How one man measured the Earth with a stick
Tonight's idea
Eratosthenes ran the library at Alexandria, in Egypt. He read that in Syene, far to the south, the midday Sun on the longest day of the year shone straight down a deep well: no shadow, Sun exactly overhead.
On the same day in Alexandria, a vertical stick did cast a shadow. He measured the angle between the stick and the Sun's rays: about 7.2°. That is \(7.2 / 360 = 1/50\) of a full circle.
The Sun is so far away that its rays arrive at both towns parallel. If the rays are parallel and the stick in Syene points straight at the Sun, then the 7.2° between the Alexandria stick and its ray is the same as the angle between the two sticks measured from the centre of the Earth. (Look at the picture below: the two marked angles are alternate angles, the two equal "Z" angles a line makes when it crosses two parallel lines.)
So the arc from Syene to Alexandria is \(1/50\) of the way round the world. Travellers said the distance was about 5,000 stadia, roughly 800 km. Fifty times that is 40,000 km. The modern value is 40,008 km round the poles and 40,075 km round the equator.
The same trick, shadows and parallel rays, measures anything tall. A one-metre stick and a tree in the same sunshine make similar triangles: the ratio of height to shadow is the same for both.
Watch one: the whole calculation
Fraction of the circle: \(7.2° \div 360° = \tfrac{1}{50}\).
So the Syene–Alexandria distance is \(\tfrac{1}{50}\) of the circumference: \(C = 50 \times 800 = 40{,}000\) km.
Radius from circumference: \(C = 2\pi r\), so \(r = 40{,}000 / (2 \times 3.14) \approx 6{,}370\) km. Modern value: 6,371 km on average.
Tree by shadow: a 1 m stick casts a 1.5 m shadow. A tree casts an 18 m shadow at the same moment. Height to shadow is \(1 : 1.5\) for both, so the tree is \(18 \div 1.5 = 12\) m tall.
Your turn
- Suppose Eratosthenes had measured 9° instead, and the towns were 1,000 km apart. What circumference would he have got?
- Two towns are 400 km apart, due north and south of each other, and the noon shadow angles differ by 3.6°. Find the circumference of the Earth from this pair.
- If the two cities had been 2,000 km apart (same Earth), what angle would the Alexandria stick have shown?
- A 1.2 m stick casts a 0.8 m shadow. At the same time a flagpole casts a 10 m shadow. How tall is the flagpole?
- Historians are not sure how long a stadion was. If it was 185 m, what circumference do 5,000 stadia and \(1/50\) give? How far off is that from 40,075 km, as a percentage?
- Why does the method need the Sun to be very far away? What would go wrong if the Sun were, say, only 1,000 km above Syene? Draw a quick sketch.
Stretch: measure your own latitude
On an equinox (around 20 March and 22 September) the noon Sun is directly overhead at the equator. Stand a stick straight up at local noon (when the shadow is shortest) and measure the angle between the stick and the line to the top of its shadow. That angle is your latitude.
Explain why, using the same picture as Eratosthenes but with the equator playing the part of Syene. Then try it: a 1 m stick, a tape measure, and a protractor or a bit of trigonometry if you know some.
Answers
- \(360 / 9 = 40\), so \(C = 40 \times 1{,}000 = 40{,}000\) km.
- \(360 / 3.6 = 100\), so \(C = 100 \times 400 = 40{,}000\) km.
- \(2{,}000 / 40{,}000 = 1/20\) of the circle, so \(360° / 20 = 18°\).
- Height to shadow is \(1.2 : 0.8 = 1.5\), so the flagpole is \(1.5 \times 10 = 15\) m.
- \(5{,}000 \times 185 = 925{,}000\) m \(= 925\) km, and \(50 \times 925 = 46{,}250\) km. That is about 15% too big. With a shorter stadion (some were about 157 m) the answer comes out close to 39,000 km. The method was excellent; the uncertainty was all in the distance.
- With a nearby Sun the rays are not parallel: they fan out from a point. Then the angle at Alexandria would depend on how high the Sun is, not just on how far round the Earth you have walked, and the alternate-angle step fails. In fact a nearby Sun over a flat Earth would also make a shadow at Alexandria, which is why Eratosthenes needed to already believe the Earth was round and the Sun very distant. Both were accepted by Greek astronomers of his time.
Stretch. Latitude is the angle, measured at the Earth's centre, between you and the equator. On an equinox the Sun's rays are parallel to the plane of the equator, so a stick at the equator points straight at the Sun. Your stick is tilted away from the rays by exactly the angle between you and the equator, by the alternate-angle argument. So the shadow angle equals your latitude. The angle has tangent (shadow length ÷ stick height); a 1 m stick with a 0.75 m shadow means a latitude of about 37°.