re: energy producing experiments

[Delburt Phend]

https://pisrv1.am14.uni-tuebingen.de/~hehl/Drehimpuls.pdf

The last two are of interest; and the labrat disproves his own theory.

You have to click on the graph to make the video play.

[Delburt Phend]

This is about a discussion on the cylinder and spheres experiments.

If you were to shoot a 1 kilogram mass moving 20 m into a 399 kilogram block; the block will not start moving 1 m/sec. This is the conservation of energy and it will not happen; as you know. But you are proposing that you put the one kilogram on a string and then it will happen; but it won't, the string will not make it happen.

You lose it all to heat; right. That is what the theory says, so stick with it.

It takes four frames to cross the black square at the start and four frames after the spheres return the motion. The spheres are much smaller than the cylinder and they can not return the energy to the cylinder. The spheres return the momentum to the cylinder, which means they received the momentum from the cylinder.

The Paul Nord experiment proves that angular momentum conservation does not work in the lab. That leaves you with only linear Newtonian momentum. And if Newton is correct this leaves us with free energy.

Circular Motion - YouTube

https://www.youtube.com/watch?v=zww3IIMRo4U

[telecom]

This is about an Atwood's machine with 199.5 kg on one side and 200.5 on the other.

Yes; the difference between M1 and M2 is 1 kg; and the sum is 400 kg. That means that the acceleration 'a' is 1/400 times 9.81 m/sec/sec.  a = .024525 m/sec²

Now the formula for time is used; d = ½ at²   With a drop distance of one meter and an acceleration of .024525 m/sec² we have a drop time of 9.03 seconds.

Doing the same for free fall we get a drop time of .4515 seconds.

9.03 sec / .4515 sec = 20

The force of 9.81 N is applied for 20 times as long in the (400 kg to 1 kg) Atwood's machine.

The final velocity of the Atwood's is found by this formula;  d = ½ v²/a;   this is just d = ½ at²  where v/a is substituted for t.

The final velocity for one meter of free fall is 4.429 m/sec; The final velocity for the 400 to one Atwood's is .22147 m/sec.

This is 4.429 (1 kg * 4.429 m/sec) units of momentum for free fall and (400 kg * .22147 m/sec) 88.589 units of momentum for the Atwood's. This is 20 times as much momentum produced; and 20 times as much time over which the force acts.

It takes 4.429 units of momentum to return the one kilogram of imbalance to the original position of the Atwood's.  And you get 88.589 out.

Now this is the quantity of momentum that Newton said was conserved and you put 4.429 units in and you get 88.589 units out. Every cycle; 4.429 in 88.589 out.

All that is needed is to transfer the motion of the 400 kilograms into the one kilogram: and that can be done with a cylinder and spheres machine.

There is also a modified Atwood's that eliminate the up and down motion; if that bother some.

Can this transfer be done by the different means?
For example using m * v = F * dt
In your case for the t= 1 sec F is equal 80 N
The piston of the hydraulic cylinder can be pressed with 80 N for 1 sec, which can be used to turn the generator...
I can be wrong, though.
Tnx

[Delburt Phend]

I did an air table experiment today were I used the center pin of the air table to interrupt the string on a puck with a 108 cm radius. After the string hit the pin; the new radius was 14.9 cm. This is a radius change of 108 cm to 14.9 cm = 7.25.   108 cm is 7.25 times larger than 14.9 cm.

Therefore if angular momentum is to remain the same the linear velocity of the puck will have to increase by a factor of 7.25.

I did a video and I placed this video on my computer; I put the video in a frame by frame mode and counted the frames needed to cross a distance of 2.86 cm on the puck. It took 21 frames to cross this distance on the puck when the radius was 108 cm and when the radius was 14.9 cm.   At 240 frames per second this is a speed of .327 m/sec for both the small circle and the large circle. There is no change in velocity when you go from the large circle to the small circle.

I did it in the opposite direction as well; going from the small 14.9 cm circle to the 108 cm circle.  There is no change in velocity when you go from the small circle to the large circle.

Now I will try to post the video.

[Delburt Phend]

https://www.youtube.com/watch?v=egP0bjVi2ss