B
BaCardi
Guest
John Retchford wrote:
> You miss two of Carl’s points. The first has to do with the nature of
> the materials. In the railway wheel analogy the steel is behaving
> elastically at the contact patch – it is being loaded and unloaded along
> the same (linear) path in the load versus elongation plot. As work is
> force times the distance the force moves, no net work is being done.
> In the case of the bicycle tyre, you are dealing with an anelastic
> material. Because its force vs elongation characteristic exhibits
> hysteresis, work must be done on it during loading and unloading. The
> rate of doing work is power and this power comes from the rider’s legs.
> The area of the contact patch of a tyre is almost solely a function of
> force on the tyre (the weight) and its pressure. This just comes from
> the definition of pressure (force/area). The force supported by the
> sidewalls is negligible for a feasible bicycle tyre. Try pressing a
> fully deflated tyre with your thumb. So the thick walled tyre will have
> essentially the same contact patch area as its thin walled counterpart
> and will absorb more of the rider’s power in hysteresis loss as its
> walls deform as it rolls along. Just like Carl said.
> John Retchford
Alright, then I must be still missing the two points you say, because
after a review I still don't understand it. If a tire's contact is
alsmost "solely a function of force on the tyre (the weight) and its
pressure" like you say, then what difference does TPI have to do with
it? That's what we've all been discussing to date.
--
> You miss two of Carl’s points. The first has to do with the nature of
> the materials. In the railway wheel analogy the steel is behaving
> elastically at the contact patch – it is being loaded and unloaded along
> the same (linear) path in the load versus elongation plot. As work is
> force times the distance the force moves, no net work is being done.
> In the case of the bicycle tyre, you are dealing with an anelastic
> material. Because its force vs elongation characteristic exhibits
> hysteresis, work must be done on it during loading and unloading. The
> rate of doing work is power and this power comes from the rider’s legs.
> The area of the contact patch of a tyre is almost solely a function of
> force on the tyre (the weight) and its pressure. This just comes from
> the definition of pressure (force/area). The force supported by the
> sidewalls is negligible for a feasible bicycle tyre. Try pressing a
> fully deflated tyre with your thumb. So the thick walled tyre will have
> essentially the same contact patch area as its thin walled counterpart
> and will absorb more of the rider’s power in hysteresis loss as its
> walls deform as it rolls along. Just like Carl said.
> John Retchford
Alright, then I must be still missing the two points you say, because
after a review I still don't understand it. If a tire's contact is
alsmost "solely a function of force on the tyre (the weight) and its
pressure" like you say, then what difference does TPI have to do with
it? That's what we've all been discussing to date.
--
