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I have a hydraulic disc road bike with a 160mm rotor on the front and a 140mm on the rear. The setup is exactly the same front to rear, except for rotor size. I'd like to know the disparity between the force I would have to apply at the brake lever for the total friction or stopping power to be equal front to rear, for a given speed over a given distance, to a complete stop, i.e. total work done by each to be exactly equal. Since the 140mm rotor is 7/8th's the size of the 160mm, I'd be tempted to say that an equal lever pull would produce 7/8th's the friction on the rear, but I'm sure there is more to it than that.

So, for example, If I'm riding at 10mph and come to a stop over 100ft, how hard would I have to pull each lever (as a ratio of front brake/rear brake) to ensure that my momentum is converted to exactly the same amount of heat by each rotor respectively?

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  • Hmm, the brake lever provides a mechanical advantage to help depress the hydraulic cylinder. I think that because of this there is not enough information to scientifically answer this. How did you want the force you exert expressed anyway? If you are looking for a number like how many newtons need to be exerted to achieve the same results from each disk, well... I'd be really curious as to how you would use those numbers to affect your braking habits... Otherwise, if you want to try to make your braking roughly equal. Maybe I'm off the mark though
    – renesis
    Commented Apr 16, 2015 at 22:25
  • @renesis I'm not looking for a force in terms of newtons, just a ratio, and how that ratio was derived. Something along the lines of 'I need to pull the rear with 8/7 the force as the front to ensure I bleed off speed equally'. Thinking more about it I'm sure angular momentum has to be factored into the answer somehow. Practically, I'm not going to consciously pull the rear harder when I'm braking, I just want to know for knowledge sake, and also to determine pad wear and replacement.
    – ebrohman
    Commented Apr 16, 2015 at 22:48
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    Gotcha. So it's more of an academic question than something to be applied. Well, if no one happens along who can answer your question, you may want to try over at physics.stackexchange.com Just a suggestion.
    – renesis
    Commented Apr 16, 2015 at 23:18
  • @renesis thanks for the suggestion. I think I'm going to take it down, and post the question in the physics community, worded a bit differently.
    – ebrohman
    Commented Apr 17, 2015 at 2:04
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    Why do you want stopping power to be equal front to rear? The front is bigger because the front wheel has more stopping power. As you brake weight is shifted to to the front wheel. Same heat is a different question.
    – paparazzo
    Commented Apr 17, 2015 at 12:23

5 Answers 5

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Don't worry about equal torque. The front wheel will do as much as ninety percent of the work on a maximum stop. You will learn to modulate the rear to prevent lock-up. Equal pressure will be fine for normal stops.

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There are some assumptions in your question that make it next to impossible to answer.

  1. Braking is not a static operation.

The friction of any wheel is directly proportional to the weight currently being supported by that wheel. As you brake the effective "weight" moves from the both wheels to almost entirely on the front wheel. The harder you brake just the rear wheel, the sooner it will lock up.

  1. Effective braking is not about absolute stopping power, but modulation.

Just about any brake can lock up a wheel, and once you lock up a wheel your stopping distance isn't going to get any shorter[1] and you've lost a significant amount of control on where the bike is going. The point of larger disks is not to increase absolute stopping power but to increase control and heat dissipation. The leverage difference between a 140mm and a 160mm rotor compared to the much larger diameter of the wheel is fairly minimal:

140/700 = 0.2

160/700 = 0.228

A larger disk allows you to get closer to the lock up point and also suffers less brake fade due to heating. In MTB jargon terms, a bigger rotor provides better "modulation", this means control of stopping power, not absolute stopping power.

The answer to your question is not a simple ratio, but a curve that is a solution to a relatively complex set of differential equations.

[1]- Depending on the surface, it may actually get longer.

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Assuming the same force applied by the cylinder and the same materials, with the same pad size, the larger rotor will have a larger "lever arm" and be more "effective", in proportion to the diameter.

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  • I know this, the larger rotor will create more torque on the wheel. I'd like to know what ratio of forces - front/rear lever pull - will cause the two different sized rotors to burn the speed equally. 160mm front, 140mm rear, or any ratio for that matter. You are going to have to pull the rear harder in this case. I'm asking how much harder.
    – ebrohman
    Commented Apr 17, 2015 at 1:18
  • @ebrohman - You can roughly assume that, above maybe 30% brake force, the force on the lever is proportional to the torque on the wheel.
    – Daniel R Hicks
    Commented Apr 17, 2015 at 1:23
  • IMHO this is correct. The rear disk is 7/8 the size of the front, so the same pressure el produce 7/8 of the torque. To get the same torque requires 8/7 of the pressure that is on the front to be applied to the back.
    – andy256
    Commented Apr 17, 2015 at 4:29
  • @andy256 that's what I was thinking, but it seems too simple. I'm looking to see that actual calculation. As maatnz said, the bigger rotor cools faster, has more air passing over it, etc.
    – ebrohman
    Commented Apr 17, 2015 at 15:53
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    @ebrohman - For most materials, in the "medium" range of operation (where brakes would likely be operating), the "coefficient of friction" is a near-constant, meaning that sliding friction is directly proportional to caliper pressure.
    – Daniel R Hicks
    Commented Apr 17, 2015 at 17:07
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Your simple answer is essentially correct. A hydraulic system is going to maintain a pretty constant mechanical advantage and frictional losses will be trifling. So 8/7 more force on the rear will provide identical torques on each wheel provided the pad friction response is linear, As pointed out, identical torques won't mean much as each wheel is under a different load. Or to put it another way, so long as both wheels are maintaining road contact they will both decelerate at the same rate, regardless of which brake you are using...

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While braking gently:

Assuming identical hydraulics, pads and rotor material, friction will be linear with applied force. However, the larger disc is moving faster past the caliper, increasing the distance on which the friction acts and therefore braking power. The ratio will be about 7/8 (with larger rotor needing less force) with some adjustment for the fact that pads are not on the edge of the disc (making the difference somewhat larger).

But:

Braking causes weight transfer - the rear wheel has a tendency to lift while the front carries more weight. This means that the rear wheel will slip under significantly lower braking power than the front (I hear it is by about 50% for common designs). This means that under some conditions some levels of braking power achievable with the front brake will be impossible with the rear.

(A detailed discussion of how to balance and combine the use of the front and rear brake is outside the scope of this question.)

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