Re: understeer/oversteer definition

["Dave Aukerman" <mhauk@ccsdana.net>]

Thanks Gross, now I've got a splitting headache.  :)

Dave

----------
> From: Scruggs, Gross <GScruggs@monterey.nps.navy.mil>
> To: 'quattro-digest@coimbra.ans.net'
> Subject: RE: understeer/oversteer definition
> Date: Monday, March 15, 1999 2:35 PM
> 
> 	Dearest Audi Borg,
> 
> 	The gauntlet having been thrown by my friend Graydon I am required
> by honor to respond.  I think there are three issues here:
> 	One... although commonly used, 'centrifugal force' is not a valid
> physics term.  (Halliday and Resnick's "Fundamentals of Physics" does not
> contain the term at all.)  What the term describes is the reaction force
to
> centripetal, or 'center seeking,' acceleration.  It is the force you feel
as
> you swing a weight on the end of a string.  By tying the weight to a
fixed
> length restraint you require that it accelerate towards your hand.  The
> magnitude of that force equals the mass times its velocity squared
divided
> by the length of the string. The faster you swing/rotate the mass, the
> greater the center seeking acceleration, the greater the reaction to that
> acceleration and the more load on the string.  Similar increases can be
felt
> by shortening the string.  It is similar to pulling that same weight
behind
> your car as you accelerate, only it is linear acceleration and we have no
> term analogous to "centrifugal force" to describe it.
> 
> 	Two... regarding the effect of a rotating tire on the amount of
> friction (and because we want this friction we call it "traction") there
is
> no effect.  It is compelling to visualize that the whirling mass of
rubber
> flinging itself toward the pavement contributes to the
friction/traction...
> but it doesn't.  For every point on a tire that is heading toward the
> pavement there is a point opposite on the tire which is heading away from
> the pavement... and their rotational effects neatly cancel one another .
> Kinetic friction/traction is the result of coefficient of friction times
the
> normal force (force which is perpendicular to the pavement).  The
> coefficient of friction is the result of the interaction of the
mechanical
> properties of the materials at the interface... steel on ice, rubber on
> pavement, etc.
> 
> 	Three... the normal force has no impact on coefficient of friction.
> They are separate and can not effect each other.  Recall that anything
that
> is a 'coefficient' doesn't have a dimension or units, it is a
> non-dimensionalized number.  A force has dimension or units and therefore
> when acting alone cannot mathematically act on a coefficient and leave it
> dimensionless.  However, multiply them together and the result is
> Friction/Traction... which will have the dimension or units of the force.
> 
> 	Amplifications?  Corrections?  
> 
> 	Regards, Gross Scruggs
> 
> 
> 		>Date: Fri, 12 Mar 1999 18:17:05 -0500 (EST)
> 		>From:	"Graydon D. Stuckey" <graydon@apollo.kettering.edu
> <mailto:graydon@apollo.kettering.edu> >
> 		>Subject:	Re: understeer/oversteer definition
> 		>
> 		>On Fri, 12 Mar 1999 FBFISH@aol.com <mailto:FBFISH@aol.com>
> wrote:
> 		>
> 			>> <<  The greater the vertical load, the >>  higher
> the coefficient of friction the greater the slip angle. >>
> 			>> Group- I can not comment on the veracity of the
> above description. I do >> remember some Physics  however, and twice in
the
> explanation it is mentioned >> that "the greater the vertical force, the
> greater the coefficient of friction" >> I think it is more proper to say
> that the greater the normal (vertical) force >> the greater the friction
> force. I suspect that the coefficient of friction >> between tire and
road
> is not a constant or simple relation. On an 
> 		>
> 		>Coefficient of friction is a constant typically, although I
> suppose it
> 		>could vary with tire design across the tread.
> 		>
> 			<<>> instantaneous basis though normal force times
> coefficient of friction is what >> determines  the force of friction
which
> is offsetting the centripetal force.
> 		>
> 		>Very good.  Friction "force" is correct, and "Centripetal"
> would please
> 		>my physic profs much more than "centrifugal."  I forget the
> explanation.
> 		>That sounds like something Gross could explain.
> 		>
> 		>Later,
> 		>Graydon D. Stuckey
> 		>'91 V8 Quattro 5-speed
> 		>
>