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Circumventing Newton's third law through Euler Inertial Forces


John2020

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6 hours ago, swansont said:

Trajectory refers to the CoM motion. It is not helical. Stop making stuff up.

Let us ignore (b) and just take our attention to (a). Please allow me to ask you the following:

1.Do the threads of the nut follow a helical trajectory? Note: screw is fixed not rotating.

2.If (1) is true could we say, because of the translation screw only helical trajectory of the threads is allowed?

3.When the nut evolves around the screw, a linear displacement is taking place, right? 

4.Is the translation screw hold between two housings and limited between their space?

5.Now from the action-reaction principle, if the nut is being displayed to the right, what is being displayed in the opposite direction in order to hold Newton's 3rd law? 

 

6 hours ago, Ghideon said:

Just to make sure you do not still claim constant velocity. The object will accelerate due to the gravity. Yes?

Yes.

6 hours ago, Ghideon said:

1: What is your definition of linear motion? 
2: What is your definition of helical motion?

1.When a real contact force applies in line with body's CoM then, the body will acquire an acceleration by following rectilinear trajectory.

2.See https://en.wikipedia.org/wiki/Helix

Edited by John2020
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8 minutes ago, John2020 said:

Does the nut follow a helical trajectory?

No. The threads outside the bolt and inside the nut have helical shape. The centre of mass of the nut follows straight line while rotating. Same as a rotating bullet exiting a rifled* barrel of a gun. Rotating, not doing any helical movement.

8 minutes ago, John2020 said:

2.If (1) is true

(1) is false

 

 

*) Helical groovings that are machined into the internal (bore) surface of a gun's barrel.

Edited by Ghideon
rifling definition
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9 minutes ago, Ghideon said:

No. The threads outside the bolt and inside the nut have helical shape. The centre of mass of the nut follows straight line while rotating. Same as a rotating bullet exiting a rifled barrel of a gun. Rotating, not doing any helical movement.

Yes, but the thread of the nut (contacts) follows a helical trajectory. It is clear the CoM of the nut follows a straight line while rotating and I am not arguing about this.

The point in my arguments is not about what the CoM of the nut does, this is perfectly clear. I admit my description was really flawed because I had in mind the threads of the nut and not the nut itself, however I used wrongly the nut as following a helical trajectory.

Anyway, could you tell me what is the answer on (5): "Now from the action-reaction principle, if the nut is being displayed to the right, what is being displayed in the opposite direction in order to hold Newton's 3rd law? "

 

Edited by John2020
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18 minutes ago, John2020 said:

1.When a real contact force applies in line with body's CoM then, the body will acquire an acceleration by following rectilinear trajectory.

2.See https://en.wikipedia.org/wiki/Helix

Thanks. Number 2 is a definition of a Helix shape, not the helical motion you describe.

Edited by Ghideon
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1 minute ago, Ghideon said:

Thanks. Number 2 is a definition of a Helix shape, not the helical motion you describe.

My description as I admitted about the nut is flawed. However, the threads of the nut follow the topology of an Helix, right?

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9 minutes ago, John2020 said:

Yes, but the thread of the nut (contacts) follows a helical trajectory.

But isn't this what I told you, @John2020?

1 minute ago, John2020 said:

the threads of the nut follow the topology of an Helix, right?

Yes, but the nut's thread is not a dynamical element of any relevance in the problem. It's rather its COM and moment of inertia.

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1 minute ago, joigus said:

But isn't this what I told you, @John2020?

Then I apologize.

@joigusCould you please tell me what is the answer on the following:"Now from the action-reaction principle, if the nut is being displayed to the right, what is being displayed in the opposite direction in order to hold Newton's 3rd law? "

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10 minutes ago, John2020 said:

However, the threads of the nut follow the topology of an Helix, right?

Seems ok, thanks for clarifying, I think that allows the discussion to progress! Looks like helical/non helical movement or its definition is not important.
The helical shape of the set of contact points along the treads is what allows your device to circumvent Newton's Third law?

 

Edited by Ghideon
better wording
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But when describing rigid bodies, points like the one you describe play a role as, say, kinematical references. An example is the instantaneous centre of rotation for a cilinder rolling over a plane. But it's not like the mass can be considered lying at that point. You must add the,

\[\boldsymbol{\omega}\wedge\boldsymbol{r}_{i}\]

to the velocity. And you get for the kinetic energy something like this:

\[K.E.=\frac{1}{2}M\boldsymbol{V}^{2}+\boldsymbol{\omega}\wedge\boldsymbol{P}_{{\scriptscriptstyle \textrm{COM}}}\cdot\boldsymbol{V}+\frac{1}{2}\omega_{\mu}\sum_{i}m_{i}\left[\boldsymbol{r}_{i}^{2}\delta_{\mu\nu}-\left(x_{i}\right)_{\mu}\left(x_{i}\right)_{\nu}\right]\omega_{\nu}\]

If V (the velocity of the reference point tracking the motion of the body as a whole) coincides with the COM of the body (nut), then you have a further simplification:

\[K.E.=\frac{1}{2}M\boldsymbol{V}_{{\scriptscriptstyle \textrm{COM}}}^{2}+\frac{1}{2}\omega_{\mu}\sum_{i}m_{i}\left[\boldsymbol{r}_{i}^{2}\delta_{\mu\nu}-\left(x_{i}\right)_{\mu}\left(x_{i}\right)_{\nu}\right]\omega_{\nu}\]

so that the energy can be expanded as translational plus rotational.

The quantity,

\[\frac{1}{2}\sum_{i}m_{i}\left[\boldsymbol{r}_{i}^{2}\delta_{\mu\nu}-\left(x_{i}\right)_{\mu}\left(x_{i}\right)_{\nu}\right]\]

is the matrix of inertia. It's not a point-particle problem.

Are you familiar with this?

Edited by joigus
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1 minute ago, joigus said:

I know of no treatment that deals with the motion of a rigid body, and a nut or a bolt, in terms of action and reaction. All the parts of the body are integrated into a whole that has 6 degrees of freedom.

No need for a mathematical description at the moment. Just by checking Fig.1-Upper or the version without the guiding bars, isn't the answer obvious?

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Every bit of the body moves as a whole, following each other by means of a velocity field:

\[\boldsymbol{V}_{i}=\boldsymbol{V}+\boldsymbol{\omega}\wedge\boldsymbol{r}_{i}\]

V is the velocity of the body as a whole from an inertial system, and r_i is the position vector from the centre to the particular place where that part of the body is with respect to the common centre. Usually the "centre" is chosen to be the COM.

IOW, every part of the body moves according to all other parts. It would be crazy to deal with this as a system of actions-reactions.

2 minutes ago, John2020 said:

No need for a mathematical description at the moment.

Really? I think it's high time we talk some maths.

Otherwise you have infinitely many points of action and reaction. Don't you?

Edited by joigus
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10 minutes ago, joigus said:

IOW, every part of the body moves according to all other parts. It would be crazy to deal with this as a system of actions-reactions.

But in Fig.1-Upper we have the ability to accelerate the nut in one direction that implies from the moment it coincides with the whole system CoM, the CoM will change. The question is, is there something that will impede the acceleration of CoM (CoM change will take place whatever the case but its acceleration is the point of interest)?

Note: Speaking about CoM and as you also pointed out the same in a previous post of yours, the CoM of the whole coincides with the CoM of the nut.

10 minutes ago, joigus said:

It would be crazy to deal with this as a system of actions-reactions.

This is what is all about this thread and my paper.

Edited by John2020
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4 minutes ago, John2020 said:

This is what is all about this thread and my paper.

Wrong method, IMO. There are infinitely many action-reaction pairings in all the radial directions, resulting in no radial net force. Except the components along the axis.

But I wouldn't care to do that sum. Too much work, I suppose, if I ever tried that.

Much better to write 4 variables and start guessing constraints.

Anyway...

Later.

36 minutes ago, John2020 said:

Could you please tell me what is the answer on the following:"Now from the action-reaction principle, if the nut is being displayed to the right, what is being displayed in the opposite direction in order to hold Newton's 3rd law? "

Displaced you mean? The screw, of course. (In free space.)

But now you seem to have a gravitational field.

Anyway...

Later.

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Thanks @joigus, nice to see some real math entering the this thread again. I'm going to hijack some of your observations and use them in the muck more naive approach I'm currently using.

23 minutes ago, joigus said:

Otherwise you have infinitely many points of action and reaction.

@John2020 To simplify case with the vertical bolt further and maybe allow for drawing a picture with forces, would it be ok to have rolling (or sliding) balls in a helical configuration instead of continuous threads? Would such a design still posses the claimed properties of the reactionless drive? No need to define an exact amount of points, we just need to agree (or disagree) that the helical configuration of contact points is the important thing to circumvent newton rather than a continuous helix-shaped thread. I imagine a few rolls or similar inserted evenly spaced in the nut threads. 

 

Edited by Ghideon
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32 minutes ago, joigus said:

Wrong method, IMO. There are infinitely many action-reaction pairings in all the radial directions, resulting in no radial net force. Except the components along the axis.

My view on this (or what I am trying to share through this thread and through my paper) is, there is nothing impeding the acceleration of the nut that implies there is no internal reaction. How can that be? Because, the only way to trigger an internal reaction in the system is to permit mass transfer in the opposite direction that means the rest of the system, which is impossible. Why is that? Because the rest of the system is hold by the housings that hold the ends of the screw. Consequently, the rest of the system cannot be displaced (as happened with the nut).

32 minutes ago, joigus said:

But now you seem to have a gravitational field.

You are confusing the proposal of Ghideon with Fig.1-Upper. I am speaking about the Fig.1-Upper.

32 minutes ago, joigus said:

Displaced you mean? The screw, of course. (In free space.)

Of course not. In case of Fig.1-Upper, the screw will rotate in the other direction without displacing any kind of mass in that direction.

23 minutes ago, Ghideon said:

To simplify case with the vertical bolt further

I would suggest to take our attention only to Fig.1-Upper, otherwise many misunderstandings and confusing statements arise and makes this discussion difficult to follow.

Edited by John2020
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1 hour ago, John2020 said:

Let us ignore (b) and just take our attention to (a). Please allow me to ask you the following:

1.Do the threads of the nut follow a helical trajectory? Note: screw is fixed not rotating.

Yes, the threads do.

1 hour ago, John2020 said:

2.If (1) is true could we say, because of the translation screw only helical trajectory of the threads is allowed?

Sure. But a nut is more than threads.

1 hour ago, John2020 said:

3.When the nut evolves around the screw, a linear displacement is taking place, right? 

Yes. I have been saying this repeatedly.

1 hour ago, John2020 said:

4.Is the translation screw hold between two housings and limited between their space?

Well, I was just considering the nut and bolt, and knowing the nut is constrained as a boundary condition. The details aren’t really important. Nut sure, that’s what it looks like.

Quote

5.Now from the action-reaction principle, if the nut is being displayed to the right, what is being displayed in the opposite direction in order to hold Newton's 3rd law? 

The rest of the mechanism would move in the opposite direction, assuming it was free to move. The center of mass of the system would remain in the same spot if there is no external net force.

 

 

 

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27 minutes ago, John2020 said:

there is nothing impeding the acceleration of the nut

That is intuitively incorrect and/or lacking detail. The nut in fig 1 has mass and hence needs a force to be accelerated horizontally. Greater thread angle* means greater acceleration along the screw and greater torque for same angular acceleration of the screw.  

I'll think of this and maybe provide a better answer tomorrow; this is my intuitive reply, means there is plenty of room for error

Note, this post is referring to Fig1 but would apply to a vertical case as well.

 

27 minutes ago, John2020 said:

I would suggest to take our attention only to Fig.1-Upper because many misunderstanding and confusing statements arise, otherwise.

I have no problem keeping them separated. I'll soon return to the vertical case if thread is still open by then. Locating the exact spot where you think current mainstream use of Newton is not giving a correct prediction is still important to the discussion. 

 

*) note again; maybe not the correct technical term

  

Edited by Ghideon
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1 hour ago, Ghideon said:

No. The threads outside the bolt and inside the nut have helical shape. The centre of mass of the nut follows straight line while rotating. Same as a rotating bullet exiting a rifled* barrel of a gun. Rotating, not doing any helical movement.

That is an excellent example, with an obvious recoil.

1 hour ago, John2020 said:

Yes, but the thread of the nut (contacts) follows a helical trajectory. It is clear the CoM of the nut follows a straight line while rotating and I am not arguing about this.

It sure seems that you are

 

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Just now, swansont said:

The rest of the mechanism would move in the opposite direction, assuming it was free to move. The center of mass of the system would remain in the same spot if there is no external net force.

I am addressing now Fig.1-Upper (or what I am trying to share through this thread and through my paper) . There is nothing impeding the acceleration of the nut that implies there is no internal reaction. How can that be? Because, the only way to trigger an internal reaction in the system is to permit mass transfer in the opposite direction that means the rest of the system, which is impossible. Why is that? Because the rest of the system is hold by the housings that hold the ends of the screw. Consequently, the rest of the system cannot be displaced (as happened with the nut).

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