Clothoid parallel to clothoid

Question

I wanted to ask the question "Can clothoids be made parallel with other clothoids" I have come across statements that this is not possible, but no proof. A previous poster "Jean Marie" states that:

"You may know that in general, parallel curves of non-polynomial curves have no explicit equation"

seems to corroborate this (clothoid is transcendental). I am using clothoids in the alignment of bridge decks. The offsets from the setting out line or "Master String" should be parallel to the Master String. I have gotten around this problem by generating coordinates along the offsets perpendicular to the clothoid and fitting a third order polynomial to the data using "Linest". This gives satisfactory results, however I would like to see a proof of "clothoids can not be made parallel to other clothoids" Can anyone supply this?

Answer 1

Let $$ s \mapsto C(s) $$ be a clothoid with $C(0) = (0,0), C'(0) = (1, 0)$, and $k_C(s) = s$, and $s$ is an arclength parameter, just to make things very explicit. Let $N(s)$ be the (oriented) unit normal to $C$ at $C(s)$, so that $N(0) = (0, 1)$. Let's look at the offset curve of $C$, along $N$, but distance $1$. That's $$ t \mapsto D(t) = C(t) + N(t) $$ where I've used $t$ to remind us that $D$ is not in general parameterized by arclength. Let's compute a few things. Letting $T(s) = C'(s)$ denote the (unit) tangent vector, and $R$ denote counterclockwise rotation by $\pi/2$, we have

\begin{align} T'(s) &= k_C(s) N(s) \\ N(s) &= R T(s) \\ R N(s) &= R^2 T(s) = -T(s) \\ N'(s) &= R T'(s) \\ N'(s) &= R (k_C(s)N(s)) = -k_C(s) T(s)\\ N''(s) &= -k_C'(s) T(s) - k_C(s) T'(s)\\ &= -T(s) - s^2 N(s) \end{align}

That should be enough to get us started. Let's compute $D'(t)$ and $D''(t)$ and see how those look. \begin{align} D'(t) &= C'(t) + N'(t) \\ &= T(t) -k_C(t) T(t)\\ &= (1 -t) T(t) \end{align} Actually, right there we have enough. For at $t = 1$, $D(t)$ reverses direction, hence cannot possibly be a clothoid (for the curvature there would have to be infinite).

To summarize: I've shown a (very simple) clothoid, and an offset curve from that clothoid that is not a clothoid. Hence in general, it's not true that every offset of a clothoid is itself a clothoid. It's possible that some offset of a clothoid is a clothoid -- for instance the zero offset! -- but in general, one cannot be confident that this will happen. Indeed, my suspicion is that a nonzero offset from a clothoid is never a clothoid, but that might take more work to prove (I think it probably follows with a few more lines from what I've already written down), and it's the answer to a different (and un-asked) question.

Answer 2

A(n attempt at a) qualitative proof.

A clothoid is a curve whose graph of curvature vs length is a straight line. Assume without loss of generality that the curvature is increasing as we go along the clothoid, and the starting curvature is $0$. The curvature at any point is finite. The curvature graph is not bounded from above.

For any point on clothoid, going the offset distance by the normal to the clothoid gets us onto the offset curve; the offset curve's radius of curvature at that point is smaller (second case, below, -- greater) in magnitude than the clothoid's radius of curvature, by the value of the offset distance exactly.

For any offset distance, go far enough along the clothoid and you get to the point where the clothoid's radius of curvature has decreased so much that it becomes equal to the offset distance.

The offset line's radius of curvature at this point is $0$. Its curvature is thus infinite. No clothoid has infinite curvature anywhere at a finite arc length, by definition.

In the second case of offset in the other direction, the curvature becomes bounded from above -- the radius of curvature can not become smaller than the offset distance.

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