On a sphere, does a rhumb line always exist between two points? If so, how to calculate it? We can calculate the bearing between two points (described by lat, lon with an elevation of 0 for simplicity) on Earth by using the Haversine formula, which yields an initial heading and a final heading. When this line is followed the distance is minimal.

But is there such a thing as a "straight line" which is not distance minimal that can be cast between any two points such that the initial heading is the same as the final heading always? How can this bearing be calculated if so? (As noted in comments, such a path is called a rhumb line or loxodrome.)

Example: If point A is some point that lies on the equator of Earth, and B is a point 200 miles to the west of point A, you could simply get from A to B by going 200 miles on a 270 heading. You could also follow a great circle path which would result in less distance but would involve an initial and final bearing which change as a function of your position on the line. Does this straight line exist for any two points on a sphere?
I am fairly sure the answer is yes, but I'd like to know how to calculate this if I am correct. What would this line be called? I've heard it described at LOS (line of sight) I think, but unsure of if that is the same idea.
 A: I expected that this question was a duplicate, but I couldn't locate another instance of it.

But is there such a thing as a "straight line" which is not distance minimal that can be cast between any two points such that the initial heading is the same as the final heading always? ... What would this line be called?

What you're describing is a rhumb line, also called a loxodrome or a path of constant heading, namely, a path on the earth's surface such that the angle between the direction of travel and the meridian (longitude line) is constant. You can produce equations for a rhumb line by parameterizing the sphere and imposing the constant-angle condition, which leaves a single functional parameter that controls how fast you move along the path; to do this it's convenient to use latitude-longitude coordinates.

How can this bearing be calculated if so?

Carrying out the above procedure (and ignoring the poles) shows that the points on the loxodrome

*

*through the point with latitude $\phi_0$ and longitude $\theta_0$ and

*making a constant angle $\alpha$ with the meridians

are characterized by the relation
$$\phantom{(\ast)} \qquad (\theta - \theta_0) \cot \alpha = \operatorname{gd}^{-1} \phi - \operatorname{gd}^{-1} \phi_0 , \qquad (\ast)$$
where $\theta$ denotes longitude and $\phi$ latitude, and where $\operatorname{gd}^{-1}$ is the inverse of the Gudermannian function: $$\operatorname{gd}^{-1} = \operatorname{artanh} \circ \sin .$$
In particular, if we fix a destination point $(\theta, \phi) = (\theta_1, \phi_1)$ and let $\alpha$ vary, we can solve for the heading $\alpha$ that will take us from $(\theta_0, \phi_0)$ to $(\theta_1, \phi_1)$. We split into two cases:

*

*If $\phi_0 = \phi_1$, i.e., the points lie on the same parallel, then $\alpha = \pm \frac{\pi}{2}$, that is, the only headings that take you between the points are due west and due east.


*On the other hand, if $\phi_0 \neq \phi_1$, then since $\operatorname{gd}^{-1}$ is strictly increasing on the set $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ of (non-polar) latitudes, $\operatorname{gd}^{-1}(\phi_0) \neq \operatorname{gd}^{-1}(\phi_1)$, and rearranging gives $$\alpha = \arctan \frac{\theta_1 - \theta_0}{\operatorname{gd}^{-1} \phi - \operatorname{gd}^{-1} \phi_0} .$$ Notice that adding any integer multiple of $2 \pi$ to, say, $\theta_1$ gives another representation $(\theta_1 + 2 \pi k, \phi_1)$ of the destination point and (since $\arctan$ is strictly increasing), giving an infinitely many headings that take you from the departure point to the destination point,
$$\boxed{\alpha = \arctan \frac{\theta_1 - \theta_0 + 2 \pi k}{\operatorname{gd}^{-1} \phi - \operatorname{gd}^{-1} \phi_0} , \qquad k \in \Bbb Z} .$$ Difference choices of $k$ correspond to rhumb lines that wrap around the sphere different numbers of times.

Does this straight line exist for any two points on a sphere?

The above computation shows that it does, at least provided neither of the points is a pole, and generically that there are infinitely many rhumb lines connecting two points; you can check readily that you can always find a rhumb line passing through a pole and a given second point, and the only headings that take you between the two are $\alpha = 0, \pi$.
Remark Like you say, rhumb lines are typically not distance-minimizing; in fact, among all rhumb lines, only arcs of the equator and of longitude lines are. Nor are they straight in a naive sense: A typical rhumb line curves away from a geodesic ("straight") path. To see this, imagine you're standing a meter from the North Pole facing east. If you walk straight ahead, you'll walk along a great circle that will pass near the South Pole. But the constant-heading path is just a small circle around the North Pole that bends to your left. This curving behavior is also what you'd expect a priori: Anything worth calling "straight lines" ought to be a feature of the geometry itself and in particular independent of the choice of coordinates, but different choices of poles give different rhumb lines.
On the other hand, one can look for a map projection that has the property that rhumb lines on the Earth correspond to straight lines on the map. But equation $(\ast)$ shows that, for any given rhumb line, $\theta$ and $\operatorname{gd}^{-1}(\phi)$ are affinely related, so the projection $$(\theta, \phi) \mapsto (\theta, \operatorname{gd}^{-1}(\phi))$$ has that property. One can check that this projection is conformal (that is, it preserves angles between vectors) and moreover that it is essentially the only projection that has these two properties; this is the classic Mercator projection.
