Rung 02 · ~150 BCE · Rhodes, Greece

The Size & Distance of the Moon

Hipparchus · two kinds of eclipse, combined like clues in a detective story

Two independent eclipse observations — one lunar, one solar — combined to give the Moon's size and distance simultaneously.

Clue 1 — Lunar eclipse: timing how long the Moon takes to cross Earth's shadow reveals the shadow is ~2.5 Moon-diameters wide at that distance.

Clue 2 — Solar eclipse: the Moon fits almost perfectly over the Sun's disc, meaning both have the same apparent angular size from Earth. This adds the crucial correction to the shadow-cone measurement.

Hipparchus used two independent eclipse observations — a lunar one and a solar one — to solve a problem that looks impossible with either alone.

Clue 1 — Lunar eclipse timing. During a lunar eclipse, Earth's shadow sweeps across the Moon's face. Hipparchus timed how long the Moon took to cross the full shadow width, then compared that to the Moon's own travel time across a distance equal to its own diameter. He found the shadow was roughly 2.5 Moon-diameters wide at that distance.

The missing correction. Earth's shadow is a cone, not a cylinder — it narrows because the Sun is larger than Earth. At the Moon's distance, the shadow is already smaller than Earth itself. This correction involves the Sun's unknown size and distance.

Clue 2 — Solar eclipse geometry. A total solar eclipse shows the Moon fitting almost perfectly over the Sun's disc. This means Moon and Sun have the same angular size from Earth: their true radius divided by their distance is equal. Substituting this into the shadow equation, the unknown Sun terms cancel beautifully, and the correction becomes exactly one Moon-radius.

Earth's radius ≈ 2.5 × Moon-radius + 1 × Moon-radius = 3.5 Moon-radii

How the unknowns disappear — the solution

Here is the elegant trick: the Sun's distance and size appear in both clues — but they cancel out completely when you combine them, leaving only the Earth–Moon size ratio.

Clue 2 reveals that the cone correction equals exactly one Moon-radius — making the unknown Sun terms cancel. Earth's radius turns out to be 3.5 Moon-radii. The true value is 3.67 — Hipparchus was remarkably close.

Actual computation

The same argument in symbols. r and d are radius and distance; the point is that r_sun and d_sun appear in both clues and divide out, so the answer never needs them.

So the Moon is about 3.5× smaller in diameter than Earth (true ratio: 3.67×). Plugging back into the Moon's known angular size of 0.5° gives its distance: about 60 Earth-radii (true: 60.3).

This rung stands entirely on Rung 1 — without knowing Earth's physical size first, all these ratios give only relative numbers, not real distances.

⚑ Assumptions that made it work

Earth's size is known — inherited from Rung 1. The Moon's distance comes out in Earth-radii, so any error there scales directly through.
Sun's rays are parallel — inherited again. The shadow-cone geometry depends on this.
A total solar eclipse means Moon and Sun have equal angular size. This is an observational fact, not derived — and it happens to be true only because we live in a cosmically lucky era. The Moon is slowly drifting away from Earth; in the distant past it was larger and would have produced annular eclipses; in the distant future it will be too small to cover the Sun entirely.
The Moon moves at a constant speed across the sky during the eclipse. Approximately true, and the resulting error is small.
The Moon shines by reflected sunlight (not its own light), so the shadow geometry is physically meaningful. Greeks understood this.

✋ Try it yourself — measure the Moon with a piece of card

Hipparchus needed two eclipses and a lot of patience. One half of his work you can redo on any clear night, with a card, a ruler and about ten minutes.

Punch or cut a small round hole in a piece of card — somewhere around 5 to 8 mm across. Measure it as accurately as you can and write the number down.
Hold the card up at the Moon, close one eye, and look through the hole.
Slide the card slowly towards and away from your eye until the Moon exactly fills it — just touching all the way round, with no sky showing at the edges and no part of the Moon hidden.
Hold still and have someone measure from the card back to your eye. That is the number that matters, so take it twice.

Now divide: eye-to-card distance ÷ hole width.

What you have just measured: you should get somewhere near 110. That is the Moon's angular size, and it says the Moon is 110 times further away than it is wide — which is the one number Hipparchus could not get from eclipses alone.

Now close the rung with it. The shadow argument above gives the Moon as about 1/3.5 of Earth's width — the true figure is 1/3.67, so Hipparchus's step is already a little generous. Rung 1 gave you Earth's width, 12,700 km. Divide it by 3.5 and the Moon comes out about 3,600 km across; divide by the true 3.67 and you get 3,470 km. Multiply either one by your 110 and the Moon lands somewhere between 380,000 and 400,000 km away.

It is really 384,400. You got the distance to the Moon to within a few per cent, from a hole in a card and the rung below — and the wobble in your answer is not sloppiness, it is Hipparchus's 3.5 being slightly off, carried all the way up. That is what it means for one rung to stand on another.

And a second measurement, for free. Do the whole thing again on a night when the Moon is rising — huge, orange, sitting on the rooftops and looking close enough to touch. Then wait a few hours and do it once more when it is high overhead and looks small and ordinary.

The number does not change. Not by a millimetre. The Moon is exactly the same size in both, and always has been — the "huge" Moon on the horizon is something your brain is doing, not something the sky is doing. Your eyes are not instruments. That is the entire reason this ladder is built out of measurements instead of impressions.

The Size & Distance of the Moon

How the unknowns disappear — the solution

When Hipparchus Measured the Moon...