Shifrin Differential Geometry Exercise $1.2.27$ -- A Differential Equation For Bikes

Question

The Question

Suppose the front wheel of a bicycle follows the arclength-parametrized plane curve $\vec{\alpha}$. Determine the path $\vec{\beta}$ of the rear wheel, $1$ unit away. As the hint explains, the goal is a differential equation involving $\theta$, the angle of the front wheel with the axle of the bike, and $\kappa$, the curvature of $\vec{\alpha}$.

This question is very interesting to me, and I haven't seen a solution written up anywhere. This is sort of shocking to me as it seems like it should be a very relevant problem for e.g. autonomous driving.

What I've Tried

I've only been able to make minimal progress.

The hint tells us to write $\vec{\alpha} - \vec{\beta}$ in terms of $\theta$, $\vec{T}$ (i.e. $\vec{\alpha}'$), and $\vec{N}$ (i.e. $\frac{\vec{\alpha}''}{\kappa}$). We obviously have $\| \vec{\alpha} - \vec{\beta}\|^2 = 1$. Differentiating, we obtain $$ (\vec{\alpha}' - \vec{\beta}') \cdot (\vec{\alpha} - \vec{\beta}) = 0 $$ That is: $$ (\vec{T} - \vec{\beta}') \cdot (\vec{\alpha} - \vec{\beta}) = 0 $$ Differentiating again, we obtain $$ (\kappa \vec{N} - \vec{\beta}'') \cdot (\vec{\alpha} - \vec{\beta}) + (\vec{T} - \vec{\beta}') \cdot (\vec{T} - \vec{\beta}') = 0 $$ Now it seems to me we should have $$ \vec{T} \cdot \vec{\beta}' = \|\vec{\beta}'\| \cos \theta $$ So we can expand $$ (\vec{T} - \vec{\beta}') \cdot (\vec{T} - \vec{\beta}') = 1 - 2\|\beta'\| \cos\theta + \|\vec{\beta}'\|^2 $$ And that's as far as I've gotten. It seems like I'm going about this all wrong. In particular, I have no idea what to do with the derivatives of $\vec{\beta}$. If only $\vec{\beta}$ were arclength parametrized I feel like I could make some progress, but I don't think there is any reason it should be. The only thing I can think is that we should have $$ \vec{\alpha} - \vec{\beta} = \lambda \vec{\beta}' $$ for some $\lambda$ that could depend on the arclength of $\vec{\alpha}$. I didn't push too far in this direction, though, since it required introducing yet another unknown.

I thought of yet another line of attack. Since $\|\vec{\alpha} - \vec{\beta}\| = 1$, we can say that $\|\vec{\alpha} - \vec{\beta}\|$ is just $\vec{T}$ rotated by $\theta$, i.e. $$ \vec{\alpha} - \vec{\beta} = \cos (\theta) \vec{T} + \sin(\theta) \vec{N} $$ With this expression I've gone as far as the hint suggests, but I don't see what to do next.

What am I missing?

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If this post summons Ted Shifrin, and he'd rather answers to his textbook questions not be given out, I'd be happy to delete this question and post it as a reference request instead. I really am shocked I haven't been able to find this problem written about anywhere. I'm guessing it's because I'm bad at searching the literature, not because it actually hasn't been written about.

Answer 1

With help from Ted Shifrin, I was able to solve the problem. Here's a sketch of the solution.

The first observation is that $\|\vec{\alpha} - \vec{\beta}\| = 1$ since the bike has length $1$. Since $\vec{\alpha}$ is arc-length parametrized, $\|\vec{T}\| = \|\vec{\alpha}\| = 1$, so $\vec{T}$ and $\vec{\alpha} - \vec{\beta}$ must be related by a rotation. The angle of this rotation is exactly the angle $\theta$ that the front wheel makes with the frame of the bike (up to a sign depending on how you define things). Applying the rotation to $\vec{T}$ gives you an expression for $\vec{\alpha} - \vec{\beta}$ in the $\{\vec{T} , \vec{N}\}$ basis in terms of $\theta$. Differentiate this expression for $\vec{\alpha}- \vec{\beta}$ to obtain an expression for $\vec{\beta}'$ in the $\{\vec{T}, \vec{N}\}$ basis in terms of $\theta$ and $\kappa$.

On the other hand, $\vec{\beta}'$ must point in the same direction as $\vec{\alpha} - \vec{\beta}$ since the rear wheel is fixed to point in the same direction as the frame of the bike. This allows us to write $$ \vec{\beta}' = \lambda(\vec{\alpha} - \vec{\beta}) $$ from which it easily follows that $$ \lambda = \vec{\beta}' \cdot (\vec{\alpha} - \vec{\beta}) $$ We have expressions for both vectors on the RHS in the $\{\vec{T}, \vec{N}\}$ basis, so we can easily calculate this dot product in terms of $\theta$ and $\kappa$.

Knowing $\lambda$, we have yet another expression for $\vec{\beta}'$. Since $\{\vec{T}, \vec{N}\}$ is a basis, we may equate the components of each expression. After just a little algebra, you should obtain $$ \kappa = \theta ' + \sin\theta $$ an expression for $\kappa$ entirely in terms of $\theta$ and its derivatives.