RE: understeer/oversteer definition

[Harry R Glesner III <hglesner@ast.lmco.com>]

Gross:
     That is a very interesting dissertation...and I find it quite over my head, 
sounds like college level double speak, and doesn't answer the original question 
one iota. For the longest time there was a theoretical limit to acceleration, 
mostly drag racers wondered were the limits of "traction", would limit e.t.'s 
and ultimate speed in the quarter mile, and the magic 1 "g" could not be 
exceeded. Well guess what, it was a long time ago what are they up to now 300 
plus miles an hour, and less than 5 seconds, Newton would be scratching his 
head. Applying straight forward physics to the situation with out looking at the 
variations of tire compounds, live mass moving around the center of gravity, 
amplification of forces at the contact patch and the resultant slip angle, you 
will have exactly what you have postulated...no answer to the original 
question...what is understeer and what is oversteer? 
     You and I should more than likely keep our day jobs, neither of us are 
headed to Jordan to be Damon's race engineer.
     Again, see Mr. Smiths "Engineer to Win" and see what a real race car 
engineer has to say on the subject, in this case I'd be only the messenger.
     
Rick Glesner
Littleton, Colorado
    
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Date: Mon, 15 Mar 1999 11:35:39 -0800
From: "Scruggs, Gross" <GScruggs@monterey.nps.navy.mil>
Subject: RE: understeer/oversteer definition
To: "'quattro-digest@coimbra.ans.net'" <quattro-digest@coimbra.ans.net>
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	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
		>
	

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