Calculating the percent of area "covered" by a vector pointing on a sphere

Question

The question is inspired by rotational dynamics and how much of the sky could a camera "cover" when it rotates in a specific way. Let's say that we have solved the equations for rotational motion of a body, and have the vector function of time $v(t)$, which is of constant length and represents the direction in which the "camera" is pointing. The camera can "see" or cover a circle projected on a sphere of radius $r$, or in other words, every point that is at an arc length´s distance of $r$ or less from the tip of the vector. Thus, the vector moves around as time goes by and "covers" a certain area of the sphere. My question then is, how could you represent the area covered after some time t as a function or some form of integral of $v(t)$ ? I thought about projecting the surface of the sphere onto a plane so that a point $O$ on the sphere is the origin, $y$ is proportional to $z$ component, and $x$ is proportional to the angle that $x$ and $y$ components of the vector form to the the vector pointing at O (angle in one direction would of course yield negative $x$, and angle in the other would yield positive $x$). This is doable, but whatever method one chooses, the problem of areas "covered" overlapping each other always exists, and I have no idea how deal with it without computing the solution; I don´t solve it as a function when overlapping is present.

Answer 1

Let $u$ be a simple smooth curve of length $L$, living on the unit sphere $S^2$, parametrized by its arc-length $s$.

$$[0,L] \ni s \quad\mapsto\quad \vec{u}(s) \in S^2 = \{ \vec{x} \in \mathbb{R}^3 : |\vec{x}| = 1 \}$$

Let $X$ be the set of points on $S^2$ at a distance at most $r < \frac{\pi}{2}$ from the image of $u$. When both $L$ and $r$ is small and the curve doesn't curved backward to make $X$ "self-intersect", $X$ can be viewed as a union of

• Two semi-spherical caps, each with area $\pi ( 1 - \cos r)$
• plus a curvy strip $Y$ of dimensions $\approx L \times 2r$.

Let $\vec{v}(s) = \vec{v}'(s) = \frac{d}{ds}\vec{u}(s)$ and $\vec{w}(s) = \vec{u}(s) \times \vec{v}(s)$. Like $\vec{u}(s)$, both of them are unit vectors and the three vectors $\vec{u}(s)$, $\vec{v}(s)$, $\vec{w}(s)$ together form an orthonormal triad along the curve $u$. Using them, we can parametrize the curvy strip $Y$ as

$$[ 0, L ] \times [ -r, r ] \ni (s,\theta) \quad\mapsto\quad \vec{Y}(s,\theta) = \cos\theta \vec{u}(s) + \sin\theta \vec{w}(s) \in S^2 $$ In this parametrization, the Jacobian for the area equals to

$$\begin{align} \vec{Y} \cdot \left(\frac{\partial \vec{Y}}{\partial s} \times \frac{\partial \vec{Y}}{\partial \theta}\right) &= (\cos\theta\vec{u} + \sin\theta\vec{w})\cdot \left((\cos\theta\vec{v} + \sin\theta\vec{w}') \times (-\sin\theta\vec{u} + \cos\theta\vec{w}\right)\\ &= \vec{v}\cdot(\cos\theta\vec{v} + \sin\theta\vec{w}') = \cos\theta - \sin\theta \vec{v}'\cdot \vec{w} \end{align} $$ It is easy to check $|\vec{v}'\cdot \vec{w}| = |k_g|$ where $k_g$ is the geodesic curvature of $u$ in $S^2$

If $u$ is not too curvy, i.e. $|k_g(s)| < \cot r$ for all $s \in [0,L]$, the Jacobian will be positive over the whole strip $Y$. Its area will be given by

$$\require{cancel} \int_0^L\int_{-r}^r (\cos\theta - \color{red}{\cancelto{0}{ \color{gray}{\sin\theta \vec{v}'\cdot \vec{w}}}}) d\theta ds = \int_0^L 2\sin r ds = 2L \sin r$$

As a result, when the curve is short and not too curvy, the area we seek has a surprisingly simple formula:

$$\bbox[border:1px solid blue;padding: 8px]{ \verb/Area/ = 2L\sin r + 2\pi(1 - \cos r) } $$

I'm pretty sure this is a special case of Weyl's formula for tube volume. Unluckily, I can find the exact statement proved by Weyl nor its derivation. I hope I didn't miss anything essential in this derivation.

Answer 2

HINT:

At any time $t$, let the tip of the radial vector make the polar angle $\theta$ with the z-axis & azimuthal angle $\phi$ with the x-axis then the elementary area covered by the tip of the vector on the spherical surface $$dS=(r\sin\theta d\phi)\times (rd\theta)=r^2\sin\theta d\theta d\phi$$ by integrating $$S=r^2\int\sin \theta d\theta d\phi$$ you need to express the angles as some function of time $t$ as $\theta=f(t)$ & $\phi=g(t)$ on the basis of rotational motion of the tip of the vector