Re: Wheel Slip Angle

David Savinkoff <[email protected]> Wed, 1 Jan 2025 19:52:47 -0800
Newsgroups gmane.games.torcs.general
Message-ID <CAN4dpXGwj_+LJ_CqTycQMssRjUH5qmGKKzv_sqvtfGvVg3RDdA@mail.gmail.com>
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On Thu, Dec 19, 2024 at 1:39=E2=80=AFPM 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 =3D 1m/s*cos(30 deg)
The vector y-component is 0.5 m/s.        , y =3D 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) =3D 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 =3D 0 deg=
rees):
longitudinal slip =3D (1m/s - 1m/s*cos^2(slip angle)) / 1m/s =3D 1 - 0.75 =
=3D 0.25

Note that all of the slip is lateral slip applied to the wheel axle.
lateral slip =3D 1-cos^2(a) =3D sin^2(a) ::  sin^2(30 degrees) =3D 0.25

                Conclusion
1) (vector longitudinal slip =3D vector lateral slip =3D 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 =3D 0.25) Shows that (1 - wheel value equals slip value).

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

4) sqrt(0.75) and sqrt(0.25) are at right angles to each other when pythago=
ras
   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 =3D sin(sa);
to:   sy =3D sin(sa) * fabs(sin(sa));

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