 PM »
Okay. poynt99,
I now know what you mean,
you wanted to say, that the first 1000 Ohm resistor
is like the resistance(impedance) of the cable
going from the battery to the circuit.

Yes, there are losses there, but you have seen,
that you also have a small positive ripple in your simulation,
that means the battery voltage rises, when the current at the shunt
is negative, i.e. it is recharging the battery.

So it would be good that if the measurements of the battery voltage
will be taken directly at the battery terminals, but you will see only
a small ripple as in your simulation there.

The only question I still have is, if the inductances of the shunts
distort the measurements, so that the mean average values of the
current shows a total negative current, also, if 6 or 40 Watts of power
are heating the heater element ?

Rosemary probably did not run the circuit 5 months contineously,
but only a few times in the 5 months during measurements, so longer
testing times are needed to see, how the battery voltage is
going up or down.

As a lead acid battery is full at about 12.5 to 12.8 Volts
and is nearly empty already at 12.0 Volts,
these voltage changes must be noted and at only 6 Watts of heating,
it should be let run for about 1000 hours, if all the 5 batteries have 100 Amphours
capacity. Only then the 5 batteries should be empty and the battery voltage
should fall to 60 Volts.

Regards, Stefan.

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #368 on: March 26, 2011, 08:51:32 PM »
Okay. poynt99,
I now know what you mean,
you wanted to say, that the first 1000 Ohm resistor
is like the resistance(impedance) of the cable
going from the battery to the circuit.

Yes, there are losses there, but you have seen,
that you also have a small positive ripple in your simulation,
that means the battery voltage rises, when the current at the shunt
is negative, i.e. it is recharging the battery.
Actually no, that's not really what I am trying to say. My point was to show that the two measurements are not the same. If the two measurements are not the same, one of them must be incorrect. I wasn't expecting anyone else to answer the question, as I had hoped Rose herself would be allowed to see the point I am trying to make.

Quote
So it would be good that if the measurements of the battery voltage
will be taken directly at the battery terminals, but you will see only
a small ripple as in your simulation there.
You will see the actual voltage across the battery, which is the required goal for obtaining the correct PVbat. A small ripple is precisely what you should see. You seem to be thinking contrary to this....why?

.99

neptune

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #369 on: March 26, 2011, 09:43:34 PM »
Here is a crazy thought if we really want to be pedantic .What is magic about the battery terminals . These are just the point at which the lead "wires" [bus bars] inside the battery change to copper wires outside the battery .Should we not really put our probes INSIDE the battery on the actual plates?

poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #370 on: March 26, 2011, 10:27:38 PM »
Here is a crazy thought if we really want to be pedantic .What is magic about the battery terminals . These are just the point at which the lead "wires" [bus bars] inside the battery change to copper wires outside the battery .Should we not really put our probes INSIDE the battery on the actual plates?

It is unfortunate that you perceive me as being pedantic.

The issue of where the battery voltage is measured is paramount to obtaining a valid battery power computation by the scope.

How was your post helpful in any way?

.99

hartiberlin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #371 on: March 26, 2011, 10:33:35 PM »
Actually no, that's not really what I am trying to say. My point was to show that the two measurements are not the same. If the two measurements are not the same, one of them must be incorrect. I wasn't expecting anyone else to answer the question, as I had hoped Rose herself would be allowed to see the point I am trying to make.

Well the second is wrong, but I don´t understand what your point is here...what you want to say with it..
hmm...a bit confusing..

Quote

You will see the actual voltage across the battery, which is the required goal for obtaining the correct PVbat. A small ripple is precisely what you should see. You seem to be thinking contrary to this....why?


No, I don´t think contrary.
The voltage will be almost constant, just only a small ripple, i.e. rising of the voltage,
when the battery current is negative and a bit falling, when the current is positive.
But the differences  will be only in the MilliVolts range, as the internal
resistance of the batteries is pretty low.

The battery works here as a big capacitor, where the voltage can not jump on it,
so only a small ripple will be seen on the DC supply voltage.

This is why we can neglect the ripple voltage and can calculate with
a "constant" battery supply voltage and just
observe the current on the shunt.

When the current trace area below the ground line , which I painted green
in the posted scope shots on the shunt, is bigger than the
area above the ground line(painted red), then we can already see, if energy
is flowing back into the battery or energy is flowing out of the
battery. Then we don´t need the battery voltage.

Hope this helps.
Regards, Stefan.


poynt99

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #372 on: March 26, 2011, 11:47:02 PM »
Stefan,

One of the two battery measurements is incorrect. That is problem number one that needs attention.

Second, now that you bring up the current measurement, that too has it's problems.

Referring to your post; your area fill-in of the csr voltage is going to come out very close to equal when comparing both halves. First inaccuracy is that cycle mean is being used, and there are multiple cycles displayed. This is not the intended way to use cycle mean. Furthermore, the scope can not know what constitutes one cycle (it simply looks for zero-crossings), and therefore it completely missed that fact that one full switching cycle includes the portion of the cycle I highlighted.
http://www.overunity.com/index.php?topic=10407.msg279211#msg279211

Second (see scope shot below), realize that you are only looking at about half of the cycle. The other half clearly shows that there is positive current sourced from the battery. I have highlighted this in a red elipse.

In summary; the shunt voltage mean value shown is of no use, and does not reflect what the real average current is for that measurement.

.99

cHeeseburger

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #373 on: March 27, 2011, 12:37:20 AM »
Now, Stefan and Poynt, we are converging on the truth finally.  The battery is a fixed DC potential with millivolts of actual ripple due to its internal resistance.  The battery voltage does not actually have the 150VAC 1.5MHz signal Rosemary is feeding the 'scope.

The importance is now properly focused on the shunt and the actual current flow there.

This demonstration shows the difference between the waveforms obtained across the inductive shunt and the resistive portion of same shunt.  The scale factors are identical (1V per division) on both traces.  Notice three super-important  things:

1)  The amplitude when we include the inductance is way higher and does not agree at all with the actual current measured just across the resistor.  The inductance allows a much larger voltage swing, fooling us into thinking the current is much larger than it really is.

2)  Look at the areas above and below zero.  In the larger (inductance included) trace, by eyeball, it looks like the areas are close to even or maybe even slightly more negative.  But in the real current trace it is clear that the area above zero is easily greater than that below zero.

3)  There is significant phase skew between the two waveforms and this will ruin the accuracy of any multiplied samples.  The true current (across just the resistive part of the shunt) does not peaqk at the same time as the false, inductor-polluted "current" trace and is in fact not always the same polarity at a given instant in time.  Notice the inductive shunt trace is approximately at its peak at the zero-crossings of the real current:  almost 90 degrees phase shift.  Basic fundamentls when the L vastly dominates the R of shunt!

So, the amplitude, waveshape and phase angle of the "current" signal Rosemary is feeding into the scope is by no means an accurate picture of the true instantaneous current flowing in the circuit. When the "battery" voltage also has an enormous misrepresentation due to series inductance inside the measuring points, and we multiply the data samples point by point, the numbers are so far from any believable reality that it boggles the mind and the results could come out anywhere and are totally meaningless, sorry to say.

Here is a challenge for Rosemary:  Submit this post to your favorite Tektronix Applications Engineer.  He or she is a certified oscilloscope measurement expert and is called on all the time to sort out these kinds of measurement questions.  Ask him or her to write a paragraph about it, agreeing or disagreeing with what I have wriiten here, attach his or her name to it, and publish it here for us.

Cheeseburger
« Last Edit: March 27, 2011, 04:35:20 AM by cHeeseburger »

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #374 on: March 27, 2011, 04:38:32 AM »
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

Poynty - Still not called off your dogs?  Shame on you.

Now.  Regarding that equation.  P never, to the best of my knowledge - is represented in any of those equations that you've put forward.  Power is ALWAYS vi dt.  Or Volts x amps x time.  THAT's it.  You can try and argue this till the cows come home Poynty.  This is the fundamental requirement for wattage analysis and this over time = POWER.  NOTHING ELSE.

SO.  Take that example that you've given us.  I'm looking at your schematic.  The amount of current discharged from your batteries will be determined by the amount of resistance in the path of that current.  Therefore resistance will be R1 + R2 + R3.  IF R1 = 1 Ohm and R2 = 1000 Ohms and R3 = 0.25 - then the total resistance determined by that circuit will be 1001.25 Ohm. 

Let us further assume that VBatt = 24 volts.  Therefore the current discharged by that supply source will be 24/1001.25 = 0.024 amps.  THEREFORE  vi dt = 24 (vbatt) x 0.024 (amps) = 0.575 watts x (say) 5 minutes would be 0.575 x 60 x 5 = 172.5 Joules.  THAT'S IT.  The ONLY correct way to determine that power.

So.  To get back to your question.  The Ohmage in the path of that P value that you refer to cannot be considered in isolation to the power over the entire circuit which will be distributed according to the resistance over the whole circuit.  You CANNOT look at one isolated part of the equation and expect it to represent a true value.

Now.  To get back to that same circuit that you drew and REPLACE R2 with a whole pile of MOSFETs in parallel.    Then.  Replace the R1 @ 1 Ohm with R = 10.86 Ohm.  NOW.  Apply a switch that the battery is ONLY connected during 20% of the time and for 80% it is disconnected and THEREFORE NO POWER IS DELIVERED. 

P = vi dt.  THEREFORE.  10.86 (R1) and 0.25 (R3) + 0 resistance at the MOSFET.  Therefore R = 11.11 Ohms.  THEREFORE if Vbatt = 24 then 24/11.11 = 2.16 amps.  24 (v) x 2.16 (i) = 51.85 watts.  Assume a 5 minute run time.  Therefore 51.85 x 60 x 5 = 15.552KJ. 

The actual question here is what happens during the period when the switch is open and the battery APPARENTLY is not able to discharge any current flow.  Hopefully you're looking at this.

Rosemary