IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[PeterAX]

To Tarsier79.
======================
The only idiot here is Tarsier79!
======================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s . Focus on the “upper� zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va� = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb� = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va�’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.
10) Vb�’ = post-zig-zag velocity of the black component = 0.1 m/s = const.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) + ((Mb) x (Vb’)) = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> ((Ma) x (Va’)) + 0 = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> (Ma) x (Va’) = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) + ((4 kg) x (0.1 m/s)) <=>
<=> 1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va�, Vb� and Vy are actually of no interest to us. Actually only the values of Va’, Va�’ and Vb�’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of Va�’ and Vb�’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va�’ = 0.5999992 m/s and Vb�’ = 0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
============================
I am asking my simple question for the 1st time: Do you have any objections against any of the above items 1 - 15? Yes or no?
I am looking forward to your answer for the 1st time. (Only one word -- either "yes" or "no"!)

[Tarsier79]

Pretty sure I am Nobel Prize Winner #1.

Let me answer your original question.

A: Neither

[PeterAX]

To Tarsier79.
======================
Asking my simple question for the 2nd time: Do you have any objections against any of the items 1 - 15 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 2nd time. (Only one word -- either "yes" or "no"!)
======================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s . Focus on the “upper� zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va� = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb� = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va�’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.
10) Vb�’ = post-zig-zag velocity of the black component = 0.1 m/s = const.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) + ((Mb) x (Vb’)) = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> ((Ma) x (Va’)) + 0 = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> (Ma) x (Va’) = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) + ((4 kg) x (0.1 m/s)) <=>
<=> 1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va�, Vb� and Vy are actually of no interest to us. Actually only the values of Va’, Va�’ and Vb�’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of Va�’ and Vb�’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va�’ = 0.5999992 m/s and Vb�’ = 0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
============================
Asking my simple question for the 2nd time: Do you have any objections against any of the above items 1 - 15? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 2nd time. (Only one word -- either "yes" or "no"!)

[Tarsier79]

There is a reason people don't have in-depth conversations with their iphone.

[ovyyus]

I am asking my simple question for the 1st time: Do you have any objections against any of the above items 1 - 15? Yes or no?
I am looking forward to your answer for the 1st time. (Only one word -- either "yes" or "no"!)

No.

I do not have any objections against any of the above items 1 - 15. What is next?

[unstable]

It seems to me that in the video, when he stops turning the crank, you can notice the pressure of a button pressed with the thumb of his right hand.

[WaltzCee]

. . .
No.

I do not have any objections against any of the above items 1 - 15. What is next?

I guess this is where we wait for a courier to bring the ransom note.
I suggest we notify the special operators.

[unstable]

But...Isn't it more effective to completely ignore his messages ? :-)

[PeterAX]

To ovyyus.
=================
Hi ovyyus,
Thank you for your reply. Good answer! You are obviously enough qualified in order to evaluate the zigzag concept in a proper manner.
------------------------------
And here is how we proceed further.
------------------------------
1) Look again at the link https://www.youtube.com/watch?v=xX14NK8GrDY.
2) From 3:45 to 3:48 we have Ma = 1 kg, Mb = 4 kg and V1 = 1 m/s. (Consider only the "upper" zigzag device.)
3) From 3:59 to 4:01 we have Ma = 1 kg, Mb = 4 kg, V2 = 0.6 m/s and V3 = 0.1 m/s. (Consider only the "upper" zigzag device.)
4) (1 kg) x (1 m/s) = ((1kg) x (0.6 m/s)) + ((4kg) x (0.1 m/s)). The last equality unambiguously shows the validity of the law of conservation of linear momentum in this particular case.
5) (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s) x (0.6 m/s)) + ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s)). The last inequality unambiguously shows the invalidity of the law of conservation of mechanical energy in this particular case.
-------------------------------
(Note. Force of friction inside the zigzag channels = 0.0000001 N = experimental result. The latter can be reduced even further. Friction does not influence in any way the validity of the mechanical effects, described in the above items 1 - 5. )
-------------------------------
Looking forward to your answer.

[ovyyus]

Looking forward to your answer.

You didn't ask a question. What is the question?

[PeterAX]

To ovyyus.
=====================
Hi ovyyus,
Thank you for your reply.
Yes, you are wright. Here is the correction. The question is: Do you have any objections against any of items 1 - 16 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
=====================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s . Focus on the “upper� zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va� = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb� = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va�’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va�’ = V2.
10) Vb�’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb�’ = V3.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) + ((Mb) x (Vb’)) = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> ((Ma) x (Va’)) + 0 = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> (Ma) x (Va’) = ((Ma) x (Va�’)) + ((Mb) x (Vb�’)) <=>
<=> (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) + ((4 kg) x (0.1 m/s)) <=>
<=> 1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va�, Vb� and Vy are actually of no interest to us. Actually only the values of Va’, Va�’ and Vb�’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of Va�’ and Vb�’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va�’ = 0.5999992 m/s and Vb�’ = 0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
16) And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) > ((0.5) x (Ma) x (Va�’) x (Va�’)) + ((0.5) x (Mb) x (Vb�’) x (Vb�’)) <=>
<=> (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) + ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s)) <=> 0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy in this particular case.
============================
============================
Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer.

[ovyyus]

Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer.

No.

I do not have any objections against any of the above items 1 - 16. What is next?

[PeterAX]

To ovyyus.
==================
Good! Very good! In one word, you admit the fact that the zigzag device severely violates the law of conservation of mechanical energy as the inlet energy is equal to 0.5 J and the outlet energy is equal to 0.2 J. The last fact is simply impossible from the point of view of standard classical mechanics if friction is either equal to zero or so small that can be neglected.
Do you accept the simple obvious fact that a kinetic energy of 0.3 J simply disappers thus breaking severely and obviously the law of consevation of mechanical energy?
Looking forward to your answer.

[ovyyus]

Yes. What is next?

[PeterAX]

To ovyyus.
====================
There is a great variety/bunch of next steps. (Some of these next steps seem to be quite obvious.) I will write to you on this topic in detail in the nearest future.