Eratosthenes and the Size of the Earth
In which inhabitants of Quaint Towns of the American Midwest can take a brief vacation and determine the size of the Earth using simple household items and a little bit of middle school arithematic without advance mathematics like trigonometry
(For a related method, click here.)
Eratosthenes measured it.
Eratosthenes: Jack of All Philosophers
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 ... well, you're the Jack of All Trades.
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 practical way to determine which numbers are primes. 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.
But north of Aswan at the Mediterranean coastal city of Alexandria, obelisks did cast shadows on that day, and the Sun was never directly overhead on any day. Eratosthenes realized the reason the shadows were of different lengths was due to the curvature of the Earth. Eratosthenes figured that from the difference in shadow lengths he now calculate the size of the Earth. Of course, this means that Eratosthenes knew that the Earth was round - which the Greeks knew anyway.
Discours de la Méthode (of Eratosthenes, that is)
Now modern scholars pointed out that Eratosthenes probably did not measure the Earth as accurately sometimes stated. But what he and the other Greek mathematicians were interested in was getting the method down and were not worrying about getting the numbers right on (to quote Shakespeare). And Eratosthenes did get the method right.
The method Eratosthenes used to calculate the size of the Earth was based on a simple rule he learned from Euclid. The rule was that if you have a straight line that intersects two parallel lines, then the alternate angles are the same. And he figured that two lines that are parallel are the rays of the Sun.
Alternate (and Equal) Angles
Yes, yes, it is true that the Sun's rays when they strike the Earth are not exactly parallel. But because the Sun is so much larger than the Earth, the rays were parallel enough for Eratosthenes.
Parallel Enough
Now if you want to calculate the actual numbers, you need to know not only Eratosthenes's method but you have to also have some empirical data. First you have to know the height of the obelisk at Alexandria and the length of its shadow at noon on June 21. That's when the noonday Sun is overhead in Aswan.
The other bit of data needed is the north-south distance from Aswan to Alexandria. Today we know that's about 490 miles.
Eratosthenes, of course, used Greek measurements, and they didn't have a mile. Instead used a unit of measurement called the stadion (στάδιον). This was the length of the stadium at Olympia where they had the Olympic games. The Greek historian Herodotus said a stadion was 600 feet but he was using Greek feet which modern scholars say was a bit longer than a modern foot. But we'll stick with a stadion being 600 feet.
So if Eratosthenes says the distance from Aswan to Alexandria was 5000 stadia that's 3,000,000 million feet or 568 miles. But since Eratosthenes was probably talking about the distance along the Nile - which wiggles back and forth a bit - we'll go with 500 miles.
OK. 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
Here r is the radius of the Earth (line segment AB), h is the height of the obelisk in Alexandria (AD), s is the shadow length of the obelisk (AE), and d is the distance from Aswan to Alexandria (AC).
First note that we represent the Suns rays with the two vertical (and hence parallel) lines, CB and DE. So the line defined by the top of the obelisk to the center of the Earth, DB, forms the two acute alternate interior angles with the Sun's rays, ∠ ABC and ∠ ADE.
As Euclid told us, these angles are equal.
∠ ABC = ∠ ADE
So all you need to do is to calculate the size of the ∠ ADE, which is formed by the base of the obelisk (A) to the top (D) and then to the end of the shadow on the ground (E). This angle will be the same as the angle formed by the well to the center of the Earth to the obelisk, ∠ ABC.
Once the size of ∠ ABC is in hand, divide that number into 360. Multiply that number by the distance from the well to the obelisk (i. e., from Aswan to Alexandria) and, hey, presto!, you'll have the circumference, c of the Earth.
So it all boils down to the trick of accurately measuring the angle, ∠ ABC, something that was tough to do in the olden days, and is not all that easy even now.
We don't have Eratosthenes' original paper (or rather, his papyrus) so we don't know the exact numbers he had. But suppose he used a 68 foot obelisk (like Cleopatra's Needle) and found that the Sun cast a shadow that was 8 feet 7 inches long.
As for measuring the height of the obelisk there are various ways to do that. One way, although not the most practical, 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. Or what is easier, you can measure your height to shadow ratio at any time, and multiply that ratio by the length of the obelisk's shadow.
At this point we need to do some fairly simple math. Although we said we can do the arithmetic without trigonometry for now we'll go ahead and use trigonometry just to get the basics.
If draw the obelisk set up a little more to scale, you'll see that the obelisk, the shadow, and the Earth form a right triangle (the curve of the Earth is way too slight to affect the calculation). That is ∠DAE is 90 degrees.
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 .
This means if the shadow is 8 feet 7 inches (8.58 feet) and the obelisk is 68 feet, the tangent of the angle ∠ ADE = 8.58/68 = 0.1262. If you look up the angle for that tangent - that is the arctangent of 0.1262 - you find the angle is 7.2 degrees.
| tan(∠ ADE) | = | s/h |
| = | 8.58/68 | |
| = | 0.1262 |
... and ...
atan(0.1262) = 7.2 degrees
But since ∠ ADE and ∠ ABC are alternate angles from of the parallel Sun's rays. So we know that
∠ ADE = ∠ ABC = 7.2 degrees
Now degrees are not absolute numbers but are relative values of how much of a 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:
| % Circumference | = |
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 | = | 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.
Εὕρηκα!
(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.)
Remember, though, that Eratosthenes was interested in getting the method right. Today that means we'd calculate the general formula for the circumference. Fortunately, that's pretty easy to do. You just run through the calculation using the letters and don't calculate the actual numbers.
Remember, the angle defined by the obelisk and its shadow is the same as the angle formed by the two cities and the center of the Earth. This angle divided by 360 fraction of the circumference spanned by the distance from Aswan to Alexandria.
And since the angle is the same as the arctangent the heights of the shadow and the obelisk height, you get:
atan(s/h)
/
360 |
= |
d
/
c |
Elementary algebra then lets us solve for c and hey, presto!, we have the formula.
| c | = |
360 × d
/
atan(s/h) |
For this formula atan(s/h) must be in degrees. But usually it's better if you use trigonometric functions to put the angles in radians. A radian is defined by:
2π Radians = 360 Degrees
... and in radians the formula is:
| c | = |
2π × d
/
atan(s/h) |
And as we'll see if you want to avoid using trigonometric functions - which is what we said we'd do - using radians makes things easier.
Eratosthenes, Yes! Trigonometry, No!
To avoid trigonometry look again at the obelisk in Alexandria and the well in Aswan. The general layout of the calculations is:
And a relook at the schematic:
The Schematic
... and at the formula using trigonometry:
| c | = |
2π × d
/
atan(s/h) |
Now remember that s/h is the tangent of the right angle where s and h are the opposite and adjacent sides of a non-right angle, α. So the arctangent, atan(s/h), is simply the angle α.
The equation then becomes:
| c | = |
2π × d
/
α |
Now there is a commonly used trick in trigonometry called the small angle approximation. In the small angle approximation, is that if the angle, α, is - quote - "sufficiently small" - unquote - we can write:
| sin(α) | = | α |
| cos(α) | = | 1 - α2/2 |
| tan(α) | = | α |
So we can replace the value of the angle α with the tangent which is s/h as long as the angle is small enough.
| c | = |
2π × d
/
s/h |
... which can be simplified a bit as:
| c | = |
2πhd
/
s |
As far as how small is small enough, you can just calculate some tangent values:
| s | h | [tan(α)] | Error | |
| 0.0 | 68 | 0.0000 | 0.0000 | 0.00% |
| 5 | 68 | 0.0735 | 0.0734 | 0.14% |
| 10 | 68 | 0.147 | 0.146 | 0.75% |
| 15 | 68 | 0.221 | 0.217 | 1.61% |
| 20 | 68 | 0.294 | 0.286 | 2.80% |
| 25 | 68 | 0.368 | 0.352 | 4.34% |
So as long as the angle is under around 20 degrees or so we should be all right.
The Errors Don't Matter (Much)
There are lots of approximations in Eratosthenes's measurement. The distance from Aswan and Alexandria is not a straight line. And the two cities are not in a direct north-south line.
Also the Sun does not shine straight down in Aswan on any day. That's because it's 45 miles north of the Tropic of Cancer. It's only in the tropics that there are days where the noon-day Sun will be directly overhead.
But the reasons the calculations still work is that none of the errors are large enough to really matter. Even if the Sun doesn't shine exactly straight down in Aswan like Eratosthenes said, it's a pretty good approximation. You can always do the wrong calculations if the errors are small enough, or (even better) if you have "compensating errors. That way if you don't get the true answer, the answer will be true enough.
What You CAN Do At Home
Trying it at home
Well, this is all and well if you live in Alexandria or Aswan, Egypt. But suppose you live in the town of, say, Gene Autry, Oklahoma? Are you forever forbidden to measure the circumference of the Earth?
Well, have no fear! It is possible for the good citizens of Gene Autry to calculate the circumference every bit as good as Eratosthenes. The key - as shown below - is all you need is the two towns separated a reasonable distance from north to south and a few household items.
As to how far they need to be separated there's no fixed answer. But the difference of the shadow lengths in the two towns 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, by taking a look at the map. There we see that Wichita, Kansas, is nearly due north of Gene Autry but also a goodly ways away.
From Gene Autry to Wichita
So we can create the schematic.
First we'll do the most accurate calculation using, yes, trigonometry.
You'll remember that we showed that the formula from Eratosthenes's experiment using the small angle approximation is:
| c | = |
2πhd
/
s
|
... where d is the distance from Aswan to Alexandria, h is the height of the obelisk, and s is the length of the shadow.
Now this equation works for any two towns where one is at a spot where on the day of the measurement the sun shines down directly overhead - it does not have to be Aswan. So we can solve the equation for the distance, d, which is the north-south distance between the towns.
| d | = |
cs
/
2πh |
Now let's calculate the distance of two different towns that are a given distance from the town where the Sun is overhead.
| d1 | = |
cs1
/
2πh |
and
| d2 | = |
cs2
/
2πh |
... where s1 and s2 are the shadow lengths. Note that we only need one h value since we make sure that whatever casts the shadows in the two towns are the same height.
Now take the difference of the two distances, d1 and d2:
| d2 - d1 | = |
cs2
/
2πh
-
cs1
/
2πh |
This can be rearranged as:
| d2 - d1 | = |
c × (s2 - s1)
/
2πh
|
And solving for c:
| c | = |
2πh(d2 - d1)
/
(s2 - s1)
|
So you can see all we need to know the length of the two shadows (s1 and s2), the height of the dowel, h, and the north-south distance between the two towns, which is d2 - d1. It doesn't matter where the towns are as long as they're in a north-south line.
Of course, the bigger the distance between the towns the better because the longer the shadow the more accurate you can measure it.
So let's mosey on down the Santa Fe Trail - or as some call it US Interstate 35 - to Gene Autry, Oklahoma. Actually you can either mosey down from Oklahoma City or mosey up from Dallas.
But what you should do first is find a hardware store and buy a 60" (ergo, 5 foot) dowel rod. While you're at the store, you might want to get a carpenter's square and make a plumb bob to help you accurately set the dowel up perpendicular.
For those not familiar with Gene Autry, you drive along I-35 and then when you reach Exit 40 you'll see the sign directing you to Oklahoma Highway 53 East toward Springer and yes, Gene Autry. If you take the exit and turn east, you can drive about 8 miles until you see the entrance to the Ardmore Municipal Airport dead ahead. Turn right on - yes - Gene Autry Boulevard and continue south past the large Dollar Distribution Center. In a little over half a mile you'll see that Gene Autry Boulevard turns left and reverts back to good old OK 53. Rather than follow this left curve, keep going straight on what is Frost Street (OK 53A). Strictly speaking Frost Street will tee in to Melton Avenue although it looks like Frost is simply bending to the right. Shortly after that, Melton itself curves left to become North Main Street. Keep heading south and soon you'll be at the corner of Main and Commerce. You're in Gene Autry.
Now if you find a convenient spot to take a welcome break, you can set up your dowel rod. You want it to be firmly set in and perpendicular to the surface. Of course, the flatter the surface, the better, and you want find a place where you can take the proper amount of time and not annoy any of the local citizens. If you have a 5' dowel, then you should be able to set it up so that at least 4.5' is above the ground. But the point is to have the dowel set up perpendicular and to accurately measure it's height above ground.
Strictly speaking you can pick any day of the year. But we'll pick June 21 for reasons we'll elaborate on later. However, whatever day you pick, the measurement should be made at solar noon. That is when the Sun is at the highest point in the sky.
Of course, you can look up the proper time of solar noon (make sure you correct for Daylight Savings Time if that's needed). And then you measure the height of the shadow.
But you can also determine solar noon by measuring the movement of the solar apex using the shadow itself. What you do is set everything set up a couple of hours before noon and make repeated measurements of the length of the shadow and note the time. Make sure you measure the length to the nearest sixteenth of an inch and the time to the nearest second or so. You keep doing this through noon time. Then you can draw a curve and pick out the time where the shadow length is the smallest.
As to what day is best, remember we want the ratio of the shadow to the dowel height to be small enough for our small angle approximation to work. So the best day is the Summer Solstice or June 21. The days are longer and so the noon shadow is the shortest of the year. So the triangles are long and tall.
At this point if everything is d'accord with proper technique and procedure, you'll find the noon shadow length of a 4½ foot dowel rod in Gene Autry on June 21 is about 10 inches 3/8 inches or 0.8646 feet.
Now, Page 2 (whoever said that).
After you've got your measurements recorded - time and shadow lengths - hop in your car. Take a note of the odometer reading once you turn onto I-35 and head north to Wichita, Kansas. Sure, it's a bit of a drive - 4 odd hours - but you can still manage to get there while it's daylight.
Then the next day - there's plenty of motels in Wichita - make a note of how far you've traveled which should be close to 245 miles. Then set up the dowel rod as you did the day before and with plenty of lead time before noon. Again make sure you note the time of solar noon which for all intents and purposes is the same time as in Gene Autry.
Measure the shadow and lo! You should find the length of the shadow is about 1 foot and 13/4 inches or 1.1458 feet.
Now admittedly the two measurements are a day apart and it's best to have both measurements the same day. You can correct the difference by remeasuring the shadow height the next day in Gene Autry and adding ½ of the difference of the two measurements to the first measurement. But it turns out this is another error that's small enough not to matter and the shadow length will be virtually unchanged. So we'll just stick with the 10 3/8 inches.
So now we can go back to our equation:
| c | = |
2πh(d2 - d1)
/
s2 - s1
|
Now the difference in the distances, d2 - d1, is just the distance from Gene Autry to Wichita. Strictly speaking it's the north-south distance, but we'll just go with the odometer distance and we'll use 245 miles.
Now we can plug in the numbers:
| c | = |
2π × 4.5 × 245
/
1.1458 - 0.8646
| miles |
.. which gives us a circumference of:
| c | = |
2π × 4.5 × 245
/
1.1458 - 0.8646
|
| = |
6927
/
0.2812
| |
| = | 24634 miles |
With the "best" modern value being 24,901 miles, that's is only 1 % off. Not too bad, considering.
Eratosthenes: Pretty Good, But Not That Good
We mentioned it's best to do the calculation at solar noon on June 21. However, if the exigencies of modern life make that inconvenient, you can do the experiment on other days, although you may have to use trigonometry.
For instance, if you happen to be visiting relatives in south central Oklahoma on December 21, you'll find the shadows are lengthier than in the summer. The shadow in Gene Autry of the 4.5 foot dowel rod will be about 7 feet and 10 15/16 inch or 7.916 feet. In Wichita, the shadow will be 9 feet and 13/16 inch or 9.073 feet.
Now if we use the non-trigonometric formula:
| c | = |
2πh(d2 - d1)
/
s2 - s1
|
Plugging in the numbers:
| c | = |
2π × 4.5 × 245
/
9.073 - 7.916
|
| = |
6927
/
1.157
| |
| = |
5987 miles
|
Hm. Not very good. So what the hey is up?
Well, look at the angles. The tangent of the angle in Gene Autry is 7.911/4.5. This is equal to 1.758. But the arctangent - that is the angle - is 1.054 radians. So the small angle approximation is in error by 66%! Adding the the error from the measurement in Wichita, and you get an error of 74%
However, if you go and use the trigonometric formula.
| c | = |
2π(d2 - d1)
/
atan(s2/h) - atan(s1/h)
|
... and the numbers are:
| c | = |
2π × 245
/
atan(9.073/4.5) - atan(7.916/4.5)
|
| = |
1539
/
atan(2.01622) - atan(1.79511)
| |
| = |
1539
/
1.11037 - 1.05388
| |
| = |
1539
/
0.05649
| |
| = |
27084 miles
|
Which again ain't too bad.
And Eratosthenes?
Actually, despite the modern writings, Eratosthenes didn't do any better than we did when we drove from Gene Autry to Wichita.
Note we said that the reference books say a stadion is about about 600 feet. But 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 traveled is 745 miles. So we'll go by land. That distance is about 670 miles. So the miles that Eratosthenes really calculated was:
Circumference = 50 × 670 miles = 33,500 miles
So he was off by more than we were driving from Gene Autry to Wichita. Still good old Beta did pretty good, even if he wasn't perfect.
References
"Measure Earth's Circumference with a Shadow", Ben Finio, Scientific American, September 7, 2017
"Eratosthenes of Cyrene", The MacTutor History of Mathematics Archive, University of St. Andrews, Scotland.
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
"Shadow Length", PlanetCalc.