IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[George1]

To ME.
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And one more question.
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1) In our initial example the mass and the inertia moment of body 2 (Earth(ball) + ramp) are much bigger than the mass and inertia moment of body 1 (the block). In this case body 2 is practically motionless and body 1 is in motion. Besides |F| = |-F| = 10 N = const.
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2) Let us reverse the situation. Let us assume that the mass and the inertia moment of body 1 (the block) are much bigger than the mass and the inertia moment of body 2 (Earth(ball) + ramp). In this second case body 1 (the block) is practically motionless and body 2 is in motion. Besides |F| = |-F| = 10 N = const.
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What is the difference between the above items 1 and 2? There is no difference.
(Please read carefully and thoroughly again our post of Sat Oct 24, 2020 11:12 am.)
Looking forward to your answer.
Regards,
George

[George1]

To agor95.
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I can't understand what are you talking about. Do you have any reasonable and mathematically supported arguements related to this topic? If yes, then please share these arguments with us.

[ME]

When a member has a bad rating then it's best to put the member into the ignore list.

That way you 'Don't Feed the Troll'.

Agor, that only applies when you actually feel trolled and outraged.
When you feel trolled then it's better to lower expectations knowing that communication will likely not improve.
You can't really blame a troll for lacking actual substance, as is proven by trolling.

In many replies on this forum I just try to reply the best I can, for my own exercise.

2A) If the situation is "flat" (that is, θ = 0), then bodies 1 and 2 do not move and the two dynamometers register forces F and - F, respectively.
2B) If the situation is "inclined" (that is, θ > 0), then bodies 1 and 2 move relative one to another (combined motion) and the two dynamometers register AGAIN forces F and - F, respectively.
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Question:what is d the difference between 2A and 2B?

You already know the difference between a "flat" situation (θ = 0) and "inclined" (θ > 0).
Either the attracting force is countered completely, or it is not.
There should be no distinction between (θ = 0) and (θ > 0) because that's covered by trigonometry.

To repeat what I wrote in post: https://www.besslerwheel.com/forum/view ... 035#176035

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Extreme situation 1:
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Let's say the plane is a horizontal table: θ=0°
This table is able to provide a reaction to the block's force "mg" IN FULL
Thus: mgcos0° = mg

Where does this block go?
mgsin 0° = 0
In other words: Absolutely nowhere!!

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Extreme situation 2:
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Let's say the plane is a vertical wall: θ=90°
This wall is unable to provide a reaction to the block's force "mg".
Thus: mg cos 90° = 0

Where does this block go?
mgsin 90° = mg
In other words: This block remains in total FREEFALL, despite that wall!!

In all these cases, including the intermediate, it accelerates down the ramp with an acceleration of "mgsin θ".
The answer: mathematically there's no difference between 2A and 2B.

[agor95]

Hi ME

I have ignore of this thread. I see only your replies.

So the question is have you found an improvement in George1 conduct?

Cheers

[George1]

To ME and agor95.
==============================
But you are not reading my posts (if reading them at all). Because this is not a discussion -- this is a monologue.
Or you both deliberately simulate lack of understanding, which is not an honest approach.
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Let me explain again. (PLEASE READ CAREFULLY AND THOROUGHLY THE TEXT BELOW.)
1) We have two bodies -- body 1 and body 2.
2) Body 1 is the block. A rectangular parallelepiped. Sizes much smaller than 1 meter. A few centimeters in the three dimensions.
3) Body 2 consists of a ball and a ramp, which are firmly attached one to another thus forming one united whole. (For the convenience of the considerations Earth is replaced with a ball, whose diameter is equal to 1 meter for example and the ramp's sizes (a few centimeters for example) are much smaller than 1 meter.
4) The two ends of a spring are attached to the centers of mass of bodies 1 and 2.
5) The spring contracts and causes the motion of body 1 and/or body 2. Besides two identical dynamometers are attached to the two ends of the spring. One of the dynamometers registers force |G| and the other dynamometer registers force |-G|, where |G| = |-G| = G = const. (Spring's characteristic 4 -- please look at our previous posts.)
6) The experiment is carried out in a space station under weightlessness conditions. Friction is negligible.
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7) CASE 1. The mass and the moment of inertia of body 2 are much bigger than the mass and the moment of inertia of body 1.
7A) Body 2 is motionless.
7B) Body 1 is in motion under the action of force G', where G' = G sin θ.
7C) Because of the last equation you assert that body 1 exerts force G on body 2, but body 2 does not exerts force - G on body 1. (This is your point of view, which you have been supporting for a long period of time.)
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8) CASE 2. The mass and the moment of inertia of body 1 are much bigger than the mass and the moment of inertia of body 2.
8A) Body 1 is motionless.
8B) Body 2 is in motion under the action of force G', where G' = G sin θ.
8C) Because of the last equation you have to assert that body 2 exerts force G on body 1, but body 1 does not exerts force G on body 2.
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Which of the above two cases 1 and 2 do you accept to be correct? Case 1 or case 2? (According to your logic case 1 and case 2 must be valid simultaneously. But this is an absurd. Besides you always ignore the readings of the two dynamometers. Why?)
Looking forward to your answer.
George1

[George1]

In the second line of my last post the sentence "But you are not reading my posts (if reading them at all)." must be replaced with the sentence "But you are not reading carefully my posts (if reading them at all)."

[ME]

In my previous post I wrote: "To repeat what I wrote in post: https://www.besslerwheel.com/forum/view ... 035#176035".

This can be interpreted as: "You are not reading carefully my replies, if reading them at all!"

You wrote: "Because this is not a discussion -- this is a monologue."
This means that you are in the wrong place.
This is a discussion board.

[George1]

To ME.
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YOU ARE NOT READING MY POSTS AT ALL! STOP IMITATE IGNORANCE! ARE YOU AN AGENT OF THE BIG MAFIA?
I AM EXPLAINING AGAIN SOME SIMPLE THINGS. KEEP IN MIND MY PREVIOUS POSTS.
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1) The experiment below is carried out in a space station under weghtlessness conditions.
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2) We can always choose a suitable combination of (1) sizes, masses and inertia moments of bodies 1 and 2 and (2) a spring, for which if the spring contracts, then an astronaut would see that (a) the longitudinal axis of symmetry of the spring is motionless, (b) the distance between the centers of mass of bodies 1 and 2 decreases, (c) each of the two identical dynamometers registers a force of 10 N, (d) body 1 rotates at an angle α and (e) body 2 rotates at an angle β.
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(Note 1. The centers of rotation of bodies 1 and 2 are their centers of mass, respectively. Besides bodies 1 and 2 rotate in the same direction, let's say clockwisely.)
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(Note 2. The force of 10 N, registered by each of the above mentioned two identical dynamometers, does not depend on decreasing of the distance between the centers of mass of bodies 1 and 2. This is because we use a spring of characteristic 4. Please refer to our previous posts.)
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3) In one word, a constant force of magnitude of 10 N is applied to the center of mass of body 1 and is directed to the center of mass of body 2. At the same time another constant force of the same magnitude of 10 N is applied to the center of mass of body 2 and is directed to the center of mass of body 1. Under the action of these two forces the centers of mass of bodies 1 and 2 move one to another in a straight line (the common line of action of the two forces coincides with the longitudinal axis of symmetry of the spring) and at the same time rotate clockwisely at angles α and β, respectively.
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4) Having in mind the text above can we conclude that body 1 exerts force G on body 2 and body 2, on its behalf, exerts force - G on body 1 as forces G and - G are equal in magnitude (10 N) and opposite in direction? Can we accept the validity of equality |G| = |-G| = 10 N = const.?
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(Note 3. Symbol G can be replaced by any of the symbols F, K, L, M, etc.; we can choose any capital Latin letter.)
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I hope that all of us here in this forum are honestly seeking for the truth and none of us here is arguing because of arguing itself.
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Looking forward to your answer.
George1

[ME]

To George,
YOU ARE NOT READING MY REPLIES AT ALL! STOP IMITATE IGNORANCE! ARE YOU AN AGENT OF THE BIG MAFIA OR JUST PLAIN STUPID?
I KEEP EXPLAINING SOME BASIC THINGS LIKE AN INCLINED PLANE. KEEP IN MIND MY PREVIOUS REPLIES.

And also... a spring does really not alter things.
Because you can't read anyway, let's see how you process animated gifs.

For those who can read let me introduce that animation.
Somewhere in space there's an incline that gets attracted via a spring by a sphere of equal mass.
A box also gets attracted by that sphere. Only that incline sits in its path.
Would something completely different happen compared to a situation on Earth where the attraction happens because of gravity?
Ehm, no.
Only that the incline is a bit large and overhanging the sphere: Thus funny, but expected, stuff happens.

Looking forward to your answer.

No you're not.

[George1]

To ME.
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I have mentioned many times already that you are not reading carefully my posts (if reading them at all). YOUR ANIMATION DOES NOT PRESENT THE SITUATION CORRECTLY. THE RAMP AND THE BALL ARE FIRMLY ATTACHED ONE TO ANOTHER THUS FORMING ONE UNITED WHOLE, WHICH IS CALLED BODY 2. SO CORRECT YOUR ANIMATION!
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Further let us split the text of our last post into several sub-steps. Sub-step 1, which is given below, is limited/surrounded up and down by double dashed lines.
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SUB-STEP 1
1) The experiment below is carried out in a space station under weghtlessness conditions.
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2) We can always choose a suitable combination of (1) sizes, masses and inertia moments of bodies 1 and 2 and (2) a spring, for which if the spring contracts, then an astronaut would see that (a) the longitudinal axis of symmetry of the spring is motionless, (b) the distance between the centers of mass of bodies 1 and 2 decreases, (c) each of the two identical dynamometers registers a force of 10 N, (d) body 1 rotates at an angle α and (e) body 2 rotates at an angle β.
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(Note 1. The centers of rotation of bodies 1 and 2 are their centers of mass, respectively. Besides bodies 1 and 2 rotate in the same direction, let's say clockwisely.)
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(Note 2. The force of 10 N, registered by each of the above mentioned two identical dynamometers, does not depend on decreasing of the distance between the centers of mass of bodies 1 and 2. This is because we use a spring of characteristic 4. Please refer to our previous posts.)
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Question: Do you accept the validity of the above written SUB-STEP 1?
If yes, then you have to correct your animation for a second time, that is,
the spring must be motionless.
If not, then why?

[George1]

An addition to our last post. More precisely, the longitudinal axis of symmetry of the contracting spring must be motionless. Correct your animation!

[ME]

I like my animation in space... it stays the way it is.
It's almost as if you want this spring to behave as is it were gravity.
If that's really the case than you're not really progressing, and you find the right reply somewhere in some earlier post.

Talking about timing:

I will do the picture in the very nearest future.

I showed mine, where's yours george?

[George1]

To ME.
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First of all I would like to apologize for being a little more rough than necessary. Please excuse me for my, let's say to some extent, inadequate behaviour. I am really sorry.
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Actually you have created a perfect animation and I would like to ask you only to make 6 small corrections, if possible. These 6 corrections are given below.
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1) The longitudinal axis of symmetry of the spring must be motionless.
2) The sliding surfaces of the red block (body 1) and the green ramp must be always in contact, i.e. the sliding contact must never be lost.
3) The green ramp and the blue ball are firmly attached one to another thus forming one united whole, which is called body 2.
4) The geometric center of the red block and the geometric center of the blue ball approach each other on a straight line, which coincides with the motionless longitudinal axis of symmetry of the spring.
5) The red block rotates clockwise at an angle α as the center of rotation is its geometric center.
6) The blue ball together with the green ramp (body 2) rotates clockwise at an angle β as its center of rotation is the blue ball geometric center.
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(Note. We have assumed that the above mentioned centers of rotation coincide with the related centers of mass, respectively.)
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Would you be so polite to realize the above mentioned 6 corrections, if possible? Let us consider the case in which the longitudinal axis of symmetry of the contracting spring is motionless.
Looking forward to your answer.
Regards,
George

[George1]

And one more correction, if possible. Only one contracting spring, which is connects the geometric center of the red block with the geometric center of the blue ball.

[George1]

And one more correction, if possible. Only one contracting spring, which connects the geometric center of the red block with the geometric center of the blue ball.