Do you think this calculation would tell you anything about the expected wear on the wheels?
In each case, the perimeter of the wheel
is given by P = d.
So for the large wheel, approximately, P = 3.1416 145 = 455.53 cm = 4.555 metres.
Hence the number of rotations is
100 / 4.555 = 21.95 rotations.
For the small wheel we have P = 3.1416 85 = 267 cm = 2.67 metres.
Hence the number of rotations is
100 / 2.67 = 37.45 rotations.
Given that there is weight distributed equally
on the large wheel and the small wheel,
we would expect the wear on each wheel
to be proportional to the
number of revolutions it makes.
Hence there should be more wear on
the small wheel than the large.