"Ivan Vegvary" wrote in
news:RVcDf.7671$Cf7.7003@trnddc06:

Watching "Classic Car Restoration" the host made the statement that
the brake caliper on the restoration car (1969 Alfa Romeo) had twice
the force on the rotor (compared to single piston calipers) because it
has a dual opposing pistons, one pushing on each of the brake pads.

Is this a myth, like 2 cars, 50 mph each traveling at each other,
creating an equivalent collision of 100 mph? (In reality it is
equivalent to one 50 mph car slamming into an unmovable (50mph
reaction) object, e.g., concrete wall, 5' diameter oak tree, etc.

For example, if a single piston caliper is able to apply 500lbs. of
push, there would have to be 500lbs. of resistance from the other side
of the caliper frame. Total 1000 lbs. 'squeeze'. All things being
equal, if in a dual piston caliper, with each piston pushing with 500
lbs. force, doesn't the rotor experience the same amount of friction
against turning?

Where is my thinking wrong?

Thanks for replies,
Ivan Vegvary



Force = Pressure x Area

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.

--
Anthony

You can't 'idiot proof' anything....every time you try, they just make
better idiots.

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