The Lee-Tseung Lead Out Theory

[Pirate88179]

@ TinselKoala:

Excellent information and explanations sir.  You know, I would love to see what the Secret Service might do to Lawrence if he was on the White house lawn with a tube shaped device containing a coil with wires coming out of it.  Here is a question:  Can you Lead out gravitational energy while being wrestled to the ground by Secret Service agents?  When you fall to the ground, are you falling faster than 9.9 cm/sec/sec due to the additional energy being introduced to the system by the 5 agents?  Maybe we will see.

Bill

[TinselKoala]

I don't know why I'm doing their research for them--unless it's because I so resent being lied to by fools and idiots.
Now, Devil and LTseung and TopGun, how is this device different, and why is Naudin's work not an exact enough replication of experiment001, and why are his conclusions incorrect?
Don't challenge me or my education, but rather confront the issues involved and refute Naudin, if you can.

From JLN's website:

(begin quote)
"Test Results  : When spring is released by the burning of the nylon wire, the Impulse Drive ( its support ) moves 5 mm from the left to the right but the internal moving mass M has shifted 6 mm from the right to the left, and we have : 84 g x 6 mm ~ 104 g x 5mm.

Conclusion : Today, I am able to confirm the effect observed in the Bull's Impulse engine with this simple experiment: After that the spring is released the device has translated 5 mm in one direction.

But, after a short analysis of the photos above you will notice that there is no motion of the center of mass in spite of the apparent motion of the device. This is only a 'one-shoot' experiment because the device needs to be resetted before each test, the net impulse in the Bull's device sums to zero over a complete cycle ( when the moving masses return to its initial position at the end the cycle ). "
(end quote)

[hansvonlieven]

OK. Lawrence, Top Gun, Devil, Forever and Whoever

Let's see how good you are with your own formula.

You say:

The lead-out-energy from a pendulum is equal to the vertical component of the tension of the string multiplied by the vertical displacement.

Since this is your own formula it should be easy for you to demonstrate the following:

Please calculate the amount of available Lead Out energy for the following pendulum.

Location : Hong Kong
Length of pendulum from fulcrum to centre of gravity of pendulum bob: 2 Meters
Weight of pendulum bob: 1 Kilogram
Gravitational Acceleration for Kong Kong: 9.790 m/s²

You can specify your own values for the Lee Tseung Pull.

Show us.

Hans von Lieven

[Top Gun]

OK. Lawrence, Top Gun, Devil, Forever and Whoever

Let's see how good you are with your own formula.

You say:

The lead-out-energy from a pendulum is equal to the vertical component of the tension of the string multiplied by the vertical displacement.

Since this is your own formula it should be easy for you to demonstrate the following:

Please calculate the amount of available Lead Out energy for the following pendulum.

Location : Hong Kong
Length of pendulum from fulcrum to centre of gravity of pendulum bob: 2 Meters
Weight of pendulum bob: 1 Kilogram
Gravitational Acceleration for Kong Kong: 9.790 m/s²

You can specify your own values for the Lee Tseung Pull.

Show us.

Hans von Lieven

Please refer to http://hk.geocities.com/winghang20022002/Presentation.htm#presentation

Start at the 10th slide and finish at slide 14 for the detailed physics explanations.

I shall use a horizontal pull at 1/10 the weight.  (Or 0.1 of the 1Kg bob)

Mathematical steps:
(1)   At equilibrium, the three forces tension T, weight Mg and Horizontal Force F will obey the Law of Parallelogram of forces.  The displaced angle a can be calculated from tan(a) = F/mg or tan(a) = 0.1.  From the trigonometry tables, a = 5.711 degrees.
(2)   The Old tension before the horizontal force = T = Mg = 1 x 9.790 Newtons.
(3)   The New tension after the horizontal force = T1 =Mg/cos(a)   
= 9.790/0.995036 = 9.838835 Newtons.  This is slightly higher than T.
(4)   The horizontal displacement = dX = Lsin(a) = 2 x 0.099511
= 0.199022 meters
(5)   The vertical displacement = dH = L (1-cos(a)) = 2 x (1-0.995036)
= 0.009927 meters

(1)   The horizontal work done = F x dX = 0.1 x 1 x 9.790 x 0.199022
= 0.194842 Newton-Meters
(2)   The vertical work done = Mg x dH = 1 x 9.790 x 0.009927
= 0.097186 Newton-Meters
(3)   The ratio of horizontal work over vertical work = 0.194842/0.097186
= 2.004844

Thus in the case of the First Lee-Tseung Pull using the supplied values, approximately 2 parts of horizontally supplied energy will lead out 1 part of vertical gravitational energy.  The Lee-Tseung lead-out theory cannot be wrong.  The Physics and Mathematics cannot be wrong.

*** The attached file shows the same thing but formatted better.  ***

[Pirate88179]

Quote from TopGun...

"Thus in the case of the First Lee-Tseung Pull using the supplied values, approximately 2 parts of horizontally supplied energy will lead out 1 part of vertical gravitational energy.  The Lee-Tseung lead-out theory cannot be wrong.  The Physics and Mathematics cannot be wrong."

Once again, I don't see any where that you account for the energy required to supply the first Lee-Tseung pull.  Maybe I missed something but I sure didn't see it in your supplied numbers.  I have said the same thing to Lawrence many times when he posts these types of numbers.

Bill