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A hypothesis for a possible reactionless drive


PeterAX

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

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Please have a look again at our post of 06/11/2021 02:50 PM

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V2 = ? V3 = ?

Looking forward to your two answers.

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To JeffreysTubes8.

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Hi dear colleague,

Thank you for your last interesting post. Need some time to consider it carefully. I will write to you in the nearest future.

Regards,

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P.S. By the way is there any video and/or drawings related to the concept that you describe in your posts? Or it is a matter of a secret project for the present?    

 

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

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

Please have a look again at our post of 06/11/2021 02:50 PM

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

V2 = ? V3 = ?

Looking forward to your two answers.

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To those 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.
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.
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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 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.

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Looking forward to your comments.

Edited by PeterAX
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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.

Looking forward to your comments. 

Edited by PeterAX
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To those members of this forum, who are interested in our zigzag concept.

Please have a look again at our post of 06/17/2021 03:53 PM.

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The “X effect” does not depend on the value of the force of friction Ffr inside the zigzag channels. No matter how small is Ffr the “X effect” still exists. According to our last experiments Ffr = 0.0000001 N. Therefore (a) Ffr, (b) the related generated heat and (c) the related due-to-friction experimental error (which is obviously much smaller than 1 %) can be neglected.

Looking forward to your comments.

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Still waiting for your comments related to our zig-zag concept. All that I can say about PART 3 of our first video is that the described-there mechanical phenomenon has been proved experimentally many (may be thousands of) times. The “X effect” exists no matter how small/big is the value of the force of friction inside the zig-zag channels. The “X effect” does not depend on the value of the force of friction inside the zigzag channels. (Please refer to our previous posts, if you like.)

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To JeffreysTubes8.

1) Well, as if there is some misunderstanding here. Our so called "X effect" has nothing to do with the strong molecular X-structure, that you mentioned in your last posts.

2) But your last two posts are really very interesting. Need some time to consider them carefully.   

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To JeffreysTubes8.

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Hi there,

Actually I did not understand your last posts very well. I am not against you, dear collegue!:) What makes you think that I am against you?:) On the contrary, we (all members of our team including me) are always ready to support and help realizing of new technology ideas. The latter seems to be clear from all previous posts of ours.

Looking forward to your answer.

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

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V2 = ? V3 = ? Any experimental results?

You can make your first model of the zig-zag construction by using simple copper or iron wire (we made this wire construction many years ago) and experiment on your kitchen table.:) It will be interesting for you, please believe me.:)

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