Delburth,


great, thank you,really  helped me.


Mike

Some data from the pulley.

The gate trip time of a 35 mm Atwood’s pulley with 60 g on each side equal to the trip times for 105 mm Atwood’s with 20 g on each side; and 20 g on the 105 side and 60 g on 35 mm side? 

I placed 80 g (77.366 g) on the left side and 60 g (59.185 g) on the right side at the 35 mm diameter of the 4 radii pulley. The trip time of the photo gate after a quarter rotation were 0571, .0565, .0561 sec.  Average .0566 sec

I placed 20.04 g on the left side at the 35 mm diameter of the 4 radii pulley. And I placed 19.8 g on the left side at the 105 mm diameter and 19.8 g on the right side at the 105 mm diameter of the 4 radii pulley. The trip times of the photo gate after a quarter rotation were .0568, .0560, .0571, .0569, .0564 sec. Average .0566 sec

I placed 78.84 g on the left side at the 35 mm diameter of the 4 radii pulley. And I placed 20.23 g on the right side at the 105 mm diameter of the 4 radii pulley.
The trip times of the photo gate after a quarter rotation were .0568, .0563, .0568 sec. Average .0566 sec

All the 20 g drive masses are at the 35 mm diameter. So the 20g at the 105 mm diameter is equal to the 60 g at the 35 mm diameter.

 I took the position of proving that it was (105 mm / 35 mm = 3   3² equal 9)  9 times harder to rotate the balanced masses at the 105 mm diameter. That would mean that  1/9 the mass of 60 g each side at 35; would be equal to 6.66 g each side at 105 mm. So I placed 6.960 grams on the right and 6.960 grams on the left side at the 105 mm diameter. I accelerated this with 20.04 g at the 35 mm diameter.

The same arrangement tripped at .0468, .0468, .0467 sec instead of the .0566 sec for the 60 g on each side at the 35 mm diameter.

So mr is off by a few 10,000th (.0566 to .0566) and mr² is off by 100 / 10,000th (.0468 to .0566).  This means that the likelihood of mr being correct is about 99.9% and the probability of mr² being correct is somewhere around zero.

And this proves that energy is not a conserved quantity.

Magbet motor  https://www.youtube.com/watch?v=shjp6uyp3ns&list=PLpuBp-WaqFobl1olMsCLo1hzwvTIewbvj&index=3

Compare NASA despin quote: “The relatively small weights have a large effect since they are far from the spin axis, and their effect increases as the square of the length of the cables”. And Prof Quote: “Why should string lengths matter---? How foolish”.

This is a quote from NASA concerning the Dawn Mission despin. The Satellite mass was 1420 kg and the released masses were less that 3 kg. I simplify this to a 400 kg rim 1 m in diameter spinning with and arc speed of 1 m/sec. This rim has two ½ kg mass unwinding from both sides. This unwinding stops and reverses the satellite’s spin.

The cylinder and spheres machines are the same experiment, and they show that the original spin of the center mass will be restored if the masses are allowed to rewrap around the center mass.

If the original arc speed of the 400 kg rim was 1 m/sec then “squaring the radius as NASA predict allows for the sphere (two ½ kg masses) masses to have a speed of only 16 m/sec (1kg * v * 12.5 m cable extended / .5 m cable wrapped).     

If the original arc speed of the rim was 1 m/sec then energy conservation would allow a sphere velocity of 20 m/sec.  ½ * 400 kg * 1 m/sec * 1 m/sec = 200 J and ½ * 1 kg * 20 m/sec * 20 m/sec = 200 J   

But we know that there is no such law as the law of conservation of kinetic energy and in fact we know that 1 kg moving only 20 m/sec can not give 400 kg a velocity of 1 m /sec. So kinetic energy conservation is not a logical option.

So: “Why should string lengths matter---? How foolish”. Why should 1 kg moving 16 m/sec have more momentum when attached to a 12.5 m cable as opposed to being attached to a .5 m cable? How foolish!!!

Let's ask it another way. What takes 1.6 N applied for 10 seconds to stop: a 1 kg mass moving 16 m/sec on a 12.5 m string or 1 kg moving 16 m/sec on a .5 m string?

A 1 kg mass moving 16 or 20 m/sec can not give a 400 kg mass an arc speed of 1 m/sec. It can not return the motion. This 400 kg at 1 m/sec is 400 units of Newtonian momentum and it will take 400 kg*m/sec of momentum to return the original motion.

When the 1 kg is moving 400 m/sec it has 400 times as much energy as the 400 kg moving 1 m/sec.    ½ * 1 kg * 400 m/sec * 400 m/sec = 80,000 J and ½ * 400 kg * 1 m/sec * 1 m/sec = 200 J

This makes the 'cylinder and spheres' experiments worth several trillion dollars.