Is there a way to parameterize a path on a sphere? Say we want a particle to travel a certain path along a sphere, always travelling a certain direction (namely an angle from the equator). For example, starting at the origin and travelling a north-east direction, would be an angle of $\pi/4$. If we let then angle be $\alpha$, then is there a way to parameterize the equation of the path? 
What I've tried: In spherical coordinates we can write the equation of a sphere of radius R as $$\textbf{r}(\theta,\phi)=<R\sin{\theta}\cos{\phi},R\sin{\theta}\sin{\phi},Rcos{\theta}>\text{where }  0\leq\theta\leq2\pi\text{ and }0\leq\phi\leq\frac{\pi}{2}$$
However, I don't think we can simply say that one of the angles is $\alpha$, as the vector from which we're measing $\alpha$ is constantly changing. If the angle is $\pi/4$ or similar, Ive tried making the argument that $\theta$ and $\phi$ would change at the same rate, therefore you can let them both equal $t$ and proceed from there, but I'm not sure if that's true. 
Any help would be appreciated, cheers! 
 A: There are several conflicting conventions for spherical polar coordinates
among mathematicians, physicists, geographers and astronomists.
In this answer, I will stick to following parametrization of the sphere:
$$[0,2\pi] \times \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \ni (\theta,\phi) \quad\mapsto\quad (x,y,z) = 
(R\cos\phi\cos\theta,\,R\cos\phi\sin\theta,\,R\sin\phi ) \in \mathbb{R}^3$$
i.e. $\theta, \phi$ corresponds to the longitudes and latitudes in geography and $\phi \in [0,\frac{\pi}{2}]$ corresponds to the northern hemisphere.
The question at hand is find a curve whose tangent is making an constant angle $\alpha$ with the circles of latitudes. In above parametrization, the metric on the sphere is given by 
$$ds^2 = R^2 (\cos^2\phi \, d\theta^2 + d\phi^2)$$
What this means is:


*

*if we move along the line $\phi = \text{const}$ for a small amount $\delta\theta$, the distance traveled $\approx R \cos\phi \delta\theta$.

*if we move along the line $\theta = const$ for a small amount $\delta\phi$, the distance
traveled $\approx R \delta\phi$.
In order for the curve to make a angle $\alpha$ with the circles of latitudes, we need to make displacement $(\delta\theta, \delta\phi)$ such that
$$\frac{R \delta\phi}{R \cos\phi \, d\theta}\approx m \stackrel{\text{def}}{=} \tan\alpha$$
This implies the curve satisfies following ODE:
$$\frac{1}{\cos\phi}\frac{d\phi}{d\theta} = m
\quad\iff\quad\frac{d\sin\phi}{1 - \sin^2\phi} = m \, d\theta
$$
If the curve start at a point $(\theta_0,\phi_0)$, we can solve above ODE and get:
$$\frac{1 - \sin\phi}{1 + \sin\phi} = \frac{1 - \sin\phi_0}{1 + \sin\phi_0} e^{-2m(\theta - \theta_0)}
$$
For example, if we start the curve at $(\theta_0,\phi_0) = (0,0) \iff (x,y,z) = (R,0,0)$, we get
$$
\begin{align}
\frac{1-\sin\phi}{1+\sin\phi} = e^{-2m\theta}
&\iff \sin\phi = \tanh(m \theta)\\
&\iff
\begin{cases}
x &= R \frac{\cos\theta}{\cosh(m\theta)}\\
y &= R \frac{\sin\theta}{\cosh(m\theta)}\\
z &= R \tanh(m\theta)
\end{cases}
\end{align}
$$
Such curves are called Rhumb line, there are more details on its wiki entry. 
At the end is a picture of $6$ rhumb lines. All of them start at $(\theta_0,\phi_0) = (0,0)$ and their $\alpha$ vary from $10^\circ$ (the red curve) to $60^\circ$ (the cyan curve). As one can see, they sort of "spiral" around the north pole as they approach it.
$\hspace0.8in$ 
