# Kinematic Bicycle Model theta units

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?

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 $$\delta = 2^\circ$$.