# Thread: Curious about rotor to wheel size

1. ## Curious about rotor to wheel size

It seems logical that the larger the rotor the more effective the leverage it will have on the wheel, and similarly, the larger the wheel, the stronger the leverage upon the brake. Which leads me to wonder if anyone has done a calculation on the effective difference in leverage between a given rotor to a given wheel size, with other factors (rotor and pad material, tires, spoke lacing pattern, fork flex, etc.) ignored. It seems like a 165mm rotor would have 'x' effect on a 26" wheel but be not as effective (powerful) on a 29" wheel, as the 29" wheel has a greater leverage. How to compare? I'm sure there is a formula to compute the difference in leverage, but I don't know enough about physics to work it out.

2. It can be solved simply using a sum of the moments around the axle. The force exerted by the brakes x the radius of the rotor is = to the force exerted by the tire x the radius of the tire. Therefore:

160mm rotor, 26" wheel: Force of Tire=Force of Brake x .25
160mm rotor, 29" wheel: Force of Tire=Force of Brake x .22

So what does that mean? The force applied at the brakes must be 4 times what is exerted on the tire for a 26" wheel and ~4.5 times for a 29" wheel. This means that 12.5% more force is required of the brakes to generate the same stopping power for a 29" wheel as a 26" wheel with the same 160mm rotor.

Hope that's what you were looking for.

*As stated by OP- neglects rotor and pad material, tires, spoke lacing pattern, fork flex, reduced rotor velocity for larger wheels, etc.

3. Nvm. Too_Fast_46 beat me to it.

4. Thanks! So, again, all things being equal...if 12.5% more force is needed to obtain the similar stopping power, then can we say that a 12.5% larger rotor will accomplish that? If you took a 180mm rotor, for example, a 12.5% bigger rotor is 22.5mm larger, making for a 202.5mm rotor. In other words, a 203mm rotor on a 29" wheel will have about the same stopping force as a 180mm rotor on a 26" wheel?

Let me look at it another way:

203mm rotor is 7.99" - call it 8" - 8 / 29 = .276 (3.62 multiplier)

180mm rotor is 7"; 7 /26 = .269 (3.71 multiplier)

So, they are roughly comparable, no? The higher the number, the lower the force required, correct? In other words, the 203mm rotor on the 29er would very slightly outperform the 180mm rotor on the 26" wheel - or would give the same performance for just a slight bit less lever force. Again, ignoring everything else. Clearly tire patch size would be different between the two wheels, as would be stiffness of the wheel for a given spoke pattern, etc.

5. Yes, the higher the multiplier the lower the force required.

160mm rotor, 26" wheel- 6.3/26=.24 ~ 180mm rotor, 29" wheel- 7.09/29=.24
180mm rotor, 26" wheel- 7.09/26=.27 ~ 203mm rotor, 29" wheel- 8/29=.28

So, a 29" wheel with one rotor size up from a similar 26" wheel will require approximately the same braking force.

6. Too Fast,

thanks for confirming and your help!

7. No problem. Good to know a Mechanical Engineering degree is good for something

8. Awesome info...Was wondering if my 160MM rotor was good enough for my 29" wheel...this says yes, real world use says not so much. Any minimal amount of moisture really takes the performance away from my brakes. I'm assuming its that way with any disc style brake though. Maybe I'll step up to a larger rotor and see what I can do then...

9. On my 29er, I found a 180mm rotor on the front was marginal in performance with the Juicy 7 brake system I used to have. I upgraded to 203mm front, 180mm rear, and Shimano Saint calipers and levers, and am now extremely happy with my braking system. Obviously, the larger rotors contributed, as did the 4-pot calipers and servo-wave levers.

10. The actual difference between a 29er and 26er is less than 12.5%. The amount depends on the size tire, but with a 52mm tire height it is 9.5%. Even the bare rim difference is only 11.3%.

(622 + 104) / (559 + 104) = 1.095

The same size rotor will produce that much reduced braking on the larger wheel because the rotor spins that much slower for a given ground speed. You don't need leverage ratios to solve the problem.

Still, the ultimate answer is to go one size up for the larger wheel.

11. It all pails in comparison to rider weight.

12. How much more rotational mass does a larger rotor add, given that the distance from the inside of the rotor braking surface to the outer rotor braking surface is almost always the same? It's just that the distance from the center of the wheel to the rotor braking surface has increased. So by the fact that a larger diameter disc is farther from the hub will mean that it will contribute, in some form, to the rotational inertial of the wheel itself, thereby making it harder to stop.

Does anybody make different braking surface width discs? that would give you more contact surface to brake against.

13. Friction is mostly independent of the area of contact. Downhill rotors have large brake tracks to increase the heat capacity of the rotor and increase the surface area to dissipate heat. Basically a large brake track trades weight for less fade.

The rotational momentum of the rotor is negligible compared to the forces the brakes have to counteract while braking, especially if one appreciates the fact that the rotor is at most 4.5" from the center of the axis of rotation. But theoretically, yes, a larger rotor does increase the rotational momentum of the wheel, not only due to the distance the braking surface's mass is from the axis, but also due to the increased weight (obviously).

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