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Thread: Hydraulic Brake Question

  1. #1 Hydraulic Brake Question 
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    Hello, everybody. I was reading about some hydraulic disc brakes for bicycles and came across a concept that threw me off. I hope people here can help me understand it.

    First, this is how the brakes in question work:

    The rider pulls a lever which moves a "master piston" into the fluid. In the caliper, two "slave pistons" move in opposing directions (toward each other), clamping a pair of brake pads onto the rotor. The master piston has a smaller area than the slave pistons. Therefore, it travels a longer distance with less force while the slave pistons each travel a shorter distance with greater force.

    This is the concept that threw me off:

    "Due to the fact that there are two slave pistons, it might seem intuitive that the combined surface areas of the slave pistons would factor into the piston ratio, effectively doubling the clamping force on the rotor. This would be true if the combined slave piston areas were moving in the same direction (as is the case with one side of a four-piston caliper). However, if the two slave pistons move in opposing directions, the total clamping force is only equal to the force produced by one of the pistons. For example, if one of the slave pistons were immobilized, and the other slave piston generated the overall clamping force on its own, the force produced would be no different than if both pistons were working together."

    This is a quote from the designers of the brakes and I assume that it's correct. But why is it so? Do the opposing pistons work against each other, thus canceling out some of their combined force?

    Thanks for reading!

    - Eric


    Last edited by ericbakuladavis; October 3rd, 2014 at 11:08 AM. Reason: Made less wordy
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  3. #2  
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    Pistons moving in the same direction (adjacent) add their area as if they were one piston.
    Pistons moving in opposite directions are effectively in-line at their individual areas and so do not add.
    In both cases the hydraulic volumes add.

    Think of two hydraulic cylinders back to back. One pressing down at the ground, and one up. Both are 1 square inch @ 1000 psi.
    Putting 1000 pounds on top loads both to capacity.


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  4. #3  
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    Thanks for your answer, G O R T

    Quote Originally Posted by G O R T View Post
    Pistons moving in the same direction (adjacent) add their area as if they were one piston.
    This makes sense to me.

    Quote Originally Posted by G O R T View Post
    Think of two hydraulic cylinders back to back. One pressing down at the ground, and one up. Both are 1 square inch @ 1000 psi.
    Putting 1000 pounds on top loads both to capacity.
    This makes sense to me also. The 1000 lb weight is acting on both pistons simultaneously.

    In the case of two face-to-face pistons moving toward each other, clamping a rotor, why is the total clamping force only equal to the force of 1 piston? I'm thinking there's more to it than the fact that they are in-line. Does it have to do with the fact that the rotor is flexible and therefore the pistons are pushing against each other? Would it be different if the rotor were perfectly stiff?
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  5. #4  
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    Yes the pistons push against each other, useful rotors cannot be made that inflexible.
    We have opposed pistons in a frame. Each piston is 1 square inch and designed for 1000 psi.
    The first piston can supply 1000 lb. force max, and is extended against an internal stop to keep it together.
    If the second (opposed) piston applied anything less than 1000 lb. against the first piston, the first piston would still apply 1000 lb.
    If the second piston exceeded 1000 lb. of force, it would just push the first piston back and increase it's hydraulic pressure.
    This would exceed the design limit.
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  6. #5  
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    OK. I think I got it now. Because the slave pistons have the same area and are powered by the same master piston, they push with equal force. So one piston acts as a backstop for the other and visa versa, Now I see how it would be the same if one piston were immobilized and it's pad were almost touching the rotor. The immobilized piston would push back with equal force, just like it would if it were moving.

    I didn't think there could be a rotor that stiff. But if there were one, then the clamping force of the caliper would be doubled, right?

    BTW: In these brakes, the pistons have no internal stops. If you take the brake pads and rotor out, you can pop the pistons right out of their holes!
    Last edited by ericbakuladavis; October 5th, 2014 at 02:11 PM.
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  7. #6  
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    Weight is critical on a bike thus I expect the rotor to be thin to the point of flexibility. A single pad would require a brace on the other side of the rotor. I suggest that it is simpler and more effective to use symmetrically counter-acting pads that will not deform a light, thin but circumferentially strong disc.
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  8. #7  
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    Quote Originally Posted by WaltBJ View Post
    Weight is critical on a bike thus I expect the rotor to be thin to the point of flexibility. A single pad would require a brace on the other side of the rotor. I suggest that it is simpler and more effective to use symmetrically counter-acting pads that will not deform a light, thin but circumferentially strong disc.
    They are indeed. However, a fair number of hydraulic (and cable disk) brakes have only one actuator (a piston) that moves against a frame. The frame wraps around the rotor so that the piston/rotor has something to act against. (There are pads on both sides.) They seem to work fine.
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    Quote Originally Posted by WaltBJ View Post
    Weight is critical on a bike thus I expect the rotor to be thin to the point of flexibility. A single pad would require a brace on the other side of the rotor. I suggest that it is simpler and more effective to use symmetrically counter-acting pads that will not deform a light, thin but circumferentially strong disc.
    As billvon said, you are right on. However, I posed the question not because I think a perfectly stiff rotor would be a good idea. It was a hypothetical question aimed at better understanding physics. If the rotor were "perfectly stiff", would the clamping force of the pistons be affected?
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    Quote Originally Posted by ericbakuladavis View Post
    if the two slave pistons move in opposing directions, the total clamping force is only equal to the force produced by one of the pistons.
    ...

    But why is it so?
    Think of someone weighing themselves on a bathroom scale. The person's weight presses down on the scale, and the floor resists by pushing up on the scale with [basically] the same weight.
    Grief is the price we pay for love. (CM Parkes) Our postillion has been struck by lightning. (Unknown) War is always the choice of the chosen who will not have to fight. (Bono) The years tell much what the days never knew. (RW Emerson) Reality is not always probable, or likely. (JL Borges)
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    Quote Originally Posted by jrmonroe View Post
    Quote Originally Posted by ericbakuladavis View Post
    if the two slave pistons move in opposing directions, the total clamping force is only equal to the force produced by one of the pistons.
    ...

    But why is it so?
    Think of someone weighing themselves on a bathroom scale. The person's weight presses down on the scale, and the floor resists by pushing up on the scale with [basically] the same weight.
    Thanks jrmonroe. This is a good way to illustrate the concept.

    Quote Originally Posted by ericbakuladavis View Post
    If the rotor were "perfectly stiff", would the clamping force of the pistons be affected?
    Thinking about it some more, I'd say a perfectly stiff rotor would not affect clamping force. The flexible rotor pushes back on each pad by means of the other pad. A stiff rotor would push back on each pad by means of it's stiffness. In both cases, I'd say the total clamping force is basically the same. What do y'all think?
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