IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[George1]

Ok, ok, I will explain again.
==========================
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) 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.
============================
Everything seems to be clear now, doesn't it? Each item of the text above (items 1 - 15) is correct, isn't it?
But please ask questions, if any. We are ready to answer.
Regards,

[George1]

Looking forward to your comments, related to our previous post.

[George1]

To JerryVolland.
----------------------------
Hi there,
1) Well, you have generated some really interesting ideas! My respect to your inventiveness! Good! But please give me some time to consider carefully and thoroughly your suggestions. Need some time. Really.
2) For the present I would like to ask you again to focus solely and only on the "X effect". Do you accept the validity of the "X effect"? (The latter has has been proved experimentally many (may be thousands of) times.)
Looking forward to your answer.
     

[George1]

To JerryVolland.
======================
But you are not reading my posts at all! This is not a dialogue! This is your monologue!
======================
Please read carefully the text below and please answer the question at the end of the text below. Please!
======================
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) 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.
============================
DO YOU HAVE ANY OBJECTIONS AGAINST ANY OF THE ABOVE ITEMS 1 - 15? YES OR NO? ONLY ONE WORD -- EITHER "YES" OR "NO"!

[George1]

To JerryVolland.
=======================
You are simply afraid of truth. You are simply rejecting obvious physical reality.
Consider carefully and thoroughly again our post of June 21, 2021, 02:06:31 PM.
I am asking you my direct and clear question for the 2nd time: DO YOU HAVE ANY OBJECTIONS AGAINST ANY OF ITEMS 1 - 15? YES OR NO?
Looking forward to your answer for the 2nd time. (ONLY ONE WORD -- EITHER "YES" OR "NO"!)