Involutes

Let $\alpha, \beta : (a,b) \to \mathbb{R}^3$ be regular curves. We say that $\beta$ is an involute of $\alpha$ when

1. $\beta(t)$ lies on the tangent line to $\alpha$ at $\alpha(t)$, and 
2. the tangents to $\alpha$ and $\beta$ at $\alpha(t)$ and $\beta(t)$ are orthogonal.

In this post, we will explore the basics of what one can say about an involute of a given curve.

Let $\alpha(s)$ be a unit speed curve. Let $\beta$ be an involute of $\alpha$. What can we say about $\beta$?

Evaluate this Sage cell to get an example of a cuve and its involute.

In this picture, the original curve is a circle (in black), its involute is in red.

From the definition, we get these conditions:

\[ \beta(s) - \alpha(s) = \lambda(s) \alpha'(s) \tag{1} \]

for some function $\lambda(s)$, and 

\[ \langle \beta'(s), \alpha'(s) \rangle = 0 . \tag{2} \]

Here the primes denote differentiation with respect to $s$. Remember that $s$ is the arclength parameter for $\alpha$, but probably not for $\beta$.

If we differentiate equation (1), we find

\[ \beta'(s) - \alpha'(s) = \lambda'(s) \alpha'(s) + \lambda(s) \alpha''(s) .  \tag{3} \]

Now, take the inner product of both sides with $\alpha' = T_{\alpha}$ and you find

\[ LHS(3) = \langle \beta'(s) - \alpha'(s), \alpha'(s) \rangle = \langle \beta - T_{\alpha}, T_{\alpha} \rangle = -1 \]

and

\[ RHS(3) = \lambda'(s)\langle \alpha'(s), \alpha'(s) \rangle + \lambda(s) \langle \alpha''(s), \alpha'(s) \rangle = \lambda' \langle T, T \rangle + \lambda \langle T', T \rangle = \lambda' \]

If we put these together, we get the equation $\lambda'(s) = 1$, which integrates to 

\[ \lambda = C - s \]

for some constant $C \in \mathbb{R}$. This gives us a very reasonable description of the involute $\beta$:

\[ \beta(s) = \alpha(s) + (C - s) T_{\alpha}(s) . \tag{involute} \]

When is $\beta$ a regular curve? We compute in a straightforward manner:

\[ \beta'(s) = \alpha'(s) + (C-s) T_{\alpha}'(s) - s T_{\alpha}(s) = (C - s) \kappa_{\alpha}(s) N_{\alpha}(s) . \]

Thus $\beta$ is regular at those values of $s$ different from $C$ that have $\kappa_{\alpha}(s) \neq 0$.

Observation: 

The quantity $||\beta - \alpha|| = ||(C-s) T_{\alpha}(s)|| = |C-s|$ serves both as the distance between corresponding points on the curves and as a measure of arclength along $\alpha$. For this reason, involutes are sometimes called string involutes.  (And this is related to the way that Christian Huygens discovered them while trying to build a better pendulum clock.) Imagine the involute as the path taken by an endpoint of a string unwound from the original curve. This is a bit easier to imagine if you view the segment $[\alpha(s) \beta(s)]$ as describing the part of the string which has been pulled off of the curve already. For our picture above, the string segments in are in blue. Unwind the string by picking up the spot where the curves meet on the right, and then go around counter-clockwise.

Some Results about Involutes:

Theorem: An involute of a plane curve lies in the same plane.

Proof: Suppose that $\alpha$ lies in a plane with normal vector $n$. Then by an appropriate translation, we can assume that $\langle \alpha(s), n\rangle = 0$. Differentiating, we find that $\langle \alpha', n \rangle = 0$. By our characterization of involutes above, we see that for an involute $\beta$, there is some constant $C$ so that $\beta = \alpha + (C-s) \alpha'$. Therefore,

\[ \langle \beta , n \rangle = \langle \alpha + (C - s) \alpha', n \rangle = \langle \alpha, n \rangle + (C-s)\langle \alpha', n \rangle = 0 \].

Thus $\beta$ lies in the same plane as $\alpha$.

$\Box$

Theorem: An involute of a generalized helix is a planar curve.

Proof: Suppose that the generalized helix $\alpha$ is a unit speed curve which has axis $\ell = P + t v$, where $v$ is a unit vector. By definition of generalized helix, $\langle T_{\alpha}, v \rangle$ is a constant. Differentiating twice, we find that $\langle N_{\alpha}, v \rangle = \langle N_{\alpha}', v \rangle = 0$.

Let $\beta = \alpha + (C - s) T_{\alpha}$ be an involute of $\alpha$. We differentiate this twice to find
\[ \beta' = (C - s) \kappa_{\alpha} N_{\alpha}, \quad \beta'' = a N_{\alpha} + b N_{\alpha}'\]
for some functions $a$ and $b$. As a consequence, we get that
\[ \langle v, \beta' \rangle = 0, \quad \langle v, \beta'' \rangle = 0. \]
This means that $\beta$ lies in a plane perpendicular to $\ell$.

$\Box$

This last theorem calls for a picture (right +Neil Martinsen-Burrell?). So let's show a string unwinding from a standard circular helix. I suggest you use FireFox to see this one, as it requires Java to run the 3D interactive visualization. Hit Evaluate to enjoy the wonders of colored lines giving a perception of space.