CooperToons HomePage Merry History Dept. of Education CooperToons Books

Eratosthenes and the Size of the Earth

(Something You Can Try at Home)

©2013 by Charles F. Cooper

 

Eratosthenes measured it.

Eratosthenes: Jack of All Philosphers

Eratosthenes got no respect. Oh sure, his friends said he was bright and knew a lot. But he could never focus on anything for long enough to become a real expert. His philosophical buddies even called him "Beta" - meaning second rate. After all, if you're the Jack of all Trades, then you're ...

But for a second rate philosopher, Eratosthenes did all right. You probably first heard about him in grade school. He worked out the famous "Sieve of Eratosthenes" which was the first efficient way to determine which numbers are primes between any two specified numbers. The sieve is still used in modern number theory. Not bad for old Beta.

But probably what Eratosthenes did that still wows people today is that he measured the size of the earth. Not only that, we read, but he got the number right on. It didn't require advanced math, either, and anyone who's made it through middle school can easily understand how he did it.

Eratosthenes was born in Cyrene, Libya, near the Mediterranean coast and lived a lot of his life in Egypt. He knew that on a certain day in Aswan - in the south of Egypt - the noonday sun shone directly down into a well. Or to put it another way, a vertical structure like the Egyptian obelisks would cast no shadow on that day and in that place.

But north of Aswan in the Mediterranean coastal city of Alexandria, obelisks did cast shadows on that day, and the sun was never directly overhead on anyday. Eratosthenes realized the reason the shadows were of different lengths was due to the curvature of the earth. Of course, this means that Eratosthenes knew that the earth was round - which the Greeks knew anyway. But he also realized that he could now calculate the size of the earth.

As smart as his method was, he might have realized that you can measure the size of the earth from any two cities and on any day as long as you know how far north and south they are from each other. But what could you expect from good old Beta? In any case, we'll not only talk about the method Eratosthenes actually used, but the one could have used as well.

Discours de la Méthode (of Eratosthenes, that is)

The original method Eratosthenes used to calculate the size of the earth was based on a simple rule we learned from Euclid. It was the rule that if you have a straight line that intersects two parallel lines, then the alternate angles are the same, and so as shown here, the alternate internal angles are equal.

Alternate (and Equal) Angles

Yes, yes, it is true that the sun rays when they strike the earth are not exactly parallel. But because the sun is so much larger than the earth, they were parallel enough for Eratosthenes.

So if you look at the drawing below, you can get the idea. The obelisk at Alexandria casts the shadow, but the noon sun at Aswan is directly overhead. We can measure both the distance from Aswan to Alexandria - which is almost exactly 500 miles in a north south direction (actually 491 miles). We can also measure the height of the obelisk and the length of its shadow.

Alexandria and Aswan

Eratosthenes, of course, used Greek measurements. They didn't have a mile per se, but instead used the unit of measurement called the stadion (στάδιον). This was the length of the stadium at Olympia where they had the Olympic games. Since different experts come up with different lengths of the stadion, it means that no one really knows. But all in all, the distance is about 1/10 of a mile. Eratosthenes said the distance from Aswan to Alexandria was 5000 stadia which, sure enough, is 500 miles.

What we need to do now is to draw a schematic showing the whole obelisk-well-earth system. Of course, you don't draw it to scale, but that doesn't matter. In fact, it helps if you don't.

The Big Picture

First note that we represent the suns rays with the two vertical (and hence parallel) lines. So the line defined by the top of the obelisk to the center of the earth, Line DB, forms the two acute alternate interior angles with the sun's rays. These are the angles ∠ ABC and ∠ ADE. which like we said (that is, what Euclid proved) are equal.

∠ ABC = ∠ ADE

So all you need to do is to calculate the angle formed by the base of the obelisk to the top and then to the end of the shadow on the ground. This angle will be the same as the angle formed by the well to the center of the earth to the obelisk. Divide that angle into 360 and multiply the ratio by the distance from the well to the obelisk (i. e., Syrine to Alexandri) and you'll have the circumference of the earth.

The trick is now to measure the angle, something that wasn't so easy to do in the olden days.

We don't have Eratosthenes' original paper (or rather, his papyrus) so we don't know the exact numbers he had. But we do know if he had used a 68 foot obelisk (like Cleopatra's Needle) it would have thrown a shadow that was 8 feet 7 inches long. As for measuring the height of the obelisk there are various ways to do that. The simplest way is to wait until the length of your shadow is the same length as your actual height. That means the length of the obelisk's shadow is the same as its actual height, too.

At this point we need to do some fairly simple math. If you look at the picture of the obelisk and the shadow, you'll see it is a right triangle.

Right Triangle

Now the definition of the tangent of a non-right angle of a right triangle is the ratio of the two sides opposite and adjacent to the angle .

tan(∠ ADE) = opposite / adjacent = s / h

This means if the shadow is 8.6 feet and the obelisk is 68 feet, the tangent of the angle ∠ ADE = 8.6/68 = 0.1265. If you look up the angle for that tangent - that is the arctangent of 0.1265 - you find the angle is 7.2 degrees. This is indeed the angle Eratosthenes calculated.

tan(∠ ADE) = s/h = 8.6/68 = 0.1265
arctan(0.1265) = ∠ ADE = 7.2 degrees

But since ∠ ADE and ∠ ABC are alternate angles from of the parallel sun's rays. So we know that

∠ ADE = 7.2 degrees = ∠ ABC

Now degrees are not absolute numbers but are relative values of how much of the circle the angle spans. For instance an angle of 36 degrees spans 100 × 36/360 = 10 % of the circle. An angle of 90 degrees spans 100 × 90/360 = 25 % of the circle. It doesn't matter what the actual size of the circle is. 90 degrees spans 25 % of a circle whether it's the size of a pinhead or - yes - the size of the earth.

So from Eratosthenes' calculation we know that 7.2 degrees spans the arc of the earth's circumference that runs from Aswan to Alexandria. So the percentage of a circle between the two cities is:

Percent of circle (circumference) spanned = 100 × 7.2 / 360 = 2 %

Now 2 % is the same than as 1/50th. So the distance from Aswan to Alexandria is 1/50th of the total circumference of the earth. Since the distance from Aswan to Alexandria is 500 miles, then the circumference of the Earth is fifty times that distance. Ergo:

Circumference of the Earth = 50 × 500 miles = 25,000 miles

According to the most recent value from NASA, the circumference of the Earth is 24,900 miles (actually 24,901 miles, but who's counting?). That's only 0.4 of a percent off.

(For what it's worth, Christopher Columbus went to the various sources and also calculated the size of the earth. He determined the earth had a circumference of about 15,000 miles - off by almost 40 %. Christopher was better at sailing than he was at math.)

If you want the general formula for the circumference, it's easy to derive. Just go through the calculation using the letters and don't calculate the actual numbers. So you get:

arctan(s/h) / 360 = d / c

Elementary algebra then lets us solve for c and hey, presto!, we have the formula.

c = 360d / arctan(s/h)

(Note: For this formula arctan(s/h) must be in degrees. If the angle of arctan(s/h) is in the units of radians - usually considered more advanced - the formula is:

c = 2πd / arctan(s/h)

But we'll stick with the simple stuff.

Alternative Calculations

You can also estimate the value simply by assuming the ratio of the shadow to the obelisk is the same as the ratio of the distance from Aswan to Alexandria the radius of the earth. So as a reminder of the general set up we can look again at our picture

Alexandria and Aswan

And look at the schematic:

The Schematic

And the ratios are:

s/h = d/r

Or the Earth's radius is:

r = hd/s

Since the circumferences is 2 × π × radius, the general formula for the circumference is:

c = 2πhd/s

And if you plug in the numbers:

c = 2πhd/s = 2 × π × 68 feet × 500 miles/8.6 feet = 24,840 miles

which is only about a quarter of a percent off.

The Errors Don't Matter (Much)

Note that these methods are not mathematically exact. The triangle Δ ADE is assumed to be a right triangle, and the shadow of the obelisk equal to the length of one of the sides. This is - strictly speaking - not correct. The shadow is actually lying along the curve of the earth an so is not really the side of a triangle.

And if you use the second method, you make an assumption that the distance from Aswan to Alexandria is a side of a triangle - in fact you assume the triangles Δ ABC and Δ ADE are similar - that is, they share the same angles and hence the sides of the triangles are in proportion to each other.

Of course, the distance from Aswan and Alexandria is also not a straight line, but is curved. And the two triangles are not similar in a mathematical sense. Δ ABC is an isosceles triangle (the two long sides are equal) and so the ratios s/h and d/r are not the same.

(Note: Mathematical purists may cite non-Euclidean geometries that the "triangle" formed by Alexandria, Aswan, and the center of the earth is isosceles, yes, but because the smallest side is a curve, then the triangle is indeed not only a right triangle, but a double right triangle, something not possible in Euclidean space. Well, this is true enough, but not really relevant to what we're doing.)

Of course in reality none of these errors are large enough to really matter. This is true for any "practical" calculation. That is, if the errors arising from an incorrect assumptions are too small to be measured (of if you have "compensating errors), then the errors don't effect the final conclusions.

What You CAN Do At Home

 

Trying it at home

Well, do you want to measure the circumference of the earth a là Eratosthenes? Well, all you need to do is have a friend in Aswan tell you when the sun is overhead, you then fly to Alexandria, and measure the length of the shadow of an obelisk whose height you know on the same day.

But you really don't need to do this. If Eratosthenes had thought things through, he would have realized he didn't need to know where the sun was directly overhead at noon on a specific day. All you need to know is that there is someplace on the earth where the sun is directly overhead at noon.

Of course the sun is directly overhead somewhere any day of the year. So you can measure the length of two shadows at noon in two different towns. They don't even need to be directly in a north-south line. One person can be in Orlando and the other in Spokane. You don't need an obelisk either. A yardstick will work fine, thank you.

The key - as shown below - is that the two towns need to be separated a goodly distance from north to south. How far must they be separated? Well, there's no fixed answer, but the difference of the shadow lengths needs to be large enough to be reliably measured. As long as you do that, we can get the circumference of the earth. But first we'll go through the proof of how this will work, and what the formula is.

You'll remember that we showed above that the formula using Eratosthenes experiment is:

c = 360d / arctan(s/h)

So we can solve the equation for the distance, d, which is the north south distance between the town and whatever spot on earth the earth is directly overhead at noon. It does not have to be Aswan.

d = c × arctan(s/h) / 360

Now let's calculate the distance of two two different towns north of the place with the sun overhead.

d1 = c × arctan(s1/h) / 360

and

d2 = c × arctan(s2/h) / 360

Now take the difference of the two distances, d1 and d2:

d2 - d1 = c × arctan(s2/h) / 360 - c × arctan(s1/h) / 360

This can be rearranged as:

d2 - d1 = c × [arctan(s2/h) - arctan(s1/h) ] / 360

And solving for c:

c = 360 × (d2 - d1) / [arctan(s2/h) - arctan(s1/h)]

So you can see all we need to know the length of the two shadows (s1 and s2), the height of the yard stick, h, (which is 36 inches, of course), and the north south distance between the two towns, d2 - d1. It doesn't matter where the towns are.

Of course, the bigger the distance between the towns the better. And the longer the shadow the more accurate you can measure it. So let's pick a day late in the year - say November 26 - and two towns quite a ways apart. We'll again pick Spokane and Orlando. So what you do is contact someone in Spokane if you live in Orlando, and if you live in Spokane contact someone in Orlando. Then you agree to measure the length of a shadow at noon.

It is important that the time be solar or local noon. Because noon by the clock occurs at the same time in time zones hundreds of miles across - that is, the noons are simultaneous - the noon by the clock is not the noon by the sun. In Orlando on November 26, local noon is 12:13 p. m. and in Spokane it is 11:37 a. m. Fortunately, in November you don't have to worry about the Daylight Savings Time nonsense.

Now once you get your yardstick, you have to make sure it is vertical to the ground. So you have to have a flat surface and make sure that the stick is 90 degrees to the surface. So a carpenter's square or an artist's triangle will help.

But if you do everything right, then the length of the shadow of a yardstick measured at solar noon (that is 12:13 p. m.) on November 26 in Orlando will be 42 inches long. Similarly the shadow at noon on the same day in Spokane will be a whopping 92 inches long.

With these numbers we can calculate the angle of the shadow at the top of the yardstick.

Orlando Angle: arctan(42/36) = arctan(1.17) = 49.4 degrees
Spokane Angle: arctan(92/36) = arctan(2.56) = 68.6 degrees

So the difference is

d2 - d1 = 68.6 - 49.4 = 19.2 degrees

Now the only thing you need to know now is that Spokane is 1329 miles north of Orlando (we are not counting the 3000 odd miles in the east-west distance). So now we can calculate the circumference of the earth as

c = 360 × 1329 / 19.2 = 24,919 miles

This is only 0.08 % off - only 19 miles.

It might be interesting to try an even more basic approach. Say you measure the length of the shadow at noon in Little Rock on December 21. You'll find a yard stick casts a shadow of 58 inches. Well, the next year you get in your car and drive due north as much as the roads allow and end up in Columbia, Missouri, which is about as close to due north of Little Rock as you can get. But it's also not as far north as Spokane is from Orlando, and you'll find the shadow of the yardstick is only 69 inches. So again calculating the angles:

Little Rock Angle: arctan(58/36) = 58.1 degree
Columbia: arctan(69/36) = 62.4 degree

So the difference is

Angle at Columbia - Angle at Little Rock = 62.4 - 58.1 = 4.3 degrees

Now on your drive you can't only go north south so you'll get some extra mileage. Even if you take some direct non-interstate roads, you'll find you'll put about 342 miles on your odometer. So based on these number you end up with a circumference of the earth as:

c = 360 × 342 × / 4.3 = 28,633 miles

Which isn't as good as Eratosthenes, but still not too bad, considering. This error is almost entirely due to the inaccuracy of using road travel as a measure of north-south mileage. The actual north-south distance in a straight line from Little Rock to Columbia is 291 miles. So if you use your angles from the shadows you find the circumference is:

c = 360 × 291 × / 4.3= 24,362 miles

Which is much better, 2 % off.

But why did Eratosthenes get so close to the actual distance? Alexandria and Aswan aren't in a straight north-south line either. But with our conversion of 250,000 stadia at 1/10 of a stadion per mile we get right smack on the 25,000 miles circumference. Why can't we do as good a number as he did?

Eratosthenes: Pretty Good, But Not That Good

Actually, Eratosthenes didn't do any better than we did when we drove from Little Rock to Columbia. And for much the same reason.

Note we said that the reference books say a stadion is about 1/10 of a mile. Actually that's what you read on the Fount of All Knowledge when someone wants you to think Eratosthenes made an incredibly close measurement. But if you turn to reference books - those non-electronic devices with white flappy things in the middle - you'll find no one really agrees on what a stadia is. Some say it's about 606 feet others, some say 616 feet. If that's the case, we'll find that Eratosthenes made an error in the distance from the two cities - an error that worked in his favor.

Eratosthenes would have measured the distance either by a land route or by travel on the Nile. Because the Nile at one point does a north/south loop the distance is 745 miles. So we'll go by land. That distance is 672 miles. So the miles that Eratosthenes really calculated is:

c = 50 × 672 = 33,600 miles

So he was off by more than we were. Still good old Beta did pretty good, even if he wasn't perfect.

References

"Eratosthenes of Cyrene", The MacTutor History of Mathematics Archive, http://www-history.mcs.st-andrews.ac.uk/Biographies/Eratosthenes.html. One of the longest lived and most accurate websites. No unnecessary flashing animations, no scripts that would glacially run even on a supercomputer, just a nice readable, useful, and comprehensive website. From the University of St. Andrews in Scotland

The Eratoshthenes Project, "http://www.physics2005.org/projects/eratosthenes/TeachersGuide.pdf". Although the math in this article was worked out by the author of CooperToons ipse, he was not so naive as to suppose similar derivations for the generalization of Eratosthenes' method had not been undertaken by others. Here is one fairly recent derivation.

Sun or Moon Altitude / Azimuth Table, http://aa.usno.navy.mil/data/docs/AltAz.php/

Measure for Measure, R. A. Young and T. J. Glover, Blue Willow Press, 1996.

"Archeological Site of Cyrene", UNESCO, http://whc.unesco.org/en/list/190