New accelerating gravity wheel ! Converted video from www.newenergymachine.com !

[LarryC]

Hi worker,

Your machine would work, but it doesn't appear to be OU, as it would be using external pump pressure to lift the fluid to the top. The energy gained by rotating to the bottom would be less than the energy used in the pump to lift the fluid to the top do to various losses.

I've included a screen capture of your drawing using a freeware package from Cnet called ScreenHunter 5.0 to change it to a jpg file. It is more effective to let people view directly, without clicking on the link at the bottom. Many people don't even notice the bottom link.


Regards, Larry

[spider4re]

Thought this might be interesting. Could this be used in assisting bobs machine?


http://www.engadget.com/2009/02/10/regenerative-shock-absorbers-developed-by-team-at-mit/

[unity2zero]

It is sad that finally, Bob's claim of his over-unity self propelling gravity wheel has come to nothing. Sigh.....!!!!!! 

[LarryC]

@All,


If Bob's machine is a self-runner then it must compress a volume of air into the receiver/accumulator which is >= the volume of air released from the receiver.

When 2 cubic feet of air is compressed into 1 then the psi is 15 or a multiple of 2. The formula is (Gauge PSI + 14.7) / 14.7 to find the multiple at different Gauge PSI. So 20 psi is (20 + 14.7) / 14.7 or a multiple of 2.36.

I will be using 15 PSI in my examples as it is easier to demonstrate.

The drawing shows the long outer cylinder and the short inner cylinder.

First the cam and spring would do the first 4 inch of lift. Then the air would be switched on for the 4 inch to 13 inch move. The air would be shut off and locked from escaping at this point, so it would still be pressuring the cylinder, until the Check Valve in can open. The 9 inch movement actually uses about 11 inches of compressed air because the 4 inch movement drew in air at Atmospheric pressure.

Then the CF slam capture will start compressing the air in the short cylinder. At 3 inch slide in, the PSI will be 15. Then the receiver check valve will open and allow the compressed air to return to the receiver.


The spreadsheet show the supporting calculations and the CF multiplying factor required, which is 4X in this case. This is break even with no losses, so it would need to be higher.

In this case if a 24" X 2" cylinder was used to compress the air. A full 24" stroke would give you 15 PSI at 12" and the rest of the stroke would recover the air lost. Since that stroke would not work for this machine a 6" X 4" cylinder with the the same Cu In is used. Shorter stroke, but four times the force is required at midpoint (3 inch). I also think that 20 PSI is more likely used with a compression factor of 2.36. It would increase the diameter of the short cylinder and the force required.

Can a CF capture force factor of 5X or greater be achieved, I only got 3X + in my previous testing? Yes, more weight, higher RPM, longer arm, etc. Some of these may have timing consequence, so only thru additional testing can it be known.

I will need a sturdier CF testbed slide to add more weight, so when I get the time???

I included the xls if you wish to see the formulas. But keep in mind that other PSI need to be adjusted for the compression factor and the distance to PSI. For example 30 PSI has a compression factor of 3.04, so 30 PSI wouldn't be reached until 4 inch on the 6 inch stroke.

Another interesting value is the Time in seconds at 12:00 and 3:00, But keep in mind it is the time with a non rotating unit.   


Regards, Larry

[Xaverius]

Anyone heard from Bob lately?  I was just wondering what was the status of his manufacturing endeavors as announced last November.