Re: [Torcs-users] Wheel Slip Angle

[David S. <dsa...@te...>]

On Thu, Dec 19, 2024 at 1:39 PM Bernhard Wymann wrote:
>
> Hi David
>
> I just skimmed through your mail, but I think it is wrong:

Dear Sirs:
Sine(slip angle) of a wheel IS a measure of slip.
All that is required is to draw a different conclusion from the derivation.

Thus, Sine(slip angle) is lateral slip, and can be shown to be in the same
terms [units] as longitudinal slip.

The derivation with the correct conclusion follows:

                Wheel Slip Angle
Imagine pushing a wheelbarrow straight ahead at 0 degrees and
sqrt(0.75) meters/second from the origin. At the same time
another person pushes straight ahead at 90 degrees and 0.5 m/s
on the wheel axle.
What results is: The wheelbarrow moves at an angle of 30 degrees
at 1 meter per second with 0% wheel longitudinal slip, and
100% wheel side slip (lateral slip).
Now you can draw a diagram with the vector: 30 degrees, 1 m/s.
The vector x-component is sqrt(0.75) m/s. , x = 1m/s*cos(30 deg)
The vector y-component is 0.5 m/s.        , y = 1m/s*sin(30 deg)
The x-component slip is 0% (wheel rotates)
The y-component slip is 100% (wheel slides)
The wheel slip angle with respect to the vector is -30 degrees.

The vector (30 degrees, 1 m/s) represents the combined longitudinal
and lateral velocity, so what is the wheel slippage with respect to
this vector? Well, this vector sees a wheel rotating at sqrt(0.75) m/s
at a slip angle of -30 degrees. The wheel is off-angle from the vector.
The component of wheel speed that is in line with this vector
is: sqrt(0.75)m/s*cos(-30 deg) = 0.75 m/s. Thus, the wheel speed is
1m/s * cos^2(30degrees) and the ground speed is 1m/s.

Now to determine the longitudinal slip for the vector (slip angle = 0 degrees):
longitudinal slip = (1m/s - 1m/s*cos^2(slip angle)) / 1m/s = 1 - 0.75 = 0.25

Note that all of the slip is lateral slip applied to the wheel axle.
lateral slip = 1-cos^2(a) = sin^2(a) ::  sin^2(30 degrees) = 0.25

                Conclusion
1) (vector longitudinal slip = vector lateral slip = 0.25) Shows that
   longitudinal slip and lateral slip are on equal terms with each other
   because they both agree on the same result.

2) (1 - 0.75 = 0.25) Shows that (1 - wheel value equals slip value).

3) (0.75 + 0.25 = 1) Shows that all of the terms are in the same units, and
   sum along the hypotenuse (hypotenuse = 1).

4) sqrt(0.75) and sqrt(0.25) are at right angles to each other when pythagoras
   theorem is applied.

Thus, lateral slip can be expressed in terms of longitudinal slip where
lateral slip is:  ground_speed*Sine(slip_angle)
and longitudinal slip is:  ground_speed - wheel_speed

** The following experiment was tried and found to be lacking:
            An experiment on TORCS

change line 275 of:
https://sourceforge.net/p/torcs/code/ci/r1-3-1/tree/torcs/torcs/src/modules/simu/simuv2/wheel.cpp
from: sy = sin(sa);
to:   sy = sin(sa) * fabs(sin(sa));