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Author Topic: Joule Thief 101  (Read 620743 times)

Offline MileHigh

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Re: Joule Thief 101
« Reply #2580 on: May 14, 2016, 04:11:14 PM »

So here is my problem with that. many times before he admits to being corrected, his stance is this.......

"You are the epic failure others claim you to be.
You are a total disaster.
Your (sic) a fraud.
You epic failure.
You are now the laughing stock of this forum."

As far as your rant about me goes, I may deal with that later.  What I don't want to see is predatory harassment from you like you did to me the other day.

The reason for this posting is to clarify the words you quote above.  Those are Brad's words, not mine.

MileHigh

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Re: Joule Thief 101
« Reply #2580 on: May 14, 2016, 04:11:14 PM »

Offline tinman

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Re: Joule Thief 101
« Reply #2581 on: May 14, 2016, 04:54:03 PM »
You are dead wrong.  So where did you make your mistake?

No mistake MH.
The results were the same on every test--> 5 tests done.

Should i make a video showing you this?
Did you watch the video i posted the link to a few post back--that will knock ya sock of when you do the calculations.

Here are mine from the video i speak of.

So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.


Brad

Offline MileHigh

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Re: Joule Thief 101
« Reply #2582 on: May 14, 2016, 05:06:19 PM »
An experiment where the final voltages are not equal is an invalid experiment and is not even applicable to the statement about half of the energy being lost.  It is as ridiculously simple as that.

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2582 on: May 14, 2016, 05:06:19 PM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2583 on: May 14, 2016, 06:02:32 PM »
An experiment where the final voltages are not equal is an invalid experiment and is not even applicable to the statement about half of the energy being lost.  It is as ridiculously simple as that.

Ah,that good old MH paradox kicks into action again lol.

Nice one MH--were all getting the hang of how you work now ;)

So what would you say if i carried out the test again,and included a parallel resistor of 5K across the LED,and i ended up with more than half the starting voltage in both caps?
What paradox inclusion would you use then?


Brad

Offline tinman

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Re: Joule Thief 101
« Reply #2584 on: May 14, 2016, 06:06:57 PM »
No mistake MH.
The results were the same on every test--> 5 tests done.

Should i make a video showing you this?
Did you watch the video i posted the link to a few post back--that will knock ya sock of when you do the calculations.

Here are mine from the video i speak of.

So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.


Brad

Perhaps you missed this one MH.
What paradox will you add to this?
You will note that the lower voltage value cap is more than half the starting voltage that was in the charged cap,and the second cap has an even higher that half voltage value than the starting cap--and he was using a little DC motor to make the transfer,and it was running the whole time :o

Cant wait to see you explain this one ;)


Brad

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2584 on: May 14, 2016, 06:06:57 PM »
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Offline poynt99

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Re: Joule Thief 101
« Reply #2585 on: May 14, 2016, 06:15:20 PM »
So ,after some testing of my own on the cap to cap transfer,by way of a 12 volt LED in series with the caps,i have found that you do not loose half of your stored energy when doing the transfer.
See diagram below for details and circuit used.

Awaiting for the MH paradox to kick in. :D


Brad

P.S--that should be end total of joules x 2= 900.422mJ,not 900.442mJ

Brad, replace your LED with a silicon diode, and repeat the experiment. Your two caps should be closer in value at the end.

Offline Magluvin

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Re: Joule Thief 101
« Reply #2586 on: May 14, 2016, 06:21:38 PM »
No mistake MH.
The results were the same on every test--> 5 tests done.

Should i make a video showing you this?
Did you watch the video i posted the link to a few post back--that will knock ya sock of when you do the calculations.

Here are mine from the video i speak of.

So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.


Brad

I can see that the inductance in the motor would act similar to the inductor with series diode to get most from 1 cap to the other. So it would be cool to try some things with this where we tried different motors for possibly better initial results.  What I was just thinking of is the difference between draining a battery(or cap) into the motor where the only thing we get out is motor action and we dont charge another cap, vs cap to cap with motor action calculated in as part of the whole energy result. First do the cap to motor only unloaded, then loaded. Not worrying about measuring the motor output yet, do it just to see the difference in time till stop.

Then do the cap to cap and do the test with the motor loaded then unloaded, and see if we lost anything along the way by loading the motor vs unloaded. Get it? like did the other cap not get charged as much when the motor is loaded, etc. I may see what I have and give it a go as it interests me. 
Cool stuff ;)

Maybe there is nothing to it. But the only way is to see. ;)

Mags

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2586 on: May 14, 2016, 06:21:38 PM »
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Offline tinman

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Re: Joule Thief 101
« Reply #2587 on: May 14, 2016, 06:35:04 PM »
Brad, replace your LED with a silicon diode, and repeat the experiment. Your two caps should be closer in value at the end.

I placed a 5k resistor across the LED Poynt,and ended with a higher than half value on each cap.

Did you see the video i linked some posts back,where the guy used super caps,and made the transfer via a small DC motor.
He ended with a lot more than half the starting energy,and ran a DC motor the whole time of the cap to cap transfer.
Results below.

 So 2.52 volts across 350F=1111.32 joules of energy in one cap,and none in the other.

The end result was
Cap 1 -1.468v across 350F= 377.129 joules
Cap 2-1.273v across 350F= 283.593 joules.

Total =660.722 joules
If we double this,we have 1321.444 joules.

So yes,he lost less than the half expected.
But there is also one other thing to take into account-->the little motor was running the whole time,and there for was also dissipating power the whole time current was flowing through it,by way of resistive heat loss.


Brad

Offline MileHigh

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Re: Joule Thief 101
« Reply #2588 on: May 14, 2016, 07:06:34 PM »

Cant wait to see you explain this one ;)

Brad

The answer is simple.  You are deceiving yourself with a pure bait and switch on yourself.  You are using an inductive type of load which by definition facilitates the transfer of energy from one cap to the other with less losses than a resistor, just like Magluvin said.  And you are not doing as complete as possible an energy transfer because the final voltages are not the same.  The less of an energy transfer you do, the less the losses are going to be.

So your experiment has done absolutely nothing to "refute" the classic example of 50% losses when using a resistor.  You are simply doing a different experiment, that's all.

There is no "gotcha," just a completely different experiment.

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2588 on: May 14, 2016, 07:06:34 PM »
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Offline MileHigh

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Re: Joule Thief 101
« Reply #2589 on: May 14, 2016, 07:36:16 PM »
author=MileHigh link=topic=8341.msg483226#msg483226 date=1462588590]

Oh,a conundrum :D -->the MH paradox lol.

Quote
Here is the simple conclusion:  It doesn't matter what value of resistor you use you always lose half the energy.  Hence when you short two caps together with "no resistor" you still lose half the energy.

Once again--very wrong.
You do not loose half of the energy--you loose less than that.
Repeated bench experiments prove this to be the case.

Nope--you dont loose half the enrgy :)

Again, in no uncertain terms, you are dead, dead wrong.  You are comparing apples and oranges, which is nothing more than a silly stunt.

The bigger question to ponder is how is it possible that you don't even realize this for yourself?

Offline Magluvin

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Re: Joule Thief 101
« Reply #2590 on: May 14, 2016, 09:02:21 PM »
yes,,

the condition set forth is just as nonsensical as saying that I have a full bucket and am going to poor it into a bucket with a hole in it,, see I loose some of what ever was in the bucket.

By the way,, it is not the external resistance that eats up the juice,, it is the internal resistance.

There is a good point in that....

In my opinion, if the ideal inductor does have a 100%efficiency where the impedance mechanism is ideal and 100%, adding a resistor in series still wont cause that ideal insuctors internal mechanism to be less than 100% eff and I think no current would flow then either.  So putting a resistor in series with an ideal inductor does not now make the ideal inductor a normal one.  That is a good point.  ;D ;)

Mags

Free Energy | searching for free energy and discussing free energy

Re: Joule Thief 101
« Reply #2590 on: May 14, 2016, 09:02:21 PM »
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Offline MileHigh

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Re: Joule Thief 101
« Reply #2591 on: May 14, 2016, 09:49:48 PM »

By the way,, it is not the external resistance that eats up the juice,, it is the internal resistance.

If you are talking about the standard setup with a resistor between two capacitors, then I don't know what you mean.  The vast majority of the energy is lost in that resistor.

Offline poynt99

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Re: Joule Thief 101
« Reply #2592 on: May 15, 2016, 12:34:56 AM »
As I mentioned Brad, go back to basic principles. Do the basic version of the experiment with just a silicon diode between.

Offline tinman

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Re: Joule Thief 101
« Reply #2593 on: May 15, 2016, 03:04:38 AM »
The answer is simple.  You are deceiving yourself with a pure bait and switch on yourself.  You are using an inductive type of load which by definition facilitates the transfer of energy from one cap to the other with less losses than a resistor, just like Magluvin said.  And you are not doing as complete as possible an energy transfer because the final voltages are not the same.  The less of an energy transfer you do, the less the losses are going to be.

So your experiment has done absolutely nothing to "refute" the classic example of 50% losses when using a resistor.  You are simply doing a different experiment, that's all.

There is no "gotcha," just a completely different experiment.

Dear MH
What kind of a load is an LED?

Oop's.


Brad

Offline tinman

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Re: Joule Thief 101
« Reply #2594 on: May 15, 2016, 08:16:43 AM »
I can see that the inductance in the motor would act similar to the inductor with series diode to get most from 1 cap to the other. So it would be cool to try some things with this where we tried different motors for possibly better initial results.  What I was just thinking of is the difference between draining a battery(or cap) into the motor where the only thing we get out is motor action and we dont charge another cap, vs cap to cap with motor action calculated in as part of the whole energy result. First do the cap to motor only unloaded, then loaded. Not worrying about measuring the motor output yet, do it just to see the difference in time till stop.


Maybe there is nothing to it. But the only way is to see. ;)

Mags


Quote
Then do the cap to cap and do the test with the motor loaded then unloaded, and see if we lost anything along the way by loading the motor vs unloaded. Get it? like did the other cap not get charged as much when the motor is loaded, etc. I may see what I have and give it a go as it interests me. 
Cool stuff ;)

Mag's
I just carried out this very test-a number of times.
I went and bought two new 55 farad caps just for this experiment.
I am using a small DC motor from a drone. The first 5 tests i carried out without a load(propeller removed from motor),the second 5 tests with the propeller on the motor as the load.

Without the load,i ended up with exactly half the voltage on each cap to that of the starting voltage on Cap A.
Start-Cap A=2v, cap B=0v.
End test-cap A=1v, cap B=1v-->half energy lost.

Motor with propeller on--loaded test
Start-cap A =2v, cap B=0v
End test--cap A =1.043v, cap B=1.037v.

As we added a load,we also increase the current flow through the two caps and motor. One would have thought that this means we would loose more of the stored energy to heat,due to resistive losses being higher,due to the higher flow of current.
But seems that is not the case here.

So the next thing to do ,was to find out why we lost less energy by applying a load to the little motor.

I wonder if MH knows the answer?
Why do you think it happened this way Mag's ?.

There is a perfectly good explanation for it ;)


Brad

 

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