Re: [Torcs-users] Wheel Slip Angle

[Bernhard W. <be...@bl...>]

Hi David

I just skimmed through your mail, but I think it is wrong:
- "The wheelbarrow moves at an angle of 30 degrees": The last time I checked for an angle of 30 degrees you would need 1:0.5. Stopped reading here.
- Gk/H*sin(a) -> a = asin(Gk/H) -> asin(0.5/sqrt(0.75)) ~ 35.3 deg
- The slip angle is calculated above, see "sa", sx and sy are the slip contributions on the respective coordinate.

Kind regards
Bernhard

On 18.12.24 22:33, David Savinkoff wrote:
> Dear Sirs:
>                  Wheel Slip Angle
> Sine(slip angle) of a wheel is not a measure of slip.
> Lateral slip = Sine(slip angle) * |Sine(slip angle)|
> Here is why:
> 
> 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
> 
> Lateral slip is positive or negative for left or right, and independent of
> forward or reverse. Thus, lateral slip has the following formula:
> 
> lateral slip = sin(a) * abs(sin(a))
> 
>              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));
> 
> Sincerely,
> David Savinkoff
> 
> 
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