re: energy producing experiments

[Delburt Phend]

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 

[Kator01]

Delburt,
wow, 12 years have passed by...you remember ?
For those here to better understand how the dynamics of your system works we go back 12 years where you posted the pictures of the evolving motion in series including description: go to reply#116, January 18th

https://overunity.com/1995/free-energy-from-gravitation-using-newtonian-physic/105/
I am still scratching my head...to find a technical solution
Mike

[Delburt Phend]

A pulley with two radius lengths can be used to make energy. As in a double Atwoods. When the ratio of the length of the long radius to the length of the short drive radius is greater than the square root of the mass ratio between the accelerated mass over the drive mass. The speed of the accelerated mass throws it beyond the distance dropped.

[Delburt Phend]

The ratio of accelerated mass to drive mass would be; as in the above experiment 1318.75 g / 8.75 g.  And then you place a 1 / 12.28 mass at 12.28 times the distance.

[Kator01]

Delburt,
found this simulation-applet, however you can not vary the radius. I myself can not run the simulation , my Linux-System misses the Java-code
https://demonstrations.wolfram.com/DoubleAtwoodMachine/
Now concerning this guy here:
https://www.asc.ohio-state.edu/durkin.2/treb2.htm
Lets do some energy calculations:
Mass of the Brick 1.5 kg, g 0.98 kg*m/sec exp2,
Mass of golfball 0.046 Kg, speed 16 m/s
From the video I only can guess the hight of the falling brick is about 0.8 m
Potential energy of the brick : m * g + h = 1.5 x 9.8 x 0.8 = 11.76 JEnergy of the golfball : 1/2x 0.046 x 16 exp2 = 5.88 J
So this system is not showing energy-increase. I assume it misses the right point in time to release the ball.He mentioned that - accoding to his observation - the brick almost stood still at times while falling.
I would really find more about this Mr. Gordon here : http://garydgordon.com/Despin/
especially how he connected the pucs at the rim so they could be catapulted tangentially.
Good find anyway

Mike