Originally Posted by

**timo888**
The Schwalbe site explains such an assertion: given the same inflation pressure, a tire of a specified diameter will be less out-of-round than a tire whose diameter is smaller. The absolute "squish" or deformation area is the same for both tires, but the tire with the larger diameter, and thus having the greater circumference, has *proportionally less* deformation, i.e. is less out-of-round than the smaller. The more out of round, the greater the resistance. The larger the circumference, the more gradual the ascending arc of the tire; the smaller the circumference, the more abrupt the arc. To state the thing in extreme terms, if you had to fashion a wheel made out of wood in a polygonal shape, would you rather ride a small pentagon or a larger duodecahedron?

Let's say the rims are the same width, and that we have two wheels, one 700mm Ø and the other 305mm Ø. We will ignore the tire wall height for the sake of simplicity. The circumference (Π * Ø) of the larger is 700 * 3.14 = 2200 (roughly) and the circumference of the smaller is 305 * 3.14 = 958. If there's a 25mm deformation along the tire surface, on the larger that is 25/2200 (.011) and on the smaller it is 25/958 (0.2), about double.

Regards

T