Is Clarke's astronautical geometry wrong? (solving triangles)

Question

I'm reading a cool book called The Exploration of Space by Arther C Clarke. In the book, on the topic of navigation within the solar system he says:

By means of a sextant, or whatever its equivalent device may be in astronautics, the navigator first measures the angle A between the Sun and the Earth. He knows, from the almanac, the position of the Earth, and hence the line Sun-Earth is fixed. Next he measures the angle B between Venus and the Sun - and since the line Sun-Venus is also known, simple geometry fixes the spaceship's position at X.

My understanding of trigonometry is at barely-remembered-schoolboy-from-15-years-ago level...but I basically think that 'aint right, right? We're basically making 2 triangles; Earth - Sun - Ship and Venus - Sun - Ship. For both of those triangles we know one side (Planet - Sun) and one angle (from the ship, between the Planet and the Sun). The closest I can get to solving is to divide the two triangles again into 2 right angles, which lets you solve all the angles for 2 of those right angled triangles, but not any of the sides for the triangles you have the angles for.

So, is this a solvable problem, or is Clarke mistaken? If it's solvable; how?

n.b. Annoyingly, it's difficult to google for information on this specific topic because NASA in their wisdom have named their newest navigation project SEXTANT. This is a backronym that I don't recall, but essentially replicates the effects of the GPS network using doppler and an almanac of the signals emitted by pulsars.

Answer 1

Given two points $A$ and $B$, and the unoriented angle $\angle ACB$, the set of all possible locations of $C$ is the union of two circular arcs from $A$ to $B$, each with central angle equal to $2(\pi - \angle ACB)$.

So if you assume Sun ($S$), Earth ($E$), Venus ($V$) and you ($Y$) are coplanar, and you know the (distinct) positions of $S$ and $E$ and $V$ and the angles $\angle SYE$ and $\angle SYV$, then the set of possible positions for $Y$ is the (finite) intersection of two unions of two arcs. Of the finite remaining possibilities you can probably decide which is which and where you are by approximating the distance to the Sun by solar intensity or something, or simply by knowing roughly where you should be according to your flight plan.

In a 3-dimensional system, two measurements are not enough to constrain three degrees of freedom (each measurement constrains you to a spindle torus surface, and intersecting two of those still yields an entire curve of possible locations), but you can deal with that remaining degree of freedom by also measuring the angle $\angle EYV$ and intersecting three spindle tori instead of two.

Note that this measurement can still degenerate in certain situations, for example when you, Earth, Sun and Venus lie on a common circle. In that case you'll need to employ another planet, or the background stars, or start measuring distances.

Edit:

In the planar case, there can be up to four distinct points in the solar system where you can measure the same two unoriented angles $\angle SYE$ and $\angle SYV$. You can orient the angles you measure by looking at the background stars to figure out ecliptic north.

Answer 2

This is the situation:

Our astronaut measures the angles $A$ and $B$. Angle $A$ fixes the spaceship $X$ to lie on the circle through $S,X,$ and $V$; and angle $B$ fixes the spaceship to lie on the circle through $S,X,$ and $E$. So $X$ lies at the intersection of these two circles.

This only works if we know that all four points lie in the same plane, which is usually the case (or nearly so) for interplanetary travel. Also, as Magma explains, it doesn't work if the two circles happen to coincide.