Generalized Helices: Curves of Constant Slope

A regular curve in space is said to have constant slope when its tangents make a constant angle with a given fixed line. Such a curve is also called a generalized helix, or even just a helix amongst the cognoscenti. The line is called the axis of the helix.


An analytic way of stating the condition is as follows: Let $A$ be a unit vector parallel to the axis $\ell$. Denote the parametrized curve by $X$ and its arclength parameter by $s$. We use standard notation for the Frenet-Serret apparatus. Then the definition means that
$$\begin{array}{}\text{(1)}& ⟨T,A⟩=\cos \alpha .\end{array}$$
Here $\alpha$ is the angle mentioned because $T$ and $A$ are both unit vectors.

Our goal is to describe the consequences of equation (1) for the shape of the curve.



The geometry of a curve is encoded by the curvature and torsion functions? Is there a way to characterize helices by using these functions?

Lancret's Theorem: A regular curve has constant slope if and only if the ratio $\kappa /\tau$ of curvature to torsion is constant.

Proof: We begin by differentiating equation (1) with respect to $s$. This gives us
$$⟨\kappa N,A⟩+⟨T,A^{\prime }⟩=0.$$
Since $A$ is a constant, we deduce
$$\begin{array}{}\text{(2)}& ⟨N,A⟩=0.\end{array}$$
Thus $A$ must be parallel to the rectifying planes of $X$. (All of them!) Though we often picture our vectors as "free" vectors, they really are all bound with their tails at the origin, so we know that $A$ lies in $\mathrm{Span}\{T,B\}$. Since $\{T,B\}$ is an orthonormal basis for the rectifying plane, we can express the unit vector $A$ as a linear combination this way:
$$\begin{array}{}\text{(3)}& A=\cos (\alpha )T+\sin (\alpha )B.\end{array}$$
We differentiate this to find
$$0=\cos (\alpha )\kappa N+\sin (\alpha )(-\tau )N=(\cos (\alpha )\kappa -\sin (\alpha )\tau )N.$$
Now, $N$ is a unit vector, so we deduce that the coefficient must vanish, from which we can deduce that
$$\begin{array}{}\text{(4)}& {\displaystyle \frac{\kappa }{\tau }}=\tan (\alpha ).\end{array}$$
a constant.

Now, suppose that $X$ is a regular curve with $\kappa /\tau =C$. We choose $\alpha$ to be the unique real number so that $\alpha \in (-\pi /2,\pi /2)$ and $\tan (\alpha )=C$. Then a little algebraic manipulation gives us that $\cos (\alpha )\kappa -\sin (\alpha )\tau =0$. Now define a vector $A$ by
$$A=\cos (\alpha )T+\sin (\alpha )B.$$
One can check that $A$ is a unit vector, and then
$$A^{\prime }={\displaystyle \frac{d}{ds}}(\cos (\alpha )T+\sin (\alpha )B)=\cos (\alpha )\kappa N-\sin (\alpha )\tau N=0.$$
Hence A is a constant vector, and the tangents make angle $\cos (\alpha )$ with $A$. Thus $X$ is a helix.

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Suppose we choose coordinates so that the axis is in the direction of our third coordinate, and hence the helix is expressed in the form
$$(x(s),y(s),\cos (\alpha )s).$$
This means that our upward drift is well understood, and what remains to be investigated is the plane curve $(x(s),y(s))$.


Theorem: Let $Y$ be the projection of the helix $X$ with angle $\alpha$ onto a plane perpendicular to the axis $\ell$. The principal normal to $Y$ is parallel to the principal normal to the helix, and the curvature of $Y$ is ${\kappa }_Y={\kappa }_X{\csc }^2(\alpha )$.

Proof: By definition of orthogonal projection, we know that $Y=X-⟨X,A⟩A$. We differentiate this with respect to $s$ to find
$$\begin{array}{}\text{(5)}& Y^{\prime }=T-⟨T,A⟩A=T-\cos (\alpha )A.\end{array}$$
It is important to remember that $s=s_X$ is the arclength parameter for $X$, and not necessarily for $Y$. So when we want to compute things for $Y$ we will need to use the chain rule.

$${\left({\displaystyle \frac{d{s}_Y}{ds}}\right)}^2=⟨Y^{\prime },Y^{\prime }⟩=⟨T-\cos (\alpha )A,T-\cos (\alpha )A⟩$$
$${\left({\displaystyle \frac{d{s}_Y}{ds}}\right)}^2=⟨T,T⟩-2\cos (\alpha )⟨A,T⟩+{\cos }^2(\alpha )⟨A,A⟩=1-{\cos }^2(\alpha )$$
Or, more simply,
$${\displaystyle \frac{d{s}_Y}{ds}}=\sin (\alpha ).$$
So we get the very useful measurement of the speed of the projection:
$$\begin{array}{}\text{(6)}& {\displaystyle \frac{d{s}_Y}{ds}}=\sin (\alpha ).\end{array}$$

Now it is straightforward to compute derivatives for the planar projection $Y$ in terms of its arclength $s_1$. The tangent to $Y$ is

$$T_Y={\displaystyle \frac{dY}{d{s}_Y}}={\displaystyle \frac{ds}{d{s}_Y}}{\displaystyle \frac{dY}{ds}}={\displaystyle \frac{1}{\sin (\alpha )}}(T-\cos (\alpha )A)=\csc (\alpha )T-\cot (\alpha )A.$$
And so we differentiate to find $N_Y$ and ${\kappa }_Y$:
$${\displaystyle \frac{d}{d{s}_Y}}{T}_Y={\displaystyle \frac{1}{\sin (\alpha )}}(\csc (\alpha ){\kappa }_X{N}_X)={\csc }^2(\alpha ){\kappa }_X{N}_X.$$
Now, $N_X$ is a unit vector, so it must also be the principal normal to $Y$, and we get the desired description of the curvature of $Y$.

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