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Overunity Machines Forum



Quantum Energy Generator (QEG) Open Sourced (by HopeGirl)

Started by madddann, March 26, 2014, 09:42:27 PM

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MarkE

Quote from: verpies on May 05, 2014, 08:35:20 PM
Well, at least we had an intelligent conversation.  I wish that there was more of that on this forum.

To tell you the truth I don't feel comfortable with the Lenz law being in conflict with the C.o.E. either.

I watched 3 videos about the Lenz law from Lasseviren1 and I like the way he expresses himself.
I noticed that he wrongly used B where he should have used Φ, because a "hoop wants to minimize" the change in total flux, not a change in flux density. 
The "Status Quo" that he mentions should refer to total flux (Φ) through the hoop - not the flux density (B) through the hoop.

I wish I had his email to talk to him, because I am not registered on YT and never will be.
Lenz' Law enforces CoE for Faraday's Law of Induction.

F_Brown

Quote from: Vortex1 on May 05, 2014, 08:48:44 PM
Would a tape wound core work in this application? I understand there are articulations in the center ring, but maybe there are ways around this. The tooling cost would probably be a lot less than a custom die or laser cut laminations. Tapewound cores are often used in large Variacs and toroidal power transformers.

How much does the core weigh?

The rotor and core laminations total 33 kgs.  The windings add another 20 kgs.

I tend to doubt a tape wound core would work.  I thought about that too.

verpies

Quote from: MarkE on May 05, 2014, 08:50:42 PM
Lenz' Law enforces CoE for Faraday's Law of Induction.
Yes it does, but sth is different in this situation when a closed coil is used with movable ferrite and KE is accounted for.

In an ideal shorted coil, any non-zero EMF (ℰ) would result in infinite current because I=ℰ/R.
So Faraday's Law of Induction (ℰ = -dΦ/dt) does not seem to apply in case of ideal shorted coils because the Lenz law always keeps dΦ/dt=0.

MarkE

Quote from: verpies on May 06, 2014, 03:57:50 AM
Yes it does, but sth is different in this situation when a closed coil is used with movable ferrite and KE is accounted for.

In an ideal shorted coil, any non-zero EMF (ℰ) would result in infinite current because I=ℰ/R.
So Faraday's Law of Induction (ℰ = -dΦ/dt) does not seem to apply in case of ideal shorted coils because the Lenz law always keeps dΦ/dt=0.
Lenz' Law only states the direction of the induced EMF that results from Faraday's Law of Induction.  Lenz' Law is the "-" sign in:  ℰ = -dΦ/dt.

In the case of a superconducting ring, there is no resistance, but there is inductance.  The current does not build to an infinite value.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2

Consider that in the familiar experiment of a strong magnet suspended above a superconducting plate, the Lorentz force is exactly equal to the acceleration due to gravity.  The magnet does not fall, and it does not go shooting towards the sky.  The induced current generates a finite magnetic flux.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/maglev.html



verpies

Quote from: MarkE on May 06, 2014, 05:57:15 AM
Lenz' Law only states the direction of the induced EMF that results from Faraday's Law of Induction.  Lenz' Law is the "-" sign in:  ℰ = -dΦ/dt.
Not only.  It also refers to the magnitude of the induced current.

Quote from: MarkE on May 06, 2014, 05:57:15 AM
Quote from: verpies on May 06, 2014, 03:57:50 AM
In an ideal shorted coil, any non-zero EMF (ℰ) would result in infinite current because I=ℰ/R.
In the case of a superconducting ring, there is no resistance, but there is inductance.  The current does not build to an infinite value.
I know.  That's why I used the conditional word "would" to illustrate the absurd proposition of non-zero EMF in a zero resistance loop.

Quote from: MarkE on May 06, 2014, 05:57:15 AM
Consider that in the familiar experiment of a strong magnet suspended above a superconducting plate, the Lorentz force is exactly equal to the acceleration due to gravity.  The magnet does not fall, and it does not go shooting towards the sky.  The induced current generates a finite magnetic flux.
Again, you do not need to prove this to me because I already know that induced flux is finite.

My point was another one.
Namely, that a closed superconducting loop maintains constant flux through its crossection.
Constant flux also means that dΦ/dt=0 and that EMF is zero.

Instead of proving the obvious to me, you should concentrate on answering this interesting question:
Q: "A magnet is pulled out of a shorted superconducting aircoil.  Does the magnitude of the final current induced in that coil depend on how quickly the magnet is pulled out ?".