IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[George1]

Hi guys,
Please have a look again at the first post of ours in this topic. Let us make the following comparison. (We will repeat some parts of our previous posts.)
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1) Please have a look at Fig. 4, Fig. 5, Fig. 5A and Fig. 6.
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1A) THERE IS NO FRICTION. (Modern technologies allow to reduce practically as much as you want the experimental error, related to friction.)
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1B) Let us assume that the black component is fixed motionless somewhere to some obstacle and is not able to move. In this case it is evident that after entering the zigzag channel section the blue component will decelerate. And it is also evident that the deceleration d (its absolute value and its mean value) will be bigger than zero, that is, d > 0.
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1C) It is evident that if the blue component enters the zigzag channel section, then the black component exerts force Fc on the obstacle, which does not allow the black component's linear motion. It is evident that the direction of Fc coincides with the direction of Vo. It is also evident that the force Fc (its absolute value and its mean value) is bigger than zero, that is, Fc > 0.
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2) Please have a look at Fig. 1, Fig. 2, Fig. 2A and Fig. 3.
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2A) THERE IS FRICTION in the linear channel s-segment section.
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2B) Let us assume that the black component is fixed motionless somewhere to some obstacle and is not able to move. In this case it is evident that after entering the linear channel s-segment section the blue component will decelerate. And it is also evident that the deceleration d' (its absolute value and its mean value) will be bigger than zero, that is, d' > 0.
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2C) It is evident that if the blue component enters the linear channel s-segment section, then the black component exerts force F'c on the obstacle, which does not allow the black component's linear motion. It is evident that the direction of F'c coincides with the direction of Vo. It is also evident that the force F'c (its absolute value and its mean value) is bigger than zero, that is, F'c > 0.
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3) Please compare item 1 with item 2. It is evident that we can always 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. Therefore it is evident that zigzags successfully imitate resistance, which is identical to friction, but without generating heat. And if we use this simple fact in the experiments, described in Figs. 1 - 6, then we can conclude again that:
a) the law of conservation of linear momentum is not correct;
b) the law of conservation of mechanical energy is not correct;
c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.
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Let us stress upon the fact that IN GENERAL both the law of conservation of mechanical energy and the law of conservation of linear momentum are absolutely true and correct. There is no doubt about this. But any rule/law has its exceptions and there is nothing special, disturbing and tragic in this fact.
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Looking forward to your answer.
Regards,
George

[ME]

Well ok, here one of the many possible remarks on your homework assignment:

I think it is obvious that formula 3 is the same as formula 7. With "the same" I mean: an exact copy.
Obviously the only difference between formula 4 and formula 8 is your factor "Q".
From this it is evident that when ignoring your "Q" --because you obviously choose to ignore friction in your turning instance-- should therefore simply imply that you need to ignore your "Q" in all sliding instances for allowing a fair comparison.
Hence from that 'observation' it seems obvious that it is evident that the set of formula 3 and formula 4 are just exactly the same as the set of formula 7 and formula 8.
Therefore it is evident that your explanation is actually just a distraction to remove that "Q". It's not even an excuse.
And from that we can simply conclude that you found no discrepancy in those conservation laws.
Also important, writing a sentence that friction is not present does not imply that force-decomposition (like in your figure 5a) becomes obsolete....obviously!

Actually there is nothing "evident" about your "conclusion" just because you keep repeating the word "evident".
Perhaps I misunderstood, so then we likely need a better explanation why friction appears at one place and not the other.
We don't learn those things from your paper. It's not that evident.

[George1]

Hi ME,
Thank you for your reply.
1) First of all please excuse me for my often using of word "evident".
2) Yes, may be I have not explained clearly enough all aspects of the situation. There is friction in the segments s in straight-line modification (Figs. 1-3) and there is no friction in the zigzag modification (Figs. 4-6). (For Figs. 1-6 and the related text please refer to the first of our posts in this topic.) Actually the zigzag generates resistance, which is identical to friction, but without generating heat. So we can always 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. It's simple.
Looking forward to your answer.
Regards,
George

[ME]

You can repeat, but still doesn't explain why it doesn't generate heat when it moves in zig-zags on the same type of rails.
I think: the same type of rails = the same type of friction.
And that's on-top of the zig-zag resistance you just left out of the equation?
It's apparently not simple.

So, I don't know what to make of that.
What happens in the intermediate situation when you replace your Sine-shaped zigzag pattern by straight line segments? Like: /\/\/\/\/
Perhaps for simplicity reasons just ignore the sharp cornering and find out what happens when it's on a slanted track.
Maybe you can find things to be a function of the slant-angle?

[George1]

Hi ME,
Yes, you understand correctly -- the same types of rails. But (a) there is friction in the straight-line rails case and (b) there is no friction in the zigzag rails case. That's the only difference.
Looking forward to your answer.
Regards,
George

[ME]

George... your replies are a bit stuck in a loop without processing new information. I hope for you it's not a too severe system bug, or a burn-in problem.
Now I can only ask the same question as an answer to your 'clarification' which didn't actually clarify anything but only reiterated what you claimed earlier.

Perhaps I should attempt protocol. One moment please.

[ME]

Hi Georg,
Inquiry: What happens to friction when your straight-line rails are set at a tapered angle?
Situation: Say, those rails move inwards towards the left-side.
Looking forward to your answer.
Regards,
Marchello E.

[George1]

Hi Marchello E.,
Thank you for your reply.
Very good question! You actually ask what would happen to friction if the zigzags are not rounded curves but sharp formations similar to saw teeth? Well, if I undestood your question correctly, then you have just created an interesting variation of the basic conception. In this case the violation-of-the-two-conservation-laws effect would be as if even much greater. (Friction is always assumed to be zero in the zigzag rails case (no matter what is the form of the zigzags); and friction is always assumed to be bigger than zero in the straight-line rails case.)
Or may be I did not understand correctly your question? If so, then please excuse me and please give some more additional explanations, if possible.
Looking forward to your answer.
Regards,
George

[ME]

You actually ask what would happen to friction if the zigzags are not rounded curves but sharp formations similar to saw teeth? Well, if I undestood your question correctly, then you have just created an interesting variation of the basic conception. In this case the violation-of-the-two-conservation-laws effect would be as if even much greater. (Friction is always assumed to be zero in the zigzag rails case (no matter what is the form of the zigzags); and friction is always assumed to be bigger than zero in the straight-line rails case.)

Hi Georg,
I don't share this idea where "friction is always assumed to be zero in the zigzag rails case".

So yes, I meant a saw-tooth formation as an assisting form. You can make the points as sharp or as shallow as you want. And as such you get a variable angle: let's call it a steepness-angle. Now just ignore the points of the saw tooth. What is of interest is to discover how friction is affected by this steepness-angle. I think it is best to avoid anticipating on negative-friction just yet, because such would simply follow from measurement!

Yet, when you are already totally convinced you will gain much greater effects then that's really super good news!!
Simply eliminating the zig-zag altogether and just use inclines will drastically simplify your proof of concept design. Glad to be of help.
Looking forwards to see the result of your experiments.
Good luck and kind regards,
Marchello E.

[George1]

Hi Marchello E.,
Hi dear friend,
Thank you very much your reply.
You are real generator of good ideas! Congratulations for this ability of yours!
1) Actually you suggest to use straight-line inclines that form the zigzag sequence? Did I understand this correctly?
2) Or to use only one single incline and thus to simplify entirely the situation? Not many zigzags but only one single incline, that is, 1/4 (the length of) one single complete saw-tooth zigzag? Something like that?
3) Yes, you are absolutely right that friction can never be equal to zero in real life. But at the same time (by using modern and not so modern technologies) friction can be reduced practically as much as you want -- by using for example permanent magnet slides and bearings or rolling friction instead of sliding friction, etc. In this way we can reduce the experimental error due to friction to a certain negilgible value (practically as much as you want).
4) Yes, it is an interesting question how friction is affected by the steepness-angle. May be it would also depend in some way on the nature of the two sliding surfaces, on their area of contact and on the lubricant's kind? Because as if the simplest method and approach is to use sliding friction, one single incline, some standard lubricant and the most suitable steepness-angle. Don't you think so? Any idea for some preliminary calculations?
Looking forward to your answer.
Regards,
George

[ME]

Any idea for some preliminary calculations?

Start with friction down an incline.

[George1]

I see. But there are many ways. Which one to choose? Wouldn't be simpler if we neglect frition (which is acceptable if friction is beyond some limit) and calculate F'c = ?

[George1]

Hi guys,
I haven't been writing in this forum for many months. Fully occupied with business and experiments.
Let us recapitulate the situation. Please look at the three links below.
1) https://mypicxbg.fil...pages_01-12.pdf
2) https://mypicxbg.fil...5/figs01-08.pdf
3) https://mypicxbg.fil...have_a_look.pdf
Experimentally proved already many times, especially the third link.
Let me remind again that the theory and the related experiments must be evaluated ONLY by highly qualified experts in theoretical and applied mechanics.
Looking forward to your answer.
Regards,
George

[George1]

And here is an abstract from the third link. "It is evident that we can always 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." DO YOU HAVE ANY OBJECTIONS AGAINST THIS LAST CLAIM? THE LATTER HAS BEEN EXPERIMENTALLY PROVED ALREADY MANY TIMES.
Looking forward to your answer.
Regards,
George

[George1]

Any opinions related to our zigzag mechanical conception?