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To @citizenschallengev3.

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Hi there,

Thank you for your reply.

I am absolutely sure solely and only about the validity of the experiments, described in PART 3 of our first video. The text below is a combination of our posts of June 15, 2021 at 5:01 am and of June 14, 2021 at 8:07 am. As if this combination clarifies entirely our zig-zag concept both theoretically and experimentally.

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1) Please look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s   . Please 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) @mrmhead wrote: “…take gravity and friction out of equation and consideration…”. Perfectly agree with this.
14) 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 = 1 kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
15) In our numerous real experiments we strongly reduce friction (force of friction = 0.0000001 N) and the mean 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.

16)  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…” as @memhead mentioned in his/her last post.) 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.

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Do you have any objections against any of the above items 1 – 16? (If no, then it’s ok. But if yes, then please specify exactly which item you do not agree with and why.)

Looking forward to your answer.