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RE: [ST] for Matt and Richard - tire width & cornering



Yes - you are correct in traction being proportional to the weight over
the surface area - same as a normal force acting upon a point.  A
diagram would help me in my point immensely here.  

What I'm saying is that the heavier object has the greater centrifugal
force during the turn.  And the only thing opposing the centrifugal
force is the tire and it's traction to the road.  With a said tire
having a coefficient of friction of "x", that tire can only oppose so
much force before it breaks its traction.  As speed increases in a turn,
so does that pseudo centrifugal force that is at odds with the tires
traction.  Let make up a number and say that a bike weighing 500 lbs
going through example turn 1 at a speed of 100mph causes that force to
be 100 units.  Another bike that weighs 300 lbs going through the same
turn at the same rate of speed might only cause a force of say 70 units.
If the tires limit is 100 units before loosing traction, that lighter
bike can go up to 100 units and thus can travel faster through the same
arc (thus increasing that outward force) until it reaches that "100
units" number.  So the lighter bike can travel through the same corner
on the same tire at a higher rate of speed due to its lighter weight and
consequent lower outward force than a heavier bike at any given
cornering speed.

Matt Heyer

And the car thing had nothing to do with lean - it was about understeer
and oversteer due to the drive wheels either pushing the car or pulling
the car through a turn.


- -----Original Message-----
From: owner-st@xxxxxxxxxx [mailto:owner-st@xxxxxxxxxx] On Behalf Of
Thomas Emberson
Sent: Tuesday, November 16, 2004 10:16 AM
To: ST@xxxxxxxxxxxxxx
Subject: Re: [ST] for Matt and Richard - tire width & cornering

Heyer, Matthew A. wrote:

> Good information.  I still believe that in response to the initial
post
> ("I didn't realize a narrower rear tyre improves handling.  Maybe that
> explains why 125's and 250's can carry more corner speed than the
> diesels."), the reason the lighter bikes "carry more cornering speed
> than a liter class bike" is the weight.  There is less weight being
> pulled out (the centrifugal force) and thus less force acting on the
> tire at the same speed as a heavier mass/bike.  Thus, a higher speed
can
> be held to bring the tire up to its traction threshold.

As far as pure Physics, if the material is not deformable, then traction

is proportional to weight over surface area. That is, more surface area,

means less weight on the given surface area, thus same amount of overall

traction.

If we removed the deformable and material limitation of the tire, a 
1000lbs bike would be able to hold the same speed as a 100 lbs bike. 
Same with dragsters, if it where no for the tires deforming and other 
wear characteristics of a tire, a 1" wide tire would provide as much
as the 30" wide monsters.

BUT, the rub comes in as rubber is only so strong, that is the rubber 
can only handle so much stress before it warms and liquefies. Some of 
which is good and designed in, but too much would obviously result in 
unacceptable wear.

This is why you have the 385mm wide tire on the back of a Viper, you 
need to distribute the stress over a wide area of rubber, not for 
traction per'se, but for material considerations.

Thus larger bike, you need more rubber to spread the force over. I am 
sure Rossi could get *a single* great lap out of a 120mm wide rear made 
of the same rubber, but the tire would be completely toast going into 
the second.

> You are correct in that the tire differential (and the physics of any
> front-steer machine, in which the rear wheel follows) causes the
> differing arc lengths and the understeer (from the differing arc
lengths
> and the fact that the power is being applied at the rear tire - e.g.
> front-wheel and rear-wheal drive cars)

Yeah, but cars lean the wrong way :-)

cheers,
Tom

- -- 
Thomas Emberson


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