Stubblefield coils (bifilar) and speculations

[jeanna]

All I am saying is you want to look at V and I together in your work and if you use the very simple approach of P=V X I you will see where many of the reading found by experimenters are about the same when it comes to P. Yes indeed V may be higher, but I is lower, then I is higher and V is lower and the product stays about the same.

RStiffler,

Thank you.

I have been looking at the power equation, but I am not sure I understand about how the meter can't really say.

Are you saying I must set it up as a true circuit in order to honestly see what I am getting?

(It is difficult sitting on the dirt keeping the wires clean etc, but),  I will certainly change my test methods if this is what I must do. I may be hitting a wall right now. I am not sure what next to do.

This may be what I need.

Please describe how you would do a circuit if you were trying to measure a Stubblefield battery.

Also, do you have any advice about how to keep the leads of this circuit in good operating condition. I have had problems with wires breaking from over-flexing them in the course of testing.

Thank you,

jeanna

[Pirate88179]

@ Dr. Stiffler:

First, your input is ALWAYS welcome here on this topic.  As you know, I have been following your work on the SEC with great interest. (and great ignorance on my part)

Second, I have to second Jeanna's question about how to properly measure E and I at the same time?  I have checked E first usually, then set the meter to mA to measure I.  So, with  say about .8 E and 80 mA's, P would equal .646?

@ Jeanna:

One thing I do that might help is to make up some jumper leads with alligator clips soldered to the ends and I just clip them on and off my cu and fe wires because as you say, they can get very brittle very quickly. (I found this out myself by snapping one off already)  This way, I don need to position the original wires much at all, I just move the jumper wires wherever I want them to be.  My jumpers are a pretty heavy gauge so I don't think/hope I am not adding much to the resistance.

Bill

[jeanna]

One thing I do that might help is to make up some jumper leads with alligator clips soldered to the ends and I just clip them on and off my cu and fe wires

heavy gauge so I don't think/hope I am not adding much to the resistance.

Bill

good idea. Thanks, Bill.

What gauge? Is it that big fat hookup wire you used in your early earth battery pics?

thank you,

jeanna

[Pirate88179]

@ Jeanna:

Yes, I believe it is 12 or 10 gauge stranded wire.  12 I am pretty sure.  Not too big to be a pain, but hopefully big enough to have fairly low resistance.

Bill

[DrStiffler]

All I am saying is you want to look at V and I together in your work and if you use the very simple approach of P=V X I you will see where many of the reading found by experimenters are about the same when it comes to P. Yes indeed V may be higher, but I is lower, then I is higher and V is lower and the product stays about the same.

RStiffler,

Thank you.

I have been looking at the power equation, but I am not sure I understand about how the meter can't really say.

Are you saying I must set it up as a true circuit in order to honestly see what I am getting?

(It is difficult sitting on the dirt keeping the wires clean etc, but),  I will certainly change my test methods if this is what I must do. I may be hitting a wall right now. I am not sure what next to do.

This may be what I need.

Please describe how you would do a circuit if you were trying to measure a Stubblefield battery.

Also, do you have any advice about how to keep the leads of this circuit in good operating condition. I have had problems with wires breaking from over-flexing them in the course of testing.

Thank you,

jeanna

@jeanna
Here is what you are up against, every generator, be it battery, solar panel, alternator, et. al. has a characteristic impedance and for maximum energy transfer the load on the source should be equal to the source.

So what does that mean? Before I get there let me address connecting meters directly across coils. A meter has a built in impedance, this can range fro 2,000 ohms for a simple vane analog meter to 10 million ohms for a DVM or DMM. So when the meter is connected across the source (your coils) with no additional load,  other than the load of your meter, lets say you have a DMM with an input impedance of 10 megohm (10 million ohms) and it reads 2 volts, so what is the power at this point in time? Calculate the current first; I=E/R..... I = 2/1E6 = 2E-6amps., P=I^2 X R so (2E-6^2) X 1E7 = 4E-5 watts or 40uW.

Now lets look at it a bit different, lets say you have a DMM with a impedance (resistance) of 0.01 ohm in current mode and you connect it across your coils and read 10mA, so what is the wattage here. Again we can say now what is the voltage; V=I X R so V=1E-2 X 1E-2 = 1E-4 or 100uW or 0.1milli-watt.

So what do you do? Well if you don't know the impedance of your coils it is not simple. But lets say you have a handful of resistors, say 10ohm, 100ohm and 1000ohm. Now you need two connect the resistors one at a time across your coils and with your DMM in current mode in series with the resistor. Measure the current and calculate the power at that point in time. P=I^2 X R

Do this for each of the three resistors and see which one is highest. From this you can find a direction for the coil impedance and match that with a resistor or correct size to allow a true power calculation.

Does this make sense or have I just thrown mud in the water???