What does a disc brake piston seal do?

pbonsalb

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Drives: 18 F90 M5, 99 E36 M3 Turbo

That is just a dust boot. The brake fluid seal is an o-ring further in. I don’t know whether the piston will cooperate with you so you can just wiggle it out far enough to replace the boot. I have pulled the calipers when replacing boots and seals. It is fine to drive in the meantime. You will have to force the piston out in order to get the dust boot seated properly in the groove. If the piston won’t wiggle out and it probably won’t you can try carefully levering from the inside but you may need to use a compressor blow gun in the brake hose port. Put a piece of wood in the caliper so the piston does not come all the way out. Give it 1/2 inch of space to come out into. If it comes all the way out you want to replace the seal as well. If you have to do that be careful with lubing the seal since it can twist and tear as you reinsert the piston and you may not know it.

Last edited by pbonsalb; 03-06-2021 at

01:38 PM

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Tim shows ample evidence that brake pistons retract - depending on the source, by 0.003" to 0.008". This is absolutely true.

I maintain that the brake pads skim the rotor lightly at all times. I maintain that this is demonstrable by simply attempting to pass a sheet of paper between the pads and rotors of any car that has used the brakes to stop. Alternately, just put your ear by the brakes while you turn the hub - you will be able to hear the friction. In fact, you should be able to feel a *light* drag as you rotate the hub. It is for this reason that I said that disk brakes don't "retract".

Both statements, although apparently contradictory, are (I believe) true. How can that be? The explanation is somewhat long, but for the few of those who might still care . . . .

All real world materials both compress and stretch under force. Of course, some things stretch more than others (rubber vs steel). But most objects behave as springs. Even steel bolts are actually very stiff springs. When you torque a head bolt, you are stretching it and establishing a spring force to hold the head to the block.

Disk brakes work by applying a force to a friction material (the pad) which presses against a rotating disk. In a simplified model, the actual braking force (that stops the wheel from rotating) is simply

Friction Force = Coefficient of Friction x Force on Pad

The actual Friction Force is then multiplied by the radius of the rotor to get the angular torque (this is somewhat simplified, since the rotor isn't a ring you would have to use calculus to find the torque - but it is fairly close). But, essentially, the Braking Torque is going to be the

Braking Torque = Friction Force x Some Constant

so all we have to deal with is the Friction Force.

Brake pads have some coefficient of friction. I don't know what it is, but I'm assuming it is fairly high, so let's assume it is 1.0 (normal materials can't get higher than that - if they are > 1.0, they are sticky).

It has been hard to find a reliable number for the actual pressure in a braking system during a panic stop, but given that tubing and components are tested at several thousand PSI, it seems reasonable to assume that braking hydraulic pressure can reach 500 to 1000 PSI. I seem to recall that the caliper pistons in an MGB are about 1.5" in diameter, so this gives an area of about 1.75 square inches. That means the pad is subjected to a force of between 883 pounds and 1750 pounds.

Now, a quick reality check to make sure these numbers seem plausible. Given the high coefficient of friction of the pads, this translates to an angular torque of somewhere around 750 foot pounds. Considering that the car weighs about 2500 pounds, but that is divided across 4 tires (625 pounds per wheel). Since the tires have a coefficient of friction of about 0.75, that means they can generate at most a friction force of about 470 pounds. Given the radius of our tires, this translates to about 500 foot pounds of torque. So, the brakes can generate slightly more torque than the tires and can lock the wheels under panic braking. Everything seems plausible.

So - what does this have to do with springs and retraction?

The force due to a spring is given by

F = kx, where k is Hook's constant and "x" is the amount the spring is compressed or stretched. The formula is for ideal springs, which our braking components are not, but as a first approximation this will be close enough.

Hopefully, by now you have run out to your car and tried to pass a piece of paper between the pad and the rotor. I just tried it. It wouldn't fit. As of 2 minutes ago, this was true (by experiment) on my pads and rotors. I suppose you could argue that I needed a thinner sheet of paper, but I invite you to rotate the wheel and listen and you will hear it dragging. So, I consider the hypothesis that the pad touches the rotor to be a proven fact.

The question is, how hard is the pad pressing against the rotor? It is surely very light, because the drag is minimal. Although I've never measured the effort required to rotate a front wheel, it is perhaps a couple of foot pounds. Even if the pad's coefficient of friction is 1.0, that means there can't be more than a couple of pounds residual force applied to the pad.

But - how can that be? Tim has shown that the piston's retract by 0.003" to 0.008". I believe the answer is in everything that lies between the pistons and the rotor. Basically the interface between the steel surface of the pad, the steel back of the pad, the glue between the steel pad and the pad, and the pad itself. Particularly the contact surface between the piston and the steel back of the pad is not perfect. It may only contact at certain points (under no load). Under hydraulic force the whole back will deform slightly and make contact. However, these are not the only surfaces that potentially flex under hydraulic pressure. Since the pad will press against the rotor, by Newton's laws it will press equally back against the caliper. The whole caliper will deform under braking pressure.

Basically, everything stretches as hydraulic pressure increases and braking force is generated. In reality, it is a bunch of different springs, all with different strengths (Hook constants) bolted together. But for simplicity, we can consider it to be a single composite spring.

What we know is that the force of the pad against the rotor goes from 1 pound (under the no load, marginal residual hydraulic pressure) to between 883 and 1700 pounds in panic stop - let's call it 1001 pounds to make our lives easier. The change in force is therefore 1000 pounds.

Since Tim has supplied us with the figure for piston movement (0.005&quot we can find the Hook's constant (k)

1000 = k (0.005&quot

so k = 200,000

Without a bunch of lab equipment (which I don't have) it would be hard to prove this figure, but I would argue that the braking system itself is the lab equipment.

We know (based on good source) the movement of the piston.
We know (based on scientific test) that the pads touch the rotors even when the brake pedal is not pressed.
We have calculated the force on the pads from first principles.

Therefore, our derivation of "k" is likely to be accurate, at least to the correct order of magnitude. Our k includes all of the deformations in all of the materials (including the flexing that takes place as microscopic gaps in surfaces close under tension).

Whew!

So, I would argue that everyone is right and everyone is a winner.


Terry Ingoldsby
terry.ingoldsby@DCExperts.com

Here is one more thought that might bring the whole discussion into alignment and agreement.Tim shows ample evidence that brake pistons retract - depending on the source, by 0.003" to 0.008". This is absolutely true.I maintain that the brake pads skim the rotor lightly at all times. I maintain that this is demonstrable by simply attempting to pass a sheet of paper between the pads and rotors of any car that has used the brakes to stop. Alternately, just put your ear by the brakes while you turn the hub - you will be able to hear the friction. In fact, you should be able to feel a *light* drag as you rotate the hub. It is for this reason that I said that disk brakes don't "retract".Both statements, although apparently contradictory, are (I believe) true. How can that be? The explanation is somewhat long, but for the few of those who might still care . . . .All real world materials both compress and stretch under force. Of course, some things stretch more than others (rubber vs steel). But most objects behave as springs. Even steel bolts are actually very stiff springs. When you torque a head bolt, you are stretching it and establishing a spring force to hold the head to the block.Disk brakes work by applying a force to a friction material (the pad) which presses against a rotating disk. In a simplified model, the actual braking force (that stops the wheel from rotating) is simplyFriction Force = Coefficient of Friction x Force on PadThe actual Friction Force is then multiplied by the radius of the rotor to get the angular torque (this is somewhat simplified, since the rotor isn't a ring you would have to use calculus to find the torque - but it is fairly close). But, essentially, the Braking Torque is going to be theBraking Torque = Friction Force x Some Constantso all we have to deal with is the Friction Force.Brake pads have some coefficient of friction. I don't know what it is, but I'm assuming it is fairly high, so let's assume it is 1.0 (normal materials can't get higher than that - if they are > 1.0, they are sticky).It has been hard to find a reliable number for the actual pressure in a braking system during a panic stop, but given that tubing and components are tested at several thousand PSI, it seems reasonable to assume that braking hydraulic pressure can reach 500 to 1000 PSI. I seem to recall that the caliper pistons in an MGB are about 1.5" in diameter, so this gives an area of about 1.75 square inches. That means the pad is subjected to a force of between 883 pounds and 1750 pounds.Now, a quick reality check to make sure these numbers seem plausible. Given the high coefficient of friction of the pads, this translates to an angular torque of somewhere around 750 foot pounds. Considering that the car weighs about 2500 pounds, but that is divided across 4 tires (625 pounds per wheel). Since the tires have a coefficient of friction of about 0.75, that means they can generate at most a friction force of about 470 pounds. Given the radius of our tires, this translates to about 500 foot pounds of torque. So, the brakes can generate slightly more torque than the tires and can lock the wheels under panic braking. Everything seems plausible.So - what does this have to do with springs and retraction?The force due to a spring is given byF = kx, where k is Hook's constant and "x" is the amount the spring is compressed or stretched. The formula is for ideal springs, which our braking components are not, but as a first approximation this will be close enough.Hopefully, by now you have run out to your car and tried to pass a piece of paper between the pad and the rotor. I just tried it. It wouldn't fit. As of 2 minutes ago, this was true (by experiment) on my pads and rotors. I suppose you could argue that I needed a thinner sheet of paper, but I invite you to rotate the wheel and listen and you will hear it dragging. So, I consider the hypothesis that the pad touches the rotor to be a proven fact.The question is, how hard is the pad pressing against the rotor? It is surely very light, because the drag is minimal. Although I've never measured the effort required to rotate a front wheel, it is perhaps a couple of foot pounds. Even if the pad's coefficient of friction is 1.0, that means there can't be more than a couple of pounds residual force applied to the pad.But - how can that be? Tim has shown that the piston's retract by 0.003" to 0.008". I believe the answer is in everything that lies between the pistons and the rotor. Basically the interface between the steel surface of the pad, the steel back of the pad, the glue between the steel pad and the pad, and the pad itself. Particularly the contact surface between the piston and the steel back of the pad is not perfect. It may only contact at certain points (under no load). Under hydraulic force the whole back will deform slightly and make contact. However, these are not the only surfaces that potentially flex under hydraulic pressure. Since the pad will press against the rotor, by Newton's laws it will press equally back against the caliper. The whole caliper will deform under braking pressure.Basically, everything stretches as hydraulic pressure increases and braking force is generated. In reality, it is a bunch of different springs, all with different strengths (Hook constants) bolted together. But for simplicity, we can consider it to be a single composite spring.What we know is that the force of the pad against the rotor goes from 1 pound (under the no load, marginal residual hydraulic pressure) to between 883 and 1700 pounds in panic stop - let's call it 1001 pounds to make our lives easier. The change in force is therefore 1000 pounds.Since Tim has supplied us with the figure for piston movement (0.005&quotwe can find the Hook's constant (k)1000 = k (0.005&quotso k = 200,000Without a bunch of lab equipment (which I don't have) it would be hard to prove this figure, but I would argue that the braking system itself is the lab equipment.We know (based on good source) the movement of the piston.We know (based on scientific test) that the pads touch the rotors even when the brake pedal is not pressed.We have calculated the force on the pads from first principles.Therefore, our derivation of "k" is likely to be accurate, at least to the correct order of magnitude. Our k includes all of the deformations in all of the materials (including the flexing that takes place as microscopic gaps in surfaces close under tension).Whew!So, I would argue that everyone is right and everyone is a winner.Terry Ingoldsby