Kinematic Bicycle Model theta units

Question

I don't understand the $\theta$ units. The bicycle model defines $\theta$ as: $$\theta=\theta +\frac{v\cdot\tan\delta}{L}\cdot$$

Where $v$ is the vehicle velocity, $L$ is the wheelbase distance, and $\delta$ is the desired steering angle.

Say I'm driving at $30m/s$ in a vehicle with $L=2.3$, want to steer $50\deg$, and the $\Delta t=0.1$, then $$\theta=0+\frac{30\cdot\tan50\deg}{2.3}\cdot0.1=1.554$$

But what are the units of $\theta$? Radians or Degrees?
Since we need $\theta$ to calculate $x$ for example, with $x=x+v\cos\theta$, when $\theta$ in degrees $x=30\cos 1.554\deg=29.9889$ (slightly changed) or $x=30\cos 1.554\text{ rad}=0.503$ (significantly changed)

The $\theta$ unit is very important but I can't find any reference for it. I guess it should be degrees. Am I right?

Answer 1

No, $\theta$ has to be in radians. If you follow through the analysis in, say, this guy, which was the first hit in a web search for "bicycle model kinematics", you see that $\omega = \dot\theta$ is the rate of rotation of the reference point (rear axle, in his section 2.1) around the instantaneous center of rotation, with radius R. Since he uses $\omega = v/R$, it's clear that it all has to be radians.

BTW, if I were peddling my bicycle with my $v$ exceeding the automobile speed limit on most roads where I would feel safe riding a bike, I might not be steering that hard. (It might be interesting to compute how many g's the centripetal acceleration is, at $30m/s$ and $\delta=50^\circ$). For your numeric sanity check, maybe try something more sedate like $v=4 m/s$ and $\delta = 2^\circ$.