Fred Seelig wondered in a comment if I could provide a scientifically derived explanation of the apparent greater stopping power in a disc brake over a rim brake. Fred’s comment got me thinking about it. How, physically, do disc brakes generate more stopping force? Today’s post is a rambling attempt to sort out in my mind the physical parameters involved in disc vs rim braking.
There is no doubt that disc brakes generate more stopping force at the tire-road (or tire-trail) interface for a given hand force and lever travel. The first time I rode a bike with disc brakes, I was surprised, and a bit scared by the raw stopping power at my fingertips. We could argue about modulation, temperature fade, performance when wet or muddy, cost, weight, rim wear, spoke forces, and on and on. But there are literally thousands of discussions on the topic out there already. Just Google “Rim vs Disc Brakes”.
Fred wondered how disc brakes can generate more stopping force from the same input. He noted that it is easier to stop a wheel with a tangential force near the outer diameter (a rim brake) than at some smaller diameter (a disc brake).
The laws of physics, unlike the laws of god and man, cannot be broken. Conservation of energy is one of those laws. I can’t get more work out of a system than the work I put in. All I can do is convert the energy from one form to another. In the case of a friction brake, I am converting energy from kinetic (rotation and forward motion, mostly forward motion energy) to heat. Something else is going on that enables the disc to generate so much stopping power.
Follow along as I sort this out in my own mind. I’ll start with some assumptions:
- We are comparing cable-actuated rim and disc brakes. (I’ll briefly consider hydraulic systems under the cable friction discussion later.)
- I assume that both systems require the same amount of lever travel for caliper actuation. They do. In fact, most cable-actuated disc brakes are designed to operate with the same levers as MTB rim brakes!
- The same hand force will be applied at the brake lever for each system.
- I am considering performance of clean, dry systems.
I can think of four five six seven parameters that determine braking power at the tire/road interface, where it matters.
- Braking Surface Diameter:
When a brake caliper grabs a disc or a rim, it is generating a tangential force opposite to the direction of motion. This results in a braking torque, which is translated to a horizontal braking force by the friction between the tire and the road. If you generate the same tangential force at a greater diameter, you will generate more braking torque and more horizontal braking force at the road.
Consider acceleration of a bicycle. You can accelerate faster on a large rear cog than on a small cog. In fact you can accelerate twice as fast (all else being equal (which it rarely is)) on a 28 tooth cog as you can on a 14. The 28 has twice the diameter. The chain applies the tangential force, and the 28 gives you a longer lever. The same principle applies to braking.
Disc brake rotor diameters range from 140 to 203 mm, with 160mm being a common size. The outer diameter of a 700c road rim is 622mm, almost four times the diameter of a 160mm disc.
So a 160mm disc brake has to generate almost four times the braking force – at the brake pads – just to break even (no pun intended) with a rim brake for stopping power.
- Lever Travel:
Lever travel and applied force is what it is for ergonomic reasons, and it is the same for disc and rim brake systems. A hand-operated brake system is designed to be ergonomically operable by a normal rider. The lever cannot start too far away from the bars or you can’t get your fingers around it. The travel cannot be too great or the lever will bottom out against the bars. And the required hand force must be in a range that is achievable by an average rider in all sorts of conditions. This is why I began my assessment with the assumption of equal lever travel and applied force for both systems.
Oh, by the way, force multiplied by distance equals energy. Hand force multiplied by lever travel is exactly the energy put into the system. Per my assumptions, we will input the same energy for rim or disc.
OK, I know that the first 20% or so of lever travel is at a very low force. You are just overcoming return springs in the lever and caliper as you bring the pads into contact with the rim or rotor. If it’s alright with everyone I’ll call that a wash. The return spring forces are similar for both systems.
- Coefficient of Friction:
I really expected to find that the coefficient of friction of a disc brake pad on a steel rotor is higher than that of a rim brake pad on an aluminum rim. There is a wide range for both, but a lot of internet resources suggest that a value of ~0.4 is a good average for either system, dry and clean. This whole discussion goes out the window when you throw in variables like wet rims, oil on rotors, mud, grit, glazing, and heat fade. And I couldn’t find any data on friction coefficient of a rim brake on a carbon rim. I believe the coefficient of friction on a carbon rim is lower than on an aluminum rim. I know I have to squeeze the brakes on my road bike harder when I am using my carbon fiber rims and carbon-specific pads to achieve the same deceleration.
Brief digression: Formula One race cars use carbon pads and rotors. It’s not carbon fiber with a resin like bicycle rims. It’s a different material, which apparently doesn’t even work well below 600C, and gets up to 1200C during braking events. But once it is hot, the coefficient of friction can be over 0.6! Here is an article on F1 brakes with a cool video of a carbon rotor exploding.
Wouldn’t night criteriums be even more exciting if the racers’ rims glowed red-hot through the corners and occasionally exploded scattering red-hot shrapnel across the peloton? No?
By the way, “Bicycling Science” by David Gordon Wilson contains an extensive discussion of rim brake friction coefficients in various conditions.
- Contact Area:
The contact area between a pair of pads and a rotor is similar to the contact area between a pair of pads and a rim, but for purposes of generating a stopping force, surface area really doesn’t matter. The equation for friction force includes only the coefficient of friction and the force perpendicular to the direction of motion. It doesn’t matter how much you spread it out.
Heat dissipation is another story altogether. A larger contact area might allow more heat to transfer into the brake assembly, decreasing the heat transfer into the rotor or rim. Hhmmm, this might be an argument for larger rim brake pads. I recall a long time ago seeing rim brake pads with cooling fins built into the brake blocks.
- Cable Friction:
My previous experiments on cable friction showed that the loss of tension along the cable is a function of the cable/housing coefficient of friction, the tension at either end of the cable, and the cumulative change in direction of the cable from one end to the other. I see two obvious differences between rim and disc systems:
- The additional cable length to reach the disc brakes down near the hubs. This should not affect the cable friction if the cable is routed without additional turns.
- The 90 degree turn at the V-brake noodle. This could be significant. It roughly doubles the cumulative direction change in the cable. A few quick calculations suggest this is a minor issue if the coefficient of friction through the noodle is as low as typical cable/housing friction measurements. But I don’t know that it is. Road rim brakes have no 90 degree turn, yet disc brakes still generate more stopping power.
I’ll call this one a parameter that may favor disc brakes over V-brakes. Maybe I’ll do some tests on it, but not today.
I promised I’d discuss hydraulic brakes, so here goes.
Note: The following paragraphs apply as much to a hydraulic rim brake as they do to a hydraulic disc brake. But not many of us have squeezed a hydraulic rim brake, while most of us have at least played with hydraulic disc brakes at a bike shop.
What makes hydraulic brakes so incredibly powerful is the lack of cable friction between the lever and the caliper. With a cable-actuated brake, the harder you pull the lever, the more force you lose along the cable. See my previous post on the topic. In theory, you will reach a situation where no additional tension reaches the calipers, no matter how hard you pull the lever. The braking force is probably sufficient to throw you over the handlebars before this happens, but the maximum cable tension at the caliper can never be as great as the cable tension at the lever.
With a hydraulic brake, the absence of cable friction means 100% of the pressure you generate at the lever appears as force at the caliper. Of course some of your energy (remember – energy equals force times distance) goes into ballooning the hydraulic tubes and compressing any air in the system.
- Caliper Travel:
This is the biggie. This is the Hokie Pokie. This is what it’s all about. You could have skipped everything above and just read this section. This, I think, is why disc brakes can generate more stopping power than rim brakes.
Shimano specifications call for 3mm to 4mm total clearance between rim brake pads and the rim. It’s got to be this much to accommodate a reasonable amount of rim lateral run-out, aka side-to-side out of true.*
Typical disc brake rotor clearance specifications call for 0.4mm to 0.8mm total clearance. (Shimano calls for 0.2mm to 0.4mm on each side. Avid recommending tightening the pads against the rotor and then backing them away until the desired pad/rotor clearance is achieved. Avid mentions no specific clearance value.)
So a pair of rim brake pads must move about five times the distance of a pair of disc brake pads just to start braking. Recall, force times distance equals work (or energy). If my braking mechanism is designed to move the caliper over a larger distance, it will not be able to generate as much force.
Consider a pair of wire cutters as an analogy. If you want to cut a tough wire, where do you position the wire? You place it deep in the crotch of the tool, close to the hinge. Study the motion of the cutters as you open and close them. Out near the tip, the blades are moving a large distance, but if you tried to cut a tough wire out there you’d find that you couldn’t generate enough force. Deeper in the crotch, the blades move less distance, but can generate more force.
So disc brakes can generate about five times the force because they need move only one-fifth the distance. Recall from discussion of braking surface diameter that a disc brake must generate four times the force to break even. It seems it can do five, giving it a 25% advantage in braking power.
- System stiffness:
There’s one last parameter that I think augments the disc brake’s performance but I don’t know how to quantify it – system stiffness. A disc brake caliper assembly is a compact chunk of metal, mounted on a pretty beefy platform. The actuation force is transmitted through a short stubby arm. There is very little flex in a disc brake caliper assembly.
In comparison, rim brake mounts are not as rigid, especially in the rear when they are mounted on slender seat stays. Rim brake caliper arms are definitely not as rigid. They must be longer to move the pads farther. If they were beefy enough to eliminate flex, the weight would be unacceptable.
There is a point to this comparison. I am not talking about a subjective feeling like softness or sponginess. Extra lever travel is required to accommodate the amount taken up flexing the calipers. There is only a certain amount of travel available. Any amount that is used to flex the caliper arms is not available to increase the force against the rim.
I may have gotten a little carried away with this conversation. In conclusion, yes, there is a scientific basis for the greater braking power of a disc brake system. Mostly it is due to the reduced travel required of the disc pads compared to rim pads. Does that make a disc brake system “better”? You’ll have to define “better”.
* Shimano’s specification for V-brake clearance is 2mm, 1mm per side! This is a bit of a cheat to reduce cable travel in my opinion. It is really difficult to run a set of brakes that close to the rims, especially brakes that rely on independent springs on each side to maintain centering. Contrast with dual pivot road brakes that maintain center via a positive stop mechanism. Throw in a little lateral wheel run-out with V-brakes set to a 1mm clearance and the brakes will rub.
3 thoughts on “How Do Disc Brakes Generate Greater Stopping Power Than Rim Brakes?”
I’ve always thought of the caliper clearance as being pretty arbitrary (as long as the wheel isn’t dragging against the brakes), but thinking more about the relationship between caliper travel and stopping power makes it clear that I should care about it. If I have, say, 6mm instead of the recommended 3-4mm of clearance on my rim brakes, it seems like it might actually cause a noticeable reduction in stopping power.
Here’s a thought – Adding to the stiffness part, along with the caliper of a disk brake being stiffer I think the rotor also might be a big contributing factor. Whereas the rotor in a disk brake is being sandwiched by the caliper/pads and is a pretty stiff and fixed system, the “brake rotor” of a rim system being the actual rim is a lot less stiff in comparison and may provide a lot more give/compliance when squeezed, thus “absorbing” some of the applied pad force? Also, any unevenness in the rim is actually trying to open the caliper as it rotates through it a lot more than unevenness of a disk rotor.
I guess all the World Cup trials riders are wrong for using rim brakes instead of discs….ya know, the discipline that requires the strongest locking brake…..
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