The Lee-Tseung Lead Out Theory

[ltseung888]

Another personal email

Hi Lawrence,

  You do not know me, so let me introduce myself:
  I am XXX a no-post dormant member. Still a Newbie, and I
  intend to stay a Newbie.

  But, would you so nice to send me the original formulation
  of the celebrated Lead Out Theory in an e-mail?
  It would be if you sent in in an attachment: pdf or word doc.

It would be be greatly appreciated.
Regards,
  XXX

Dear XXX,

Please refer to the two documents in the post by M@rcel (reply 1906)

(1)   http://www.wipo.int/pctdb/en/wo.jsp?wo=2006077451
This is our patent application document.

(2)   http://www.wipo.int/ipdl/IPDL-IMAGES/PATENTSCOPE/74/32/bd/0032bd.pdf
This is our reply to the patent examiner.  You can also click on the document tab in (1) to get this document.

Do not be intimidated by the debunkers.  Ignore their insults and jeers.  Some of them are doing their paid jobs.

[exnihiloest]

...
(2)   http://www.wipo.int/ipdl/IPDL-IMAGES/PATENTSCOPE/74/32/bd/0032bd.pdf
...

"If we give a horizontal push or pulse force to a stationary pendulum at the pendulum.
weight, this pulse force can be treated as the initial force that supplies the initial
energy. This horizontal pulse force immediately increases the tension of the string."

It's right.

"This increased tension will do work against gravity to lift the pendulum up. In other
words, the pulse force immediately causes the tension of the string to do work against
gravity."

It's wrong. The tension of the string does no work because there is no motion along the string.
F.dl=0 as dl=0 in the direction of the force vector (tension) of the string.

[Tribune]

Lee.

So to look at from an alternate perspective logically it could be said that you have not gained anything but infact manipulated gravity to save you something; Gaining a better than expected result given the current rules for the input.

Does that not seem similar to the Gyroscope. given the mass of the object; the force required to keep it upright normally would be greater than the force required (to spin it) for the duration that it remains upright.

Still the fun bit is how to use the re-directed gravity for something practical. And what percentage of this "Gravity" is re-directed by various experiments.

Trib.

[ltseung888]

...
(2)   http://www.wipo.int/ipdl/IPDL-IMAGES/PATENTSCOPE/74/32/bd/0032bd.pdf
...

"If we give a horizontal push or pulse force to a stationary pendulum at the pendulum.
weight, this pulse force can be treated as the initial force that supplies the initial
energy. This horizontal pulse force immediately increases the tension of the string."

It's right.

"This increased tension will do work against gravity to lift the pendulum up. In other
words, the pulse force immediately causes the tension of the string to do work against
gravity."

It's wrong. The tension of the string does no work because there is no motion along the string.
F.dl=0 as dl=0 in the direction of the force vector (tension) of the string.

Dear exnihiloest,

We finally have some good physics discussions. 

You correctly stated that there is no motion along the string or that the string itself does not stretch. However, there is movement of the string from its original to its new position.

The Tension of the string itself can be resolved in the vertical and the horizontal direction. The vertical component is T1cos(a) which is equal and opposite to the weight Mg.

The horizontal component is T1sin(a) which is equal and opposite to the horizontal force F.

The displacement of the string or bob can also be resolved into the vertical and the horizontal directions.  The vertical component is dH and in the same direction of the vertical force component of the string.  Thus vertical work must be done.  The value is T1cos(a) x L(1-cos(a)) or T1 x L x cos(a) x (1-cos(a).  If we substitute the actual values where M = 60 kg; g = 9.8 m/s/s; F=1/6 Mg; angle a = 9.46 degrees; L = 1 meter; we get the value (as in the post by Top Gun, reply1906), 0.8133 Joules.

The horizontal component of the displacement is dX and is in the opposite direction of the horizontal component of the force (tension) of the string.  In Physics, that is regarded as negative work done.  Negative work done can be viewed as energy stored in the system.  That negative work is 1.610728 Joules.

[Cap-Z-ro]

As much as I do not wish to make this report...it appears the head of Mr.Tsung has been photo shopped on to the body of another speaker at the conference.

If you look closely at the photo you will notice that whoever did the job inadvertently left the head that was once where Lawrence's is now situated in the frame in amongst the flower arrangement over to the left of Lawrence's head.

*Always clean up once you complete your task, and a similar faux pas will not occur to you.

Regards...