A Perpetual Motion device

[Art]

Basically the output cross sectional area for these calculations (for determining the height of water the weight on the piston can sustain against the atmosphere ) can be anything you want .However if say the outlet pipe contains more volume of water than the "piston tank" then there won't be enough water available to raise the water in the outlet to the maximum height that the force on the piston will sustain . But if you were to fill this large outlet with water from another source , then the piston (with that particular weight applied to it) would hold it at the height calculated against the atmosphere .

The "approximate" part though of this method of calculating the height of the water though , is assuming that the outlet cross section is small in relation to the piston cross section.

[helloha]

Art

1000 square centimeters is only one tenth of a square meter but it will raise the water in the outlet by 100 centimeters . If you make the "piston " one square meter the 100 Kg force will only rise the water against the atmosphere by one tenth of a meter (ie 10 centimeters).

If 1000 square centimeters of input area changes to one square meter, do remember to take into account the 1m height of 100kg water also have to change.

That means adjusting the input area will affect the height of the water input force,
which also apply to the output area,
adjusting the output area will determine the height of output water rises,
so it's important that output area need to be taken into account

... more on that output area, need time to prepare all the diagrams

[helloha]

A side note:

When changing the input area, there are three options for the water input force
(1) adjust the height, without changing the weigh of water
(2) increase/decrease the weigh of water, without changing the height
....
(3) just randomly mix both

[Tarsier79]

But any variation or combination isn't going to give you what you need to get a gain.

I think I see what you originally wanted to do... release a valve to overflow the water on the output, then float the wood back up for reset. Not going to happen.

[Art]

Helloha ,

Once you set the input piston cross sectional area , then with any given weight you have a fixed pressure on the water . It is this pressure per unit area that acts on the height of the water column of the output .

If the area of the piston is one square meter and the weight is 100 Kg, the pressure per unit area is only enough to support a column of 10 centimeters . It doesn't matter (as far as height of water in the output is concerned) what the cross sectional area is in the output.That height is predetermined by the input piston area , the weight on the piston , and atmospheric pressure.

The only options you have to increase the height of the outlet column is to decrease the area of the input piston or increase the weight on the piston .

Unfortunately there doesn't look as if there are any other options unless you can find some way of manipulating the atmospheric pressure .

[helloha]

Tarsier79

float the wood back up for reset

You got the idea, but do know that wood is not a good buoyancy, there are better and heavy, yet still can float buoys... or custom design a new type.
Also the bigger the input area, the more buoys added,
but the bigger the valve that need to open (not good),
also more friction on the edges (thinking of a hybrid solution, but that's for later).

Raising the water is no magic,
but getting the wood/buoy, in one moment standing on top of the water below,
in another moment totally submerged under water, that's magic.
This is making use of the characteristic of water, like using hydrogen gas to float up in the "height" concept

Whether can there be any gain.... absolutely no idea

[helloha]

Art

Still have no idea what you're describing, might take some time to figure that out.

Anyway for the description of the image below
- assume the input piston move down just 1cm (a very little moving down)
- since the input area is 1000 square centimetres, and with 1cm, that's 1 litres (or kg) volume of water being pushed to the output
- so if the output area is also 1000 square centimetres, that's easy, the output water raise up by 1cm
- but what if the output area is 1 square centimetre, what's the height of output water ?
- How will the 1 litres volume of water being distributed at the output area of 1 square centimetre ?

[Art]

The problem is that you are thinking in terms of volume displacements . You have to think in terms of Pressure changes , ie forces being applied through the lever system of the piston and outlet .

How high the water rises in the outlet in a system at equilibrium with atmospheric pressure is dependant only on pressure changes on the water not on the volume changes . The volumes are a separate issue .

The calculation of the height of the water in the outlet can only be made when the system is at equilibrium with the pressure of the atmosphere .

In your last drawings above , only allowing the piston and weight to fall 1 CM means that you have to hold the piston back and that in actuality you don't have 100 kg acting .Doing that to the system means that you don't have a system in equilibrium.

When the system is in equilibrium , the height of the water is calculated OVER the level of the top of the piston and this is how high the water will rise in the outlet .It doesn't matter how big the outlet tube is , providing you allow the system to reach equilibrium with atmospheric pressure , - and the system is such that you allow enough water for this to happen .

There are some very simple demonstrations that will give you a good understanding of the effects of atmospheric pressure which can be easily done in the kitchen . All you need is a few meters of soft flexible see through plastic tubing and some suitable plastic bottles with airtight twist on caps that you can drill holes in . If you drill the holes so that you can connect the tubes to the bottles with no leaks (which is easy if you have the right tubing and drills) then you can make some very simple equipment that will make atmospheric pressure a lot more understandable .

I have to say that before I did many of these experiments I really didn't fully understand the subject even though I had a very good formal academic training in the standard theory and I thought I knew it .

Its the old story of you don't know what you don't know until you know it , and then you only really know you know it when you have to do it !

[helloha]

Art

Still have no idea what you're saying,
anyway is the image attached fit your description, I'm using this information of
- the water level on input and output is the same due to atmospheric pressure
- the raise of water will not exceed the height of the water load/weigh ?

only allowing the piston and weight to fall 1 CM

of course that piston might (or might not) continue to fall further, that value is an indication of "at that moment". How far the piston can fall, I have no idea, although I think (if output area is 1 square centimetre, the rise of water can easily go up to few meters) the piston might actually only fall less than 1cm.... just guessing.

[Fletcher]

Just for a good refresher try ...

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Go to Mechanics (top left) > Fluids (mid right) > Pascal's Principle down thru Static Fluid Pressure and also down thru Buoyancy to get an good understanding and overview.

Especially Hydraulic Press.

Cheers

[Art]

Good Helloha , those diagrams are about right

But :-

Your Quote "only allowing the piston and weight to fall 1 CM
of course that piston might (or might not) continue to fall further, that value is an indication of "at that moment". How far the piston can fall, I have no idea, although I think (if output area is 1 square centimetre, the rise of water can easily go up to few meters) the piston might actually only fall less than 1cm.... just guessing."

Thats not quite right , - The piston with the 100 Kg weight can only fall to the point where the outlet height will reach the one meter level over the piston and then it will stop .

Even if the outlet area is one square centimeter the water won't rise above the one meter mark unless you add more weight to the piston .

[Art]

Thanks Fletcher for the link , thats a good site !

I think when we're finished here we should be able to offer them some stuff about the interaction of fluid pressures and gravitational interactions : )

[helloha]

Art

Just asking, if the input force is a solid weight, instead of water weight (same 100kg, 1m, 1000 square centimetre), will the solid weight also be subjected to the same water level and height raise cannot exceed 1m ?

[Art]

Yes that is correct Helloha ,

converting the weight to the equivalent water volume is just a visual method of calculating the height the water can go to in the outlet .

That weight could be a block of lead but the height of water it can raise in the outlet will still be the same .

[helloha]

Art

So ask again, if the input area is doubled to 2000 square centimetres, while the solid/water weigh is still 1m high and now having more than 100kg, will the weight also be subjected to the same water level and height raise cannot exceed 1m ?