1200% Motor

[mark australia]

Hi Will,
I do wish you the best, the maths is my week point but I have forwarded it on to a couple of people to check the maths out for you.
Is it possible to build a small working device to test some of the principles before commiting to a full blown prototype?
Just some thoughts..
Kind Regards
Mark

[will1green]

Hi Will,
I do wish you the best, the maths is my week point but I have forwarded it on to a couple of people to check the maths out for you.
Is it possible to build a small working device to test some of the principles before commiting to a full blown prototype?
Just some thoughts..
Kind Regards
Mark

Yes and mostly with off the shelf components and a few machined parts...
Try to lookup Steven Manthy inventer of OX2 engine...

[oouthere]

Well, I'm tired and it's very late for me but looking over the numbers I see extreme math errors.

This is how I figure it:

Total surface pressure area is 37.68 sq inches

37.68 sq inches * 450 psi= 16,956 lbs overall

16,956 * (12"/.5") stroke is 706.5ft/pounds torque

I don't know the formula for the torque curve but you only achieve 705.5ft/lbs of torque for a small fraction of 180 degrees of power development, this has to be averaged for the true constant torque value.

And still there are the losses involved in pumping the fluid out of the cylinder as it doesn't move itsefl nor does the valves move themselves.

Anyway, I'm tired and don't think it will work.....want to know what I do think may work?

Many of the claimed motionless o/u generators/transformers have a 90 degree magnetic offset between the flux of the input and induced current coil.  What about a transformer simply wrapped 90 degress out?  It probably wouldn't work either as it's too simple and I'm merely missing something.......

Rich

[will1green]

Well, I'm tired and it's very late for me but looking over the numbers I see extreme math errors.

This is how I figure it:

Total surface pressure area is 37.68 sq inches

37.68 sq inches * 450 psi= 16,956 lbs overall

16,956 * (12"/.5") stroke is 706.5ft/pounds torque

I don't know the formula for the torque curve but you only achieve 705.5ft/lbs of torque for a small fraction of 180 degrees of power development, this has to be averaged for the true constant torque value.

And still there are the losses involved in pumping the fluid out of the cylinder as it doesn't move itsefl nor does the valves move themselves.

Anyway, I'm tired and don't think it will work.....want to know what I do think may work?

Many of the claimed motionless o/u generators/transformers have a 90 degree magnetic offset between the flux of the input and induced current coil.  What about a transformer simply wrapped 90 degress out?  It probably wouldn't work either as it's too simple and I'm merely missing something.......

Rich

Hello Rich

Looks like you missed a very important part, that is the crank redious which is 3.5" to arrive at the proper torque.
A = Cylinder Head Area = pi*(r^2) = 3.14*(1*1) = 3.14159 sq. in.
Ta = Total Area = A*active cylinders = 3.14*3" = 9.42477 sq. in.
P = Pressure = 450psi
F = Ta*P = 9.42477*450 = 4,241 lbs Force
T = ft-lbs = (crankshaft radius / 12)*4239 = (3.5/12) =. 2916*4,239 = 1,236.7 ft-lbs.

You may want to try it again later

Will


I do know how to make an OU/ Electric motor but at only about 120-150% efficent along with supproting math...

[FreeEnergy]

http://www.ox2engine.com