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Author Topic: Centrifugal guns...  (Read 4091 times)

Offline telecom

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Re: Centrifugal guns...
« Reply #15 on: September 17, 2019, 06:41:19 PM »
If it hits a funnel like target, it should go up and then overboard

Free Energy | searching for free energy and discussing free energy

Re: Centrifugal guns...
« Reply #15 on: September 17, 2019, 06:41:19 PM »

Offline Delburt Phend

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Re: Centrifugal guns...
« Reply #16 on: September 19, 2019, 02:56:18 PM »
No mater what the length of the acceleration the steel sphere can not be moving at a velocity greater than the speed of the stick that is pushing it. With a radius of 25 cm the circumference is 1.57 meters. If the drill is up to 1200 rpm then the spheres are moving; 1200 rpm / 60 sec * 1.57 m = 31.4m/sec.

But when you transfer the momentum of a rim mass wheel to the spheres, by using a weighted string, then the spheres can be moving much much faster.

If the sphere on the end of the string has 1/40th the mass of the rim then it will be moving 31.4 m/sec * 40  = 1256.6 m/sec when released. 

I think you could use the weighted string concept to achieve speeds to put things in orbit; but you have to get them past the atmosphere.

Offline Delburt Phend

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Re: Centrifugal guns...
« Reply #17 on: September 20, 2019, 12:18:54 AM »
A one kilogram mass that is on the end of a string draped over a 99 kilogram rim will produce 44.29 units of momentum after the one kilogram has dropped only one meter.

A one kilogram mass with 44.29 units of momentum will rise 100 meters.

Transfer the momentum cause by an unbalancing mass to the mass that caused the imbalance and you create very large amounts of energy.

Free Energy | searching for free energy and discussing free energy

Re: Centrifugal guns...
« Reply #17 on: September 20, 2019, 12:18:54 AM »
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Offline Delburt Phend

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Re: Centrifugal guns...
« Reply #18 on: November 01, 2019, 01:33:55 AM »
https://www.asc.ohio-state.edu/durkin.2/treb2.htm


This is a 50 to one mass ratio between the accelerating mass (brick) and the thrown mass (golf ball). The brick is stopped even though it is still being accelerated by gravity.

This could be an Atwood's with 1250 grams on one side and 1300 grams (1250 + 50) on the other side; that is accelerated down for .1 meters.  Only the 50 grams (golf ball) is lowered .1 meters. The center of the 2500 gram (balanced) mass remains in the same position. At the end of a drop of .1 meter (for the 50 grams golf ball) the 2550 grams would be moving .19614m/sec.

That would be 2.550 kg *.19614 m/sec =.5001 units of  Newtonian momentum.

A 50 gram golf ball with .5001 units of momentum, would have to be moving 10.003 m/sec.

A golf ball moving 10.003 m/sec will rise 5.1 meters; it was only dropped .1 meters.
   
The 2.550 kg Atwood's moving .19614 m/sec would have .04905 joules of energy.     ½ mv²

A 50 gram golf ball moving 10.003 m/sec will have 2.50155 joules. 

 

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