IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[George1]

To nix85.
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Hi nix85,
1) About the first link. A small question: The brake is used in the wheels, that support the platform, isn't it?
2) About the second link. A small question: If put on a toy-boat, then wouldn't the platform (together with the toy-boat) move in a direction opposite to the direction of motion of the ball?
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Please note -- NOT REJECTING ANYTHING, only asking small (and to some extent may be stupid ) questions. But this is because this device/concept is entirely new for us for the present.
Regards, 

[George1]

We give below again a copy of our post of February 26, 2021, 04:01:00 PM.
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Our simulation/animation is given below.
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Firstly, please always keep in mind and please always refer to our first post of July 21, 2018, 02:11:37 PM and to the related two links
https://mypicxbg.files.wordpress.com/2018/04/pages_01-12.pdf
https://mypicxbg.files.wordpress.com/2018/05/figs01-08.pdf
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Secondly, our simulation/animation presentation is subdivided into three consecutive separate parts for an easier understanding.
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PART 1. Please look at the link https://youtu.be/9cBGurYSryw
1) The zigzag device and the straight-line device are put together vertically one to another and are fixed motionless to a horizontal motionless plane.
2) The two blue T-shaped components start free falling together and simultaneously. Friction is negligible.
3) In the straight-line case the T-shaped blue component falls freely without any interruptions and obstacles.
4) In the zigzag case however after the blue balls enter the zigzag channel the blue T-shaped component slows down its vertical downward motion and decreases its downward vertical velocity.
5) In one word, the straight-line modification blue T-shaped component covers the distance between the highest position and the lowest position much faster than the zigzag modification blue T-shaped component.
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PART 2. Please look at the link https://youtu.be/aVOfWLDrYwA
1) How to slow down the vertical downward motion (that is, how to decrease the vertical downward velocity) of the straight-line modification blue T-shaped component?
2) The answer is simple. The straight-line segment "s" is made rough (inside the channels) as the related force of friction is chosen in such a manner so that in the lowest position the linear downward velocities of the two blue T-shaped components are one and same and equal one to another.
3) In one word, the zigzags generate mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat.
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PART 3. Please look at the link https://youtu.be/pPGPktU_kpo
The last link simply repeats the experiment, described in our first post of Sat Jul 28, 2018 12:41 pm and in the related two links
https://mypicxbg.files.wordpress.com/2018/04/pages_01-12.pdf
https://mypicxbg.files.wordpress.com/2018/05/figs01-08.pdf
The zigzags generate again mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat.
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In one word, the text above and the related links above unambiguouly show that (a) either the law of conservation of linear momentum is not correct or (b) the law of conservation of mechanical energy is not correct or (c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.
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Looking forward to your answer.

[George1]

We give below again a copy of our post of March 01, 2021, 03:35:45 PM.
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Let us shorten our previous explanations (of February 26, 2021, 04:01:00 PM) by jumping directly to PART 3 (and having a brief glimpse at a small section of PART 2).
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So let us start our shorter explanations.
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1) Firstly, please have a look for a while at PART 2 and at the related link https://youtu.be/aVOfWLDrYwA from 00:00 to 00:03. This is only for getting a notion about the limits of the segment "s", that is, how this segment "s" is situated in relation to (relative to) the zigzag section.
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2) Now let us focus on PART 3 and on the related link https://youtu.be/pPGPktU_kpo . The experiment is carried out in a space station under weightlessness conditions. Friction is negligible as the only exception is the friction inside the two straight-line channels of the segment "s". (The inside surfaces of the straight-line channels of the segment "s" are made rough thus able to generate friction (and heat, respectively).)
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3) The mass of each blue component is Ma.
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4) The mass of each black component is Mb.
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5) There are four couples blue ball/blue rod. Each blue ball is firmly attached to the related blue rod thus forming one united whole.
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5A) The mass of each blue ball is negligible (if compared to Ma or to Mb), but not equal to zero.
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5B) The mass of each blue rod is negligible (if compared to Ma or to Mb), but not equal to zero.
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6) From 00:00 to 00:03 the two blue components move simultaneously and uniformly. Each blue component's linear velocity is V' as V' = const. The two black components are at rest.
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7) At 00:03 the four blue balls enter simultaneously (a) the "upper" black component's smooth zigzag channels and (b) the "lower" black component's rough straight-line channels of the segment "s", respectively.
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From 00:03 to 00:15 the four blue balls move (a) inside the "upper" black component's smooth zigzag channels and (b) inside the "lower" black component's rough straight-line channels of the segment "s", respectively.
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9) At 00:15 the four blue balls exit simultaneously (a) the "upper" black component's smooth zigzag channels and (b) the "lower" black component's rough straight-line channels of the segment "s", respectively.
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10) The force of friction inside the two rough channels of the segment "s" is chosen in such a manner (we could use for example a variable roughness and the related variable force of friction, respectively) that:
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a) the blue components decelerate in one and same manner, that is, their decelerations are one and same and equal one to another;
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b) the black components accelerate in one and same manner, that is, their accelerations are one and same and equal one to another.
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11) From 00:15 to 00:17 the two blue components move simultaneously and uniformly. Each blue component's velocity is V" as V" = const.
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12) From 00:15 to 00:17 the two black components also move simultaneously and uniformly. Each black component's velocity is V"' as V"' = const.
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13) Therefore for the "upper" zigzag modification we can write down that
(Ma) x (V') = ((Ma) x (V'')) + ((Mb) x (V''')) (1)
(1/2) x (Ma) x (V') x (V') = ((1/2) x (Ma) x (V'') x (V'')) + ((1/2) x (Mb) x (V''') x (V''')) (2)
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14) And for the "lower" straight-line modification we can write down that
(Ma) x (V') = ((Ma) x (V'')) + ((Mb) x (V''')) (1)
(1/2) x (Ma) x (V') x (V') = ((1/2) x (Ma) x (V'') x (V'')) + ((1/2) x (Mb) x (V''') x (V''')) + Q (3),
where Q is the heat, which is generated while the two blue balls move inside the two rough channels of the segment "s" in the "lower" modification.
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15) It is evident that (a) the system of equations in item 13 and (b) the system of equations in item 14 cannot be true simultaneously.
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16) And it directly follows from the previous item 15 that either (a) the law of conservation of linear momentum is not correct or (b) the law of conservation of mechanical energy is not correct or (c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.
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NOTE. Please refer, if necessary, to our first post of Sat Jul 28, 2018 12:41 and to the two related links
https://mypicxbg.files.wordpress.com/2018/04/pages_01-12.pdf
https://mypicxbg.files.wordpress.com/2018/05/figs01-08.pdf
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Looking forward to your answer.

[George1]

Any comments, related to our last two posts?

[nix85]

To nix85.
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Hi nix85,
1) About the first link. A small question: The brake is used in the wheels, that support the platform, isn't it?

No, brakes must be put on flakes to prevent the bakes from sakes in case of fakes.

2) About the second link. A small question: If put on a toy-boat, then wouldn't the platform (together with the toy-boat) move in a direction opposite to the direction of motion of the ball?
-------------------------------
Please note -- NOT REJECTING ANYTHING, only asking small (and to some extent may be stupid ) questions. But this is because this device/concept is entirely new for us for the present.
Regards,

Professional athletes move where they want.