rec.autos.simulators

This is a question which has been bugging me for a while now:

For cars which derive a significant proportion of their overall grip from
aerodynamic effects, does the following scenario ever arise for a given
corner?

1) There is a flat-out maximum speed at which the car can negotiate the
corner, employing mechanical and aerodynamic grip (I know this is the case)

AND

2) There is another maximum speed (slower than the speed in 1) at which
mechanical grip alone is sufficient to keep the car on the black stuff,
unaided by aerodynamics (a trip to the grocery store in my own vehicle shows
this to be true as well)

AND (here's the tricky bit)

3) There is a range of speeds SLOWER than the upper-maximum speed, yet
FASTER than the lower-maximum speed, wherein the car's aerodynamic grip is
not sufficient to collaborate with the physical grip to keep the car on the
racing surface, and yet the mechanical grip alone cannot do the job either.
(this is the bit which I *THINK* should come into play, at least in theory,
but I don't know if it's ever an issue in the real world)

...so, in essence, can a driver overcook a corner by driving TOO SLOWLY,
while a quicker pace would allow the aerodynamics to provide the necessary
adhesion?

I suppose whether or not this is an issue IRL would be mostly due to size of
the window between the upper- and lower-maximum speeds, which I would think
would be fairly significant, due to the exponential rate at which the
aerodynamic grip falls off with decreasing speed. Of course, any racing
driver worth their salt would be pushing up against the upper-maximum speed
all the time anyway, unless forced to slow down by unusual circumstances, so
perhaps the point is moot...except for duffers like me, who aren't quite as
well-seasoned. :)

Well, thanks to anyone with the patience to read through this lot -- I look
forward to your responses!
------------------------
Doug :)

  Interesting question, I was wondering about this last week too :0)  

  I'll try to delve into some "back of the napkin" math here and see what might
happen.  I'm making this up as I go, so bear with me, as there's probably a
really short, simple answer to this that I'm unaware of. :0)

  Let's look at some basic calcs for an Indycar with a maximum downforce setup.
 They can pull about 4.5g's at around 170mph, I think, so I'll use that as a
starting point in my rant.  With no aero downforce at all, I think these cars
pull somewhere around 1.7-1.8g.  If not, I'll assume 1.7 anyway for this
discussion ;-)

  Let's take a look at a 500 foot-radius corner and see what would happen with
no aero-downforce.

  The quickest a car could negotiate the corner pulling 1.7g is:

  Velocity = SquareRootOf(CornerRadius *
LateralAccelerationInFeetPerSecondPerSecond)

  LateralAcceleration (feet/sec/sec) = 32.16 * 1.7 = 54.7 feet/sec/sec

  Velocity = SquareRoot(500 * 54.7)
  Velocity = 165 feet/sec = **112 mph**

  Aero downforce is a function of velocity squared.  For simplicity, the
"mechanical grip" or "friction coefficient" we're talking about here (1.7) is
multiplied by any amount of force pushing downwards at the tires to find the
lateral force, which in turn is divided by the vehicle mass to find lateral
acceleration.

  The car weighs about 1600 lbs, which converts to 49.75 (lets say 50) "slugs"
mass.  

  How much force would it take to accelerate the 50 slug car laterally at 4.5g,
or 144.7 feet/sec/sec?

  Force = Mass * Acceleration
           =  50  *  144.7
           =  7235 lbs

  So, to corner at 4.5g's would require 7235 lbs of force laterally.  The
amount of force pushing down on the tire if the grip stays constant at 1.7,
(which it doesn't because of tire load sensitivity, I'll get to that later)
would need to be at least:

  Downforce = 7235 / 1.7
                  = 4256 lbs

  This includes the weight of the car itself (1600 lbs), so the actual
aerodynamic contribution would have to be:

  4256 - 1600 = 2656 lbs

Assuming we get this amount of downforce at 170mph (249 feet/sec), giving us
our 4.5 g lateral acceleration, we could calculate a "downforce coefficient"
like this:

  Downforce coefficient = downforce / (velocity * velocity)

Downforce coefficient =  2656 / (249 * 249)              
Downforce coefficient = .043

  What's the point?  This lets us seperate the aerodynamic and mechanical
contribution at any speed we want.  We can pick a speed, and find the
approximate aerodynamic downforce using this "downforce coefficient."

  For instance, if the car is moving at 100 feet/sec, the aerodynamic downforce
would be about:

  Aero downforce = DownforceCoeff * Velocity * Velocity
                         = .043 * 100 * 100
                         = 430 lbs

  Now, we can add the car weight to this:

  Total downforce = 430 + 1600
                          = 2030 lbs

  What's the lateral force available?  The tires can produce 1.7 times as much
sideforce as downforce, so:

  Lateral force = 1.7 * 2030
                     = 3451 lbs

  What's the lateral acceleration?

  Acceleration = Force / Mass
                     = 3451 / 50
                     = 69.02 feet/sec/sec

  If we want it in "g's", we divide by 32.16:
                  g = 69.02 / 32.16
                     = 2.14g

  What good does all this do?  We can now run this car through a corner at a
few different speeds to find out if there's a point where you'll overcook a
corner by going too slow.  

   Let's try a 400 foot radius corner.  We'll try it at 50 mph, 100 mph, 150
mph, and 200 mph, and see if we can stay within the required lateral
acceleration at all these speeds.

   First, what's the required lateral acceleration to stay on track at 50 mph
(73.3 feet/sec)?    

  Required lateral accel = Velocity * Velocity / Corner Radius
                                  = 73.3 * 73.3 / 400
                                  = 13.43 feet/sec/sec

  Dividing by 32.16 will give us the acceleration in "g's", which is more
intuitive to us race freaks :-)  
 g's = 13.43 / 32.16
      = **.42g**

  Well, we already know that without any downforce at all, we can pull up to
1.7g, so we can skip the aero-downforce stuff entirely.  The car will easily
make this corner regardless of aero-downforce.

  Ok, at 100 mph (146.7 feet/sec), we must accelerate at 53.8 feet/sec/sec, or
1.67 g.  This is still below 1.7g, so even without our aero-downforce, we'll
make the corner (barely ;-) )

  At 150 mph (220 feet/sec), we must accelerate at 121 feet/sec/sec, or 3.76g.
To make the corner at this speed, we'll need quite a bit of downforce.  Let's
see if we have enough:

  Aero downforce = DownforceCoeff * Velocity * Velocity
                         = .043 * 220 * 220
                         = 2081 lbs

  Total downforce is aero downforce + car weight = 2081 + 1600 = 3681 lbs

  Our tire friction/grip coefficient is 1.7, so at this speed, including weight
and aero downforce, how much lateral force will we develop?
   3681 * 1.7 = 6257 lbs

  Skipping a couple calculations, this gives us 125.15 feet/sec/sec
acceleration, or 3.89g.  We needed 3.76g's to make the turn, so we'll make it,
but again, it'll be close.

------
  This is probably about the speed where we're wondering what the trade off
will be as we go a little faster or slower.  If we slow down somewhat, will the
aero downforce drop off enough to make us overcook the turn?  What if we go
faster?
------
  Slowing to 130 mph (190.7 feet/sec), we must accelerate at 90.91
feet/sec/sec, or 2.82g.  This is quite a bit below the 3.89g's we needed to
make the turn at 150mph, but if we crank out the numbers again to make sure:

  Aero downforce = DownforceCoeff * Velocity * Velocity
                         = .043 * 190.7 * 190.7
                         = 1563 lbs

  Total downforce is aero downforce + car weight = 1563 + 1600 = 3163 lbs

  Our tire friction/grip coefficient is 1.7, so at this speed, including weight
and aero downforce, how much lateral force will we develop?
   3163 * 1.7 = 5377 lbs

  Skipping a couple calculations again, this gives us 107.54 feet/sec/sec
acceleration, or 3.34g.  We needed 2.82 g's to make the turn, so we'll make it
by a long shot once we slow down only 20mph.  From before, at 150 mph, we
needed 3.76g's to make the turn, and had 3.89g available.  

So, at 150 mph we had 103% of the lateral acceleration we required to make the
turn (barely making it!), while at 130mph we had 118%.  It looks to me like the
slower we go, the easier it'll be to make the turn.  
----
  Going up to 200 mph now (293 feet/sec), we must accelerate at 215
feet/sec/sec, or 6.69 g.  Will we make the turn?

  Aero downforce = DownforceCoeff * Velocity * Velocity
                         = .043 * 293 * 293
                         = 3691 lbs

  Total downforce is aero downforce + car weight = 3691 + 1600 = 5291 lbs

  Our tire friction/grip coefficient is 1.7, so at this speed, including weight
and aero downforce, how much lateral force will we develop?
   5291 * 1.7 = 8994 lbs

  Skipping a couple calculations, this gives us 179.88 feet/sec/sec
acceleration, or 5.59 g.  We needed 6.69 g's to make the turn, so we'll crash
at 200 mph.  We can only generate 83% of the required lateral acceleration at
200 mph.

    >3) There is a range of speeds SLOWER than the upper-maximum speed, yet

We have similar calculations in Race Car Vehicle Dynamics, see the
last section in Chapter 15 (pages 570-577).  Some of these do include
tire load sensitivity, but they are still very simple compared to real
life.  We came to the same general conclusion that you did, there is
only one limiting speed for any corner (with fixed aero configuration).

Book is available from:
  <http://www.sae.org/servlets/productDetail?PROD_TYP=BOOK&PROD_CD=R-146>
     ....my standard shameless plug!

-- Doug
                Douglas Milliken  <bd...@freenet.buffalo.edu>
                Milliken Research Associates Inc.

On 26 Feb 2001, J. Todd Wasson wrote:

Depending on what you assume for the load sensitivity of
the tires, you can have easily have a situation where the
minimum path radius eventually gets worse at higher speeds.

Doug Olson:

This condition would require that the minimum turning radius would be
reducing with increasing speed in some speed regime.
In flat cornering, this doesn't happen even with zero tire load sensitivity
because the required lateral acceleration goes up with speed squared
but not all the terms of the available lateral acceleration are proportional
to speed squared (the weight term). So the best you can do in flat cornering
with aero is to have the minimum turn radius go up only very slowly as
you increase speed.

I think you _might_ find this for cases where the tire load sensitivity is low
and with banking. (I haven't done or seen an analysis of this, though,
with banking and simplistic downforce and tire load sensitivity effects.
It's too late here for me to try to do it without making some silly mistake. ;)
Zero load sensitivity isn't enough by itself, as above,
but it may help with banking because generally the tire normal forces
are higher in a banked case so any penalty would be worse.

To give a sim "experience"  for that question:
I once "drove" a model on an oval and found I could
take a particular corner at 180 mph that I couldn't
take at 140 mph. The downforce should increase by
about 65% from 140 to 180 mph, so in that particular case
(and model) going too slow was bad, and proper procedure
would then be to go in faster and trust to the downforce ;)
However, I suspect that this model had low or no load sensititivity
programmed in and was at about 10 deg of banking.
And I couldn't reproduce it now, so don't flame me if it was just
my bad driving at 140 mph that caused this result ;)

--
Matthew V. Jessick         Motorsims

Vehicle Dynamics Engineer  (972)910-8866, Fax: (972)910-8216

(Where is a graduate student when you need one? ;)

--
Matthew V. Jessick         Motorsims

Vehicle Dynamics Engineer  (972)910-8866, Fax: (972)910-8216

Mercedes-Benz have a vertical bank at their proving ground, and many county
fairs have the "Wall of Death" (or whatever it's called -- usually a
motorcycle driven in a vertical cylinder, spectators look in from the top).

-- Doug

                Milliken Research Associates Inc.

And then there's the Pants Ring.  :-)

        Marc

...

But make sure to keep the throttle on in loopings. :)
Ooh, one of the things I would love to try is to create a small upside
down track (with a 180 degree screwdriver in the middle) and see if I
can get the car to stick to the road driving upside down, hehe!

Ruud van Gaal, GPL Rank +53.25
Pencil art    : http://www.marketgraph.nl/gallery/
Car simulation: http://www.marketgraph.nl/gallery/racer/

The Race Drivin' car didn't have aero downforce...so you have to be going
fast enough or you will fall off the road, when you are upside down.
Please tell us if you can get your model to "stick to the ceiling" with
aero forces, when you are going fast enough!!

I would say with banking it will be very necessary to include the
banking angle to get it right (esp. at Nascar-type tracks, where
banking effects make a lot of difference, I think). And if you have
banking included, well, the rest is for free (you have to rotate the
downforce vector to match the car's orientation, so whatever the
orientation, you always have to do the same amount of work).
But I would create the track for my own sim, and test the aerodynamics
with that. Easier to create, because I need not too much more than the
3D model (and I know the track format obviously ;-).

Ruud van Gaal, GPL Rank +53.25
Pencil art    : http://www.marketgraph.nl/gallery/
Car simulation: http://www.marketgraph.nl/gallery/racer/

I won't have to do too much; been reading much about aerodynamics but
the formulas for producing a force are not really clear in my head
yet.
I can do something like:
F_downforce = relative_air_velocity * area * Cdf

where Cdf is the aerodynamic coefficient, like 0.3 for a whole car.
Then just apply the force in the car's coordinate system. However, I
don't know good values for Cdf for a typical F1 wing (or perhaps an
airdam like those in Nascar), and I haven't a clear idea of
implementing the angle at which the air strikes the wing. In other
words, if I turn the wing from 2 to 12 degrees for example, how would
I include the effect in the formula? (probably another curve)

When I have that cleared, adding wings will be quite easy; it's just
the parameters need to be right to feel realistic.
And there's another minor detail, namely I'm detecting the road
surface I think by always shooting a ray from the wheel center
downwards (in world coordinates). This needs to be done in car
coordinates.

Will be fun though. A little upsizing of Cdf will help too to get it
stick upside down. ;-)

Ruud van Gaal, GPL Rank +53.25
Pencil art    : http://www.marketgraph.nl/gallery/
Car simulation: http://www.marketgraph.nl/gallery/racer/

I'm listed on the credit screen as a "Test Driver" -- I asked to be listed
under "Vehicle Dynamics" but Atari management wouldn't allow it...they
didn't want the competition to figure out that the game actually used a
math model of "real car physics".
 ....I think the French say, "C'est la vie", don't they??

-- Doug

Let's try to put it in a simpler form : the lateral force required to
negotiate a corner, as you said, depends on velocity squared :
F_req = k1 * v^2,
where k1 is a constant for a given vehicle and corner radius

Force available from downforce is :
F_av  = k2 + k3 * v^2
where k2 and k3 are two other constants that include the effect of the
coefficient of friction, the weight, the aero effect, etc ...

The ability of the car to negotiate the turn without loosing control is
given by the sign of :
F_av - F_req.
For v = 0,
F_av - F_req > 0, of course.
The variation with speed of this quantity, denoted Q, is given by its
derivative with respect to speed :
dQ/dv = 2*k1*v - 2*k2*v
Since v remains positive, this is positive if k1>k2, and negative otherwise.
So, if k1>k2, Q will be positive for all v : means that whatever your speed
may be, the increase in aero downforce will be greater than the increase in
required lateral force.
If k1<k2, Q will decrease until 0, and there will be _only one_ critical
speed for which grip will be lost

If you take tire load sensitivity into account, you have :
F_av = (k4 - k5*v) * (k6 + k7*v^2)
where (k4 - k5*v) is the coefficient of friction and (k6 + k7*v^2) is the
downforce.
So, Q is a function of v, v^2 and v^3. It is left as an exercise to the
reader :-) to show that it is possible to find a combination of constants
that lead to the kind of situations Doug was talking about.
Well, believe me or not, it IS possible :-)
Theoretically ...

Eric. Had I said "simpler form" ?

Relative velocity squared would be better ...

For the Cx of a passenger car, yes ... But it is about three times higher
for an F1 IIRC.
And the Cz, which leads to aero downforce, is quite different.
Values are in RCVD, right ?

Eric