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Long before steam engines powered factories, a teacher in Alexandria described a ball that spun on its own when water boiled beneath it. The same man wrote down the triangle-area formula you meet in Class 9. Hero of Alexandria is a good person to study because his work sits right between physics and mathematics, and because he used toys to teach. This post covers what he built and wrote, and works through the physics and maths behind two of his most familiar results.
Who was Hero of Alexandria?
According to the MacTutor biography of Heron of Alexandria from the University of St Andrews, he lived from about AD 10 to about AD 75. His dates were disputed for a long time, with guesses from 150 BC to AD 250. They were settled when the historian Otto Neugebauer found, in 1938, that one of Hero's works refers to a recent eclipse. That eclipse can be identified as one seen in Alexandria at 23:00 on 13 March AD 62.
MacTutor says his writings look like lecture notes, so he probably taught at the Museum in Alexandria. His surviving books cover surveying, mechanics, war machines, mirrors, measurement and pneumatics, the study of devices worked by air, steam and water.
What is the aeolipile?
It is described in Hero's Pneumatica, a work in two books, the first with 43 chapters and the second with 37. An English translation from 1851, edited by Bennet Woodcroft and translated by J. G. Greenwood, is freely available on Project Gutenberg. Chapter 50 is titled "The Steam-Engine" and begins: "Place a cauldron over a fire: a ball shall revolve on a pivot."
The design: water boils in a closed cauldron. Steam rises through a bent tube into a hollow ball that can turn on a pivot. Two bent pipes stick out of the ball at opposite ends of a diameter, bent in opposite directions. Steam rushes out of both pipes, and the ball spins.
MacTutor quotes a description of it as "the first known device to transform steam into rotary motion". The Pneumatica also includes, by MacTutor's count, descriptions of over 100 machines, among them a fire engine, a wind organ and a coin-operated machine. In the Gutenberg translation, that machine dispenses water when "a coin of five drachms" is dropped into a sacrificial vessel.
Why does the aeolipile spin? The physics
This is Newton's third law. As each nozzle pushes steam out backwards, the steam pushes the nozzle forwards with an equal and opposite force.
The two nozzles point in opposite directions on opposite sides of the ball. So their forces are equal in size, opposite in direction, and act along parallel lines a distance apart. That pair is called a couple. Its net force is zero, so the ball does not move off its pivot, but its net torque is not zero, so the ball rotates.
Worked example with illustrative numbers. Each jet pushes with a force of 0.05 N, and each nozzle is 0.10 m from the axis.
Torque from one jet: τ = F × r = 0.05 N × 0.10 m = 0.005 N·m.
Both jets turn the ball the same way, so total torque = 2 × 0.005 = 0.010 N·m.
Net force = 0.05 N − 0.05 N = 0. The pivot carries no sideways load from the jets.
The ball does not speed up forever. Friction at the pivot and air resistance grow with speed until they balance the jets' torque, and the speed levels off.
Common mistakes:
"The steam pushes on the air." The jet does not need air to push against. Rockets work in a vacuum for the same reason.
Forgetting the zero net force. In a couple, forces cancel but torques add. Exam questions on couples test exactly this.
Our post on Newton's three laws in his own words shows how the third law was first stated. For how steam later became real power, read our post on James Watt and the steam engine.
What is Heron's formula, and how do you use it?
In Book I of his Metrica, MacTutor reports, Hero proves the formula for the area A of a triangle with sides a, b and c. Let s = ½(a + b + c), the semi-perimeter. Then A² = s(s − a)(s − b)(s − c).
Worked example: the 13–14–15 triangle.
s = ½(13 + 14 + 15) = 21.
s − a = 8, s − b = 7, s − c = 6.
A² = 21 × 8 × 7 × 6 = 7,056.
A = √7,056 = 84 square units.
Check it another way. Drop a perpendicular from the top vertex to the side of length 14. It splits the triangle into two right triangles with sides 5, 12, 13 and 9, 12, 15. The height is 12, so area = ½ × 14 × 12 = 84. The answers match.
Our Class 9 guide to Heron's formula has more practice of this kind.
How did Hero find square roots?
MacTutor quotes the Metrica's method for √720, which it notes was known to the Babylonians long before:
The nearest square is 729 = 27².
Divide 720 by 27: 26⅔.
Add 27 and halve: (26⅔ + 27) ÷ 2 = 26⅚.
Check: (26⅚)² = 720 1/36, so the error is only 1/36.
Repeating the step with 26⅚ in place of 27 gives an even better answer. It is the same idea as Newton's method for roots, which you may meet in higher maths.
What can a student take from Hero?
Teach with objects. MacTutor suggests Hero used his mechanical toys to teach physics. Building or sketching a device makes a law concrete.
Check an answer two ways. Heron's formula and base × height agree on 84; that agreement is your proof you did not slip.
Know what a demonstration shows. The aeolipile proves action and reaction; it does not prove a practical engine.
When torques and couples get confusing, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
Try it now: use Heron's formula on a triangle with sides 5, 5 and 6, then check your answer by dropping a perpendicular to the side of length 6.
