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Eratosthenes measured the Earth using two cities and the Sun. Abu Rayhan al-Biruni, more than 1,200 years later, needed only one hill and one angle. His method is a favourite in olympiad-style geometry because it is short, clever and full of places where a student can go wrong. This post gives you the method, a full worked example and the mistakes to watch for.
Who was al-Biruni?
According to the MacTutor biography, al-Biruni was born in 973 in Kath, in Khwarazm (now in Uzbekistan), and died in 1048 in Ghazna (now Ghazni, Afghanistan). He was taken to Ghazna after Sultan Mahmud conquered Khwarazm in 1017, and from around 1022 he travelled with Mahmud's campaigns into northern India.
Michio Yano's entry in the Biographical Encyclopedia of Astronomers notes that al-Biruni wrote 146 works, of which only 22 survive. One of them, the Tahdid al-amakin (Determination of the coordinates of cities), was written on his journey to Ghazna in 1018. His study of Indian science and culture, India, was completed in 1030, and Yano calls him the first Indologist in the modern sense.
How did al-Biruni measure the Earth from a hill?
Amelia Sparavigna's review of al-Biruni's science describes the two steps. First he found the height of a hill by measuring the angles it made from two points a known distance apart. Then he climbed it and measured how far the horizon dipped below the horizontal. He did this near the fort of Nandana in the Punjab.
The geometry is one right-angled triangle. Your line of sight to the horizon just touches the Earth, so it meets the radius at the horizon at 90 degrees. The three corners are the Earth's centre, the horizon point and you on the hilltop. The side from the centre to you is R + h. The side from the centre to the horizon is R. The angle at the centre equals the dip angle θ, because both are measured from lines that are perpendicular to each other. So:
cos θ = R / (R + h), which rearranges to R = h cos θ / (1 − cos θ).
Worked example: what radius do the numbers give?
Take a hill of height h = 500 m and a measured dip of θ = 43 minutes of arc. These are illustrative numbers, not al-Biruni's own, chosen so you can check them on a calculator.
Convert the angle: 43′ = 43/60 degrees = 0.7167°.
cos 0.7167° = 0.99992177. Keep all eight decimal places.
1 − cos θ = 0.00007823.
R = 500 × 0.99992177 / 0.00007823 ≈ 6,391,000 m, or about 6,391 km.
That is within half a per cent of the modern volumetric mean radius of 6,371 km given by NASA. A quick check uses the small-angle form: for small θ in radians, 1 − cos θ ≈ θ²/2, so R ≈ 2h/θ². Here θ = 0.012508 rad, θ² = 0.00015645, and 2h/θ² = 1000/0.00015645 ≈ 6,392 km. The two methods agree.
What mistakes do students make with this method?
Rounding cos θ too early. If you round 0.99992177 to 0.9999, then 1 − cos θ becomes 0.0001 and R comes out as about 5,000 km, wrong by more than 20 per cent. When you subtract two numbers that are almost equal, keep every digit.
Mixing units. Put h in metres and you get R in metres. Do not mix a height in metres with a radius in kilometres halfway through.
Using degrees in the small-angle formula. R ≈ 2h/θ² only works with θ in radians.
Forgetting how sensitive the result is. With the same 500 m hill, a dip of 42′ gives about 6,699 km and 44′ gives about 6,104 km. One minute of arc, one-sixtieth of a degree, shifts the answer by roughly 5 per cent. Sparavigna writes that his self-built instrument could measure angles to about 10′ of arc, and that a precise sine value was the key to his precision.
Our guide to error propagation in measurements shows how to estimate this kind of sensitivity before you calculate.
How accurate was al-Biruni's answer?
MacTutor gives his radius as 6339.6 km. Sparavigna reports that, with an Arabic mile of 1.225947 English miles, his value was 3928.77 miles, about 2 per cent above the mean radius of curvature at the latitude where he measured. Converting medieval units is itself uncertain, so treat any modern percentage as approximate.
There is also a physical effect he could not correct for. Air bends light near the horizon, so you see slightly beyond the geometric horizon and the measured dip is a little smaller than the true one. A smaller θ gives a larger R, so refraction tends to push this method's answer up.
What can you learn from al-Biruni's way of working?
Al-Biruni recorded his own observations, compared Greek and Indian sources and said which he trusted. For your own problem solving, copy three habits: draw the geometry before writing a formula, estimate how sensitive the answer is to each measurement, and check a result two ways, as we did with the exact and small-angle forms. Compare his method with how Eratosthenes measured the Earth and note which measurements each one needed.
If you set up a geometry problem like this one and get stuck turning the picture into equations, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Try it now: redo the worked example for a 1,000 m hill and work out what dip angle you would expect to measure.
