The Lee-Tseung Lead Out Theory

[utilitarian]

Are there any relevant constructive comments or suggestions that I missed?

To your credit, you can be very methodical, as demonstrated here.

Since you are asking, I have two comments. 

(1) You never did respond to my analogy.  The analogy is this:

Instead of a pendulum, we have a hollow pipe (or even half-pipe, like the skaters use), fixed in place so it does not roll.  Ball rests at bottom, and you give it a push, and the ball rises some distance.  In this situation, has the pipe done work?

(2)  Another analogy, and I am sure this was raised earlier, given 2000 posts, but bear with me, as I am not clear on this. 

Scenario A:  You take a pendulum, start it horizontal, let go, and let it drop and then rise on the opposite side.  What is the difference during the upswing between this and:

Scenario B:  Take a pendulum at rest (ball is at bottom) and give it a push exactly equal to what it would experience at that point had you done Scenario A.

I hope you see what I am getting at.  If there is no functional difference for the ball during the upswing, then it would be trivial to test whether or not energy is lead out.  If the ball gets to a higher point then where you dropped it from, then energy is lead out.  Otherwise, no energy is lead out.

I welcome your comments.

[ltseung888]

Are there any relevant constructive comments or suggestions that I missed?

To your credit, you can be very methodical, as demonstrated here.

Since you are asking, I have two comments. 

(1) You never did respond to my analogy.  The analogy is this:

Instead of a pendulum, we have a hollow pipe (or even half-pipe, like the skaters use), fixed in place so it does not roll.  Ball rests at bottom, and you give it a push, and the ball rises some distance.  In this situation, has the pipe done work?

(one point or one analogy at a time)

I welcome your comments.

Dear utilitarian,

It is always difficult to consider a new scenario and try to compare it with an existing one that one has done months or years of research.  Let me focus on your analogy of the half pipe.  The mathematics of a circular path is complex.  So I am using the inclined plane as a simplification.  One can give the ball a horizontal push and hope to keep it up or move it up.  However, there is a mathematical requirement that the horizontal force F must be large enough so that Fcos(a) is greater than Mgsin(a).  Please see the attached figure.

This condition is not required in the case of the Pendulum.  Any small horizontal pull at rest position or at the point of maximum swing can make the Pendulum rise farther.

Thus the two scenarios are different.  Different mathematics and different analysis are required.  Without doing the detailed analysis of the Pulsed Incline Plane or half pipe, I would not and should not comment on the actual outcome. I believe that the curvature of the half pipe may be an important factor in the analysis.  I shall repeat that the two scenarios are different.  The same mathematics cannot be applied in these two different scenarios.  Thus, whether the pipe does work is irrelevant.  (The Pulsed Incline Plane or the half pipe might not be an OU device at all.  The pipe will not do work.)

This is good physics discussion and should be encouraged in this forum (instead of the abusive insults).  We can all have our brains ?stretched? and learn something.

[utilitarian]

Are there any relevant constructive comments or suggestions that I missed?

To your credit, you can be very methodical, as demonstrated here.

Since you are asking, I have two comments. 

(1) You never did respond to my analogy.  The analogy is this:

Instead of a pendulum, we have a hollow pipe (or even half-pipe, like the skaters use), fixed in place so it does not roll.  Ball rests at bottom, and you give it a push, and the ball rises some distance.  In this situation, has the pipe done work?

(one point or one analogy at a time)

I welcome your comments.

Dear utilitarian,

It is always difficult to consider a new scenario and try to compare it with an existing one that one has done months or years of research.  Let me focus on your analogy of the half pipe.  The mathematics of a circular path is complex.  So I am using the inclined plane as a simplification.  One can give the ball a horizontal push and hope to keep it up or move it up.  However, there is a mathematical requirement that the horizontal force F must be large enough so that Fcos(a) is greater than Mgsin(a).  Please see the attached figure.

This condition is not required in the case of the Pendulum.  Any small horizontal pull at rest position or at the point of maximum swing can make the Pendulum rise farther.

You did a good job showing why an inclined plane is different from a pendulum, but you have not at all shown why a perfect half-circle halfpipe is different from a pendulum.  The only assumption we would need to make is there is zero surface friction.  Just like any horizontal force at all is sufficient to move a pendulum, so is any horizontal force sufficient to move a ball in a halfpipe, assuming zero surface friction.

The pendulum and the halfpipe are identical in all respects but one.  All they do is change the direction of the force.  They do no work.  The only difference between them is that the pendulum changes the ball's direction from the top, while the halfpipe does so from the bottom of the ball.

And, I assume you will get to my second analogy later?

[ltseung888]

You did a good job showing why an inclined plane is different from a pendulum, but you have not at all shown why a perfect half-circle halfpipe is different from a pendulum.  The only assumption we would need to make is there is zero surface friction.  Just like any horizontal force at all is sufficient to move a pendulum, so is any horizontal force sufficient to move a ball in a halfpipe, assuming zero surface friction.

The pendulum and the halfpipe are identical in all respects but one.  All they do is change the direction of the force.  They do no work.  The only difference between them is that the pendulum changes the ball's direction from the top, while the halfpipe does so from the bottom of the ball.
And, I assume you will get to my second analogy later?

Dear utilitarian,

You pointed one important element - friction.  In the pendulum, we can assume that is almost zero.  In the case of the half pipe, we cannot have a true frictionless surface in practice.  Friction implies that energy will be "lost" as heat.

In the Lee-Tseung mathematical analysis of the pendulum, we can have a tiny Horizontal Force F to start and repeat it to add energy to the system.  If you use the spreadsheet to do the analysis, you can find out that the higher efficiency is obtained for the smaller angle (or smaller horizontal force).  That implies the best Lee-Tseung Pull mechanism should be tiny but frequent.  In the inclined plane, we need a minimum Horizontal Force F to start motion even if we assume no friction.  In the case of the half pipe, without the detailed mathematical analysis, can we say that the analogy is closer to the pendulum than to the inclined plane.  Or would it be somewhere in between?  How would the radius of the curvature affect the calculation?

Can you help to do the mathematical analysis?  It is much more complicated than the pendulum or the inclined plane.  However, if that is not done and we just assume the half pipe is identical in theoretical treatment as the simple pendulum, we cannot call ourselves true scientists.

To repeat, I am worried about the ?minimum force? element in the case of the inclined plane.  Will the same ?minimum force? element be required in the case of the half pipe?

Let us settle one issue after another.  There is no hurry.  This type of scientific analysis or reasoning needs much more thinking.

In order for gravitational energy to be Lead Out, a simple pendulum will not do.  We need a pulsed (or more exact, a Lee-Tseung Pulled) pendulum.  We had discussion on why a simple elastic collision is NOT a Lee-Tseung Pull some weeks ago.  I know how to apply a Lee-Tseung Pull to a pendulum but I have not identified the mechanism to apply a Lee-Tseung Pull to a half pipe yet.

[Top Gun]

I cannot resist in joining the fun.

Utilitarian may be focusing on the comparison between a simple pendulum and a frictionless half pipe.

Tseung may be thinking of the Pulsed or the Lee-Tseung Pulled pendulum and how can that be implemented or compared in the frictionless half pipe.

I want like to add that a subsequent Lee-Tseung Pull after the first Displacement will Lead Out gravitational energy.  However, a subsequent Lee-Tseung Push with the same magnitude as the Lee-Tseung Pull will NOT Lead Out gravitational energy. 

If we examine the equations outlined in Slides 10-12 in detail, the Lee-Tseung Pull is used to increase tension of the string by pulling it further from the vertical.  That will make the string do more negative work (or store more energy in the pendulum system.)  If the pull becomes a push, the tension of the string will be reduced because the effect is to move the string closer to the vertical.  This will have a totally different effect.

The caution from Tseung is justified.  If we assume that the ?Pulsed or Lee-Tseung Pulled pendulum? is identical in theoretical analysis as the ?Pulsed half pipe?, we are not true scientists.  That assumption is definitely suspect according to the above analysis.  Let us do the vigorous scientific analysis before making such an assumption.  We may learn more in such an analysis.