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Posted to rec.crafts.metalworking
Tom
 
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Default Hydraulic force question

Nick Müller wrote:

Anthony wrote:

Force = Pressure x Area


ACK

So, If you have 1 1.5" piston, and a supply pressure of 500 PSI, the
force is 500x1.7671 = 883.55, now divide that by 2, since half the
pressure will go to the other side, and you have 441.775 lbs of clamping
force on each pad, minus any friction losses in the caliper slider.
Now, say you have 2 opposed 1.5" pistons, and the same 500 PSI supply
pressure, the force is still 883.55 lbs per piston, but you do not divide
by 2, since the cylinders are opposed, thus your clamping force is now
883.55 lbs per pad and you do not subtract anything for caliper slider
friction. So, yes, the host is correct.


NAK

To put an end to this funny discussion (I have put oil into the fire):
With one piston, you have a reacting force on the other side of the
bracket of exactly the same force coming from the piston (actio =
reactio; or summ of vectorial addition of forces must be zero). Now if
you introduce a piston on the other side, that piston can only have the
same force as the first piston (same pressure, same diameter) and thus
just will compensate the actio. One piston actio, the other reactio.

Vector addition! Force has direction!

Think of having a C-clamp with two opposed spindles. Close one spindle
with all the force you can bear with your hands. Now close the other
spindle with all the force. Will there be a difference? Will you be able
to turn the other spindle? No.

So it makes no difference regarding the forces wether there is one
piston or two opposed pistons (same diameter).

Nick

LOl. What you're saying then, is that 1 x one ton jack will lift
as much as 2 x 1 ton jacks? Interesting hydraulic philosophy, I
wonder what Bramah would say.

Why is it then, that comparable capacity single piston brake
calipers have a far greater piston diameter than the equivalent
2 piston calipers?
Brake calipers work on one simple formula:

total piston area x working pressure ÷ pad area = pad pressure psi.

Tom