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Overunity Machines Forum



SMOT! - (previously about the OC MPMM)

Started by rotorhead, October 03, 2007, 11:01:31 PM

Previous topic - Next topic

0 Members and 17 Guests are viewing this topic.

Omnibus

@tinu,

You'll find the answer to your question in @modervador's post.

modervador

Quote from: Omnibus on January 19, 2008, 12:03:37 PM
Correct. Despite the fact that in both cases that follow the hand lifts the ball from A to B we have:

1) Ea(final) = (mgh1 + mgh1 + Kc) = (Ma + other energies)
This is observed in the discussed case.

OK, we have agreement! This is quite a step for me.
More later.

Omnibus

Quote from: modervador on January 19, 2008, 01:02:33 PM
Quote from: Omnibus on January 19, 2008, 12:03:37 PM
Correct. Despite the fact that in both cases that follow the hand lifts the ball from A to B we have:

1) Ea(final) = (mgh1 + mgh1 + Kc) = (Ma + other energies)
This is observed in the discussed case.

OK, we have agreement! This is quite a step for me.
More later.

This quotation is incomplete. My post should be quoted in full:

QuoteCorrect. Despite the fact that in both cases that follow the hand lifts the ball from A to B we have:


1) CoE violated:

Ea(final) = (mgh1 + mgh1 + Kc) = (Ma + other energies)

This is observed in the discussed case.



2) CoE obeyed:

Ea(final) = (mgh1 + Mb) ? (mgh1 ? (Ma ? Mb)) = Ma

This is not observed in the discussed case.

There should be agreement with the second part of my post as well:

Low-Q

Quote from: Omnibus on January 18, 2008, 04:45:35 PM
@tinu,

Focus on the fact that at C the ball has energy of the amount (mgh1 + mgh2 + Kc) which is available to be transferred (and is actually transferred when the ball it at A) into other kinds of energy. It is this energy which the ball has at C which the CoE forbids the ball to have available because all the energy that was imparted to it was only the amount (mgh1 - (Ma - Mb)).

What the exact energies are which the energy (mgh1 + mgh2 + Kc) turns into when back at A, including energy Ma, is a detail and isn't part of the argument.
What is the practical difference between mgh1 and mgh2? Why are you excluding mgh2 when talking about the energy in point C?
I ask because I'm a little confused. Shouldn't the energy imparted to the ball also include mgh2 and be something like this: Mc = [mgh2 - mgh1 - (Ma - Mb)] , as the ball in fact is ending up in point C?
Or is it so that mgh2 is the excess energy in the path A, B, C? If so, why?

Vidar

modervador

Quote from: Omnibus on January 19, 2008, 01:12:01 PM

This quotation is incomplete. My post should be quoted in full:
. . .
There should be agreement with the second part of my post as well:

I agree with the substance of both parts:

Ea(final) = (mgh1 + mgh1 + Kc) = (Ma + other energies) is observed in the discussed case.

Ea(final) = (mgh1 + Mb) ? (mgh1 ? (Ma ? Mb)) = Ma is not observed in the discussed case.

I wished to delay consideration of CoE issues (more later). I was trying to be brief and didn't think that trimming part of the quote, which was there further up the page, altered the meaning or impact of the quote. Sorry for the confusion.