Hydro Differential pressure exchange over unity system.

[johnny874]

  see3d,
not on a pc but am asking this based on your answers, does one side have 10% more surface area ? if so, that could equalize force.
you know, psi x surface area woulc show the potential force of h1 & h2.

[derricka]

...On another issue... I have been looking for the patent. I see some drawings from it, but I've been back through this entire thread and over both your mrwaynesbrain website and the PESN page and I can't find a listing for the patent. I probably just missed it... I am literally half blind these days, my left eye is failing... so could you please link to it again? And it is an actual, fully granted patent, right? Not "just" an application, I hope.

Found the patent, er, application:
http://www.google.com/patents?id=BfIPAgAAEBAJ&printsec=frontcover&dq=wayne+travis&source=bl&ots=klAmREbeB3&sig=A1L1SPYXJ-do-TsHrO-VPGu2xPw&hl=en&sa=X&ei=gQ0vULLoBITHrQG7nYHwBg&ved=0CEYQ6AEwBw

Just call me "document boy" from now on...
P.S. Seeing as the other forum is down at the moment, how's the frequency counter troubleshooting going?

[mrwayne]

MrWayne: thank you again for answering.  No, you are not really in my "OUIdiots" file... you are still one level above that in the directory tree.


Crossheads are used whenever a straight push needs to have some out-of-line motion at the end of travel. Perhaps the most familiar place we see crossheads being used are in railway steam locomotives, where the straight push of the piston needs to be sent off-axis to the driving wheel. It is sort of like an external wrist pin; it acts to relieve the side loads on the piston and cylinder and conrod packings by moving them to easily lubricated sliding guides.
Another improvement in geometry I was thinking might be useful for approximately the same purpose would be to adopt the shape of an oil pumpjack's head, which serves to change the straightline travel of the pumpshaft going into the well, into the arc motion of the jack's rocker arm without putting side loads on the pump shaft. I know you have plenty of pumpjacks around so you know what I mean here.


OK, so the hydraulic reservoir that is at or near ambient pressure is the small white one at the top, and its rams that pump fluid from it back into the pressure side are the two rams at the very top, also seen in that same image?
And that big tank at the bottom is a water tank, and there are no other reservoirs (except the pressurized hydraulic accumulator)?

And how does the hydraulic fluid get from the motor's outlet back up to that reservoir? Is there enough pressure in what is being supplied to the motor to drive it back up there or is it assisted in some way to get back to the reservoir which is much higher than the motor?

On another issue... I have been looking for the patent. I see some drawings from it, but I've been back through this entire thread and over both your mrwaynesbrain website and the PESN page and I can't find a listing for the patent. I probably just missed it... I am literally half blind these days, my left eye is failing... so could you please link to it again? And it is an actual, fully granted patent, right? Not "just" an application, I hope.

Hello TK,
This is the reservoir - above all the hydraulics - With the sticker on it - it is not square because it used to be mounted inside our 7 foot wide 12 foot tall test model.

You can see a little white oil filter just below.

If someone has not linked you to the patent - Search under Travis Wayne S, It was released under the PCT.

Thanks Wayne

p.s.

Thank you for the Crosshead info - That would help - I will bring it up when the design team meets.

[LarryC]

Component comparison of pneumatic cylinder to Travis pneumatic risers in first picture.

Regards, Larry

[see3d]

  see3d,
not on a pc but am asking this based on your answers, does one side have 10% more surface area ? if so, that could equalize force.
you know, psi x surface area woulc show the potential force of h1 & h2.

Very good Johnny,

You are on the right track.  By trial and error last night I found a factor I could use to make the internal PSI check consistent.  It was Factor = H2Area / H1Area.  Unfortunately, that violates my first assumption that the PSI must counter the input force: Air PSI = InputForce / H0Area.  So, I must reject the answer of  PSI = InputForce / ( H0Area * Factor ).  However, with the clue that it is off by that ratio, I will look to the formula further down stream that uses that same ratio in finding H1 and H2.

Thanks for the suggestion.