Gravity Mill - any comments to this idea?

[tbird]

hi prajna,

did you get the answer to my quiz?  of course the first time i heard it, i failed.

one question and we'll put this to bed.  according to your default example, if you took your shuttle to pressurize it at the gas station, what pressure would you fill to, according to their gauge?

houston, i think we have a problem.  i'm not sure about this, but the amount of weight to recompress you come up with is different (default example) than what i figured.  being a simple arkie, i just figured the square cm of the pumping pipe diameter, then muliplied it times the pressure needed.  your example being in metric, i couldn't follow how you came up with your number.  the big reason i have to wonder about the way i did it is if you made the diameter of the pipe smaller, the weight would be less.  i kinda said that was ok, it would just travel further.  the thing that is on my side (in my head) is if i use psi (i know what this means), that's pounds per square inch.  since the only place we have to apply weight is the top of the shuttle, i figured this area would have to have the ratio (whatever it is) times (in inches) 14.5 (or there abouts) per square inch.  i think i used 1kg for each square cm in your example.  houston, do we have a problem?

i think your 'Introduction to ELSA' is pretty cool.  i didn't see mentioned the type of pressure (g or a) you were using.  might be helpful.

i read through your description of the system again and have a suggestion.  at the bottom of the pipe where you expand the shuttle, make the pipe so it looks like a donut is around it, but without the inside wall.  this way if for some reason the shuttle spins, it won't miss the transfer pipe.  for that matter you could have more ports for a faster transfer and wouldn't have to work the valve at the top.  now if you use a magnet arrangement at the top to hold it inplace while it fills, this would elimate another operation of the valve.  to make sure it didn't release too soon, the ports for intake and exhaust could be different sizes.  leaving just one thing it had to do, allow water to fillin behind shuttle's descent.  this could be controlled with a check valve.  now no need at all for slector valve.  now everything would work automatic with less to go wrong.  just a thought.

tbird.

[2tiger]

Hi Stefan

@2tiger,
your calculation is a bit hard to understand:
1. What case do you want to calculate,
a)the lifting of the shuttle and pumping up the water
or
b) the downward movement of the shuttle ?

I was analysing only the down stroke of ELSA. I mean the downward movement of the shuttle and pumping water up throug smaller tube.

2. You seem to forget, that it is done in big watercase, where the
outer water must be calculated in and which has a big weight.
This is also the reason, why so much water is lifted,
if you look at it, if you disconnect the exit pipe and just look
how much water is lifted to the same height of the seawaterlevel
via the shuttle.

Do you mean to disconnect the exit pipe so that it is on sealevel? In that case the shuttle will sink and all the water of the "pistontube" will flow out right over sealevel. But then you have no usefull work, as energy is W=F x distance (lifting height) and distance is cero, then you get 0 Nm of energy.

where is exactly your 1 cm^2 and 5 cm^5 area ?
Seems you have drawn this in the same spot...
This is confusing...

I mean the area of the cross-section of the tube, so 1 cm^2 = 0,564 cm radius. The tube will have a diameter from  1,128 cm.

So this is the hydrostatic paradoxon at work making our device
less efficient, as we can only pump the water as high, as the shuttle
itsself is high !

This is wrong. If the shuttle is 50% heavier than te same volume of water, then you will lift the water on the other side ONLY HALF a height of the shuttle.
As prajna pointed out before, in an U-tube (ration 1:1, dimension 1liter per 1m height) you have to pour 1 liter water to increase the waterlevel on both sides 50 cm.
In U-tube (ration 1:2) you will have to pour 2 liters to reach the same height.

Stefan, if have not understood my calculations or my drawings compairing water with marbles, here I have another example. This times with boxes/cubes and without pressure and water. Just by using the lever rules.

In drawing No. 2 you see on the balance 0,1 N versus 0,20 N. To be conform with the hydrostatic paradoxon I moved the axis of the balance in a ratio 2:1 (NOT 1:2), so that it is balance out. You have to put twice weight on the shorter lever to lift the same amount on the longer one.

Now look at drawing No. 3. There you have on the balance 0,1 N versus 0,21 N (plus 50% weight of the shuttle).
On the right side you have 0,01 N more than on the left side. But because of the lever rules, you will only lift 0,005 N on the left side.
Remember 1 box/cube = 0,01N =1cm^3   there for 0,005 N = 0,5 cm^3 ist a HALF box/cube.
Volume = area x height   =>   height = volume / area,
so  height = 0,5cm^3 / 1 cm^2
               = 0,5 cm
               = half the height of the shuttle

And you will not be able to lift the water higher since you don?t change the weight of the shuttle.

Do you still want me to test this out with my "barbiefamilytube" ?

SeeU
2Tiger

[prajna]

Yes, tbird.  I did get it and I did understand why.  I must admit that I did have to look twice.

I will change the calculator to read:

So the pressure inside our shuttle will be 4.41  kg/cm²  (air pressure at sealevel is taken to be approximately 1kg/cm².)
[total compressed volume / shuttle volume]

I think that is unambiguous since I state what I consider normal air pressure to be as well as the shuttle pressure.

If I take the compressed shuttle to the garage the tire gauge will read 3.41 kg/cm² or 3.41 bar or 49.4578686 psi (according to google's wonderful convert function.)

being a simple arkie, i just figured the square cm of the pumping pipe diameter

Did you figure the suare cm by taking half the diameter and multiplying that by pi or by multiplying it by itself?  I hope the former since
we are dealing with the area of a circle rather than the area of a square.

The way that I figure the pressure is to take the combined weight of the shuttle and the head, convert that to a volume and divide it by the volume of the shuttle.  That figure is multiplied by the pressure at whatever depth you selected (10m so 2 times in the default example).  That tells me the neutral bouyancy pressure at the required depth. I then double that pressure so that the air will expand to twice its volume at the required depth.  Previously I indicated that pressure as gauge pressure by subtracting 1 bar but now I indicate it as you see above.

Gotta go out so I'll continue this later....

[2tiger]

@tbird
If you want to win the bottle of champaigne that Stefan offered you, make the shuttle 110% heavier than the same volume of water. Volume of water for example 1000 cm^3 = 1kg = 10 N, then the shuttle must have 2,1 kg = 21 N of weight.


bye
2Tiger

[hartiberlin]

Hi 2tiger and Tbird, I meant to have the shuttle itsself only to have a few grams of weight, so it could be avoided to have any impact. So only the volume of water it is displacing counts. In my view you can still only get the water up over sealevel with an exit tube as high as the shuttle height is. Maybe we then really have to go via a nozzle at sealevel to get it higher or use a vacuum inside the upper reservoir to suck it up.