# Eratosthenes measures the Earth - simplify in simple English, please!?

It is an article about how Eratosthenes measures our Earth and I don't understand a single word. Please help me:

In a remarkable feat, the Greek astronomer and geographer Eratosthenes estimated the size of Earth in about 240 B.C. He did it by comparing the altitude of the Sun passed directly overhead in Syene on the summer solstice. He also knew that in the city of Alexandria to the north the Sun came within only 7 degrees of the zenith on the summer solstice.

He therefore reasoned that Alexandria must be 7 degrees of latitude to the north of Syene. Because 7 degrees is 7/360 of a circle, he concluded that the north-south distance between Alexandria and Syene must be 7/360 of the circumference of Earth.

Eratosthenes estimated the north-south distance between Syene and Alexandria to be 5000 stadia.

Thus he concluded that:

7/360 x circumference of Earth = 5000 stadia

From this he found Earth's circumference to be about 250000 stadia. Today, we don't know exactly what distance a stadium meant to Eratosthenes. Based on the actual sizes of Greeks stadiums, it must have been about 1/6 kilometers. Thus, Eratosthenes estimated the circumference of the Earth to be about 250000/6 = 42000 kilometers - remarkably close to the modern value of just over 40000 kilometers.

Your help would be much appreciated it. Thanks for your time.

### 3 Answers

- poornakumar bLv 77 years agoBest answer
First, how did he guess the angle (those days they probably didn't do it in degrees, but by the effect of that angle like what is the 'tangent', that is easily measured).

At Syene, he looked into a well in which the noon-Sun didn't cast a shadow on the walls. A well is normally constructed, aligning with a plumb-bob so as to have vertical walls. If Sun is at the 'zenith' a ground object placed vertically, doesn't cast a shadow of noon-time Sun. It implies Sun is on the same angle as the latitude of that place (but Eratosthenes must have used a different language). He concluded that Syene is located on the Tropic of Cancer (as we know it, or perhaps he too knew it that way). Alexandria, supposed to be on the same longitude (it is easy to check with a compass, but he might not have had one then) but beyond the tropics & tropics is the land beteen the 'tropics of Cancer' & 'Capricorn' where at every place Sun appears vertical (at zenith) on noons of two days in Sun's transit (one south to north & one north to south).

Anyway on the summer solstice day, he must have observed the shadow in a well in Syene. By the ratio of the shadow's length on ground, to the height (from the base where the shadow is, to the 'top' that causes the apex of the shadow) that we call 'Tangent', For a 7°, it should be 0.1227645. This as a fraction of the semicircular arc (180°) that we call (here)

f = (7/180).

If this fraction is used to divide the arc distance (on the Great Circle, every great circle is Equator-like) from Syene to Alexandria, or its reciprocal (1/f = 180/7) is multiplied with the distance, that gives Earth's half Circumference. They knew the value of Pi ('π') then, that is the ratio of Circumference to the diameter (alternately, ratio of semicircular arc to the radius), But of course they were interested in the total circumference of Earth. Even Columbus and his friends based their voyage on this figure.

At 7°, one can say the tangent almost equals the angle. Actually it shoud be, with small angle approximation, is "distance/[2 Tan(7/2)]"

2 Tan(7/2)≈ 2[(7/2)] = 7° (= 0.122173 Radian; see how remarkably close to the "Tan7°" above).

Well, I used the degree measure in 7°, just for demonstration but that should never enter Mathematical calculations, which should be in Radians. Radian measure (though cumbersome to handle for someone not well-versed in calculations) directly gives the true ratio, needing no other conversion (degrees need that). In Radian measure the circumference is π times Diameter. Instead we say the angle is π Radians. If Syene to Alexandria distance is also factored as a ratio in 'Radian measure'; we get circumference by scaling it up 2π times,

distance X [2π/Radian measure] = circumference.

Now we know from the calculator the value of 'π'. But they used to do it then, with 'fractions'; the best fit for 'π' is not 22/7 but

π = 335/113.

Easy to remember too: write pairs of first odd numbers in a sequence,113355. Cut it into two, the left part (lower integers) is put as denominator or divisor and the other (right) part goes to the top as numerator, that gets divided by the denominator.

And he even has factored the error (percentage) and to know how far he can be off the mark. I think as a civilised race, by now, educated humans should figure it out without going into esoteric details like, 'degrees', 'tangents' & all that, that take us further away from what we see on ground. Things like these tell the laymen apart from Scientists; but everyone needn't be a Scientist and talk in their language. It simply cuts off people from indulging in Science. They yearn to unravel the mystery but a gulf separates them from reaching 'undersatnding'.

- LoganLv 57 years ago
I'm assuming its a problem of knowing what words mean what.

The zenith is the part of the sky that is directly overhead. The Sun passed directly overhead Syene on the Summer Solstice (which happens on the same date every year).

North of Syene, in Alexandria, the Sun only came within 7 degrees of the zenith. From the very reasonable assumption that the Earth was round he could then calculate Earth's size.

And stadia is the plural of stadium.

That's about as clear as I can imagine it. If you still don't understand you should specify which parts.

- John WLv 77 years ago
He knew that on the Summer Solstice, the Sun will be at it's highest point so that gave him a reference. He knew that it didn't seem to rise as much in a city north or him and he knew how far that city was therefore he could calculate the curvature of the Earth from the two different readings of how far the Sun got in the sky. The angles being parts of 360 and the circumference being prorated from that is based on an assumption that the Earth was spherical and he was just trying to work out how large the sphere was.

I would've used three distant points probably sails of ships at sea but that would've taken three people to do, his method just took one person and two summer solstices.