IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[eccentrically1]

https://www.besslerwheel.com/forum/faq.php#37

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Using Eccentrically1 as a wedge, let's see what he has to say:

How many more steps must we endure? I'm sure this is a waste of time. Once the experiment is in space, there aren't any means to initiate motion; it's effectively isolated from its environment. The zigzag concept is nullified.

My original post questions were never addressed?

It isn't a reactionless drive or a PMM. How does the blue component start moving? You'd have to include that in your theory. How does the blue component get back through the segment S to reset for another cycle without cancelling the initial motion? Also important to your theory.

And then:

this example of two machines, one with a frictive linear section, and the other with a non-frictive zigzag channel.

The claim is the reactive forces from the black component would be the same in both machines (Fc = F'c), because the zigzag channel would generate resistance without thermal losses, and proving your claim. In the thought experiment, the ideal conditions stated there were no electromagnetic forces. (4th paragraph)

https://mypicxbg.files.wordpress.com/20 ... _01-12.pdf

Given that, the experiment fails at that step. There would be no resistance in either machine if there are no EM forces present. Both blue components would move through without encountering resistance. The linear s section would behave as though it were a zigzag channel, i.e., without friction.

The other inconsistency I note is in the above paper, the black component is allowed to move. In the 3rd link, containing the now-infamous question, the black component isn't allowed to move.

Given just these two issues with your thought experiment (there are others),
the answer to your question is no, you can't choose a suitable combination of (a) magnitude of force of friction, (b) length of segments s and (c) number and shape of zigzags, for which Fc = F'c, Fc > 0, F'c > 0, d = d', d > 0 and d' > 0?"

Given your responses it would seem that you are the one who doesn't understand theoretical or applied mechanics.

Was about to reply too, but agreed with ecc1.

The clock on your reputation is ticking, George1.

[ME]

I don't understand Ecc. I agree on all those replies. Thought it was best to quote your conclusion as a counterpoint to George's attempt to make some point.

Was about to reply too, but agreed with ecc1.

That was page 11. I already replied as in-depth as I thought was feasible for George on page 5 (correction page 8)

A reminder to those concerned about my reputation: Everyone can hit my red dot.
Perhaps revoke the given dot first by hitting that cross-icon first.

<--- around here in the sidebar

[eccentrically1]

I was talking about George’s rep clock ticking.
Art replied first and I thought that would have been the end of it.
Then Julio made an animation that should have really ended it.

[ME]

:-)

Yup I agree, they both made excellent efforts too!
In retrospect I hit Julio's green.

[agor95]

Yes agreed

In retrospect I hit Julio's green.
(En retrospectiva, golpeé el verde de Julio.)

Regards

[George1]

To eccentrically1. (All other participants in this discussion are simply uneducated amateurs in the field of theoretical and applied mechanics (and as if in physics as a whole).)
-------------------------
Hi eccentrically1,
1)What are you talking about? Your attempt for sumulating ignorance and stupidity is quite unskillfull. Are a BIG OIL/BIG MAFIA agent?
2) V' = V" and Vi = Vr. Do you have any theoretical (ONLY THEORETICAL!) objections against the validity of these two equalities? Look again at our post of Apr 05, 2020, 3:48 pm.
Looking forward to your answer.

[eccentrically1]

Yes, I am. I’m a big oil mafia agent.

[George1]

OK. It's your choice. But you didn't answer my question. Do you have any theoretical (ONLY THEORETICAL!) objections against the validity of the two equalities V' = V" and Vi = Vr?

[ME]

It has been a while since I read that homework assignment... please remind us the quantities and units of V', V", Vi and Vr .
-- Described in new words please !! --

While awaiting for the specifics of your formulae, here's my preliminary conclusion:
As far as I recall you tried to compare something one might label as 'Lost energy' and 'Rationed energy'.
While you could seemingly make a match, there is still a theoretical difference.
One will be lost, while the other can (theoretically) be recovered.
Neither situation is reactionless.

[George1]

To eccentrically1.
-----------------------------
I am asking you four times in a row one and same question and four times in a row you avoid answering my question.
You obviously have no arguements against our concept and this follows directly from our 4-steps discussion whose final result (please look again at our post of Apr 05, 2020, 3:48 pm) are the equalities V' = V" and Vi = Vr. These two equalities unambiguously show the invalidity of the law of conservation of mechanical energy and/or the law of conservation of linear momentum.
Looking forward to your answer.

[George1]

To ME.
--------------------
Hi ME,
Thank you for your post.
Please look at our 4-steps discussion with eccentrically1. The final result of this discussion (please look again at our post of Apr 05, 2020, 3:48 pm) are the equalities V' = V" and Vi = Vr. These two equalities unambiguously show the invalidity of the law of conservation of mechanical energy and/or the law of conservation of linear momentum.
Looking forward to your answer.
George

[ME]

The explanations you received throughout this do not compute, so there must be a reason you keep getting stuck.

Thus, could you please give us a story where you explain things in completely new words?
Start from zero.
A picture with arrows is appreciated too.
All I see here is something like "a=b", and "c=d". As if that would mean something - and it doesn't.

So I ask again, please give a specific example with appropriate quantities and units so we can apply some comparative calculus.
It's the best way to see how things get violated and by how much.

Looking forward to your answer.

[George1]

Hi ME,
Thank you for your reply.
We (our team) gathered together and carefully considered our so far correspondence in this forum. And we concluded that it is actually our fault that we did not explain the things in a proper manner. So we decided to use another approach. We would use a step-by-step explaining procedure (step 1, step 2, step 3, ....etc.) as we would not procede to the next step without reaching an agreement on the present step. And here it is.
----------------------------
STEP 1.
Please look at the link https://www.knowledgeuniverseonline.com ... -plane.php
This is the basic example for a motion of a body on a smooth (friction is negligible) inclined plane. The drawing is very illustrative.
The the body moves without friction (as it is a matter of a smooth inclined plane) and the inclined plane is motionless and firmly attached to the ground. The body exerts a vertical downward force G on the inclined plane. At the same time however the inclined plane on its behalf exerts a vertical upward force -G on the body. Besides G=-G, that is, the two forces G and -G are opposite in direction and equal in magnitude.
Do you accept the validity of this STEP 1?
------------------------------
Looking forward to your answer.
George

[ME]

A liter is sometimes a very crowded space. Glad you reached consensus.

A picture with arrows is appreciated too.

Actually any picture showing vector decomposition would have been a great start.

I agree with the math of the inclined plane as shown on that website.
I don't agree with the validity of your STEP 1- addition.

I usually like to motivate my yes/no (I hope you do to), so here it is:

That object on an incline plane is attracted straight down by a gravitational force with the magnitude of m·g.
The structure of the inclined plain is able to withstand that force and reacts by pushing back with its surface.
When you consider that force as being a pressure (Force per Area [units: Pascal]) then it's perhaps easier to understand the direction of this push-back.

The direction it pushes back is perpendicular to this surface - also known as the normal.
Normally the letter 'N' is used to indicate the Normal force, but on that website they use 'R' to indicate a Reaction force.
- as a side note: That's why indicators like V', V'', and Vi are basically meaningless when not properly defined and only vaguely mentioned.

Because this Normal force is only partially countering the gravitational attraction, enough for the object not to sink into the incline, there will be some force left.
So there's no "G=-G" because that would mean that gravity was countered in total by the plane itself. And that's incorrect.

Force decomposition shows that gravity still acts on this object tangent/parallel to the plane with a magnitude of mg·sin(&#952;).
When this is more than static friction of the plane then the object is able to accelerate down the plane.

Please try step 1. again by motivating your yes/no/maybe.

[George1]

Hi ME,
Thank you for your reply.
But you are not reading carefully my posts.
1) Firstly, we are not talking for the present for V', V", Vi and Vr at all! Please pay special attention to this fact!
2) Secondly, please have a look again at the link https://www.knowledgeuniverseonline.com ... -plane.php
3) Thirdly, R is not equal to mg, this is clear. We are not argueing about this fact.
4) Fourthly, G = mg = weight of the body.
4) Fifthly, if you accept the validity the considerations in the above link, then you have to accept the validity of the equality G = - G. The body exerts a vertical downward force G on the inclined plane. At the same time however the inclined plane on its behalf exerts a vertical upward force -G on the body. Besides G=-G, that is, the two forces G and -G are opposite in direction and equal in magnitude.Do you accept the validity of the considerations in this item 4?
Looking forward to your answer.
George