<p>We know, precisely, the ratio between the Earth-Venus and Venus-Sun distances. Kepler's laws give us this.</p>
<p>We know, in terms of apparent size in the sky, the precise size of the Sun in degrees - just by looking at it.</p>
<p>Placing Venus between the Sun and the Earth, we can look at it from one point in the Earth (call it <span class="math-container">$N$</span>) and see Venus against the Sun at a point I'll call <span class="math-container">$n$</span>. Repeating from another place (call it <span class="math-container">$S$</span>) at a different latitude, we get another point, <span class="math-container">$s$</span>. If we know how far apart these two points are in the sky, we can use the <em>relative</em> distance of Venus to tell us the angle, at Venus, of the lines <span class="math-container">$N$</span>-Venus-<span class="math-container">$n$</span> and <span class="math-container">$S$</span>-Venus-<span class="math-container">$s$</span>. Since we know the distance in kilometres between <span class="math-container">$N$</span> and <span class="math-container">$S$</span>, that means we can solve the triangle and get the distance in kilometres between Venus and <span class="math-container">$N$</span> or <span class="math-container">$S$</span>. (I am tacitly omitting a few steps here, such as the fact that the &quot;distance&quot; between <span class="math-container">$N$</span> and <span class="math-container">$S$</span> is a north-south distance and not a great-circle distance along the surface of the Earth). Having the distance between Venus and <span class="math-container">$N$</span> or <span class="math-container">$S$</span>, we can multiply that up (since we know the ratio accurately) to the distance between the Sun and <span class="math-container">$N$</span> or <span class="math-container">$S$</span>.</p>
<p>So it remains to find the distance in the sky between <span class="math-container">$n$</span> and <span class="math-container">$s$</span>. Direct measurement cannot be done accurately enough. Accordingly we use <em>transit time</em> 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 <span class="math-container">$s$</span> or <span class="math-container">$n$</span> is travelling.</p>
<p>Three transit times are actually needed, not two. To see this, suppose that the times for <span class="math-container">$s$</span> and <span class="math-container">$n$</span> 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 <em>calculate</em> the duration of a hypothetical transit along the diameter, and that figure can then be combined with just two actual observations.</p>
<p>To summarise:</p>
<ul>
<li>Transit durations give latitudes on the Sun's disc.</li>
<li>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.</li>
</ul>

CC BY-SA 4.0
2020-11-14T07:35:49.420Z
How do you go from transit durations to latitudes?
usernumber
CC BY-SA 4.0
2020-11-13T17:31:22.450Z
2020-11-13T17:31:22.450Z
Martin Kochanski
<p>In the early 18th century, <a href="http://www.astronomy.ohio-state.edu/%7Epogge/Ast161/Unit4/venussun.html" rel="nofollow noreferrer">Halley described a method</a> 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?</p>

CC BY-SA 4.0
2020-11-13T11:08:29.007Z
Different question, but somewhat related: [Before the 1761 transit, what was our best estimate of the distance to the Sun?](https://astronomy.stackexchange.com/q/39233/7982)
uhoh
<observational-astronomy><distances><mathematics><astronomical-unit>
How do you go from transit times to Earth-Sun distance?
usernumber
# How do you go from transit times to Earth-Sun distance?

(Asked by: username_0 on 2020-11-13)

In the early 18th century, [Halley described a method](http://www.astronomy.ohio-state.edu/%7Epogge/Ast161/Unit4/venussun.html) 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?

**Tags:** observational-astronomy, distances, mathematics, astronomical-unit
> **Comments:**
> **username_1**: Different question, but somewhat related: [Before the 1761 transit, what was our best estimate of the distance to the Sun?](https://astronomy.stackexchange.com/q/39233/7982)  

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## 1 answers

### Answer by username_2 on 2020-11-13. Score: 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.

> **Comments:**
> **username_0**: How do you go from transit durations to latitudes?  

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