Knitel's InfinityPump

[sushimoto]

@all

<SNIP>

So it is called Knitel's InfinityPump...because 60%-70% and closing the loop is done by that fabulous, wonderfull, brilliant, careing, goodlooking, civilized, magicall, explorer, brain stormer, extraordinary mind, protector of weak, justfull, leader, strong carrisma,with prodigy, dexterity, agility,...what to say then perfect...

the very same Knitel Igor

Wiz

Okay Wiz,
Whatever, However.

I love it. And I love you too, if you want.

....And what do you think about of adding a spring to store some of the gravitational energy
in order to support the buoyancy - upward motion?

best,
sushi

[tbird]

Hi,

When working with hydraulics you must use the correct formulas to base your answers on.

5 meter of head = 0.499 871 996 66 kilogram-force/square centimeter.

This may help  http://www.onlineconversion.com/pressure.htm

Wayne

i knew i should have checked his work, considering 1 cm3 weights only 1 gram. in 5 meters you have 500 cm3, so 500 x 1 = 500 grams.  pretty close numbers.  thanks for the correction.

stefan, as you can imagine, this will change the values, but the principle still is the same.

tom

[xnonix]

hi stefan,

i think i found where you are going wrong.


this is not a 100% true.  from your first set of givens, you said the tube would have 1 cm2 (not 1 cm diameter).  so for your above statement to be true, the swimmer can not exceed 6.369 cm2.  did that turn on a light?

the 100kg isn't applied to just the 1 cm2 of the tube unless the swimmer is only 1 cm2.  it is divided over the intire area evenly.  so if you have a 100 cm2 swimmer, your weight would have to be 100 x 15.7kg (tube area 1 cm2) = 1570 kg.  much more volume (and mass) than 100kg.

i'm sure you want to tell me i'm wrong here and i don't need to go on so i will stop.  but this point is what hans was trying to get you to learn by studying.

tom

Hi all, nice thread...

Then tbird,

are you saying that if you put the 5 meters tube coming from ocean you need a weight in the other side equivalent to the ocean's weigth for sucking the water?

Just trying to understand the 2 versions.

Thx,

xnonix

[hartiberlin]

hi stefan,

i think i found where you are going wrong.


this is not a 100% true.  from your first set of givens, you said the tube would have 1 cm2 (not 1 cm diameter).  so for your above statement to be true, the swimmer can not exceed 6.369 cm2.  did that turn on a light?

the 100kg isn't applied to just the 1 cm2 of the tube unless the swimmer is only 1 cm2.  it is divided over the intire area evenly.  so if you have a 100 cm2 swimmer, your weight would have to be 100 x 15.7kg (tube area 1 cm2) = 1570 kg.  much more volume (and mass) than 100kg.

i'm sure you want to tell me i'm wrong here and i don't need to go on so i will stop.  but this point is what hans was trying to get you to learn by studying.

tom

Sorry, I meant the tube to have a 1 cm diameter.

So the height of 5Meter will have 15.7 kg of water in it.
As the air pressure is pressing on the water surface,
it is just as a syringe, which needs to pull 15.7 Liters of water up 5 Meters
high.
So you need to overcome 15.7 Kg of weight....

Tom do you mean, the dimensions of the 100 kg swimmer body play any role
in it ?  because the 1 cm diameter pipe is changed into a bigger cylinder
at the top at the V1 valve ?

Okay, than make the swimmer weight body 2000 Kg, where is the problem ?

[tbird]

Hi all, nice thread...

Then tbird,

are you saying that if you put the 5 meters tube coming from ocean you need a weight in the other side equivalent to the ocean's weigth for sucking the water?

Just trying to understand the 2 versions.

Thx,

xnonix

hi xnonix,

i think most of the water weights are using fresh water, so if you use sea water, the weights will have to be a little heavier.  you don't need the weight of the sea, just a little more than in the tube per square unit of area of tube diameter.  did i say that well enough?

tom