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Author Topic: re: energy producing experiments  (Read 40622 times)

Offline Delburt Phend

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Re: re: energy producing experiments
« Reply #180 on: January 20, 2020, 04:37:31 PM »
In an Atwood's 8.75 grams can accelerate a balanced mass of 1310 grams to a velocity of .09128 m/sec after the 8.75 grams has dropped .064 meters. This occurs when the accelerated mass of  1310 grams is at the same radius as the 8.75 g drive (the mass that provides the force) mass. In this experiment the drive mass is at a 10 mm radius; and the 1310 grams of balanced mass is at the same location.

In an Atwood's 8.75 grams can accelerate a balanced mass of 135 grams to a velocity of .885 m/sec after the 8.75 grams has dropped .064 meters. This occurs when the accelerated mass of  135 grams is at a radius 97 mm while the 8.75 g drive mass remains at 10 mm. The 8.75 grams is still moving .09128 m/sec and the 135 g is moving 9.7 times faster.

The 1310 grams has a momentum of 1.310 kg * .09128 m/sec = .11958 units

The 135 grams has a momentum of .135 kg * .885 m/sec = .11948 units

The 8.75 grams has dropped .064 m; To restore its original position you would have to throw it upwards .064 m. A velocity of 1.12 m/sec is needed to throw a mass upwards .064 m. So .00875 kg * 1.12 m/sec is the needed momentum to restore the original position of the drive mass.

The momentum needed to restore the original position of the 8.75 grams is .00875 kg * 1.1206 m/sec =  .0098005 units

The quantity of momentum available in the 1310 grams or the 135 grams is 12.19 times that which is necessary to recycle the system.  .11952  / .009805

This is possible because the time over which the force acts in a free fall of .064 m is .114 sec.; and the time over which the force acts to produce the motion of the 1310 grams or the 135 grams is 1.4023 seconds.    1.4023 sec  / .114 sec = 12.33 

The 1310 is at a radius of 10 mm. And 1/9.7ths of that mass (135 g) is at 9.7 times that radial distance (at 97 mm).

What would prevent us from placing 1/12.28ths the mass (106.67 g) at 12.28 times (122.8 mm) the distance?

The 106.67 grams would be moving 12.28 times faster than the 8.75 grams moving at .09128 m/sec.  You would have 106.67 grams moving 1.12 m/sec.

106.67 grams moving (.09128 m/sec * 12.28) 1.12 m/sec would have .10667 * 1.12 m/sec = .11947 units of momentum.

The 1.12 m/sec velocity is sufficient to reload the system and you have 106.67 grams instead of 8.75 grams.

Or how about 26.2 grams at a radius of 500 mm?    50 * .09128 m/sec = 4.564 m/sec 

Free Energy | searching for free energy and discussing free energy

Re: re: energy producing experiments
« Reply #180 on: January 20, 2020, 04:37:31 PM »

 

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