The Lee-Tseung Lead Out Theory

[Rosphere]

@ltseung888,

Maybe because it is past my bedtime, but I am having a hard time making solid connections between these two steps:



Slide 4. shows the 2 units of feedback, implying a dynamic system.  If so, from what bounds; is it starting from rest at a=0, up to some point a, and back to rest at a=0, where the velocity at the start and end of the cycle is zero?

Or, is it one cycle in a continuous movement where initial and final velocities do not equal zero?

Maybe a good nights sleep will help me see it tomorrow. 

[ltseung888]

@ltseung888,

Maybe because it is past my bedtime, but I am having a hard time making solid connections between these two steps:



Slide 4. shows the 2 units of feedback, implying a dynamic system.  If so, from what bounds; is it starting from rest at a=0, up to some point a, and back to rest at a=0, where the velocity at the start and end of the cycle is zero?

Or, is it one cycle in a continuous movement where initial and final velocities do not equal zero?

Maybe a good nights sleep will help me see it tomorrow. 

Note that 2 parts of horizontal pulse energy leads out 1 part gravitational energy. (In diagram labelled 3)

The energy equation (In diagram labelled 4) is effectively:

   2 parts Input horizontal pulse energy
      + 1 part Lead Out gravitational energy
= 3 parts Output Energy (if there were no loss)

Then from the Output, 2 parts are fed back as Input horizontal pulse energy

The above equation repeats.

The presentation slides assumes someone knowledgeable is presenting them.  May be the description file is better.  It is also updated to show the best pulse force (not horizontal but perpendicular to radius).
   

[tinu]

@ltseung888,

I began reading about your Lead-Out Theory this evening.  I started to read some items that you had posted way back in July, (when I was not active on this forum.)  I followed your link to http://www.energyfromair.com/beijing/Taiwan2a.files/frame.htm.

On this page I found some equations and values.  So, I generated a spreadsheet using this data, (and some stuff I picked at Lawrence Tech.,) in order to better consider the two energy terms.  As you can see, I have the spreadsheet calculating values at each one degree.

I then inserted a couple rows at the 9.48 degree value given at the web site; the top of which showing what the force would be at this given angle, 9.882 kg, and the one below is the value also given at the web site, 10kg.

...

As I wrote, I only started reading/studying this evening.  So far, so good.  Your numbers in this example jibe with my spreadsheet,... mostly.  I would not bring this up if the ratio had been given as 2.000.  However, as you can also see from the spreadsheet, this ratio approaches 2 as a limit going back to the zero angle.  I know it is not much above the limit, but it is the fact that it is above the limit that compels the anal retentive nature in me to point this out to you.  That is all.  Do with it what you will. 

I plan to continue reading more about your theory over the next few days.  What I have read, so far, is making sense.

Thank you for sharing with this forum. 

Rosphere--better late to the party then never

Excellent work, Rosphere!
I refer mostly to the chart, which is eloquent for why there is no subject to be seriously discussed into this thread.

Please take a look much earlier into this thread, close to its beginning: I?ve explained that what Mr Tseung calls ?Horizontal energy? is, in fact, the total spent energy. (That?s a very elementary fact.) It is total energy because there is no way one can horizontally deviate the weight from vertical without lifting it. Lifting <is achieved because of> / <it is a consequence/ a constraint of using> the rigid string BUT the energy comes from the pulling person.

After accepting that 'Hori energy' equation is wrong, notice that the values in the chart makes no physical sense. The computed ratio means total energy spent divided by the energy that goes into vertical lift (increase in potential energy). No physical significance whatsoever. No overunity either, just misuse of equations.

But if one reverses the ratio presented in the chart, now the new values would mean potential energy gained versus total energy spent. This may be useable for mental exercises. So, as example let?s take the angle =9.48. One has to spend 1.628 Joule to bring the weight from vertical to that angle, by pulling with a CONSTANT force of 9.882 Newton. By the end of the movement he finds that only 0.819 Joule were actually used for lifting the weight (this amount goes into potential energy). Well?! Is some energy being lost?! Where did the other half of the energy go?! The answer is very simple: it went into kinetic energy. And the pendulum will further move up by itself due to its stored kinetic energy.

Note the word ?constant? above written in capitals. It is important because the simple equations Mr. Tseung uses are no longer valid if the force is not kept constant. Because of that initial assumption already made, any distractions derived from pulsed forces and similar ?improvements of theory? can not be discussed under the above mathematical treatment (which, I must stress again, it is physically wrong.). But the required mathematical analysis never came and I suspect it will never do.

Hope I made myself clear,
Tinu

[RD Edwards]

Linda:
Thank you! You are eloquent, and right to the point. Thank you for your well thought out questions to "Dr' Tseung.

"Bunkers" with zero Credibility get that way by Leading Out Bunk in Too Many Simultaneous Conflicting Statements and Directions

freaking hilarious, thanks:)

[ltseung888]

Please take a look much earlier into this thread, close to its beginning: I?ve explained that what Mr Tseung calls ?Horizontal energy? is, in fact, the total spent energy. (That?s a very elementary fact.) It is total energy because there is no way one can horizontally deviate the weight from vertical without lifting it. Lifting <is achieved because of> / <it is a consequence/ a constraint of using> the rigid string BUT the energy comes from the pulling person. ***This is where you were taught wrong.***

*** Vertical Work cannot be done by a Horizontal Force without a machine to convert its direction. ***

Dear Tinu,

I finally know where you have the misunderstanding.  In Physics, we have the following Newtonian Equations:

(1)   Work done = Force x Displacement (Vector Mathematics)

(2)   Both Force and Displacement are vector quantities meaning that they have direction. 

(3)   If the Displacement is not in the same direction as the Force, no work is done.  This means a horizontal force cannot do work in the vertical direction unless a machine is used to change the direction of the Force.

(4)   Work done is associated with Energy.  Work is a scalar quantity.  So is Energy.

(5)   In the case of the first pull of the pendulum by a Horizontal Force, there are three Forces on the Pendulum. Namely Horizontal Force, Tension in the String and Gravitation Force or the Weight.

(6)   In Physics, we have to consider the effect of these Forces on the Pendulum.  The Weight can be considered constant.  The Horizontal Force is externally applied.  I assume that we can be extremely clever and keep it perfectly horizontal and constant.  The Tension of the Rigid String will vary.  It will definitely be greater than the Weight. 

(7)   When three forces act on an object in a two-dimensional plane as in the simple pendulum with a pull, we can use the Parallelogram of Forces to calculate the resultant force.  We can examine the work done by each of these forces.

(8 )   If we can apply a perfectly Horizontal Force, we can never do vertical work without some kind of a machine as required by the Laws of Physics.

(9)   When we examine the ?machine elements? in a simple pendulum, we can easily see that the increased tension of the String is responsible in lifting the Weight.  This is the Lead Out Energy.

(10)   Such Physics Laws and their applications cannot be wrong.  One can improve the analysis by considering the best pull force.  The best pull force is not horizontal in all cases.  In fact it is perpendicular to the Radius or Tangential to Motion.

Let us focus on the Physics and drop the Insult Training.

Lawrence Tseung
Vertical Work cannot be done by a Horizontal Force without a machine to convert its direction.   A machine can be a pulley, an inclined plane etc.