Effect of wheel diameter on bike disc brake performance?

[speedub.nate]

Greetings all!

There is a bit of a debate in the world of mountain bikes, concerning the conventional 26" wheel diameter versus the rising popularity of the 29" wheel size (nominal diameters).

One of the popular beliefs is that the 29" wheel requires a larger brake rotor than a 26" wheel.

At face value, this appears true, given that the larger wheel has greater leverage over the brake pads clamping the rotor. We're talking roughly 10% difference in diameters.

However, for the most part, rider weights are not changed, bike weights are roughly comparable, and average speeds are equal (many riders will say they are faster on one size or the other, but no study has clearly concluded that either is a "faster" wheel size).

My speculation is as follows: Two bikes rolling downhill (riders attached), at similar speeds and weights. It appears to me we have a similar amount of kinetic energy in each mass as we begin to apply the brakes to slow and/or stop them.

If the braking system's job is to convert kinetic energy to heat, then my impression is we're looking at two identical tasks.

While the 29" wheel has ~10% greater leverage over the pads clamping the rotor, the 29" rotor is also spinning ~10% more slowly at any given ground speed (compared to the 26" wheel).

For what it's worth, a typical rotor diameter is 160mm / 15mm brake track width, your average bike may weigh around 30 pounds, and an average downhill speed may be 30 MPH.

Is my approach too simplistic? Should I take other factors into consideration?

One area a 29" bike will always be heavier than a 26" bike is in the rim and tire weight -- this typically accounts for 150g increased tire weight, and 50g rim weight (per wheel).

I'm more interested in a discussion of the forces at play and am not (yet) looking to work the problem to "solve" it in the form of a numerical answer.

 

[iceman99]

I’m only a junior in high school; so I am really not qualified to give any 100% correct answers. I would like to see the results to this as I am actually quite the mountain biker. I myself am thinking about "upgrading" to a 29", as they are more expensive... Sorry I can not give any suggestions.

 

[binzing]

I don't think it should really matter. I mean if the disc is the same it shouldn't matter and you shouldn't need a bigger disk just for a bigger wheel, I would think.

 

[russ_watters]

Isn't it possible to apply enough force to lock-up the brakes either way?

 

[Rogerio]

If the 29" wheel had the same mass, and the same mass distribuition as the 26" wheel (as it was a scaled 26" wheel) , and the same tangencial speed, then its rotational kinetic energy would be the same.

However, the 29" whell is heavier. So, theoretically, ...

 

[rolls]

Old thread but I googled it just then.

Anyway if using disc brakes and ignoring the inertia/weight issue (negligible) the larger the rim the faster the wheel is passing the brake for any given road speed hence the more energy that is expended for a given braking force.

If using disc brakes then it is identical (as the disc size stays the same).

I'm not sure whether the extra braking capacity due to the speed of the tyre would swamp the extra inertia/weight of the rim or vice versa, I imagine both are negligible and you could just simply squeeze the brake lever harder if needed.

 

[rcgldr]

One of the popular beliefs is that the 29" wheel requires a larger brake rotor than a 26" wheel.

Assuming the same weight, same amount of deceleration, the amount of braking force at the contact patch is the same, but the torque produced with the 29" wheel is greater than the torque produced by the 26" because of the greater radius (torque = contact patch braking force x radius). To keep the friction force on the brake rotor the same for both wheels, the brake rotor on the 29" will would need to be larger than the rotor on the 26" wheel by the same ratio.

 

[Lsos]

Assuming the same weight, same amount of deceleration, the amount of braking force at the contact patch is the same, but the torque produced with the 29" wheel is greater than the torque produced by the 26" because of the greater radius (torque = contact patch braking force x radius). To keep the friction force on the brake rotor the same for both wheels, the brake rotor on the 29" will would need to be larger than the rotor on the 26" wheel by the same ratio.

X2

You don't need a larger disk due to kinetic energy (which should be almost the same), you need a larger disk due to a larger torque required.

If the brake disk sizes are the same, you will need to press the brake harder if you have a larger wheel.

 

[K^2]

Isn't it possible to apply enough force to lock-up the brakes either way?

It is, unless you didn't tune brakes correctly. So yeah, it's effectively independent on wheel diameter. The braking distance will be given by type properties of surface and tire thread.

Next concern is brake overheating. This is also independent of wheel size, since the energy dissipated is exactly the same.

Final concern is brake wear, and that might actually be different with different wheel sizes, but I couldn't tell you which would result in worse wear. One would require greater force on the pads, the other requires pads to travel longer distance over the disk. Which causes greater wear, I don't know.

 

[R Power]

I agree with Lsos.
The wheel with larger diameter will definitely need more braking torque to counter it's own torque. Hence, bigger the disc, more is the friction between disc(rotor) and brake pads and more is the braking torque.

 

[rolls]

I agree with Lsos.
The wheel with larger diameter will definitely need more braking torque to counter it's own torque. Hence, bigger the disc, more is the friction between disc(rotor) and brake pads and more is the braking torque.

However for a given amount of force on the brake pads, the bigger the wheel the more effective torque the brakes will apply to the wheel!

Its a double whammy, the brakes now work more effectively for a given force, but they also require more force to stop the wheel!

edit: Wait I have this wrong, I was coming from a vehicle where the wheel size is the same but the rotor changes in size.

If both the 29 and 26 are going the same effective speed the 29 wheel will be spinning ever so slightly slower not faster, my bad.

So yes the 29 will always need more braking force.

 

[thopsy]

The amount of torque required to stop the rotation of the wheel itself is negligible for any reasonable speed. Think about spinning the tire really quicky, and you can just grab it with your hand to stop it. The majority of the torque comes from stopping the linear momentum of the rider. This is the same, independent of the wheel size! Therefore, the same size rotor has the same size stopping power, no matter if your 26er or 29er. If you have rim brakes, then the 29er has more stopping power due to the increased effective braking radius.

 

[rcgldr]

The amount of torque required to stop the rotation of the wheel itself is negligible for any reasonable speed.

No one was considering the angular inertia of the wheel. As mentioned in post #7, asuming the same linear braking force at the contact patch, then the torque equals this force times the radius of the tire, so the radius is a factor.

 

[Naty1]

At face value, this appears true, given that the larger wheel has greater leverage over the brake pads clamping the rotor.

this premise is backwards...the reverse is true

One of the popular beliefs is that the 29" wheel requires a larger brake rotor than a 26" wheel.

"requires" for what purpose...what is the objective... equivalent stopping torque or equivalent stopping distance...or something else?

posts 7 and 8 are just wrong...later posts get a bit closer..but none are really correct...

Torque is the product of radius times the force of the brake (pad).

So for a given force, that is, for an equal size brake pad [force] applied on two different diameter wheels, the same force at the greater radius provides the greater stopping torque.

But the open question is how that brake pad force actually translates considering the different rotational speeds of the wheel rims...the slower speed of the larger diameter wheel might pretty well offset the greater torque via reduced friction...I don't know about that portion. A related practical consideration is that hot brake pads are LESS effective than cooler ones...one reason for the colling fins on automobile disc brakes.

 

[rcgldr]

Torque is the product of (disc) radius times the force of the brake (pad).

With an equal and opposing torque which is the product of tire radius times the linear braking force at the contact patch between tire and pavement. (Nearly equal and opposing torques, since the wheel experiences angular deceleration, there is some net braking torque). If the linear braking force is the same, then the larger the tire radius, the larger the torques involved.

 

[Lsos]

Torque is the product of radius times the force of the brake (pad).

So for a given force, that is, for an equal size brake pad [force] applied on two different diameter wheels, the same force at the greater radius provides the greater stopping torque.

You seem to be talking about the force applied by the brake pad, but at a radius of the tire...which is just wrong.

There are two torques at play here, one applied by the pad at the radius of the pad, and one applied by the ground at a radius of the wheel. If you want to be able to lock up the wheel, the pad force*pad radius must equal ground force*wheel radius.

Obviously if you increase the wheel radius, you will have to increase either the pad force or the pad radius in order to balance them out.

 

[Naty1]

I can't figure out what post # 15 is trying to say..

But this makes a lot of sense:

There are two torques at play here, one applied by the pad at the radius of the pad, and one applied by the ground at a radius of the wheel. If you want to be able to lock up the wheel, the pad force*pad radius must equal ground force*wheel radius.

That's a good way to look at it..I agree...

 

[rcgldr]

I can't figure out what post # 15 is trying to say..
But this makes a lot of sense:

There are two torques at play here, one applied by the pad at the radius of the pad, and one applied by the ground at a radius of the wheel. If you want to be able to lock up the wheel, the pad force*pad radius must equal ground force*wheel radius.

Same as post #15, you have equal and opposing torques (ignoring angular inertial of the wheel).