Joule Thief 101

[Magluvin]

Ive thought of that many times. Even argued it a few times.


If we had 2 air tanks, 1 full at say 100psi and 1 empty.
Then we dump the full into the empty till they are equal. What did we lose?  We lost pressure.

Say each tank is 1cu ft. The one thats full is able to accomplish a certain amount of work over time, depending how you release it. But if we dump that full tank into the other till both are equal, we now have the total amount of air in 2 cu ft. 50psi each? Now if we calculated the total amount of work that can be done with the total of what is in the 2 tanks, we will find that we could do more work with the full one cu ft tank than we can with 2 1cu ft tanks described above after the dump and equalization.

Now we could introduce an air motor, motor driven with air that can also pump, and we put a flywheel on the motor shaft, that would enable us to release the full tank into the empty tank through the air motor/flywheel and be able to transfer most of that air in the first tank into the empty tank. Like having a full cap and dumping it into another empty cap through an inductor and a diode and get most of the full caps energy into the empty cap, minus minimal losses.

They say that if we dump a cap into an empty cap that the reason for the 50% loss is due to resistance in the transfer. I still fail to believe that. I believe we have just wasted the full caps energy by dividing its voltage/pressure into a cap/container that is twice as large. Where we lost it is in reduced voltage/pressure, not in resistance losses. We lost it in a stupid way. We didnt do anything with the energy transfer.   

Example.....

We have one full tank and one empty. We dump the full tank into the empty tank and the pressures equalize.

Now, we have our full tank and dump it into the empty tank, but we use the air motor to make the transfer. Now we can do work with the air motor until the 2 tanks are equalized.  And we still have the same amount of pressure in each tank as when we did it the stupid way.

Im still trying to figure out the 'reasoning for blaming' the loss on resistance/heat when dumping a full cap into and empty one of the same value.    There is something wrong there.

That would be an interesting debate thread.

Mags

Back when I argued this, I had done simulations where if I have 2 10uf caps, 1 with 10v and one with 0, then made the transfer with an inductor, and diode, and timed the switch on, then switch off, I could get just above 7v in each cap.  Now the 2 caps in parallel at 7v is just about equal in energy as the 1 cap with 10v.  But if we make the comparison of the caps vs the air tanks, 10v in one cap direct dump into the empty cap, we get 5v each, and the air tanks, 100psi of 1 tank dumped into the empty tank we get 50psi in each. 

So how do we get the 2 tanks to be just above or at 70psi each from the single 100psi tank? 

Using the air pump with a flywheel, we dump the full tank till we have 70psi in the source tank then shut of the source tank and let the flywheel pump in air from the outside into the second tank.  And the second tank should be near 70psi, considering losses. The same with the caps. We disconnect the source cap at 7v and let the inductor with a diode finish charging the other cap to right around 7v

So doing the direct dump is just releasing pressure/voltage in a way that we didnt do anything with the transfer action during the dump. The energy wasnt lost in heat. It was lost by releasing pressure into a container that was twice as large. I call it stupid losses.

Mags

[Magluvin]

Trump might indeed have the "football" so let's hope being the most powerful man on Earth chills him out a bit.

I saw a show the other night saying that Trump will be like Biff Tanem from back to the future 2. Pics of the 2 were a bit uncanny.

Mags

[Pirate88179]

I saw a show the other night saying that Trump will be like Biff Tanem from back to the future 2. Pics of the 2 were a bit uncanny.

Mags

Good.  Maybe Putin will stop buzzing our carriers and China will quit stopping our ships because they might be afraid of what he might do.  We will get some respect back...hopefully.  We will see.

Bill

[tinman]

I am just posting this to make it clear that I am no liar as has been alleged many times.




.

MileHigh

Me:  It's just like you saying something about your coil being sealed in epoxy therefore you couldn't see the direction of the windings therefore you couldn't add the dot convention to your schematic.  Thus we are led to conclude that you are incapable of inventing a way to do your own test?  Another jaw-dropping moment.

And that right there is a lie.

Me--  Another lie,as i clearly stated as to how i could define the dot convention with that post.
Please stop your lies MH.

Brad, you said this on April 18th:

I cannot determine the winding direction of the primary winding of the transformer,as it is taped and lacquered,and the wire very fine. Guess i could use a compass and DC current to work it out.

Proving beyond doubt that you lied when you stated--->Thus we are led to conclude that you are incapable of inventing a way to do your own test,when i clearly stated how it could be done.

I did not see any clear statement for how you could "define" the dot convention.

Either you need new glasses,or you are lying once again
The quote you posted of my reply !once again!-->Guess i could use a compass and DC current to work it out

I am no liar, but based on what you are stating above I am forced to conclude that you are unable to come up with a ridiculously simple test to determine the dot convention for a transformer.  Assuming that I am correct, that's a real shame.

Yes you are,and you have now been upgraded to a pathological liar.
If you think supplying the coil with a DC voltage and current,and using a compass to see the field produced at each end of the coil is hard,then you need help--and fast.

Yes, you do indeed mention a compass, and I suppose that if push comes to shove you could use a compass to determine the dot convention for a transformer that puts magnetic flux in the open air, although it will not work for a closed-flux-loop transformer.

What do you mean if !push comes to shove! ?. It's a very simple way of working out winding direction of a coil.
And before you go off into your own little world,and try to twist things around again to save your sorry ass--we were not using a closed-flux-loop transformer,we were using air core coils,and so my way of determining the winding direction of the coil was very correct.

Considering that there is a pretty obvious and self-evident way to determine the dot convention that is so easy and so quick that will work for any type of transformer with standard bench equipment, your compass remark did not stick with me.

I dont give a rat's ass about what sticks with you. What i stated was correct in every way.
Please show us an easier way to determine the winding direction of the wire on an air core coil.
I have to place two clip lead's onto the coil from my power supply,and place a compass at one end of the coil--thats really hard-isnt it MH

 

What I really remembered was the shock in reading you say, "I do not know the dot convention,as the primary is sealed in apoxy (sic) resin."

That's !epoxy! by the way.
Why so shocked MH,as i never bothered to check it out.

I only made this posting to make it clear that I am no liar.

Well you failed at that -didnt you. You posted exactly what i said,and that proves that your statements are lies.

I suppose that technically Brad is correct, and you really can use a compass to determine the dot convention for an air-core-type transformer if you really want to go that strange route.

Technically right  lol. It was spot on MH,and there is nothing technical about using a DC current and compass to determine turn direction of a coil.
And yes--you know just as well as everyone else,that i was using an !!AIR CORE!! coil.

   

Or Brad will have a miraculous alternative explanation that clears up the misunderstanding

There is no misunderstanding MH--you just lied through your teeth again,and then made this post which pushed you into the very hole you dug your self.

Once again you tried to make me look like a fool,and once again you failed to do so. The biggest hoot is,you just proved to everyone here that you are indeed a liar--a pathological liar.

Pathological Liar--The individual may be aware they are lying, or may believe they are telling the truth. Sometimes however, the individual may be lying to make their life seem more exciting when in reality they believe their life is unpleasant or boring


Get help MH.


Brad

[tinman]

Ive thought of that many times. Even argued it a few times.


If we had 2 air tanks, 1 full at say 100psi and 1 empty.
Then we dump the full into the empty till they are equal. What did we lose?  We lost pressure.

Say each tank is 1cu ft. The one thats full is able to accomplish a certain amount of work over time, depending how you release it. But if we dump that full tank into the other till both are equal, we now have the total amount of air in 2 cu ft. 50psi each? Now if we calculated the total amount of work that can be done with the total of what is in the 2 tanks, we will find that we could do more work with the full one cu ft tank than we can with 2 1cu ft tanks described above after the dump and equalization.

Now we could introduce an air motor, motor driven with air that can also pump, and we put a flywheel on the motor shaft, that would enable us to release the full tank into the empty tank through the air motor/flywheel and be able to transfer most of that air in the first tank into the empty tank. Like having a full cap and dumping it into another empty cap through an inductor and a diode and get most of the full caps energy into the empty cap, minus minimal losses.

They say that if we dump a cap into an empty cap that the reason for the 50% loss is due to resistance in the transfer. I still fail to believe that. I believe we have just wasted the full caps energy by dividing its voltage/pressure into a cap/container that is twice as large. Where we lost it is in reduced voltage/pressure, not in resistance losses. We lost it in a stupid way. We didnt do anything with the energy transfer.   

Example.....

We have one full tank and one empty. We dump the full tank into the empty tank and the pressures equalize.

Now, we have our full tank and dump it into the empty tank, but we use the air motor to make the transfer. Now we can do work with the air motor until the 2 tanks are equalized.  And we still have the same amount of pressure in each tank as when we did it the stupid way.

Im still trying to figure out the 'reasoning for blaming' the loss on resistance/heat when dumping a full cap into and empty one of the same value.    There is something wrong there.

That would be an interesting debate thread.

Mags

Referring to the two air tanks Mag's-you are very correct,in that you can end up with more stored energy than what you started with.
I did such testing with MarkE working out the numbers as we went through each stage of testing.
At the last test,i ended up with 16% more stored energy in the two tank's,than what i started out with in the one tank.
This was(and can be again)proven beyond doubt,and it can be done with a simple venturi setup.

Doing it your way,you can indeed power a load without loosing or using any of the stored energy in the tanks.


Brad