Question about water

ScumLord wrote: »

In hydraulics there's something that says that you only need a fraction of the tube diameter at the control end to drive a much larger tube diameter at the work end, it's some sort of natural multiplier.

I'm sure someone can explain it in more sciency words.

Pressure (P) is equal to force (F) per unit area (A):

[latex]\displaystyle P = \frac{F}{A}[/latex]

In a hydraulic system, it is assumed that the liquid is essentially incompressible, and so the pressure is constant.

So, suppose you have such a hydraulic system, consisting of two cylinders, one with a relatively large diameter ([latex]D_L[/latex]) and one with a relatively small diameter ([latex]D_S[/latex]). Both cylinders are "floating" in a reservoir of hydraulic fluid. The cross-sectional area of each cylinder is obviously given by:

[latex]\displaystyle A = \pi\frac{D^2}{4}[/latex]

Now, let's say you apply a force ([latex]F_S[/latex]) to the smaller cylinder, pushing it down into the fluid. This force is equal to the pressure in the fluid times the area of the cylinder ([latex]A_S[/latex]):

[latex]\displaystyle F_S = P \times A_S[/latex]

This results in a force ([latex]F_L[/latex]) being applied to the larger cylinder:

[latex]\displaystyle F_L = P \times A_L[/latex]

Now, we know that the pressure in the fluid remains constant, so:

[latex]\displaystyle \frac{F_S}{A_S} = \frac{F_L}{A_L}[/latex]

So, the “gain” in force achieved is equal to the ratio of the cross-sectional areas of the two cylinders:

[latex]\displaystyle \frac{F_L}{F_S} = \frac{A_L}{A_S}[/latex]