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

[triumph1@xxxxxxxxxxxxx]

I wish I had paid attention in school now :(

Mick in Aus.


>>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. 
<<


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