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Overunity Machines Forum



Gravity Mill - any comments to this idea?

Started by ooandioo, November 03, 2005, 06:13:20 AM

Previous topic - Next topic

0 Members and 4 Guests are viewing this topic.

prajna

Sorry Stefan, you are confused about this.  The large diameter pipe is below sea level and we do not need to consider the weight of water in that pipe we need only consider the weight of water in the small pipe that is above sealevel.  Whatever volume of water that is above the sealevel we must consider as being the weight of water that our shuttle must lift.

You even did the experiment to see this effect and the whole of our effort is based on the principle that we need only account for any water we lift above the water level.

I think the confusion for you arrises in that the pressure from the water in a column 1m high will exert the same pressure per area whether the area is large or small.  So a column of water 1cm by 1cm by 1m will exert a force of 1 kilo per cm2 and will weigh 1 kilo.  A column of water 10cm by 10cm by 1m will also exert a force of 1 kilo per cm2 but will weigh 10 kilos.  The boyancy required to lift the water in the second tube is 10 times that required to lift the water in the narower tube.

ooandioo

I can confirm with tbird. Never mind you are taking a 1m high, 10cm diameter pipe above sealevel or a 2m high, 5cm diameter pipe - the pressure on the shuttle beyond is is always the same.

I still think, using a pipe for transporting water in the reservoir bassin or a sprinkling noozle will have the same results, as the pressure is also the same.

tbird

hi all,

another note to remember is reguardless of the shape or size of the shuttle, the pressure we get is only the difference between what the shuttle weighs and the weight of the water it displaces.  example:  shuttle weights 1, 125 kg (pretty big, right?) and it displaces 1,126kg, then the positive force (pressure) is 1 kg. divided over whatever surface area the shuttle has.

i think it is easy to lose track of this fact.

tbird

ps still don't see my question addressed
It's better to be thought a fool than to open your mouth and prove it!

hartiberlin

Quote from: prajna on August 26, 2006, 08:09:43 PM
10m we will require 3 litres of air in our 0.5 litre reservoir so it must be at 6 times atmospheric pressure - if my logic is correct. That should be 14.7 * 6 = 88.2psi = approx 6.2 kilos/cm2.



Okay, so far.

Quote
We can discount 1 litre of air when it comes to recompression of the reservoir because we still have that from the previous cycle. This leaves 2 litres to recompress = 2000cc. If we have a plunger with a cross section of 1 cm2 and a length of 20m then a weight of 6.2 kilos will compress the air in the cylinder to that pressure.  If we only have 0.5m to compress in then we will have a cross section size of 40cm2 and require a weight of 248 kilos.


Okay, that would need
W= Force x distance= 6,2 Kilo x 9,81 x 20 Meter= 0,3379 Watthours

Quote
The quantity of water we can pump depends on the area of the top of our shuttle.  If it is 20cm2 then 1 litre will be 50cm high in the pipe and we can therefore pump 20 litres 0.5m above the water level. Hmmm....

Is my maths wrong or did I miss something or do we have less energy than we thought?

No, your math is not wrong, it is the 0.5 Meter exit pipe height, that destroys your design.
You only get as output
W= m x g x h = 20 Kg x 9,81 x 0,5 Meter= 98,1 Wattseconds/3600 = 0,02725 Watthours, so
you are very much underunity !

You must use a nozzle at seawaterlevel and then you can redesign your
shuttle to be in a different height ! the height should be much smaller then !
Also it could be, that just with 1 Liter it does not work yet !
Hope this helps.

Please all,also if you repeat yourself, please post always all the details
in ONE posting, so to better see your design.
Don?t leave things out, just copy and paste from old postings.
This way it is much easier to see the whole design and one does not need
to go back pages and pages...
Thanks.
Regards, Stefan.
Stefan Hartmann, Moderator of the overunity.com forum

tbird

hi andi,

QuoteI still think, using a pipe for transporting water in the reservoir bassin or a sprinkling noozle will have the same results, as the pressure is also the same.

the advantage of the pipe is you can control where the water goes.  if you use leverage for recompression, it would be hard to spray it there accurately.  it's cheap too.

tbird
It's better to be thought a fool than to open your mouth and prove it!