From: tsuit@gmuvax2.gmu.edu (Tom Suit)
Subject: Re: downforce and cornering g's (was which car handles the best?
Summary: Why is this so difficult?
Organization: George Mason University, Fairfax, Virginia, USA 
Date: Wed, 30 Sep 1992 19:42:33 GMT

In article <selpJCq00iUyM5oFRp@andrew.cmu.edu>, "Daniel M. Deright" <dd36+@andrew.cmu.edu> writes:
> In regards to Dave Paper's comments on downforce
  
    
> >Ummm I hate to clarify something..
    
> >but if you have taken a basic level physics course, you would know that a
> >car can never obtain more than 1g, in actuality, .9 is VERY good, as i
> >think the dodge viper gets umm .93 or something..
    
  GM engineers have gotten the Corvette to exceed 1.0 g. F1 cars
  regularly pull upwards of 3 g's laterally. This can be easily
  proven by simply observing the speed of the car through a known
  radius curve and then calculating the centripetal acceleration.
  Centripetal acceleration is defined as Ca = V^2 / R, where V
  is velocity and R is the radius of the curve.

> >1 g would pick the car up.. and 2g' just doesn't exist, not in lateral
> >motion anyways..

  You are saying that 1g of lateral acceleration would pick a car up?
  Which direction would it go? How did the car suddenly become immune
  to the pull of gravity? What drugs are you on?

  And 2g does not exist in lateral motion?? Somebody shake his head;
  I'll bet it rattles! By "lateral motion" I assume you are thinking
  "side to side". However, there is no difference between side to side
  and front to back...both are perpendicular to the vertical axis.
  Top fuel dragsters CERTAINLY exceed 2 g's on forward acceleration.
  (That one guy is now breaking the 300 mph mark in the quarter mile
  fairly consistantly.) If acceleration is defined as A = 2*D/T^2,
  then according to you, no ground vehicle can get equal to or less
  than a 6.423 second quarter mile. (Did I do my math right, guys?)
  By pure coincedence, I happen to have the July 1992 issue of C&D
  and there is a short blurb on Kenny Bernstein being the first
  drag racer to top 300 mph. Speed was 301.7 at the end of the quarter
  mile and time was 4.823 seconds. Acceleration was 5 g's.

> >sorry guys..
   
> >-dave
  
> Perhaps you should have stayed awake in Physics a bit more.  Downforce
> modifiers, ie. chassis mounted wings, drop tubes and pressure slides
> utilize some of the energy used in forward motion (sacrifice speed) to
> effect a low pressure area under the car and a drastically higher
> pressure zone above the car, effectively doubling, tripling or more the
> cars weight.  This increase in the cars weight allows the car to corner
> at much higher levels of stress.  While the car in truth cannot ever
> exceed one gravity of lateral force, the measurement of "g" force is a

  BZZT! Wrongo! Lateral acceleration is a complete separate force from
  gravity. It has nothing to do with gravity other than the fact that
  lateral acceleration is often measured in units called g's which are
  equal to the acceleration of the pull of gravity on an object. This
  is roughly equal to 9.8 meters/sec^2. Regardless of what your lateral
  acceleration is, gravity will ALWAYS be exerting that force of 9.8 
  m/s^2. Gravity comes into the traction management picture since it is
  what gives the tires their firm contact with the road, thus giving
  that sticky compound you spent lots of money on a chance to resist
  sliding sideways. To get a clear picture of what is going on, think
  of the forces involved in terms of vectors. To avoid the complication
  of aerodynamic downforce, let's use the Ferrari F40 as our example
  since it was designed to have 0 net lift. (same with Porsche's 959)
  No matter how fast the car is being driven, gravity will always be
  applying a downward force vector of 9.8 m/s^2. Let's say you are 
  on your favorite mountain twisty and go through a curve with a radius
  of roughly 100 meters at a speed of 69 mph (Bill & Ted official
  speed limit). You would be experiencing a horizontal force vector of
  9.8 m/s^2. So now there are 2 vectors of equal force acting on your
  car but since they are in different directions, you have to average
  their angles to find the direction of the net vector. The net vector
  would be 1 g applied downwards through the center of the mass of the
  car, at a 45 degree angle to the side from straight down. Since the
  net vector is still pointing below the horizontal plane, it is
  completely impossible for the car to leave the ground via either of
  the two forces being considered. 

  Ultimately, what it boils down to is what net force is required to
  to get the tires to break traction sideways given the current pressure
  that's pushing them against the road surface.

  BTW, I realize that above I am expressing forces in terms of
  acceleration which is a no-no but I can't remember the equation
  for force. (mass*acceleration?) If so, that's about 11,136 newtons,
  assuming the car weighs around 2500 pounds.

 Anyway, enough said.

Tom Suit