re: energy producing experiments

[Delburt Phend]

To make free energy from gravity you need to utilize two simple machines. We use the Atwood's    https://www.youtube.com/watch?v=4ovhEkSIqV0    to increase the quantity of momentum; this is done by increasing the quantity of time over which the force acts. This momentum is then transferred to a small object by using the despin event as seen in a satellite despin procedure   https://www.youtube.com/watch?v=uMQnXig2hrg     and in the double despin cylinder and spheres machine   https://www.youtube.com/watch?v=YaUmzekdxTQ.      Or     https://www.youtube.com/watch?v=KgT70vSIUgA

https://www.youtube.com/watch?v=w-7d66JscI8     

[Delburt Phend]

I set up an Atwood's machine that has two pulley radii on the same rotational shaft. The one radius is 10 .035 mm; half of the  20.07 mm shaft diameter.  The second radius is a wheel on the shaft with a radius of 97.2 mm.   97.195 mm / 10.035 mm =  9.6856

A 650 gram mass suspended on the shaft could be balanced with 68 grams suspended on the wheel.

I placed 650 grams on both ends of a string and draped it over the shaft. I accelerated this Atwood's arrangement with a certain mass on the shaft and had it trip a photo gate after one rotation. The average trip time of the photo gate was .0433 seconds.

I placed 68 grams on both ends of a string and draped the string over the wheel. I accelerated this same Atwood's with the same certain mass on the shaft and arranged it to trip the same photo gate, in the same manner, after one rotation. The average trip time of the photo gate was .04325 seconds.

This proves that a two radius Atwood's produces an F = ma outcome; as long as the longer radius has a proportionally smaller mass.  The 68 grams is moving 9.68 time faster than the 650 grams.  They have about the same momentum .068 kg times 9.68 is roughly equals to .650 kg times 1.

The two Atwood's arrangements do not produce the same quantity of energy: .068 kg at 9.68 has about 10 times as much energy as .650 kg at 1. But they had the same input energy.

Atwood's can make energy. 

[sm0ky2]

They did not have the same input energy.
One had a larger mass to be initially accelerated.


What was this quantity of " input energy"?

[Delburt Phend]

The quantity of mass being accelerated has nothing to do with the input energy.

The Atwood's was accelerated by the same mass suspended from the shaft. It was only about 62 grams.  The photo gate recorded a little less than one rotation. The flag was released just after the 2nd trip gate and it came around one rotation. So the 62 grams dropped (20.07 mm * pi) * 9.81 N/kg = .038 joule for input energy.

The same mass was used to accelerate the two 68 gram masses and the two 650 gram masses.  But the exact quantity of input force is not important; because the motion of the system without the added masses is the same for both the 136 grams and the 1300 grams.

The quantity of force needed to overcome bearing friction would be the same for both the 136 gram run and the 1300 gram run.

The quantity of force needed to set the shaft and wheel in motion (to .75 RPS) would be the same for both the 136 gram run and the 1300 gram run.

The quantity of force remaining to set the 136 grams, on the wheel, in motion would be the same as the quantity of force remaining to set the 1300 grams, at the shaft, in motion.

This remaining force accelerates the 1300 grams to .04769 m/sec. Or this same force accelerate 136 grams to .46189 m/sec. The 1.300 kg has .001478 joules of energy; and the .136 kg has .0145 joules of energy.  ½ mv²

.0145 J / .00148 J = 9.814 times as must energy for this portion of the Atwood's.

The quantity of momentum is the same 1.300 kg * .0477 m/sec = .062101 units:    .136 kg * .46189 m/sec = .06281 units of momentum     .06281 /.062101 = 1.01

[Delburt Phend]

Take a one meter tube and mount a quality bearing on one end: arrange it so that the tube rotates in a vertical plane. We will start from a horizontal position for the tube.

We place a force gauge at one decimeter from the point of rotation: the gauge reads a downward force.

The experimenter then places 10 kg at the one decimeter position. The scale or force gauge will read 10 kilograms or 98.1 newtons.

The experimenter then removes the ten kilograms and places a one kilogram mass at the 10 decimeter position. The scale or force gauge will read 10 kilograms or 98.1 newtons. The lever arm multiplies the force at the 1 decimeter position.

Now lets place the bearing and tube into a horizontal plane; and place it just above a frictionless plane so that the masses rest on the frictionless plane. In this arrangement gravity no longer has any affect upon the system; the tube is now rotating in a horizontal plane. The one and ten kilogram masses no longer exert force they only have inertia.

Now lets place ten kilograms of mass at the one decimeter position. Now lets apply ten newtons of force, in a clockwise direction, at the one decimeter position. After one second the ten kilograms will be moving 1 meter per second. 

Now lets place a one kilogram mass at the ten decimeter position. Then lets apply ten newtons of force in a clockwise direction at the one decimeter position.   After one second the one decimeter position will be moving 1 meter per second. The inertia of the one kilogram mass has been multiplied by ten at the one decimeter position just as the force had been multiplied by ten in the vertical experiment above. After one second the ten decimeter position will be moving 10 meter per second.

A 10 kilogram mass moving 1 meter per second has 5 joules of energy and a 1 kilogram mass moving 10 m/sec has 50 joules. But the same amount of force * time can be used to make 5 or 50 joules.

Conclusion: It takes the same amount of force to rotate a tube having different quantities of inertia along the length of the tube: as long as the length of the radius of rotation is inversely proportion to the differing quantities of inertia.

Example: ten kilograms at one decimeter rotates as easily as one kilogram at ten decimeters;

An experimental proof: is that an Atwood's finds it no more difficult to accelerate 136 grams at a 97.195 mm radius than it does to accelerate 1300 grams at 10.035 mm. And the 136 grams has 9.81 times as much energy as the 1300 grams.

Any suggestions on a corporation that would like to make use of this?