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

here is an easier way to think about it.
lets say there is 100 psi pressure in the brake line. if we double the
size of the piston, we double the pressure on the brake pads.
if we have 2 piston calipers, and convert to 4 piston, but the total
area is the same, we have the same pressure squeezing.

HOWEVER, with 4 pistons, we might have more consistent, and even
pressure. Which means we could get a higher clamping force before the
brakes lock up.

"Ivan Vegvary" wrote: (clip) Where is my thinking wrong?
^^^^^^^^^^^^^^^^^^
Your thinking is not wrong. How much pressure is a diver subjected to,
considering that it is being exerted from all angles? Having two hydraulic
cylinders pressing on opposite sides of the brake disk is a way of
eliminating deflection without making the disk super heavy. Also, the way
the caliper is constructed, the opposing forces balance each other, so you
don't need a super heavy bracket.

"Leo Lichtman" wrote in message
...

"Ivan Vegvary" wrote: (clip) Where is my thinking wrong?
^^^^^^^^^^^^^^^^^^
Your thinking is not wrong. How much pressure is a diver subjected to,
considering that it is being exerted from all angles? Having two
hydraulic cylinders pressing on opposite sides of the brake disk is a way
of eliminating deflection without making the disk super heavy. Also, the
way the caliper is constructed, the opposing forces balance each other, so
you don't need a super heavy bracket.

Yes, a lighter bracket makes a lot of sense.
Ivan

Ivan Vegvary wrote:

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

People say the darndest things. The only thing dual pistons do is to
keep the whole caliper from moving. The forces on each pad are the
same. Its the same as pushing against the wall or pushing against
another person, the pressure is the same.


John

"Leo Lichtman" wrote in message
...

"Ivan Vegvary" wrote: (clip) Where is my thinking wrong?
^^^^^^^^^^^^^^^^^^
Your thinking is not wrong. How much pressure is a diver subjected to,
considering that it is being exerted from all angles? Having two
hydraulic cylinders pressing on opposite sides of the brake disk is a way
of eliminating deflection without making the disk super heavy. Also, the
way the caliper is constructed, the opposing forces balance each other, so
you don't need a super heavy bracket.



There's no deflection of the disc. The caliper floats, providing the pins
aren't corroded.

Garrett Fulton

"Ivan Vegvary" wrote in message
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
I think you are forgetting there are two reactions, the force of the rotor
against the near brake pad and the force of the other side of the rotor
against the brake pad on the other side, translated through the frame of the
caliper. In this case it is the rotor that supplies the equal and opposite
force. The force of a single piston would be distributed on both sides of
the rotor, assuming no friction on the sliding mechanism. A second piston of
the same size would double the "squeezing force", however the fallacy in the
commentator's statement lies in the fact that the dual piston design would
use correspondingly smaller pistons. The reason for the design is to make
sure force is equal on both sides without relying on the sliding frame, not
to double the braking force, as another poster pointed out.

"gfulton" wrote: There's no deflection of the disc. The caliper floats,
providing the pins aren't corroded.
^^^^^^^^^^^^^^^
There is no deflection of the disk with caliper brakes. What I was trying
to say is that if, instead of a caliper, you had a piston pushing on one
side of the disk, there would be a deflection.

Ivan Vegvary wrote:

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?

You can think about an experiment that will give you a clear anser:

You have one caliper with one piston that pushes onto the disk with say
1000N. Now you take a second caliper with the same piston diameter and
slip it over the disk. You will get double the force, no matter on what
side of the disk the piston of that second caliper is. You can now make
a single caliper out of the two, without changing anything compared to
having two calipers.


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

"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.

Remove sp to reply via email

People say the darndest things. The only thing dual pistons do is to
keep the whole caliper from moving. The forces on each pad are the
same. Its the same as pushing against the wall or pushing against
another person, the pressure is the same.

......... But the brake pad area is doubled. Twice the stopping force,
which is what we're talking about.
Bugs

Bugs wrote:
People say the darndest things. The only thing dual pistons do is to
keep the whole caliper from moving. The forces on each pad are the
same. Its the same as pushing against the wall or pushing against
another person, the pressure is the same.

........ But the brake pad area is doubled. Twice the stopping force,
which is what we're talking about.
Bugs

At least absent context, this sounds incorrect. Doubling the area w/o
changing the normal force would not change the (dry) friction force.
Don't buy it? Remember that doubling the area (again keeping the normal
force constant) results in a lower pressure between the surfaces.

Bill

"Bugs" wrote in message
ups.com...
People say the darndest things. The only thing dual pistons do is to
keep the whole caliper from moving. The forces on each pad are the
same. Its the same as pushing against the wall or pushing against
another person, the pressure is the same.

........ But the brake pad area is doubled. Twice the stopping force,
which is what we're talking about.

No.... there is a "passive" pad on the non-cylinder side. Where have you
seen a caliper brake system that had only one pad pressing on only one side
of the disc?

LLoyd

When the caliper float is essentially becomes a double ended piston. So
your extra 833 pounds is imposed on the closed end of the piston, forces
the entire caliper against the rotor from the other side. Host was
incorrect.

Anthony wrote:
"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.

The great majority of OEM disk brakes are single piston and there is no
deflection of the disk because the caliper body is free to move sideways on
lubricated slides.

Randy


"Leo Lichtman" wrote in message
...

"gfulton" wrote: There's no deflection of the disc. The caliper floats,
providing the pins aren't corroded.
^^^^^^^^^^^^^^^
There is no deflection of the disk with caliper brakes. What I was trying
to say is that if, instead of a caliper, you had a piston pushing on one
side of the disk, there would be a deflection.

Nick Müller wrote:

You have one caliper with one piston that pushes onto the disk with say
1000N. Now you take a second caliper with the same piston diameter and
slip it over the disk. You will get double the force, no matter on what
side of the disk the piston of that second caliper is. You can now make
a single caliper out of the two, without changing anything compared to
having two calipers.

Not if in making the single caliper out of two you effectively put the
cylinders in series.

The simple question would be, will the system still work if the
pressure line to one cylinder is removed and capped, leaving the
cylinder vented but preserving system pressure for the other one?

"Anthony" wrote in message
...
"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.


Ummm..No.

Remember ol' Newton said about equal and oppostite reactions? With only 1
piston to force is as you describe. The first force is from the piston, and
since the disk is free to move axially on the shaft, the reaction comes from
the solid leg of the caliper opposite the piston (with the disk between the
piston and the leg). Now if you replace the caliper leg with another
piston, the force doesn't change in magnatude. Both pistons are "putting
out" 882 lb (in your example) and each is reacting the force from the other.

Think of it like this:
Stand with you hands against the wall. Move your legs back away from the
wall until you will surely fall over if not for leaning against the wall.
Feel the force in you hands? It's coming from your body weight. Now repeat
this except lean against someone elses' hands (kind of like forming an arch
for the kiddies after a soccer game). Did the force you feel in your hands
double? After all, the force comes from your body weight (like the wall
experiment) as well as the other person.

Kelly Jones wrote:

Remember ol' Newton said about equal and oppostite reactions?

actio = reactio


Think of it like this:
Stand with you hands against the wall. Move your legs back away from the
wall until you will surely fall over if not for leaning against the wall.
Feel the force in you hands? It's coming from your body weight. Now repeat
this except lean against someone elses' hands (kind of like forming an arch
for the kiddies after a soccer game). Did the force you feel in your hands
double? After all, the force comes from your body weight (like the wall
experiment) as well as the other person.

Now if you replace the wall with a leaf spring (the wall is actually a
very stiff one), it will deflect with one person pressing against it. If
you put the leaf spring between the two hands, it will not deflect,
because equal force is coming from two opposite sides. - Double the
force.


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

""Nick Müller"" wrote in message
...
Kelly Jones wrote:

Remember ol' Newton said about equal and oppostite reactions?

actio = reactio


Think of it like this:
Stand with you hands against the wall. Move your legs back away from the
wall until you will surely fall over if not for leaning against the wall.
Feel the force in you hands? It's coming from your body weight. Now
repeat
this except lean against someone elses' hands (kind of like forming an
arch
for the kiddies after a soccer game). Did the force you feel in your
hands
double? After all, the force comes from your body weight (like the wall
experiment) as well as the other person.

Now if you replace the wall with a leaf spring (the wall is actually a
very stiff one), it will deflect with one person pressing against it. If
you put the leaf spring between the two hands, it will not deflect,
because equal force is coming from two opposite sides. - Double the
force.


So, Nick, try this: Put a brittle spring between your hands (I suggest an
egg). Press both hands together real hard. According to your note, the egg
will not break (deflect) "because equal force is coming from two opposite
sides".

Ivan Vegvary wrote:
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


You are right. There is no force advantage to opposing cylinders.

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
inches.
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.

In article .com,
says...
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
inches.
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.



That'd be true if the single piston caliper were not free to center
itself. But it can, and as a result applies the same pressure to the
pads on both sides of the disc.

Ned Simmons

Bugs wrote:

People say the darndest things. The only thing dual pistons do is to
keep the whole caliper from moving. The forces on each pad are the
same. Its the same as pushing against the wall or pushing against
another person, the pressure is the same.

........ But the brake pad area is doubled. Twice the stopping force,
which is what we're talking about.
Bugs

Every single piston brake I've seen had pads on both sides of the rotor.
The pressure that is produced by the single piston is applied to both
pade. It's just like a vice, each jaw pushes the same amount, even
though only one moves.
If you have two pistons, they push against each other with the rotor
in between. The force remains the same as long as the area of the
piston is the same, but instead of pushing against the other half of the
caliper with the other fixed pad mounted on it, the pad is being pushed
by the piston.

John

Kelly Jones wrote:

So, Nick, try this: Put a brittle spring between your hands (I suggest an
egg). Press both hands together real hard. According to your note, the egg
will not break (deflect) "because equal force is coming from two opposite
sides".

You forgot to put a :-) behind your nonsense^Wstatement.

Or do you really not know the difference between inner and outer forces?


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

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

--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

Doug wrote:
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
inches.
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 ****e 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
pads

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 **** 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 ********

In article .com,
says...
Jesus there's a lot of ****e on this thread.


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
ways.

Perhaps you'd prefer a thread on WMDs or the circumstances of Vince
Foster's death g.

Ned Simmons

On 30 Jan 2006 20:07:28 -0800, "Doug" wrote:

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
inches.
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.

When they make "Floating" calipers with the pistons on one side,
they use one piston with 10 square inches area, or two pistons with 5
square inches of effective working area each. The clamping effect
goes through the caliper body, and the caliper moves laterally on the
slide pins or rails to apply clamping pressure to the pads on both
sides of the rotor.

When they make a four piston fixed caliper, they put four pistons
with 2.5 square inch working area each, two on each side. They have
the same effective area (10 square inches total), so the clamping
forces applied to the brake pads are exactly the same.

The big advantage to a fixed caliper design is that the caliper
doesn't have to move laterally to apply and release, it can be bolted
down firmly. There are no large caliper slides (that have to transmit
braking forces) to get rusty and stick, and cause dragging shoes and
other brake problems.

But the four-piston designs cost more to make initially, so when car
makers try to save a buck they don't get used. And the Big Three in
Detroit wonder why they're going bankrupt...

-- Bruce --

--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
5737 Kanan Rd. #359, Agoura CA 91301 (818) 889-9545
Spamtrapped address: Remove the python and the invalid, and use a net.

Ned Simmons wrote:
In article .com,
says...
Jesus there's a lot of ****e on this thread.


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
ways.

you're prolly right, I tend to skim rather than read in depth


Perhaps you'd prefer a thread on WMDs or the circumstances of Vince
Foster's death g.

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.

On 31 Jan 2006 09:57:20 -0800, Guy Fawkes wrote:

Ned Simmons wrote:

Perhaps you'd prefer a thread on WMDs or the circumstances of Vince
Foster's death g.

dunno who vince foster is,

Oh, just some guy the Clintons had killed. Unless you believe that he
moved his own body after being dead long enough for post-mortem lividity
to set in, I suppose.

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"

....or people with selective memory, like the current crop of democrats
who pretend that they didn't agree about the WMDs before the war.

Mr. Bergman hit it exactly right. In order to maintain the same
clamping force, you must maintain the same square inches of piston
area.

A couple of other comments. The goal is to provide maximum moving
friction without locking up the rotor. In order to do that, it is
important to keep things "consistent."
Even loads across the brake bad, lack of flexing in the caliper, lack
of cocking, angling, digging, or otherwise flexing of the caliper and
its loads in relation to the disk are important.

When all done, race cars take advantage of every little bit they can.
Braking is very important. Race cars spend a lot of money and weight to
keep their calipers and pads true to the rotors.

Flexing is a bad thing, and floating calipers by definition must move
about to do their jobs. Single piston calipers are not as good as fixed
calipers.

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

Ivan Vegvary wrote:
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.

No. Energy of a moving object is 1/2 mass times velocity squared. Total
energy disspiated in a two car crash is the sum of the energy of each
car. A 2 ton car slamming into a wall at 50 miles per hour dissipates 1/4
of the energy of a car going 100 into a wall. Two cars that compact the
same way driving into each other is really like two cars slamming into a
wall. If you consider the rate of deceleration the only measure of a
crash, the original statement is sort of true, but still misleading.





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

Tom wrote:

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.

I didn't say this. If the jacks work in parallel, they lift 2 tons. But
if you stack two jacks, will they lift 2 tons?

hint: The answer is "no"


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

Nick Müller wrote:

Tom wrote:

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.

I didn't say this. If the jacks work in parallel, they lift 2 tons. But
if you stack two jacks, will they lift 2 tons?

hint: The answer is "no"

Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

Exactly, but that is not the case in a 2 piston caliper,
the forces aren't acting in tandem. Still I notice, you
tend to ignore what you don't understand..

Tom

Tom wrote:

Exactly, but that is not the case in a 2 piston caliper,
the forces aren't acting in tandem.

What do you mean by "in tandem"? Parallel or serial?


Still I notice, you tend to ignore what you don't understand..

Ignorance is often a subjective attribute. :-)


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

On Wed, 1 Feb 2006 10:26:24 +0100, (Nick Müller)
wrote:
Tom wrote:

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.

I didn't say this. If the jacks work in parallel, they lift 2 tons. But
if you stack two jacks, will they lift 2 tons?

hint: The answer is "no"

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
overloading.

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.


RIGID SURFACE
|
Jack
|
Pad
--------
Rotor - Measure Force Here
--------
Pad
|
Jack
|
RIGID SURFACE

Know what I mean, Vern?

-- Bruce --
--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
5737 Kanan Rd. #359, Agoura CA 91301 (818) 889-9545
Spamtrapped address: Remove the python and the invalid, and use a net.

"Bruce L. Bergman" wrote in message
...
On Wed, 1 Feb 2006 10:26:24 +0100, (Nick Müller)
wrote:
Tom wrote:

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.

I didn't say this. If the jacks work in parallel, they lift 2 tons. But
if you stack two jacks, will they lift 2 tons?

hint: The answer is "no"

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
overloading.

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.


RIGID SURFACE
|
Jack
|
Pad
--------
Rotor - Measure Force Here
--------
Pad
|
Jack
|
RIGID SURFACE

Know what I mean, Vern?

-- Bruce --
--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
5737 Kanan Rd. #359, Agoura CA 91301 (818) 889-9545
Spamtrapped address: Remove the python and the invalid, and use a net.

"Lloyd E. Sponenburgh" wrote in message
.. .
Huh? I _read_ the response, and disagree. But I didn't intentionally
respond. Sorry.
LS