IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[lancaIV]

https://www.youtube.com/watch?v=JB4RMIWroMY


Happy Sabado,aeh ,weekend and sunny days wishing (if not desert-common )


Shalom,shalom aleikum


Oxala


OCWL

[George1]

To lancaIV.
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Hi lancaIV,
Nice video! Thank you for it! It is always a pleasure to correspond with you!
Regards,

[George1]

To Jerry Volland.
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1) Ah, this is already another song! Your last post is reasonable and smart. Please give me some time to consider it carefully. I will write to you in the nearest future.
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2) By the way, why don't you make a simple wire model of the zigzag device (PART 3 of our first video) and experiment on your kitchen table? Reduce friction as much as possible by using in the beginning an ordinary soap mash for example. Try different shapes and number of zigzags. It will be interesting for you, I am sure. One of our first zigzag-concept-related experiments many years ago was just of that kind  -- wire zigzag (and other forms of) frames, which were sliding over a horizontal kitchen table.       

[George1]

To Jerry Volland (and to all other members of this forum, who are interested in our zigzag concept).
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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.
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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.
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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.
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.
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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.
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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 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.
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Please ask your questions, if any. We are ready to answer.

[George1]

The yellow head with the black spectacles in our last post must be replaced by number 8 followed by the bracket symbol ")". The system keeps generating some constant defect related to the writing of number eight.