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).
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 χ pad area = pad pressure psi.
On Wed, 1 Feb 2006 10:26:24 +0100, email@example.com (Nick Mόller)
Yes, in the case of lifting a car two stacked one-ton jacks would be
limited to one ton. When you start pumping the jack on the top and
reach the one ton pressure limit, even if you jimmy the relief on the
upper jack the relief valve on the bottom jack would open to avoid
But NOT when you're talking about a brake caliper, that's not the
right way to look at it - The object of the pressure would be BETWEEN
the two stacked jacks, and it WILL see two tons of pressure - one ton
from each side. They are in parallel, but the forces are opposing.
Rotor - Measure Force Here
Know what I mean, Vern?
--<< Bruce >>--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
In other words, if you stacked your two one ton car jacks nose to nose,
rather than nose to base, you'd be able to lift 2 tons??
In your ASCII diagram, no matter where in the stack you insert a force
sensor, it will register the same value. If this were not the case
there'd have to be some additional reaction to ground between the rigid
They are in series. Sorry, but if you don't know the difference between
serial and parallel, further discussion seems quite useless.
Um, ah! So if you stack two jacks and pump one with 1ton you will have 1
ton pressure between the two jacks. Then when you pump the second jack
to 1 ton (disregarding it has already a "pressure" of one ton) the force
will not increase, but the pressure between will be 2 tons?
This is my last attempt to explain:
First, I ask you to read back a bit and try to understand my posting
with the "two opposed spindles C-clamp" example. Thanks.
Now, you have to accept a fact of physics. "actio = reactio". Or, in
plain english, "every force has a counter-force (or reacting force)".
More verbose: Every force has a reacting force of the same value, but
opposed direction and colinear (in line) with that force.
Or, the vectorial summ of the forces must be zero.
Low lets look at the single piston bracket:
DDD is the disk
R is the reacting force (actio = reactio, remember?)
F is the force
[I have left out the piston and the bracket]
Now the reacting force "R" is exactly the force "F" from the piston. If
it would not be that way, you will move the disk. That doesn't happen.
The resulting force is "conducted" (don't know your word for that, but
you know what I mean) through the bracket. If F gets smaller, R will get
smaller an have the save value (but still the opposite direction).
Now to the two pistons:
Rl Fl Fr Rr
Rl: reacting force left side
Fl: force left side
| is the piston (I needed to draw it now, to show where each vector
Fr: Force right side
Rr: reacting force right side
Now if you look at the drawing, you have two reacting forces (Rl and
Rr). They do have opposed direction, same value and are colinear. They
eliminate themselves. So I leave them out in the next drawing:
Uuuups! Damned, that looks exactly like the drawing with a single
piston. Just the names are a bit different. But names don't matter. I
could have used my cats' names. :-)
And, believe it or not:
**** IT IS EXACTLY THE SAME! ****
I can't believe this thread is still going. In your sketch, the vertical
_force_ on the rotor is ZERO. That's what you _want_, unless your rotor is
on a mighty strong bearing. That's what you _get_, whether you use two
"rigid surfaces" and two jacks, or whether you use two "floating surfaces"
rigidly connected to each other, and replace one jack with a spacer.
The _horizontal_ force on your rotor is your braking force. It is your
friction coefficient times _two_ tons, because there are _two_ sliding
interfaces. But there would be two interfaces even if you replace one jack
with a spacer and let the caliper float.
Replace one of your "pads" with a (frictionless) _bearing_, and you cut your
braking force in half. But the net vertical force _on_ the rotor is still
zero, and the net vertical force _through_ the rotor is still one ton.
Umm, hydraulic pistons have two important characteristics: force and
stroke. Two identical cylinders with equal pressures placed in parallel
double force, the same two in series double stroke. Sort of the same
way identical batteries in parallel double current, the same in series
double voltage. Likewise adding links to a chain makes it longer, not
The discussion was about stationary brake calipers with opposed
pistons--effectively hydraulic cylinders in series. So you would get a
potential increase in stroke, but no increase in force. But since a
fixed brake caliper needs no more total stroke than a floating caliper
it's more comparable to replacing one long cylinder with two short ones
stacked on top of each other.
Someone did mention earlier in the discussion a caliper with parallel
cylinders. He also mentioned that the cylinders are reduced in
size--that keeps down the force on the pads and reduces the amount of
fluid the master cylinder would need to move to apply the brakes. But
that's a different arrangement.
You mentioned two bottle jacks, but did not say if they were in
series or parallel. So, yeah, under the right circumstances two, ten,
or a hundred one ton bottle jacks push no harder than a single one ton
The formula is more correctly written as Force = piston cross
sectional area X pressure - friction. Pad psi = force / pad cross
sectional area. You'll note that pads follow the same rule: two pads in
series have the same psi as one pad with the same area. But you get
more friction since friction force = total pad area X pad psi.
B.B. --I am not a goat! thegoat4 at airmail dot net
again, stop and think about it for a second.
we have 100 psi. that is 100 pounds per square inch.
if we double the square inches, we double the force.
another way to think about it.
assume we have a single piston, and 1000 psi of force being applied to
from teh piston, and further assume that the brake pads are 10 square
with a single caliper, there are 2 pads, so that 1000 psi gets spread
over 20 square inches, for 50 PSI on the pads. with 2 pistons, you
would have 100 psi, and double the friction.
You are not doubling the area because the force is colinear and shares a
common ground. Because of this the force cannot exceed that generated by
one piston. If it did it would drive the other piston back into its
bore, but it cant because it is in equalibrium.
Jesus there's a lot of shite on this thread.
1/ imagine a motorcycle front wheel and brake, it makes things easier.
2/ in the old days iy would be cable and shoes, now it's hydraulic and
3/ hydraulics give a mechanical advantage, the smaller "operating"
cylinder vs the larger "operated" cylinder, but the flipside is you get
less travel on the larger diameter cylinder
4/ if the "operating" cylinder displaces 1 cc of fluid, this doesn't
necessarily mean there is any pressure of note in the system
5/ After you take up the slack on "off" or "released" brakes, then,
apart from compressibility of the pad and friction material, generally
speaking each extra amount of fluid displaced by the "operating"
cylinder will inclrease pressure (yes, fluids are supposed to be
incompressible, but brake lines expand etc etc
REALLY IMPORTANT from some of the shit mentioned above.
psi or any other pressure measurement is NOT a force measurement and is
nothing like a force measure ment
single piston hydraulic press type system versus opposing (identical)
pistin brake caliber type system, with the same pistons, piston area
and pressures YOU DO NOT get double the force, you do not even get the
force evenly applied to both sides cos it already is, what you do get,
and all that you get is a floating caliper. this means you get a disk
that is being gripped instead of deflected, this means a hard cast disk
can be MUCH thinner = much lower unsprung weight
to the OP, the television person was talking complete bollocks
I think if you were to read the responses carefully you'd find only a
few glaring errors in reasoning. Most everyone reached the same
conclusion, though some in roundabout and not always clearly expressed
Perhaps you'd prefer a thread on WMDs or the circumstances of Vince
Foster's death <g>.
you're prolly right, I tend to skim rather than read in depth
dunno who vince foster is, as far as wmd's go I have never had any
illusions, if fact the only people who seem to be disconnected from
reality are those that rely on the mainstream media H^H^H^H propoganda
for their "facts"
the next 20 years sure are going to be interesting times.
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