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Overunity Machines Forum



Big try at gravity wheel

Started by nfeijo, May 03, 2013, 10:03:04 AM

Previous topic - Next topic

0 Members and 32 Guests are viewing this topic.

MarkE

Quote from: webby1 on February 07, 2014, 01:04:17 PM
Small problem, you are not putting in enough energy, aka air.

The air needs to be maintained down at the bottom of the cylinder during lift, this means that more air is pumped in while the cylinder is moving up.  Then after that part of the cycle, that cylinder becomes the "source" of air for another one that is in the start condition.  Just recovering the energy is fine but it will be used again :)
Webby, no that is not correct.  Let's review:
The cylinder has no mass, no volume and an SG of 1.0.  So it does not contribute or require energy to move from one position to another.  It is a platform for the payload.  Our "air" is massless and incompressible.  Whatever volume we insert at whatever depth, it displaces a like volume of water at that same depth, and if not restrained would return to the surface.  The payload mass has an SG of 10.0 and a dry mass of 2.22gms.  In water it has an effective weight of 2.00gms.

Now, we pump 2.00cc of "air" into the bottom of the tank underneath the cylinder.  That "air" displaces 2.00cc = 2.00gms of water.  The water around that "air" pushes it up the annular ring until it is stopped by the underside of the top of the cylinder.  The displaced water continues to transmit 2.00gms * 9.8m/s/s = 19.6mN force through the "air" against the underside of the cylinder.  The payload with an SG of 10 occupies 0.22cc.  The total displaced water volume is now 2.22cc, meaning that we have a total uplift of 2.22gms exactly balancing the payload dry weight.  The cylinder plus payload weight is at this point neutrally buoyant.  In order to get the cylinder to rise, we pump in another 0.1cc of "air".  Now the upward force of displaced water exceeds the gravitational force on the payload weight by 0.98N.  The cylinder rises until it is restrained by the stop.

Now, we could pump even more "air" in.  As long as we don't overrun the cylinder movement we can eventually pump in as much as:  5.04cc.  We won't be doing any extra work on our chosen payload.  We will be putting more work in pumping in "air".  In the best case we can recover all of that extra energy and it will improve our net efficiency somewhat.

We can also increase the maximum payload.  We are limited to a maximum submerged payload weight equivalent to the weight of the water in the annular ring of 2.39gms.  For a payload with an SG of 10.0 that means a maximum dry weight of 2.65gms.  Anything heavier than that at that SG and the cylinder will not rise.  The annular ring will just fill and then any additional "air" will just spill out into the surrounding water and surface by itself.

MarkE

Quote from: webby1 on February 07, 2014, 01:12:49 PM
Yes, well sure,, formulas,, :)

The drag is both a push and a suck,, so the front car only sees the push it needs to supply to move the air out of the way and the rear car only sees the suck from the air pulling it backwards,, it is not a complete removal but it does reduce costs for both, or many, depending on the length of the line, so that they can travel at a higher velocity and or reduce fuel consumption.

Theoretically, if there are 2 independent systems where the losses for one is a gain for the other and vice-verse, then the two working together get a better bang for the buck.  It is like multitasking, so the toilet is on the other side of the kitchen from where you are sitting watching TV and you need to go use the bathroom,, and you have some dirty dishes next to you because you had a snack while watching TV,, if you pick up the dishes and deposit them in the kitchen on the way to the bathroom you have saved yourself the extra journey than if you were to do each item by itself.
Well, this is not a case of concatenating two lossy processes.  It is a case of reconfiguring two instances of lossy process into a different single lossy process. N*M <> P.

minnie




   Webby,
              I really would like to see a good result here, however I have been studying what
       MarkE is saying as he's working through this with you.
           In the end I'm seeing a bucketful of cold water and I've got to imagine a way of
       bringing it to life. It just isn't that easy!
           Look at this planet and compare it with say the moon. Nothing much happens on the
       moon but on this planet with liquid water and sunlight the magic happens.
                       John.

minnie




      MarkE,
               reply 1022. Very well explained. I could well have fallen into a trap there with the
   submerged weight. One has to correct by taking off the buoyancy of the submerged weight,
   with obvious consequences if one didn't.
                              John.

MarkE

Quote from: webby1 on February 07, 2014, 04:58:36 PM
What is the total volume of displaced water under the cylinder, it is not how much air that has been pumped into it, it is the total volume of the cylinder.

The filler, the displacement replacement, takes care of the rest of the water does it not?
Webby if we had started with an "air" bubble under the cylinder and over the piston that extended down over the annular section for any amount then the buoyant force transmitted to the cylinder would be of the total weight of equivalent volume displaced above the bottom of the "air" pocket.  IE it would include the 14.23mm diameter of the piston that would have extended above the bottom of the air pocket.  That is pretty much the condition that they start with in the misleading Video #5.  If you would like to change the problem so that we start with an "air" bubble I am happy to do that.  That will necessitate a preload weight to hold the cylinder down in the Starting State.

Here is a picture that is more like Video #5 where we start with an air bubble:
Left most, the air bubble just fills the 15mm space allowable by the stop.
Middle, the air bubble extends down 15mm over the piston.
Right, the air bubble extends down 135mm over the piston.

The calculations of work in take into account the movement of the cylinder, unladen.  The applied work we have to input from left to right scales as:  1.00, 1.10, and 1.88, even though the force increases by a factor of 10:1.  This may seem interesting and the HER people try to make it seem interesting.  But force has no direct connection to work.  Energy or work is the integral of force applied through a distance. 

Now, what you can do is specify how you want to do work:  For example you could propose that we start with the left most case, plus an added weight that is enough to compress the cylinder by some amount up to the full 15mm clearance.  Then we could add a payload weight at that height and pump in "air" until until the whole thing rises as in Video #5.  We can then remove the payload, and vent as in the earlier analysis to return to the starting condition of the compressed bubble and see what the energy balance looks like.