Quote from: AmoLago on September 11, 2012, 10:34:03 AM

Hi Fletcher,

Hope you don't mind, but I've modified your second posted image to indicate what I'm trying to explain as I thought a same starting base would be better.... hopefully it'll do at least a thousand words!

I'm not sure that the final PE is correct, but from what I've gleaned from other posts, and TK's "Virtual Water" video, this is what is going on in my head as to what could happen. Am I wrong?

If not, then from Step 3 in the image, I guess you can....

   1. Unlock whatever is holding the pod in place. If the pod is buoyant, I assume it's movement could be used to do work???

   2. Lower the currently empty blue tank on the left hand side back to it's starting point. Would this take all that much work???

   3. As the blue left tank lowers back to it's starting point, the water would naturally flow back until the water is back at the    starting point.

   4. The "pod" would also sink back to it's starting point as the water level dropped and could be locked again and the process restarted.

I guess the question is, that is if this in any way reflects any kind of reality, is can you get more output more from the releasing of the pod than is required to lift the water in the left container.

Also, if the layering adds head height, then does that mean the the blue water tank on the left of a layered system would have to be lifted as far? If not, then that I guess that reduces necessary input too.

Amo

Hi Amo .. sorry to get back to you so late - busy morning.

I have included an updated pic [worth a thousand words apparently] with the essence of what you are asking - this time I used the right scale so I could calculate correct geometry & masses & actual optimal PE's etc to look at your proposal rather than a representation - I also used increment of 50 kg as per yesterday to keep consistency.

What we have here is a cost of lifting the left tank with blue water to a height such that all of it will run across & it will empty - this could be either into the top of the right tank or up thru the bottom via a tube, overcoming pressure etc.

We are just looking at the water & the buoyancy potential to do work from the POD later when released - net situations.

Before I go on, I said yesterday that there is a real energy loss when transferring water - this is made up of drag & viscosity losses - I should elaborate further & remind you that water running down hill, thru a tube for instance, must be accelerated & then decelerated to fit thru an opening - it also has a tendency to surge or slop, a bit like a ball bearing running down the inside of a U tube i.e. it oscillates dissipating energy & these account for the major energy losses.

The raising or lowering of the tank is irrelevant as it can be counter balanced so is not considered.

Back to the pic ...

1. at start the twin tank system has water PE of 70.5 J - 2. after lifting high enough to transfer all water the PE is 624.6 J - 3A. after transfer to around the POD [locked down] the system PE is 352.5 J - 3B. the last system on the right with purple water is the combined blue & green water volumes & masses after the POD is removed [to show something I going to discuss] - the PE is 168.6 J.

The reason I have found the PE of the combined masses after the POD is removed is because buoyancy is volume dependent - so, as the POD is released in pic 3 it rises [yes, it has a lot of force, INITIALLY re acceleration = f / m (inertia)] - as it rises the water volume up the sides giving the head rapidly flows under the POD - so it has a very short stroke where the force diminishes rapidly as head reduces rapidly.

The purpose of the purple water is to show the final position of the 50 kg of initial blue water [total 62.5 kg] - at all times the PE is way less than the step 2 'raising the system work done'.

OK .. now we want to know what work the POD could do to add that energy back in ? - well, the work the buoyancy force can do when the POD is released is exactly the same as the f x d of 50 kg raising 0.2 meters IINM, except this is a constant force as opposed to a variable force so its easier to work out work done joules.

N.B. you could float a battleship in a bath tub if the bath tub were big enough.

The upshot is that 50 kg rising 0.2 meters gives about 98.1 J - when you add that to the 168.6 J we get somewhere around 266.7 J, well short of the initial work input done.

This is the optimal situation where the buoyancy force equals the upthrust force by allowing the POD to have NO mass [for the exercise] - if we assign mass to the POD then the net Upthrust Force will reduce & so will its lifting ability to raise a mass 0.2 m & the subsequent joules of PE that could be added back, for the exercise.

P.S. it makes sense IMO if you consider just how high water must be raised [CoM/CoG position] to flow across & completely fill the other tank as a few here have been consistently saying.

..........

My apologies, had to push this analysis out pretty quickly so I make no claims of accuracy - hopefully someone else can check the figures & logic against your findings.

Hmmm... looks like fletcher and MT don't agree. I wish I could see MT's spreadsheet, but it won't open in my Linux system using LibreOffice... very unusual, it's the first spreadsheet I haven't been able to open. I downloaded it twice.... apparently seamus10n can read it? Is it possible to convert to an older format maybe, that I could read? Sorry, but Koalas generally only use free software or shareware; we avoid Micro$haft products like the BSOD they are.

Meanwhile....maybe this is the missing ingredient that fletcher needs to make up the difference:

http://www.youtube.com/watch?v=WpQej7M_HXw

Hi TK .. knock yourself out - I'd like to see your breakout of his spreadsheet - I don't have time today.

P.S. for those wondering - you can tun the left hand tank raised up on its side & make it very wide & not very tall [like a shallow sea] - that will reduce the PE input requirement.

Quote from: TinselKoala on September 11, 2012, 10:40:15 PM
Hmmm... looks like fletcher and MT don't agree. I wish I could see MT's spreadsheet, but it won't open in my Linux system using LibreOffice... very unusual, it's the first spreadsheet I haven't been able to open. I downloaded it twice.... apparently seamus10n can read it? Is it possible to convert to an older format maybe, that I could read? Sorry, but Koalas generally only use free software or shareware; we avoid Micro$haft products like the BSOD they are.

Meanwhile....maybe this is the missing ingredient that fletcher needs to make up the difference:

http://www.youtube.com/watch?v=WpQej7M_HXw

http://www.viewdocsonline.com

Quote from: fletcher on September 11, 2012, 11:37:46 PM
Hi TK .. knock yourself out - I'd like to see your breakout of his spreadsheet - I don't have time today.
P.S. for those wondering - you can tun the left hand tank raised up on its side & make it very wide & not very tall [like a shallow sea] - that will reduce the PE input requirement.

Goeie morgen Marcel,
Goed om jij op deze forum te hebben, ik hoop dat we die Americaners in toog kunnen houden, ps, ik kom van Belgisch Limburg.

The worksheet is interesting, a floating pod filled with water. Buoyancy neutral, very straight forward.
I always wondered if nature can be fooled to believe something different to the reality. Seamus will say "impossible" because the book says so.  Marcel's worksheet gave the first impression to do just that, it gave me the inspiration for a mind twister that i am preparing,  should have it later today.

Marcel, 
When doing calculations, it is important to follow the natural sequence, we can not build the second floor before the ground floor is completed to support the second floor. In you calculation on line 48 to 61 where you calculate the stroke
1.. Displacement water, your height should be the height of the top of the water (top of precharge or head),
2.. If you want to inject at 0.1mtr, that you need to bring the pressure of the water height into the picture (that is the head), which is the same as step 1

When looking at buoyancy it is worthwhile to remember,
1..  Water column height or head is always a LOSS, it is only a mediator to set up the conditions for work.
2..  Stroke displacement volume is the REAL ENERGY to feature in work done

Goeie dag, Michel