> The horizontal speed may be the same, but there is also the vertical

> speed to take account for.

Yes, and that is exactly what I am doing.

> I understand that the outer surface of the wheel will be travelling

> at a constant speed of 10km/h relative to the hub, but the hub is

> also moving, so the outer surface of the wheel must take this into

> account to caculate the overall velocity of the outer surface of the

> wheel.

The forces that act on the wheel because of its motion compared to the ground should be extremely small compared to the forces resulting from the rotation. And the rotational velocity is exactly the same as the speed of the car. Also, any bump or stone on the ground that hits the tire does this with the same speed as the car moves, and this would be the only competitor when it comes to forces on a free-rolling tire. The only force that goes into the calculation that has to do with the ground is the gravitational force, and apart from being a constant regardless of the speed, it should also be of neglectible magnitude compared to the other forces acting on a tire on a fast moving car.

The power working on a given point of the tires surface due to the rotation (also called centripetal force) can be calculated using the formula:

F = m*v^2/R

For a 17" (or 0,4318m) wheel weighing in at m kg on a car moving at 100km/h (or 27,8m/s) this would yield:

F = m*27,8^2/0,2159 = m * 3579,6

As the weight is somewhat distributed over the radius of the wheel this is not entirely correct, but with most of the weight located near the surface of the wheel I think we are getting close.

The constant gravitational force is calculated as:

F = m * 9,81

As you see, the impact of gravitation is not nearly as big as the force induced by the rotation.

The force of a bump hitting the wheel is a bit more tricky to calculate as you have to take into consideration that the tire is not completely compact and also is hung up in springs, an attempt for an easy calculation would look something like:

F = 10 * m * 0,05m / 0,01s^2 = m * 5000

The first 10 is because the wheel is carrying some weight from the car through the springs (approximated to about ten times the weight of the wheel). 0,05/0,01^2 m/s^2 is a very course approximation of the acceleration involved.

These calculations are extremely simplified, and if someone has the time to make better approximations I urge them to do so, even if it proves me wrong.