re: energy producing experiments

[Delburt Phend]

I used a larger diameter (129 mm) cylinder and spheres that was a  16.36 to one mass ratio. I used the new formula for determining tether length. That is (mass ratio) / (2 * pi) * r = tether length for a perfect stop at 90° to tangent. 

That is 16.36 / 6.28 = 2.60 * 64.5 mm = 168 mm for tether length

By using this length of tether I got a perfect stop when the spheres are at 90° to tangent; first try. Awesome!

[Delburt Phend]

In these models: an additional tether length of 15.9% of a radius (of the cylinder) will stop an additional 132 grams of cylinder. It does not matter if the 15.9% is added to a 1 radius length tether; or if the 15.9% is added to a 6 radius tether length.

Each 1/6.28th of a radius tether length stops 132 grams in these models. This is 15.9% of a length of radius of the cylinder. Six radius lengths of tether stop 4973 grams; 6.159 radius lengths of tether stop (4973 +132) 5105 gram. One radius lengths of tether stops (1 r * 6.28 * 132g) 829 grams; 1.159 radius length of tether stop (829 +132) 961 gram.

NASA predicts that extra length added to a longer tether has a greater ability to stop larger quantities of mass: wiki "and their (sphere mass) effect grows as the square of the length of the cables." But this is not true. The same added length of tether stops the same added mass no matter what the length of tether.

It should be noted that this is true of the restart as well. An additional tether length of 15.9% restarts another 132 grams. When the tether has twelve units of 15.9% of the radius it restarts a 1584 (12 * 132 g total mass) gram cylinder. The thirteenth unit of 15.9% of radius length, added to the tether, will restart an additional 132 grams.
 
For the 129 mm diameter model this 15.9% is 10.27 mm; for the 88.89 mm diameter model this 15.9% of the radius would be 7.08 mm.

The tether length to mass stopped (at 90° to tangent) is a linear relationship not a square relationship.

[telecom]

In these models: an additional tether length of 15.9% of a radius (of the cylinder) will stop an additional 132 grams of cylinder. It does not matter if the 15.9% is added to a 1 radius length tether; or if the 15.9% is added to a 6 radius tether length.

Each 1/6.28th of a radius tether length stops 132 grams in these models. This is 15.9% of a length of radius of the cylinder. Six radius lengths of tether stop 4973 grams; 6.159 radius lengths of tether stop (4973 +132) 5105 gram. One radius lengths of tether stops (1 r * 6.28 * 132g) 829 grams; 1.159 radius length of tether stop (829 +132) 961 gram.

NASA predicts that extra length added to a longer tether has a greater ability to stop larger quantities of mass: wiki "and their (sphere mass) effect grows as the square of the length of the cables." But this is not true. The same added length of tether stops the same added mass no matter what the length of tether.

It should be noted that this is true of the restart as well. An additional tether length of 15.9% restarts another 132 grams. When the tether has twelve units of 15.9% of the radius it restarts a 1584 (12 * 132 g total mass) gram cylinder. The thirteenth unit of 15.9% of radius length, added to the tether, will restart an additional 132 grams.
 
For the 129 mm diameter model this 15.9% is 10.27 mm; for the 88.89 mm diameter model this 15.9% of the radius would be 7.08 mm.

The tether length to mass stopped (at 90° to tangent) is a linear relationship not a square relationship.

I think there are two types of action - when the force is momentary vs the force is
constant.
The work calculations are not applicable to the momentary force action,
but they are applicable to the constant force action.
But by sending the projectile up against the gravity with a momentary force,
it becomes possible to calculate work on the way down, when the gravity force is constant.

[Delburt Phend]

I thought I was loading a smaller picture; how do you get the big one back off?

[Delburt Phend]

I think you have the concept; but I prefer to look at it as a function of time; telecom.

A one kilogram missile moving 19.81m/sec will rise 20 meters.  From d = ½ v²/a. This will take 2.019 second.  From d = ½ at².

A kilogram mass applies 9.81 newtons of force. This is 9.81 newton applied for  2.019275 seconds.  = 19.81 N seconds; N*s

Form a chain of twenty one kilogram masses; they are one meter apart, 20 meters high. This vertical chain could be dropped one meter. The original configuration can be restored if only one kilogram is raised 20 meters. 

Connect this 20 kilogram chain to a 80 kilogram flywheel. It will take 1.009637 seconds for the chain to drop one meter while it spins the wheel. But this is 20 kilograms dropping one meter for (20 kg * 9.81 N/kg)  196.2 N applied for 1.009637 seconds for 198.09 N seconds.

The output momentum is ten times that of the input momentum.

The output momentum is 100 kilograms moving 1.9809 m/sec for 198.09 kg*m/sec.

The input momentum was 1 kg moving 19.81 m/sec for 19.81 kg*m/sec.

The input energy is 196.2 joules.

The output energy is 19,620 joules.

In the 20 kilogram; twenty meter; vertical chain arrangement one kilogram applies its 9.81 newtons for 1.0096 seconds 20 times before it needs to be returned to its original vertical position at the top of the chain. The other 19 kilograms in the chain always assist in the application of force but each one kilogram applies its 9.81 N for 1.0096 seconds 20 times. 

The descending mass is 9.81 N applied for 20.19274 seconds. This is 198.09 Ns. 

The ascending mass is only 9.81 N applied for 2.0193 second for 19.809 Ns.

This time difference multiplies the momentum by 10 times and the energy by 100 times.