rec.autos.simulators

Slip Ratio and Tration Force

Abdo Haji-Al

Slip Ratio and Tration Force

by Abdo Haji-Al » Fri, 29 Sep 2006 07:56:32

Greetings,

Searching for a slip ratio equation and its relation to the traction
force, I found two seemingly different formulas:

slip_ratio = (angular_velocity * object_radius -
longtitudinal_velocity) / longtitudinal_velocity
F = slip_ratio * traction_coefficient

And,

slip_ratio = (angular_velocity * object_radius -
longtitudinal_velocity) / (angular_velocity * object_radius)
F = engine_torque * wheel_radius * slip_ratio

First question that comes to mind is which one is right? Second, I've
used the first equation, however I didn't know what would the slip
ratio be if the velocity is 0.0 (Starting up). If it would be zero then
the car would never start in the first place. Also what value does the
traction coefficient usually have?
Now, if the second formula was the right one then the wheel would never
stop spinning, because when there's no engine torque it would mean that
the traction force would be zero, so the acceleration would never go
negative (When going forward). So how can I solve this problem?

As a side question, is there any other torque (that I need to worry
about) exerted on the wheel besides traction, engine and brakes?

Thanks,
Abdo Haji-Ali
Programmer
In|Framez

Jeff Rei

Slip Ratio and Tration Force

by Jeff Rei » Fri, 29 Sep 2006 09:16:56

This one can't be right. Force would be relative to rear wheel torque
divided by tire radius.

You need a different formula. Some formulas treat a tire as if it
were a spring.

Note that these are "idealized" formulas to make modeling a car simpler.
The actual real world formulas for a tire are more complex.

For a peformance tire, around 1.0. For a road racing slick, 1.5 or
more. For a drag racing slick, over 4.0. You can't make tires out of
table tennis *** (way too soft), but it has a coefficient around
7 or 8.

As mentioned, treat the tire as a spring.

Traction just puts a limit of the torque. The engine/gearing, and
brakes generate a torque. If the car is on an incline, then gravity
applies a force to the car and tires.

Jeff Rei

Slip Ratio and Tration Force

by Jeff Rei » Fri, 29 Sep 2006 09:25:03

The complication of traction, the loading does make a difference, especially for
smaller objects (smooth surfaces tend to attract each other, which is probably
the main reason that the smaller block seems to have much more grip).

Look at the latter part of clip # 2:

http://www.gyroscopes.org/1974lecture.asp

The relationship between slip angle and side force is not linear with the load
factor, look at the graph of force versus slip angle and note that the curves
differ depending on the load (this is slip angle, not slip ratio):

http://www.smithees-racetech.com.au/ackerman.html

db_erickso

Slip Ratio and Tration Force

by db_erickso » Sat, 30 Sep 2006 02:11:37

The first equation for slip ratio is actually almost correct. The SAE
definition of slip ratio is slip_ratio =
(angular_speed_tire/ABS(speed_over_ground)) - 1. If you recall from
math
class you can convert angular velocities into linear velocities with
the
equation V= R*w where R is the radius or the wheel/tire and w is the
angular
velocity. The only flaw in the equation is that you need to use the
absolute
value of the denominator or you will not maintain the sign when going
in
reverse. I would stick to using angular velocities, there is really no
reason to convert then linear for the calculation. As a side note,
remember
that the slip_ratio is a percentage. But a percentage of what? I'll
tackle
that next....

The equations for force are lacking. The maximum tractive force, in
very
simple terms, is the traction coefficient of the tire times the
vertical
load on the tire. What the slip ratio is, is a percentage of this total
possible force. So the equation for force would be F = slip_ratio *
vertical_load * traction_coeff. Of course the actual force generated by
the
tire is much more complex. Once the tire has passed 100% of it's
possible
force it enters a state of sliding and the force will fall off. To get
accurate  forces you will need to implement an equation such as
Pacejka's
"magic formula".

To complicate things further, you will also have to take into account
the
lateral forces at the same time as the longitudinal forces (the tire
only
has so much traction to go around).

As pointed out by Jeff in another post, you will need to handle the low
speed case in a different manner (especially at 0 velocity where there
is a
singularity). The spring method he mentioned will work just fine. It is
also
interesting to point out that you need to update the tire forces at a
fairly
high frequency or you will get jerky response.

Also, the engine torque does not have any direct influence on the tire
force
other than to increase the angular velocity of the wheel which then
affects
the slip_ratio calculation.

I hope this ramble helps.

-Brent


> Greetings,

> Searching for a slip ratio equation and its relation to the traction
> force, I found two seemingly different formulas:

> slip_ratio = (angular_velocity * object_radius -
> longtitudinal_velocity) / longtitudinal_velocity
> F = slip_ratio * traction_coefficient

> And,

> slip_ratio = (angular_velocity * object_radius -
> longtitudinal_velocity) / (angular_velocity * object_radius)
> F = engine_torque * wheel_radius * slip_ratio

> First question that comes to mind is which one is right? Second, I've
> used the first equation, however I didn't know what would the slip
> ratio be if the velocity is 0.0 (Starting up). If it would be zero then
> the car would never start in the first place. Also what value does the
> traction coefficient usually have?
> Now, if the second formula was the right one then the wheel would never
> stop spinning, because when there's no engine torque it would mean that
> the traction force would be zero, so the acceleration would never go
> negative (When going forward). So how can I solve this problem?

> As a side question, is there any other torque (that I need to worry
> about) exerted on the wheel besides traction, engine and brakes?

> Thanks,
> Abdo Haji-Ali
> Programmer
> In|Framez


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