Big try at gravity wheel

[MarkE]

Well MarkE,

By your "take " on things I should be able to boil water in a heart beat,, I can't,, why is that IF you are correct,, I know,, it is because you are wrong.

By your numbers 70 percent straight into

Webby I see you are building straw men instead of buoyancy machines.  It's your claim that you made an 83% efficient device.  You offer no evidence that you did.  It was your promise to describe your device and your analysis.  In more than two weeks, all you have done is dribble out a crude and incomplete description bit by bit with no analysis.  When faced with the horrific inefficiency of what you have described you have countered with nothing more than hand waving that with some unspecified engineering the efficiency problems could be solved.  Yet, you claim you already built to 83% efficiency.  It's pretty obvious why you aren't describing what you built and how you measured or analyzed it.

You can protest and claim that I am wrong all that you like.  I have shown the factual basis for the efficiency problem that your described apparatus faces.  The longer you go without describing what you built and how you either measured or analyzed the efficiency, the more certain you make it that your claims are so much smoke.  You'll just be joining Red_Sunset and Wayne Travis with their equally empty claims.

[MarkE]

For MarkE's solution to be the *ONLY* solution it *MUST* hold true for all conditions.

2 weights a string and a pulley, nope it does not hold true, a see-saw, not there either.

What does that mean, it means that MarkE's solution is *A* solution but not the *ONLY* solution.  I choose a solution that does not require the destruction of all that potential.

Haste makes waste, and MarkE jumped on the first solution he saw, I took a little bit of time and came up with a different solution.

Webby you are still welcome to post the method that you claim that you used to reach 83%.  In order to get there you need to show that your claimed method does not suffer the problems of the methods that I have described that include the method that you dribbled out using the transfer pump.  Like Red_Sunset and Wayne Travis you keep claiming that you have some marvelous thing behind a curtain.  When pressed for evidence of the marvelous thing, you wave your hands and show nothing.

Just in case you've forgotten here again is your original sketch, and your sketch with the transfer pump you described several days after posting your sketch.

[MarkE]

MarkE,

The method I employed is simple, I used the stored potentials as a store of potentials, there are 2 of them at the end of the cycle.

Webby, there is:  The stored energy at the start, the work added, the work removed and the stored energy at the end.  We know that the A cylinder ends up at the end of the half cycle with the same amount of energy as was in the B cylinder at the start of the same half cycle.  We know that we extract the work of lifting the 26.5g weight by 15 mm.  What you steadfastly refuse to show is the work added via the pump.  I have shown that the means that you specified to move the "air" from B to A, a transfer pump loses ~50% of the energy stored in B at the start, which is in turn about twice the energy that is extracted lifting the weight.  You have failed to show that you in fact used some other method that does not suffer that loss to heat that then has to be subsequently made up by the pump.  You have failed to show by any analysis or measurement the energy that you claim to have put in over the course of the cycle.  You have failed to support your claims.

One store is within the air contained inside the cylinder under pressure, and the other is the force manifested by the cylinder moving up in the water column.

All the dogs that pull the sled assist in the sled moving, it is not like the smaller dogs pull against the larger ones.

I have agreed with your numbers and have no issue with them *IF* I were going to use the stored potentials in the method or fashion that you are choosing to use them.

I found another solution for use of those potentials that does not require the destruction of a large portion of those potentials, I would not be surprised if someone else could find a better solution than I did, I stopped looking when I found a system that met my needs.

Here we go back to your appeal that you have some magic method hidden behind a curtain.  I call BS.  You were surprised when I showed the loss and its root cause.  Since then you have flailed about claiming that you used some method that behaves magically better than what you described up to the point that I showed you the losses.  Despite the fact that if you had some means and were to show it, it would clear you from playing the fool, you refuse to identify this method that you claim that you used.  By far the most likely reason for your refusal is obvious:  You are blowing smoke just like Red and Wayne.

[MarkE]

Wrong MarkE.

I not only pointed out the loss I pointed out why it is a loss when done using your method of transfer.

After the empty cylinder is first brought up to neutral buoyancy and then lifts the 26.5g, all of that potential can be freely off-loaded external to the system without your catastrophic loss, as well as the cylinder can be taken from empty all the way up to end of lift without all of the catastrophic losses you keep insisting must be there.

You are still requiring that the 2 independent systems can only work as one interactive system, this is not what my solution requires.

Webby this is just getting sad.  You specified the two cylinders.  You specified the transfer pump.  This is the configuration that you identified.  Referring to your post 1035:

I use a straight transfer pump connected between the top of the 2 cylinders, a transfer pump is a sealed chamber with a piston in it so that when the piston is on one side the other side has enough volume to hold the medium of one unit, then when slid over to the other side it pushes that volume out and into the unit it is connected to and at the same time will pull in the medium from the unit connected to the other side of the pump. simple.

You can figure out the weight that the cylinder can lift at full fill, place this weight on the starting condition cylinder and transfer the potential from the state2 cylinder into the starting condition cylinder, this is the cost of cycle.

I come up with an 83 percent efficient transfer this way and a 33 percent efficient transfer for an open cylinder, that is no filler.

Only after I pointed out the loss in post 1091 over a week ago did you go off on this mystery behind the curtain method suggestion.

You have completely failed to show any configuration that will deliver your claimed 83% or anything close to that in net cycle by cycle efficiency.  You just keep claiming that you can shuffle energy around with little or no loss.  That's a big no sale webby.  Either you can show the work or you can't.  So far you have worked very hard for days now doing all you can to avoid showing any configuration that performs as you claim.  For all intent and purpose you are talking through your hat.

[MarkE]

An interesting note.

When the cylinder makes its lift there is a volume of water that is moved, that volume is the 15mm diameter by the 15mm height, but the volume of added air is only 90 percent of that volume.

Where did the other 10 percent go?  It was pr-paid for during the process of bringing the cylinder up to a neutral buoyancy, that 10 percent fill of air moves up with the cylinder.

Want to spend less on the lift? reduce the filler to 80 percent and only pay 80 percent of the volume of added air at its highest cost.  You will be adding in more air to bring the cylinder up to neutral buoyancy, but is there a balance

Using the transfer pump there are five interesting states:

State2, B is fully charged, A contains only water
State3, The meniscus under B is at the same absolute depth as the meniscus under A
State4, The meniscus under B reaches the top of the piston under B
State5, The meniscus under A reaches the point that the A cylinder plus payload become neutrally buoyant
State6, The B cylinder is empty, the A cylinder is fully charged and the payload weight is removed.

For each of the four transitions you can work out how much work has to be applied to the pump to get from the starting state to the ending state.  Do you think that reducing the diameter of the piston helps in the end?  Try the problem out changing from a 14.23mm diameter piston to a 13.42mm piston and do the math.  Have you ever played "Whack-A-Mole"?

Go ahead webby:  Show your work.