A hypothesis for a possible reactionless drive

[PeterAX]

To all members of this forum, who are interested in our zig-zag concept.

------------------------------------------------

V2 = ? V3 = ? Any experimental results? Did you make the simple wire-construction experiment, described in our previous post?

[PeterAX]

To all members of this forum, who are interested in our zig-zag concept.

------------------------------------------------

V2 = ? V3 = ? Still no experimental results? The zigzag experiment is quite simple and can be modified in many ways. For example in the very beginning you can make only one or two large and "flat" zig-zag channels, reduce friction as much as possible and see what will happen.  

[PeterAX]

I am absolutely sure solely and only about the validity of the experiments, described in PART 3 of our first video. And these are as follows. (Let me only remind them again.)

<hr />

1) Please look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s and at our posts of 06/15/2021 12:23 PM and 06/17/2021 03:35 PM. Please focus on the “upper” zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.

4) mass of each couple blue rod-blue ball = m = 0.0001 kg (and even smaller)

5) V1 = Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.
6) Vb’ =  pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.

7) Va”’ = V2 = post-zig-zag velocity of the blue component = 0.6 m/s = const. = theoretical value

7A) Va”’ = V2 = post-zig-zag velocity of the blue component = 0.5999992 m/s = const. = mean experimental value

8/ Vb”’ = V3 = post-zig-zag velocity of the black component = 0.1 m/s = const. = theoretical value

8A/ Vb”’ = V3 = post-zig-zag velocity of the black component = 0.0999997 m/s = const. = mean experimental value

9) Number of zigzags = 4

10) Force of friction inside the zigzag channels = 0.0000001 N = mean experimental value

Edited June 28, 2021 by PeterAX

[PeterAX]

To those members of this forum who are interested in our zig-zag concept.

====================================

Here is another detailed explanation of our zig-zag concept.

====================================

 

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.

=================================

Please pay special attention to last item 16. This is what the above experiment (and the related theoretical consideration) has been already proved.

Looking forward to your answer.

[PeterAX]

To those members of this forum who are interested in our zig-zag concept.

====================================

Here is another detailed explanation of our zig-zag concept.

====================================

 

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.

=================================

Please pay special attention to last item 16. This is what the above experiment (and the related theoretical consideration) has been already proved.

Looking forward to your answer.

[PeterAX]

To those members of this forum who are interested in our zig-zag concept.

====================================

Here is another detailed explanation of our zig-zag concept.

====================================

 

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.

=================================

Please pay special attention to last item 16. This is what the above experiment (and the related theoretical consideration) has been already proved.

Looking forward to your answer.

[PeterAX]

To those members of this forum who are interested in our zig-zag concept.

====================================

Here is another detailed explanation of our zig-zag concept.

====================================

 

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.

=================================

Please pay special attention to last item 16. This is what the above experiment (and the related theoretical consideration) has been already proved.

Looking forward to your answer.

[PeterAX]

To those members of this forum who are interested in our zig-zag concept.

====================================

Here is another detailed explanation of our zig-zag concept.

====================================

 

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.

=================================

Please pay special attention to last item 16. This is what the above experiment (and the related theoretical consideration) has been already proved.

Looking forward to your answer.

[PeterAX]

To those members of this forum who are interested in our zig-zag concept.

====================================

Here is another detailed explanation of our zig-zag concept.

====================================

 

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.

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Do you have any objections against any of the above items 1 -16 and if yes, then why?

Looking forward to your answer.

Edited June 29, 2021 by PeterAX

[PeterAX]

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Edited July 2, 2021 by PeterAX

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