IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[PeterAX]

To those members of this forum, who are interested in our zigzag concept.
====================
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.
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 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.

[PeterAX]

To those members of this forum, who are interested in our zigzag concept.
-----------------------------------
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 you mentioned in your 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.
-----------------------------------
Looking forward to your answer.

[rocky]

Is this a PM machine? Does he use his left hand fingers to slow it down?

https://fb.watch/68xMvDJyVX/

[Georg Künstler]

Hi Rocky,
it is a very clever oscillation of weights design.
It will make impacts on the sides, so it is loosing energy and still it turns.
So where does the energy come from, it is only gravity included.
Looks like the class of designs raj is designing, all directions x,y.z is used.

The video is showing a parasitic oscillation, not a parametric oscillation.
It is a swinging on the top of a swinging, or in other words a wave is carrying a wave.

It is a function Bessler has used in his flail comparison. An oscillation on an oscillation.

Thanks for finding this video. https://fb.watch/68xMvDJyVX/

With his hand he regulate the speed, in my opinion.
If he does not do this the speed will increase and then the weights will no longer slide from one side to the other.
He has to stay in the frame of the natural frequency.

[PeterAX]

To rocky and to Georg Kunstler.
=======================
The link https://fb.watch/68xMvDJyVX/ seems to be interesting. Need some time to consider it carefully.

[PeterAX]

To Rocky and to Georg Kunstler.
-----------------------------------------------------------
About the link https://fb.watch/68xMvDJyVX/
-----------------------------------------------------------
May be some automatic computer control of the mechanical resistive counter-torque, whose value would have to follow a certain frequency? What do you think about this?

[PeterAX]

To those members of this forum, who are interested in our zigzag concept.
============================
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,

[PeterAX]

Any comments about the "X effect", described in our last post?

[PeterAX]

To Rocky and to Georg Kunstler.
-----------------------------------------------------------
About the link https://fb.watch/68xMvDJyVX/
-----------------------------------------------------------
May be some automatic computer control of the mechanical resistive counter-torque, whose value would have to follow a certain frequency? What do you think about this? What is your opinion? (The design would become a little more sophisticated, of course.)
Looking forward to your answer.

[PeterAX]

To Rocky and to Georg Kunstler.
-----------------------------------------------------------
What about the discussion, related to the link https://fb.watch/68xMvDJyVX/ ? Still no comments from you? Looking forward to your answer.

[PeterAX]

Still no comments related to the "X effect", which is described in our previous posts? Let me remind only again that it is a matter of a many-times-repeated real experiment.
Looking forward to your answer.

[PeterAX]

Any comments about the "X effect", which is described in our previous posts?

[MrTim]

PeterAX continues to...

Let me remind only again that it is a matter of a many-times-repeated real experiment.

You keep repeating yourself, but you haven't shown us anything other than what squirts out it's dead butt. What next, will you put it in a sugar cone and tell us it tastes like chocolate ice cream...?
There is no need to further respond to you.

[PeterAX]

To MrTim.
------------------------
Hi Nobel prize winner number 2,:)
It's always a great pleasure to correspond with you!:) Could you formulate for all of us here in this forum the third Newton's law, please?:)
Looking forward to your answer.
Regards,:)

[Tarsier79]

Hi Nobel prize winner number 2,:)

Is that your best retort?

George, you think you are smarter than all of us, why don't you do it. It isn't our problem you can't calculate inertial interactions correctly. Build it and prove us all wrong.

Idiot.