Author Topic: Gravity Mill - any comments to this idea? (Read 101418 times)

hartiberlin · « **Reply #330 on:** August 31, 2006, 06:05:12 PM »

@2tiger,
you always calculate the case, where the shuttle goes down to its
own weight.
We are discussing always the other case over here,
where the shuttle itsself has no weight ( only a few grams for the plastic case that can be
overlooked) and pushes the whole water column above it up due to its buoyance forces.

Please only calculate for this case.
Also you use only the pressure forumula P= F/A for calculation.
If you have a nozzle, then you need to calculate in water streaming and
water velocities in it and before it , etc.. and also the exit profile of the nozzle counts in,
only then you can calculate how high the water will rise.

So please calculate to have a good nozzle directly at sealevel= 0 Meter
and have a shuttle weight= 0 Kg and have for instance a shuttle volume
of 10 Liter = 10 Kg and have an area of 100 cm^2 surface area of the shuttle
at its top and also the same at the bottom and have the
shuttle started at 10 Meters deepth to go up.

So the shuttle is 1 Meter high and has 100 Liters of water above it.
the main pipe will then have a diameter of 11,28 cm that is about 4,429 inches.

So how high will the water rise up after the nozzle output ?
Thanks.
Regards, Stefan.

hartiberlin · « **Reply #331 on:** August 31, 2006, 06:08:23 PM »

P.S: With the calculation of the hydrostatic pressure paradoxon
to my opinion, if you take any diameter 1 Meter high exit tube instead
of the nozzle ,then the shuttle would be in equilibrium forces and would not
rise and stay in 10 Meters deepth.

hartiberlin · « **Reply #332 on:** August 31, 2006, 06:17:04 PM »

Quote from: tbird on August 31, 2006, 03:20:15 PM

why do you think this will make the water go higher? will you attach a pump to the nozzle?

Well, maybe if will use a small diameter exit tube instead of a nozzle,
we might try to evacuate the upper water reservoir to 1/2 bar pressure,
before starting the experiment and see, if this needs less energy, than we
gain by sucking this way the water up maybe also a few meter more ?

I have to calculate this with the gas pressure laws and see, if this will
bring in any gain ?

hartiberlin · « **Reply #333 on:** August 31, 2006, 06:28:17 PM »

Well, I had another thought onto how to use the Hydrostatic paradoxon
not negatively as in the Gravity Mill but positively.

Well, if you look again at the Cartesian diver:

(http://upload.wikimedia.org/wikipedia/commons/thumb/2/2b/Cartesischer_Taucher_animiert.gif/180px-Cartesischer_Taucher_animiert.gif)

(http://upload.wikimedia.org/wikipedia/en/thumb/c/cf/Cartdivr.jpg/180px-Cartdivr.jpg)

Then you can see, that the upper opening of the bottle,
where the blue cap is, is pretty small.
If we need now to pressure the bottle to sink the diver
we could just place a weight onto the opening of the bottle at the water surface.
The smaller the opening of the bottle is, the smaller must be the weight to
get the same pressure inside the water as:

P= F / A and
F= mass x 9,81

So if we make the opening of the bottle very small, say only 0.1 cm^2 we can use
a small weight to get a huge water pressure inside the water bottle.

So if we want to sink and rise the diver, which is connected to a generator for output
generation, we only need to lift a small weight up and down a few Millimeters
and thus do for the input only a very small lifting work of the weight that puts
pressure onto the water.

With the Cartesian diver we just convert a STATIC Weight force
into a moving work force=
usefull energy-work output !

What do you think ?

Regards, Stefan,

tbird · « **Reply #334 on:** August 31, 2006, 06:52:40 PM »

hi guys,

in for lunch and had to ask. did you guys lose sight of the fact the elsa is just a pump?

ooandioo · « **Reply #335 on:** August 31, 2006, 08:18:35 PM »

Quote from: hartiberlin on August 31, 2006, 06:28:17 PM

With the Cartesian diver we just convert a STATIC Weight force
into a moving work force=
usefull energy-work output !

What do you think ?

Regards, Stefan,

Isn't the deep the diver can sink in relation to the pressure you apply on the top of the bottle?
I think this will net out.

Andi.

tbird · « **Reply #336 on:** August 31, 2006, 10:43:37 PM »

Quote

[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.)
/quote]

hi prajna, i'm a bit concerned about continuing to try to explain to you why you are still doing this wrong. i'm worried i will offend you and you'll be put off the project all together. my hope is you will figure out on your own so i won't have to sound negative any more. i would suggest you figure this same example using guage pressure and see if it comes out the same pressure. if it doesn't, you'll know something is wrong. if you can get it to equal, explain to me.

Quote
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.

i believe i did that part right. our answers were within 10 square cm (i used rounded pi). the numbers are; 25 x 25 x 3.14 = 1962.5 square cm. this then would be mulitied by 1kg (i think that's what you say 1 square cm has at sea level) times 3.41, the number of bar you need at 10m (just for arguments sakes we'll say it's right). this gives us this; 1962.5 x 1 x 3.41 = 6692.125kg. your answer was 1731.2 kg. did i miss something because of the metric system?

if i know your heart is in the right place (i think yours is), you can be pretty ruff with me. i do have thick skin when i know you are trying to help me and not just be mean. if anything in that last post didn't set right, don't worry about telling me.

if we keep up the struggle, we'll get there.

all parts for test are made. just have to wait on the glue to cure and clean them up. hopefully test tomorrow pm.

tbird

tbird · « **Reply #337 on:** September 01, 2006, 12:41:05 AM »

Quote

Quote
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.

i believe i did that part right. our answers were within 10 square cm (i used rounded pi). the numbers are; 25 x 25 x 3.14 = 1962.5 square cm. this then would be mulitied by 1kg (i think that's what you say 1 square cm has at sea level) times 3.41, the number of bar you need at 10m (just for arguments sakes we'll say it's right). this gives us this; 1962.5 x 1 x 3.41 = 6692.125kg. your answer was 1731.2 kg. did i miss something because of the metric system?

hi prajna, just had a light go on in my head. maybe we both are right. when i did the calc, i used the size of the shuttle with only 5cm height. this only requires a compression stroke of 2.5cm. you use external compressing and feed it to the shuttle. the size of the compression container could have a smaller compression area, but would need a longer stroke. is that right? it's been a long time since i actually worked with air compressors and the like.

tbird

prajna · « **Reply #338 on:** September 01, 2006, 12:45:24 AM »

tbird,

Quote

i'm worried i will offend you and you'll be put off the project all together.

No chance. I'll be put off the project only when it is proved to my satisfaction that it doesn't work. Today I proved to my satisfaction that it does work. I have posted the proof at http://declarepeace.org.uk/elsa/ if you would care to check my work. It does tell me that you need both strokes, up and down, and that you need to use a leverage to get it to work. I haven't used any pressure units except kg/cm².

Sorry it is in metric but all the units are easy and I will add an example in imperial as soon as I get a chance.

I will have a think about calculating using gauge pressure and the light may dawn. Thanks. Oh, and my calculator has been set on hexidecimal rather than decimal for days too! Duh!

Quote

size of the compression container could have a smaller compression area, but would need a longer stroke.

Yes, absolutely. It is like leverages: if you have a 2:1 ratio you apply twice the force by moving half the weight over twice the distance.

tbird · « **Reply #339 on:** September 01, 2006, 02:11:41 AM »

hi prajna,

you silver tongued orator!! great job!! this is what Mr. Herring has needed from day one. to put all that on paper so anyone with half a brain can understand it is nothing short of genius!! great job! great job! if it doesn't fly now, no one really wants free energy.

I DECLARE YOU KING OF THIS THREAD!

how may i serve you?

tbird

prajna · « **Reply #340 on:** September 01, 2006, 02:28:14 AM »

phew!

Quote

how may i serve you?

well actually, how about converting it to imperial for me. You don't need all the blurb, just 1' x 1' x 3' shuttle, 32.808399' pumping cylinder etc. i.e. just choose some sensible sizes in imperial units that work. Then I could add it all to the page and others who are not blessed with a familiarity with metric would have an intuitive approach to understanding it.

I am about to completely rewrite ELSACALC based on a double-acting pump with leverage. And sure, having worked out the intro page I have seen that your suggestion re choosing just the shuttle size and pumping depth is sensible. They are certainly the first things that one would need to consider.

Thanks for the accolades. I kinda feel that if Mr Herring had presented such a page to the patent office (even the US one) they could hardly have failed to grant him a patent, regardless of their position on perpetual motion.

tbird · « **Reply #341 on:** September 01, 2006, 02:43:22 AM »

Quote

well actually, how about converting it to imperial for me.

you mean any place i find a metric number, convert it (adjust size as logical) to imp and label it? be happy to.

Quote

I am about to completely rewrite ELSACALC based on a double-acting pump with leverage. And sure, having worked out the intro page I have seen that your suggestion re choosing just the shuttle size and pumping depth is sensible. They are certainly the first things that one would need to consider.

if it's anywhere near what you have alredy set up, it'll be wonderful.

Quote

I kinda feel that if Mr Herring had presented such a page to the patent office (even the US one) they could hardly have failed to grant him a patent, regardless of their position on perpetual motion.

i think when he gets back to his internet and reads your essay, his heart will be lifted.

i better get busy, unless you want to chat about something.

tbird

prajna · « **Reply #342 on:** September 01, 2006, 02:46:19 AM »

Quote

think when he gets back to his internet and reads your essay, his heart will be lifted.

Nothing I like better than lifting hearts.

Quote

i better get busy, unless you want to chat about something.

Please do. Me too.

hartiberlin · « **Reply #343 on:** September 01, 2006, 03:23:07 AM »

Hi Pranja,
wouldn?t your new 10 x 10 cm shuttle not pump 98 Liters when it goes down ?
Why do you say 90 Liters ?
When you set it into the water in compressed state it has 1 Liter volume and
if it goes down and decompresses, it has 1,5 Liter displacement,
so it would lift down 99 Liters and lift up 98,5 or 95,5, dending on how deep you let it go.
(if you use the upper surface or the lower surface of the shuttle to count the deepth)

Also you assume, that you can press with just 0,05 Kg of weight water through a small
pipe to 40 cm over the sealevel , when the shuttle goes down ?
This is not very realistic.

Also it depends if the hydrostatic paradoxon can really be neglected in this
experiment.
If your shuttle is only 30 cm high at the top, I think it can only press the
water 30 cm high in the header tube and when it is at 10 Meters, it can only
press the water 15 cm high inside the header tube due to the
hydrostatic paradoxon.?

So let?s see, what TBird will report with his setup.
Maybe TBird you can already post a picture of your setup,
not yet in water,how it looks like, if you still need to wait until the glue dries ?
Many thanks.

prajna · « **Reply #344 on:** September 01, 2006, 03:46:25 AM »

Quote

ouldnÃ‚Â´t your new 10 x 10 cm shuttle not pump 98 Liters when it goes down ?

Think about it for a bit longer, Stefan. 1. If the shuttle is in the tube then it occupies at least 1 litre of the space in the tube. 2. The shuttle is always in the tube. 3. When the shuttle reaches the bottom of the tube it is still compressed and only occupies 1 litre. 4. When the shuttle decompresses there are exactly 99 litres above it.

Ahh, you have pointed out one error I have made though; it does pump 99 litres in each direction and not 90 (my brain was a bit slow.) I'll correct the page.

With all due respect, Stefan, the hydrostatic paradox is just confusing you. Just let it go because it is not important in understanding this system and has already been taken into account. It tells us that the water level is equal across containers that are connected. You really do not need to be concerned that any aspect of it will change the calculations I have included in this example. The height of the shuttle has absolutely no bearing on the height to which it will lift water in this system. The ONLY things that matter in this respect are the displacement, the weight and the cross sectional area of the header tube. Sorry, but that is it.

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