Is this design possible to get an axcess energy?

[selamatg]

[HiggsBoson]

Your diagram is a basic hydraulic/pnuematic pressure intensifier, you will indeed get a gain in pressure in the B chamber but when this pressure is applied to the A chamber which has a larger volume the best you can expect is an equalisition of pressure.
Force = Pressure * Area
There is no gain in this configuration.

[selamatg]

@HiggsBoson

Thanks for your comment.

You are right for the on the equalisition of pressure.
Since, Force = Pressure * Area, so that mean the force down of big cylinder much bigger than force up of small cylinder.

In my calculation (if my formula are correct) then the lowest pressure of both cylinder (when joinned) is 133 PSI, it will about 500kg push down excess.


selamatg

[selamatg]

It is possible ????

Please advice....


selamatg

[Harvey]

How to raise the 2 tons up . . .

[Harvey]

Mouse Power


Click For Larger Picture


So the mouse eats and sleeps. And when he does, it shifts the weights

[HairBear]

Maybe a Hydraulic ram could help you?

Hydraulic ram - Wikipedia, the free encyclopedia

Home-made Hydraulic Ram Pump

[Harvey]

*Removed to prevent BIG misunderstandings*

[HairBear]

That's my signature Harvey, it's not a part of my post.

[Harvey]

*Removed to Prevent BIG misunderstandings*

[HairBear]

Is there something I wrote to offend you Harvey? Is my original post about a ram pump not pertinent? I was simply giving possible advice to the person asking. In no way was I replying or even insinuating anything towards your post and now you seem to be attacking me with... Dull wit?

So yeah, Harvey, either your confused about my intentions or your looking to pick an argument with me. Which is it?

[selamatg]

@Harvey

How to raise the 2 tons up . . .

Just increase the pump pressure.

Force = Pressure * Area
The efective area of small cylinder is 2.36 inch square
The efective area of big cylinder is 11.79 inch square
=2.36 x 2100 (psi) x 0.9 (power factor) = 2,025 kgs

On stroke down :

If my calculation is right then the pressure of both cylinder when joinned is 420 PSI.
The force down will be ;
The load weight + the force down of big cylinder - the force up small cylinder
= 2 tons + (11.79 x 420 x 0.9) - (2.36 x 420 x 0.9)
= 2 tons + 2 tons - 405 kgs
= 3.62 tons force down

am i right?

Regards,
Selamat

[Harvey]

Is there something I wrote to offend you Harvey? Is my original post about a ram pump not pertinent? I was simply giving possible advice to the person asking. In no way was I replying or even insinuating anything towards your post and now you seem to be attacking me with... Dull wit?

So yeah, Harvey, either your confused about my intentions or your looking to pick an argument with me. Which is it?

You clearly have a problem - and no, I do not wish any ill will toward any member here.

You may wish to choose your signatures more wisely prior to offending people with them - I see you have found it proper to change it.

[Harvey]

Just increase the pump pressure.

Force = Pressure * Area
The efective area of small cylinder is 2.36 inch square
The efective area of big cylinder is 11.79 inch square
=2.36 x 2100 (psi) x 0.9 (power factor) = 2,025 kgs

On stroke down :

If my calculation is right then the pressure of both cylinder when joinned is 420 PSI.
The force down will be ;
The load weight + the force down of big cylinder - the force up small cylinder
= 2 tons + (11.79 x 420 x 0.9) - (2.36 x 420 x 0.9)
= 2 tons + 2 tons - 405 kgs
= 3.62 tons force down

am i right?

Regards,
Selamat

[selamatg]

Hi Harvey,

sorry i forgot to write divide by 2.2 to convert lb to kg. but the result are include with divide 2.2.

My understanding the basic formula as below :

Force (lb) = Area (in2) x Pressure (PSI) x Mechanical Efficiency (0.9)

Please correct me if i'm wrong.

Another thing to consider is the volume differential when Port A and Port B are connected. The pressure in the smaller cylinder will drop as the overall volume increases which it will as the top piston drops opening up that space. So the downward force needs to be integrated for that volume differential.

You right, both cylinder will keep same pressure when connected. That point in my understanding, cause the top cylinder much bigger the effective area than small cylinder will add the downward force.

The pressure will come down gradually.

Again...please correct me if i'm wrong.


selamatg