How this was done in 1821.....

[Omnibus]

@Dingus Mingus,

On the contrary. The term [kinetic + rotational + energy losses] accounts for that. Like I said, read what I wrote, try to understand it and don't fantasize.

[Dingus Mungus]

Fantasize about what???

I never said the SMOT or your equation doesn't work,
I was only offering a more standard physics equation...

I also asked two questions: Is the "ball" pulled in to the gate by flux and accelerated out of the elevated exit gate? Do you have a good link to pictures and explanation of your measurements? I would rather read your published notes or data, not a random arguement in a thread.

~Dingus Mungus

[Omnibus]

I was only offering a more standard physics equation...

As I said, the equation you gave is accounted for in the term [keintic + ?] in my analysis. Your equation isn?t the solution to the problem, it describes only part of the output energy (specifically, part of the excess energy).

I?d like to have my arguments published but it?s impossible. Publishing is out of the question because it would open the floodgate in a direction which absolutely goes against the core of the current agenda in Physics. So, that?s out.

The arguments I gave in this thread should be enough.

Also, to answer your question, it seems that the part of +Mb which at C turns into kinetic energy (in addition to +mgh2) allows the ball to escape the sticky spot and to get ?recharged? at A. That is something to be had in mind in practical engineering applications.

[Dingus Mungus]

(specifically, part of the excess energy)
No... I was describing Joules required to maintain velocity up a slope. No excess energy described. Excess would require a new equation and the addition of a second variable for exit velocity. I compulsively want to write it... But its not worth the time.

(I?d like to have my arguments published but it?s impossible.)
I only require a link to a site that shows your measurements and device.

(The arguments I gave in this thread should be enough.)
Well I'll pour over it again, but it was only an equation, no measurements.

(Also, to answer your question, it seems that the part of +Mb which at C turns into kinetic energy (in addition to +mgh2) allows the ball to escape the sticky spot and to get ?recharged? at A.)

ok...
At what speed does it exit the gate though?
Faster... slower... same velocity?

[Omnibus]

(specifically, part of the excess energy)
No... I was describing Joules required to maintain velocity up a slope. No excess energy described. Excess would require a new equation and the addition of a second variable for exit velocity. I compulsively want to write it... But its not worth the time.

Never mind what you were describing because it?s beside the point. Read my analysis and, as I said, don?t fantasize and don?t try to confuse the issue.

(I?d like to have my arguments published but it?s impossible.)
I only require a link to a site that shows your measurements and device.

No such link. As I said, the analysis presented in this thread shoud suffice.

(The arguments I gave in this thread should be enough.)
Well I'll pour over it again, but it was only an equation, no measurements.

What measurements? What I?m presenting is a conclusive theoretical analysis. Anyone can measure heights h1, h2 and m.

(Also, to answer your question, it seems that the part of +Mb which at C turns into kinetic energy (in addition to +mgh2) allows the ball to escape the sticky spot and to get ?recharged? at A.)

Huh ok...
At what speed does it exit the gate though?
Faster... slower... same velocity?

The speed doesn?t matter in this analysis. What matters is that there is kinetic energy which is part of the excess energy. Everything else is just engineering details of no consequence for the main point of the analysis.