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cat0020 03-18-22 09:35 AM

Smaller contact patch for smaller diameter tire, therefore less traction available.
If braking force is increase with less available traction, do you know what happens?

sjanzeir 03-18-22 09:49 AM

Yeah, you spill out (or worse.) but your talk of the contact patch being smaller for a smaller wheel is a gross oversimplification. Again, it has all to do with how you brake and little to do with how much you - as in you, the rider - brake.

cat0020 03-18-22 09:54 AM

You asked.. I answered. What's with all the hostility?
Did I say that disc brake is bad for everyone?
Learn to read.

Fentuz 03-18-22 01:56 PM

Strictly speaking if you were to fit the exact same disc brake system on 700c and 20”, the brake “power” will be greater on a 20” because of the smaller mechanical advantage (radial lever).
Between 50-662 and 50-406, the contact path is greater for 700c (even if considering 40psi of 700 and 26psi on 20).

So, the mechanical load generated by a 700c on a said 160mm disc is going to be greater than what a 20” is going to produce. With the same braking system, if you apply the same load on the lever, the 20” will skid.


Rough calculations would said that for equal caliper, lever and pads, you should get similar performance between a 20” fitted with 140m disc and a 29” fitted with 203mm disc

Schwinnsta 03-18-22 07:26 PM


Originally Posted by cat0020 (Post 22443012)
Smaller contact patch for smaller diameter tire, therefore less traction available.
If braking force is increase with less available traction, do you know what happens?

This wrong by pure physics. I am not talking about or getting into the argument of disk brakes.

The contact patch does not really enter into it as such. The braking force is the friction FORCE, which is the pressure force on the tire times the coefficient of friction of the tire. The pressure force is the pressure times the area of the contact patch. If you have a wider tire, the tire patch area increases, but the pressure will be lower as it a linear relationship to the weight divided by the area of the patch. Not trying to be a know-it-all. If I am wrong about this, one of you physicist types correct me. That said there are advantages to wider tires but braking is a function of the coefficient of friction of the tire.

2_i 03-19-22 09:41 AM


Originally Posted by Schwinnsta (Post 22443550)
This wrong by pure physics. I am not talking about or getting into the argument of disk brakes.

The contact patch does not really enter into it as such. The braking force is the friction FORCE, which is the pressure force on the tire times the coefficient of friction of the tire. The pressure force is the pressure times the area of the contact patch. If you have a wider tire, the tire patch area increases, but the pressure will be lower as it a linear relationship to the weight divided by the area of the patch. Not trying to be a know-it-all. If I am wrong about this, one of you physicist types correct me. That said there are advantages to wider tires but braking is a function of the coefficient of friction of the tire.

Independence of the patch size (Coulomb law) is certainly a good first approximation, but going into more detail a larger patch for tires will less likely skid than small. In winter or on sand, we go with lower pressure. It may have to do with the persistence of contact or static vs kinematic friction in grains. The Coulomb law/model is itself empirical and more questionable at the extremes such as going into a skid.


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