Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Pirate88179]

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]

.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]

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]

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.

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]

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?