How do you go from transit times to Earth-Sun distance?

Question

In the early 18th century, Halley described a method to determine the value of the astronomical unit by observing a transit of Venus from multiple locations on Earth. How do you go from the transit time at two different locations to the distance between the Sun and the Earth? I found many descriptions in rather vague terms of this method, but what is the math behind it?

Answer 1

We know, precisely, the ratio between the Earth-Venus and Venus-Sun distances. Kepler's laws give us this.

We know, in terms of apparent size in the sky, the precise size of the Sun in degrees - just by looking at it.

Placing Venus between the Sun and the Earth, we can look at it from one point in the Earth (call it $N$) and see Venus against the Sun at a point I'll call $n$. Repeating from another place (call it $S$) at a different latitude, we get another point, $s$. If we know how far apart these two points are in the sky, we can use the relative distance of Venus to tell us the angle, at Venus, of the lines $N$-Venus-$n$ and $S$-Venus-$s$. Since we know the distance in kilometres between $N$ and $S$, that means we can solve the triangle and get the distance in kilometres between Venus and $N$ or $S$. (I am tacitly omitting a few steps here, such as the fact that the "distance" between $N$ and $S$ is a north-south distance and not a great-circle distance along the surface of the Earth). Having the distance between Venus and $N$ or $S$, we can multiply that up (since we know the ratio accurately) to the distance between the Sun and $N$ or $S$.

So it remains to find the distance in the sky between $n$ and $s$. Direct measurement cannot be done accurately enough. Accordingly we use transit time as a proxy. A line going through the middle of the Sun is longer than a line slicing across the top or bottom of it. So transit time tells us how far north or south of the Sun's diameter the point $s$ or $n$ is travelling.

Three transit times are actually needed, not two. To see this, suppose that the times for $s$ and $n$ were equal. All we would then know, strictly speaking, is that they were symmetrically disposed about the diameter - not how far north or south of the diameter they were. But since we know how fast Venus travels across the face of the Sun we can calculate the duration of a hypothetical transit along the diameter, and that figure can then be combined with just two actual observations.

To summarise:

• Transit durations give latitudes on the Sun's disc.
• Latitudes on the Sun's disc give an angular distance between two transit points. – The ratio of orbit sizes transforms this into the angle of a long thin triangle based at Venus and joining the two observation points on the Earth. – Trigonometry gives the Venus-Earth distance in terrestrial units. – The ratio of orbit sizes, again, gives the Sun-Earth distance in terrestrial units.