Strange women lying in ponds distributing swords is no basis for a system of
Government....
however in physics, Randall you are correct.
Fluids exert an equal pressure when pushed - in all directions i.e. non
compressible. Therefore 10 pounds force on a 1" square (actuator pedal)
when transmitted through the brake lines to a 3" square area (caliper
piston) will give a 10 psi push through the lines, hence 30 pounds force for
initial 10 lb force - not bad. (or you could put the 150lbs it takes to stop
my tub on the pedal) Now there's volume to deal with, so you can't get over
zealous with the brake piston size, otherwise as someone pointed out, your
foot will have to travel through the radiator before you stop. Those
wishing to change to new calipers should investigate the volume in the
caliper vs. stock, or also change the reservoir/piston of the "pusher" as
well as the "pushee".
My change for your 2c
Physics and British cars .... who knew the twain met?
;-)
Ed
-----Original Message-----
From: Randall Young [mailto:ryoung@NAVCOMTECH.COM]
Sent: Wednesday, June 20, 2001 3:07 PM
Cc: fot@autox.team.net; Triumphs@autox.team.net
Subject: RE: GT-6/Spitfire brakes
William G Rosenbach wrote :
>
> If you take the same setup and apply the 10 psi to a 0.5 square inch
> piston you are now getting 10 lbs per 1/2 square inch which is 20 psi and
> you get 20 lbs. of force out of the system.
No, that's wrong. If you apply 10 psi to a 0.5 square inch piston, you only
get 10 psi per 1/2 square inch which is 10 psi. The force exerted on the
piston is the pressure times the area, or 10 times 0.5, which is 5 pounds of
force.
Randall
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