rec.autos.simulators

Tire friction with the road surface? Higher vehicle weight = more road
friction and bearing friction. Drivetrain friction? Even air resistance to a
minimal degree. A good example of decreasing rolling resistance is when
Nascar crews pry the brake pads 1/2" away from the rotors before
qualification at Daytona and Talledega and the driver is always reminded not
to touch the brake pedal. Higher tech F1 braking systems keep the pads a
good distance from the rotors when not braking and even take into
consideration slight warping of the rotors over the course of a race. Taking
into consideration your slip ratios, your main consideration is forward
rolling resistance and there are many factors to consider. Even the
viscosity of lubrication fluids is a factor.

Ed

I would have thought that the rolling resistance would be increased
dramatically if there was any degree of lateral slip or is it totally
independent of it? The equation for rolling resistance taken from
Pacejka's Tyre & Vehicle Dynamics is:

My = -Vertical Load * Unloaded Tyre Radius * (scalar value * arctan
(Forward Rolling Speed / Reference Velocity) + scalar value * Fx /
Adapted Vertical Load) * scalar value

which would suggest that it is independent of lateral slip

Thanks for you help Ed

I think they would be totally independent of each other. Both would affect
momentum, but in entirely different directions.....I guess that is the
correct word.....:-). I hope you get my drift..:(..

Ed

In real life it isn't. Most of rolling resistance is due to
deformation at the contact patch, combined with the fact that
hysteresis is involved when *** is compressed or stretched
and then returns to it's former state. The force during the
deformation is greater than the force during the recovery,
which is why *** is good for reducing vibration.

Lateral slip increases the *** deformation, and it's enough
to slow down a race car pulling a high g turn. In the case
of a Formula 1 car, top speed at full throttle might be around
190+ mph on a straight, but this gets reduced to about 160mph
in a 4 g turn.

Yea, I was kind of "toungue in cheek" on that one.....:-). Where you can
really see these effects is at Indy where the IRL cars are flat out all the
time and the lateral force slows them in the four corners. They do a lot
with camber where on the straights, there is a much smaller contact patch
from the tires and thus causing less friction there. The by product of this
is more wear and heat on the inside of the right tires and outsides of the
lefts. This causes a much larger contact patch in the corners due to the
distortions from the lateral Gs which will place the entire tire into
contact with the track surface. There is also the effect of the extra Gs
causing compression and therefore more friction. How to determine and
calculate the extent of both the Gs and added tire friction have with
forward rolling resistance is a tough one. I am sure the F1 and IRL (and
even Nascar today) engineers have some sort of formula for this.

Ed

Correction, the speed loss is there, but it's not 30mph, more
like in the range 10 mph to 20 mph (perhaps some CART cars
back in the 1990's that went over 255 mph on the straights).

The 160mph 4 g turn was a comment in this video, but the driver
slowed before entering "pouen":

David Coulthard in F1 McLaren, 2002 (remember automatic shifters?):

http://www.racesimcentral.net/

Onboard lap from 1998 CART car, you can hear the engine rpms drop,
but it isn't a lot.

http://www.racesimcentral.net/

Just to make sure I understand:  You're talking about simply moving
the tire sideways (90 degree slip angle)?  The tire should not spin
indeed of course.  The force that will cause this is just the
longitudinal force from slip ratio.  Tire rotates forward, it gets a
backwards torque.  Tire rotates backward, it gets a forwards torque.
That should keep the angular velocity at or near enough to zero in
most tire models, although without a relaxation approach of some kind
it may still jitter a little bit.  There isn't any "missing force"
here.  What's happening wrong in the simulation?  Do the wheels
continue to spin even at 90 degrees slip angle?

Concerning the other posts:  There is indeed a change in rolling
resistance with slip angle.  Also with slip ratio.  The example of an
IndyCar losing speed doesn't have much to do with that, though.
That's induced drag caused by the component of the lateral force
pointing rearwards because there's a slip angle, which slows the car
down.  Two different things, and I suspect that's not really what Zach
was asking about.  Correct me if I'm mistaken..

I assume aerodyanmic drag is also increased by the overall slip angle.
Perhaps this is why the tires used on Indy and the former Champ and
Cart race cars are so stiff compared to the tires used on other racing
cars. The "working" (max g turn) slip angle is about 2 degrees (I
think "working" slip angle means the total yaw angle of the car versus
the direction the car is going, essentially the slip angle of the
rear tires).

Yes, the drag will change with vehicle sideslip angle, but I'm not
sure in which direction it goes on an Indycar over that first few
degrees.  I saw one aero CFD test that showed a very slight increase
in front downforce on a formula style car (I was late to the lecture
so don't know what kind of car it was exactly).  The vertical fins in
the front wing were designed to produce that increased downforce with
slip angle, interestingly enough, by getting rid of a little low air
pressure bubble that (simulation-wise at least) developed behind and
to the outside of the front wing.  What happens to drag I don't know.
It's possible it increases a bit on some cars and decreases on
others.  On a GT car with a spoiler you can wind up with quite a bit
less downforce, but it's probably not safe to say that the drag would
also decrease along with it.  Different cars, different effects...