Bike tires
JW
In article <5a9ip7$b...@agate.berkeley.edu>
Erik de Leede <ele...@xs4all.nl> writes:
> why ask it here. The question was:
> If you sit on a bike with tires that have a lot of pressure inside them,
> than these tires will dent a bit.
> Why do they only dent a bit:
> a) because of the increasing elasticity of the tyre
> b) because of the bigger surface of contact
> c) Because of volume reduction
> I thought the right awnser was b, because it's pressure in the tire
> times contact area that gives the upward lift.
> The quizmaster said it was c,
> Anyone got an idea.
> I truely enjoyed trying to solve the problems they had in the quiz,
> with these two questions though, I feel they are wrong and I am right.
> I'ld be real gratefull if someone would awnser my questions.
> jero...@worldaccess.nl
I think that none of the answers is good, but (b) is closest
If we neglect the bending stresses in the rubber (a reasonable
approximation for thin tires) then applying Newton's laws to the flat
rubber in contact with the road tells us that the (gauge) pressure in
the tire equals the force per unit area applied by the road to the
outside of the tire. Now the deformation of the tire will require only
a very small volumetric change (especially for a bike tire) so to a
good approximation the pressure in the tire is unchanged by the load on
the bike.
Subject to the above approximations, the area of rubber in contact with
the road is equal to the weight of the bike and rider divided by the
pressure in the tires. The approximations are best when the tire walls
are thin and the pressure is high.
Best wishes
Joe
BillyFish
J.W...@unsw.edu.au (JW) wrote:
>>>>>
[Moderator's note: Quoted material deleted. -TB]
I think that none of the answers is good, but (b) is closest
If we neglect the bending stresses in the rubber (a reasonable
approximation for thin tires) then applying Newton's laws to the flat
rubber in contact with the road tells us that the (gauge) pressure in
the tire equals the force per unit area applied by the road to the
outside of the tire. Now the deformation of the tire will require only
a very small volumetric change (especially for a bike tire) so to a
good approximation the pressure in the tire is unchanged by the load on
the bike.
<<<<<<<
Here is the real answer:
Because of equal pressure around the tire, a symmetrical wheel does not
get supported by air pressure. It gets supported by hanging from the tire
rim. For a wheel not in contact with the ground (supporting no weight),
the bead gets pulled by the tire "membrane" symmetrically. When the tire
supports weight, the tire gets partially squished. Near the contact
point, the membrane in contact with the bead gets to be more parallel to
the ground. This reduces the vertical (downward) component of the pull on
the bead. Thus, the tire hangs from the rim with equal outward force per
unit length of rim everywhere except near the contact point where
distortion modifies the forces.
William Buchman
J. J. Lodder
In article <5aokgi$j...@agate.berkeley.edu>, bill...@aol.com (BillyFish) wrote:
[Moderator's Note: Unnecessary quoted material deleted. Please trim your
quotes, folks. -WGA]
> Here is the real answer:
>
> Because of equal pressure around the tire, a symmetrical wheel does not
> get supported by air pressure. It gets supported by hanging from the tire
> rim.
[further correct explanation deleted]
This is not really the real answer. The bike is really supported
by the contact forces with the road, pressure times area. (As stated first)
What happens at the rim is about the transmission of the forces,
ultimately by way of the saddle springs to rider.
The next step is that the axle gets supported by an increase
in the tension in the upper spokes of the weel,
with corresponding decrease below, etc.
Best wishes,
Jan