Re-Inventing The Wheel-Part1-Clemente_Figuera

[Aaron]

posting

UFO, all working good now?

[Ufopolitics]

Too bad. Looks like it works for me. I'll give it a try with this.

bi

edit: I sent a message to admin on your behalf because you may not be able to do that yourself due to the problem.

Thanks!...seems to be my end connection.


Ufopolitics

[bistander]

can't post...

Too bad. Looks like it works for me. I'll give it a try with this.

bi

edit: I sent a message to admin on your behalf because you may not be able to do that yourself due to the problem.

[Ufopolitics]

can't post...

[bistander]

Coil power

However, you mentioned above that "All voltage is dropped across the resistance" as a "One time operation" and converted into heat...is this so?...or is it a continuous renovating process...as to keep both coil terminals with the required voltage...right?

Hi Ufo,

I did say "All voltage is dropped across the resistance". I did not say "One time operation". I do not understand that.

Voltage is an instantaneous value, as is power. In the DC static or steady state case, the voltage is the same value at each instant. Assuming the coil resistance is constant, then the power is constant or the same value at each instant. And for each and every instant in time when that voltage is present across the coil, it is dropped across the coil resistance and the resulting power is converted to waste heat in the coil resistance for the duration. If in fact the voltage and resistance were constant for that duration, then the duration time multiplied by the power yields the energy (or heat quantity) converted in the coil.

All of that has nothing to do with the magnetic field.

I hope that helps,

bi

[marathonman]

I Agree

12 volt 10 amp = 1.2 ohm
120 volt 100 amp = 1.2 ohm

can't argue with that.

Solarlab;

There is No Bemf in the Figuera device what so ever nor is it a resonant system as magnetic resistance is dynamic. (Dynamic Inductor in constant movement).

MM

[Ufopolitics]

Hi Ufo,

You didn't acknowledge that your example of applying 12 volts to a coil and applying 120 volts with both times getting 2 amps was impossible so I am guessing you still don't understand.

Let me try a different way of explaining that. Let's say you have a 1/2 inch garden hose 100 feet long. Now we have a pump supplying water at 50 psi to one end of the hose. When we measure the flow of water coming out the other end we get 10 gallons per minute. Now what happens if we raise the water pressure? The flow will go up. If the size and length of the hose is not changed then any change in pressure will change the flow. And conversely if we go to a smaller size hose our pressure will have to go up to get the same 10 gallons per minute of flow.

This is exactly the same as your coil. If you want a certain amount of current to flow then you have to apply a certain amount of voltage. Changing the voltage will change the current. So you can't have the same coil and apply 12 volts and get 2 amps of current and then apply 120 volts and still get 2 amps of current. If you want a coil that takes 120 volts at 2 amps then you have to make another coil. You can't use the one that takes 12 volts at 2 amps.

And Bi is still correct. The current flowing through the turns of the coil are all that determine the field strength. But it is understood you have to apply voltage to get the current to flow. And the amount of voltage needed is again determined by the coil.

Respectfully,
Carroll

Hi Citfta,

I would say same as I wrote to Bistander...excellent explanations...elegant!

But I know all of that...what I do not understand is to why leave V out of the Field Strength dependency, when we are considering wire number of turns which equals wire length...then voltage.

Bistander says "voltage drops down as becomes heat across coil resistance"...fine and dandy!...but this is not just a "crash and burn" one time operation...am I right?...it is a constant renewing process isn't it?

Anyways, I wanna thank you both for taking your time and your patience explaining to such plain ignorant and stubborn guy like me...really appreciate it.


Regards


Ufopolitics

[Ufopolitics]

Hi Ufo,

No, not insignificant. Exactly what we're takling about here. I know it defies first glance logic, but all the voltage across the coil is dropped across the resistance or I*R = Coil Voltage. And I^2*R = Power dissipated in the coil resistance or converted to heat, 100%. So if all the voltage is dropped across the resistance and all the power goes into waste heat, what's left to produce the magnetic field or mmf? Only the current. And only the current matters to the magnetic field.

The paragraph above addresses DC or steady state. When the coil is first energized there is power or energy transferred from coil and its excitation source into and stored in the magnetic field. This energy is equal to one half the inductance times the current squared. Again, independent of voltage.

Regards,

bi

Hello Bistander,


Thanks, that was a nice and quite elegant explanation

However, you mentioned above that "All voltage is dropped across the resistance" as a "One time operation" and converted into heat...is this so?...or is it a continuous renovating process...as to keep both coil terminals with the required voltage...right?

In Spanish terminology (related to EE) Voltage is referred to as "Tension" between two points on a conductor (not specifying the "wire sculpting"...if curled or squared out in a coil shape) ..as amperage/current relates to conductor thickness, gauge or cross section...simple terms...simple understanding.

Even though I clearly understood your excellent explanation...thanks

Still I find some empty and loose explanations, where so simple concepts get into confusions.

I see voltage exactly as the "driving force" of currents, and currents as the density of the electric flow...where one depends upon the other at all times no matter the functions both get "assigned".


Thanks and regards


Ufopolitics

[citfta]

Hi Ufo,

You didn't acknowledge that your example of applying 12 volts to a coil and applying 120 volts with both times getting 2 amps was impossible so I am guessing you still don't understand.

Let me try a different way of explaining that. Let's say you have a 1/2 inch garden hose 100 feet long. Now we have a pump supplying water at 50 psi to one end of the hose. When we measure the flow of water coming out the other end we get 10 gallons per minute. Now what happens if we raise the water pressure? The flow will go up. If the size and length of the hose is not changed then any change in pressure will change the flow. And conversely if we go to a smaller size hose our pressure will have to go up to get the same 10 gallons per minute of flow.

This is exactly the same as your coil. If you want a certain amount of current to flow then you have to apply a certain amount of voltage. Changing the voltage will change the current. So you can't have the same coil and apply 12 volts and get 2 amps of current and then apply 120 volts and still get 2 amps of current. If you want a coil that takes 120 volts at 2 amps then you have to make another coil. You can't use the one that takes 12 volts at 2 amps.

And Bi is still correct. The current flowing through the turns of the coil are all that determine the field strength. But it is understood you have to apply voltage to get the current to flow. And the amount of voltage needed is again determined by the coil.

Respectfully,
Carroll

[bistander]

Coil voltage

Yes, in the non-superconductoring world, the current in a coil which produces the mmf is dependent on the voltage applied to the coil per Ohm's Law. However the voltage to mmf or flux relationship is purely coencenedence. Easy proof is to maintain constant current in the coil as it heats up. You must increase applied voltage as the resistivity of the copper increases. But the field remains unchanged with this voltage change.

Bistander, the changes according to temperature in a coil related to resistance increase, therefore voltage increase are completely insignificant compared to what we are talking about here.

Hi Ufo,

No, not insignificant. Exactly what we're takling about here. I know it defies first glance logic, but all the voltage across the coil is dropped across the resistance or I*R = Coil Voltage. And I^2*R = Power dissipated in the coil resistance or converted to heat, 100%. So if all the voltage is dropped across the resistance and all the power goes into waste heat, what's left to produce the magnetic field or mmf? Only the current. And only the current matters to the magnetic field.

The paragraph above addresses DC or steady state. When the coil is first energized there is power or energy transferred from coil and its excitation source into and stored in the magnetic field. This energy is equal to one half the inductance times the current squared. Again, independent of voltage.

Regards,

bi

[Solarlab]

F. Y. I.

Just a quick jog/comment...

- BEMF (back electromotive force) => efficiency < 1;

- Asynchronous system (eliminate BEMF) => efficiency = or > 1;

- Figuera method => mechanical losses = 0.

And, - high Q resonance, like momentum (e.g. a swing), will remain "in motion" with
very little, or no, input for a considerable time ~~ motion is "potential" turned "kinetic"...
And If - the system resonance remains undisturbed (no BEMF - load is isolated).

Two cents worth...!

((system LR delay [filter] ~ ~ somehow uses BEMF to advantage [timing] ???))

FIN

[Ufopolitics]

Hi Ufo,

What you are suggesting is impossible. If you are using the exact same coil you can't supply it with 12 volts and get the same current through it as you would if you supply it with 120 volts. That is clearly impossible. If you supply it with 120 volts the current will be 10 times the current it would have at 12 volts. No matter what kind of supply you use you can't change that.

Also they are correct that voltage has nothing to do with the field strength of an electromagnet. The formula clearly shows only the number of turns of wire and the amount of current going through the wire will affect the field strength. Voltage only enters the picture as that is what it takes to get a certain amount of current to flow through the coil. Of course if you vary the voltage you vary the current but only the change in current affects the field strength.

Respectfully,
Carroll

Hello and thanks for answering my question to Bistander...

See, honestly I find zero logic in taking off voltage completely out of the Field strength.

If I understand clearly...Number of Turns Equals Length of Wire...and so, Length of any given conductor relates directly
to Voltage as the difference between two points in length.

So here we give another concept...just because the distance is converted into "turns" no longer applies to Voltage?


Regards



Ufopolitics

[citfta]

Bistander,

Please let's forget about "Laws" and let's use a simple Black and White example.

Say we have a Coil-Core which wire gauge and number of turns configuration allows a Max Input rating of 175V and 2 Amps during operation ...and so it produces a measurable X Magnetic Field...and so based on what Hanon and you are saying...is that since only Amperage (Current) is what generates the Magnetic Field Strength (understood Volume as well) and not Voltage whatsoever...

Resuming, what you two are saying is that the same Coil above...just using 12 Volts and the same current (2 Amps) would produce exactly the same "X" field strength and volume as when it was fed by 120V/2 Amps?

And yes, we are talking about straight DC here...no pulsed DC, no AC.

Is that what you are saying Bistander, by backing up whatever Hanon wrote?

About previous question...just forget about it...you will see soon what I meant.


Regards


Ufopolitics

Hi Ufo,

What you are suggesting is impossible. If you are using the exact same coil you can't supply it with 12 volts and get the same current through it as you would if you supply it with 120 volts. That is clearly impossible. If you supply it with 120 volts the current will be 10 times the current it would have at 12 volts. No matter what kind of supply you use you can't change that.

Also they are correct that voltage has nothing to do with the field strength of an electromagnet. The formula clearly shows only the number of turns of wire and the amount of current going through the wire will affect the field strength. Voltage only enters the picture as that is what it takes to get a certain amount of current to flow through the coil. Of course if you vary the voltage you vary the current but only the change in current affects the field strength.

Respectfully,
Carroll

[Ufopolitics]

I tried to help and tried to be polite, but when insults appear there are no other way that say the truth.

Ufopolitics: you have no idea of Physics principles as you are continuously showing post after post. That's why you hate anyone who speak in those terms. You are completely ignorant and all your test with your Figuera are doomed to fail as consequence of your limited theretical knowledge. You are trying to control current and voltage at the same time. Even a highschool boy know that this is impossible. That's the reason of your results. You are not able to apply Ohms Law to a real system. I know my limitations, but you dont know them and you do not want to know them.

Instead of investing in expensive PSUs that you do not know how to use it I recommend you to invest in your brain version , and update it from version 0.1 to version 0.2. You can do that easily, because I am afraid that your huge huge ego is impossible to change and it will be with until the last day. I guess you do not get along well with people, even if they try to help you.

I wont answer to any further insult. I just came here again to show interesting ideas not to argue with you. Keep on trying to control voltage and current at the same time with your marvellous PSU. Good luck. Ahh, my last suggestion: use batteries in your Figuera setup.

Hanon,

You obviously have a very deep and profound misconception about "insult" meaning.
Do you really consider an insult what I wrote in your quoted text?
Maybe it is all about your very poor understanding of the English Language....let me think of it that way...



Keep going on...you surely find Figura's "code"


Good luck



Ufopolitics

[hanon1492]

I knew that...that is the main reason why all your writings are only based on airy (inflated) mathematics...since your practical experience bolts down to zero...Oh, excuse moi...I forgot about the moving piece of clear paper...and the Variac and two bulbs deal...

I tried to help and tried to be polite, but when insults appear there are no other way that say the truth.

Ufopolitics: you have no idea of Physics principles as you are continuously showing post after post. That's why you hate anyone who speak in those terms. You are completely ignorant and all your test with your Figuera are doomed to fail as consequence of your limited theretical knowledge. You are trying to control current and voltage at the same time. Even a highschool boy know that this is impossible. That's the reason of your results. You are not able to apply Ohms Law to a real system. I know my limitations, but you dont know them and you do not want to know them.

Instead of investing in expensive PSUs that you do not know how to use it I recommend you to invest in your brain version , and update it from version 0.1 to version 0.2. You can do that easily, because I am afraid that your huge huge ego is impossible to change and it will be with until the last day. I guess you do not get along well with people, even if they try to help you.

I wont answer to any further insult. I just came here again to show interesting ideas not to argue with you. Keep on trying to control voltage and current at the same time with your marvellous PSU. Good luck. Ahh, my last suggestion: use batteries in your Figuera setup.