Gravity Mill - any comments to this idea?

[tbird]

hi prajna,

that is a REALLY cool tool!!  your mom would be proud of you!

i haven't converted the numbers to what i'm familiar with yet, but at first glance the weight of the shuttle seems a bit high.  one thing that did impress me was how simply you have made things.  EXCELLENT JOB!!!!

tbird

[hartiberlin]

Pranja,
great javascript, but there the shuttle volume is 10 times too big.
Also you can also introduce the stored potential energy
in the upper reservoir via
Energy= watermass in kg x 9,81 x height in meters= Energy in Wattseconds or Joule
now if you divide it / 3600 you get energy in Watthours.

Also please include the required pumping energy which is

Energy for recompression in Wattseconds =
(P1 in Pascal - 100000 Pascal atmospherical pressure) x V1 in qubic meters x ln (Volume2 / Volume1)

If you recompress from 2times the volume back to 1time the volume, we get
ln (1/2)

"100000 Pascal atmospherical pressure" = 1 bar must be subtracted from the Pressure
P1, cause we are doing it not in a vaccuum, but at sealevel with 1 bar outdoors pressure.

also for the volume
1000 cm^3  = 1 m^3 =  1 meter x 1 meter x 1 meter = 1 qubic meter

Hope this helps.
Looking forward for the complete Javascript.
Many thanks.

[tbird]

hi prajna,

me again.  been playing with your neat program.  i have a thought or 2.  the first thing you do is "head of water in the feed tube".  this weight is nice to know, but i think just as (if not more) important would be to know what the max diameter of that pipe could be for a given height.  and maybe if given a pipe diamter, what max height could be.

shortly after you get to the shuttle.  it's a bit unclear if you are referring to expanded or compressed state.  don't know if you read my previous post

i attached a paint drawing (not very good) to show how a shuttle design could start (more details later).  the black part at the bottom and the rest of the container can weigh 1.5 times as much as it displaces.  not sure what material would work, but lots of stuff heavier than water, but don't know if anything weighes that much in that little volume.  this might be a slight flaw with Mr. Herrings example.  the worse result is a higher compression like you guys were trying to use or less buoyancy for a given displacement.  the shuttle on the left is in the compressed mode with 2 bar (30psi+or-). that's the white part inside the shuttle.  since pressure and volume are an exact ratio, if we reduce the pressure by half, the volume will double.  the picture to the right shows this state.  not sure if you can tell from the drawing but that is suppose to be twice what it was in the compressed state.  since the outside water pressure is the same at 1 bar, the shuttle would not expand any more at this depth.  if the retainers at the top of the shuttle walls  were not there and the walls extended up, as the shuttle went up, the air inside would expand the volume further.  but it is there so when the shuttle reaches the surface, there will still be 1 bar (15psi+or-) inside.

i was trying to find out how much 1 cubic foot of lead weighs.  couldn't find anything i really understood, but did find the atomic weight of lead and water.  lead= 207.2  water=18  does that mean for the same volume lead would be 11.5 times heavier?

and you may need to wait to see the drawing, but if we made lead (if i really did figure it right) the standard (least volume for weight, common) to use for weight,  then we could get a better picture of the shuttle.  Mr. Herring's drawings would lead you to beleave (as i did for awhile) the paper thin housing could provide this weight.  not quite true to life.  don't know if i put this part right.  hope you get my drift.

from your program "We know that a depth of 10m will compress the air 2 times its volume at the surface.
height / 10 + 1".  not sure if this is worded right.

that's it, just some thoughts.

i have to say again GREAT JOB!!!

tbird

[tbird]

Also please include the required pumping energy which is

Energy for recompression in Wattseconds =
(P1 in Pascal - 100000 Pascal atmospherical pressure) x V1 in qubic meters x ln (Volume2 / Volume1)

oh yes, good idea.  i'd prefer knowing the weight needed to recompress (helps those who use leverage to do this).

tbird

[tbird]

If you recompress from 2times the volume back to 1time the volume, we get
ln (1/2)

stefan what does the "ln " mean here?

tbird