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



  For highly deformable matterals such as tires, surface area is a much larger component of traction/friction then the normal force. The friction being proportational to normal force (weight) is only an approximation, that works best for hard matterals on smooth surfaces. For something like tire rubber on a rough surface, it is a very poor approximation.

 To make things more difficult to figure, the surface area is not the X by Y square inches of the contact patch, but includes all the little up and down paths as the rubber molds itself around and into every nook and cranny of the road.

  So a 500 lb bike has more traction then a 250 lb bike, on the same tires. But it is no where near  the 2 times as much traction that the simple (Newtonian) friction equations predict.

  College engineering text books don't come very close to the tire - traction problem. Tire companies spend lots of time and money looking for better solutions. And they keep the results a secret.

David W. Funk
'00 Sprint ST
Pleasanton, CA


> The idea of a tire holding only so much force is obviously dependent on
> chemistry (the material properties of all substances depends on this,
> but the material science can be excluded from this debate).  The
> DIFFERENCE in the two bikes cornering ability comes down to how they
> differ from each other with the same tires.
> 
> Traction is proportional to the normal force or gravity pulling down on
> the object.  The thing trying to break this traction while cornering is
> the centrifugal force (just like my pushing a box along the floor is
> what is trying to overcome the friction between the box and the carpet,
> which is proportional to the box's weight and dependent upon the
> coefficient of friction between the two materials).  Obviously, the
> heavier bike has a greater normal force, and thus more traction; but at
> any given cornering speed, the heavier bike is also going to be exerting
> more force centrifugally.  
> 
> Now we really may need to break out our college engineering text books
> to "prove" the matter, but my experience and knowledge leads me to the
> additional weight being more detrimental because of the increased
> centrifugal force than beneficial because of the increased normal force.
> 
> To demonstrate, take a passenger with you on your next ride (I have
> ridden VERY spirited with and without my fiancé on the back of the
> bike).  Her weight (oh - she'll kill me for even referring to her
> weight...) is more detrimental in cornering due to the increased force
> fighting the turn, than it is beneficial in increasing the normal force.
> 
> Matt Heyer
> 
> 
> -----Original Message-----
> From: owner-st@xxxxxxxxxx [mailto:owner-st@xxxxxxxxxx] On Behalf Of
> Thomas Emberson
> Sent: Tuesday, November 16, 2004 11:23 AM
> To: ST@xxxxxxxxxxxxxx
> Subject: Re: [ST] for Matt and Richard - tire width & cornering
> 
> Heyer, Matthew A. wrote:
> > 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.
> 
> We are close to talking the same language. The traction of the tire is 
> proportional to the weight imposed by gravity pulling the bike down. So 
> you have 2 forces, one centrifugal and one gravity. Traction is 
> proportion to that of gravity that is the weight of the bike. Thus a 
> bike weighing 300lbs is going to have less traction than that of the 
> bike weighing 500lbs, when you take traction as meaning the amount of 
> raw grip the tire has.
> 
> So with you numbers 500 lbs bike has 100 units of traction as, thus it 
> can take 100 units of centrifugal force, the 300 lbs bikes will have 
> *70* units of traction and less gravity acting on the tires, thus only 
> 70 units of centrifugal force can be applied.
> 
> The big difference is the 100 units of traction is going to cause more 
> heating and chemical stress in the tire than the 70 units. If the 
> chemical nature of the tire can only handle 85 units is sheer force 
> before the heat and molecular bonds get torn, the 300lbs bike is going 
> to kick some serious 500lbs bike a$$ as the 500 lbs bike would 
> experience tire failure.
> 
> If that magic tire number was say 200units instead of 85, both bikes 
> would be equal.
> 
> Assuming you can separate Chemistry from Physics, tires are all about 
> chemistry and little about physics, unless Rossi can truly change 
> 9.8m/s^2  :-)
> 
> (see, I got some moto content in there :-) )
> 
> -- 
> Thomas Emberson

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