An experiment to test the hypothesis that the Earth is flat.

(Spoiler alert: it’s not flat.)

Unless otherwise noted the artwork and photographs in this slide show are original and © by Burt Carter. Permission is granted to use them for non-commercial, non-profit educational purposes provided that credit is given for their origin. Permission is not granted for any commercial or for-profit use, including use at for-profit educational facilities. Other copyrighted material is used under the fair use clause of the copyright law of the United States.

We are going to do a thought experiment, something that has served a lot of famous scientists well. Ours has to be a thought experiment because we can’t afford the time or cost to do it in fact. There are too many of us and it would require at least two or three days to accomplish, never mind the logistics and the expense. Even though this is just a thought experiment, I am keeping it accurate. When I say that the results at some place would look like so-and-so, they would look like that. If you wish to check my assertions, on your own time and at your own expense, be my guest! That is the nature of scientific skepticism – the reason that we demand that our experiments and those of others produce reproducible results – so they can be checked. We take a standard scientific approach to this problem: propose a hypothesis and test it. We select the hypothesis: “THE EARTH IS FLAT” and seek to disprove it. One point of the exercise is to see how this hypothetico-deductive process works. Proof that Earth is not flat is the other. Notice that we are not attempting to figure out what the shape is, only what it isn’t. We will address the actual shape at the end of the process, but will not prove it.

~Lo

ngi

tud

e 9

2°W

Latitude 0° (equator) North peak, Albemarle I.,

Galapagos Is., Ecuador

Latitude 30° N New Iberia, LA

Latitude 45° N Chippewa Falls, WI

(equator)

We will be comparing observations

at 3 locations along the 92nd

meridian west. These places are

pretty close to directly north/south

of each other, so the distances

between are easy to estimate or

measure and our observations can

all be made at essentially the same

time.

Map from Google Maps

The class will divide into thirds. I will accompany one group to their station and a research assistant will

accompany each of the other groups to theirs.

The sun is somewhere that-a-way

(locally) level ground

Stick, 1m long

The setup at each location is very simple. We drive a stake into level ground, taking care that it is exactly plumb (perfectly vertical). We also take care that exactly 1m of the stick projects above the ground. We also make sure the ground is perfectly level around the stick. The sun is part of our experiment, so it is relevant to note that it is above us, but not necessarily directly overhead! In Americus, for example, the sun is never directly overhead, but off to the south. So it is in all the northern hemisphere. We do the experiment at local high noon on the fall or the spring equinox. This is when the sun has reached zenith (local noon) and when it passes directly overhead at the equator (the equinox). We do this at exactly this time so that the setup at Albemarle will have a peculiar geometry, which will become obvious pretty soon.

N

Albemarle (I’ve never been, so this is where my group goes.)

Path

s o

f so

lar

rays

Because Albemarle is on the equator, and because we are observing on the equinox (currently about September or March 21) at local noon, the sun will be directly overhead here. This means that at exactly noon the stick will cast no visible shadow. More precisely, the shadow will be directly at the base of the stick, and will be exactly the same size as the stick’s diameter. Both the stick and the path of incoming solar rays are perpendicular to the ground.

Map view N

Stick N

Stic

k, 1

m

lon

g

The sun is exactly that-a-way

Chippewa Falls

Map view N

Stick Shadow

The group at Chippewa Falls, with the exact same setup and at the exact same time, will see something completely different. The shadow they see on the ground indicates that the sun is NOT directly overhead, but is some distance to their south. The solar rays are coming in at an angle A with respect to the ground. This is called the “angle of incidence”. The rays, the shadow, and the stick form a right triangle. B

A

The sun is exactly

that-a-way

Let’s take a minute to see what the setup is intended to model. Notice that there are two similar right triangles. Our model seeks to scale the characteristics of the small one we’ve made to the large one defined by the geometry of the Sun/Earth system, assuming a flat Earth. 1) The flat ground where we’ve driven the stick is intended to model the (hypothetically)

flat Earth. The length of the shadow cast by the stick will model the distance from Albemarle to Chippewa Falls. (The bottom legs of the triangles).

2) We have driven the stick vertically to model the fact that the sun is directly overhead at Albemarle. The stick’s length (1m) therefore becomes a model for the elevation of the Sun above Albemarle. (The right-hand legs of the triangles.)

3) The angles, particularly the angle of incidence of the solar rays (angle A) are the same in both triangles because the triangles are similar.

Make sure you understand this model before proceeding!

Ray

pat

h t

o A

lbem

arle

an

d d

ista

nce

to

Su

n

Stic

k 1

m h

igh

Shadow 1m long

Millions of meters

NOT TO SCALE!!!

B

A

*

Let’s take a minute to see what the setup is intended to model. Notice that there are two similar right triangles. Our model seeks to scale the characteristics of the small one we’ve made to the large one defined by the geometry of the Sun/Earth system, assuming a flat Earth. 1) The flat ground where we’ve driven the stick is intended to model the (hypothetically)

flat Earth. The length of the shadow cast by the stick will model the distance from Albemarle to Chippewa Falls. (The bottom legs of the triangles).

2) We have driven the stick vertically to model the fact that the sun is directly overhead at Albemarle. The stick’s length (1m) therefore becomes a model for the elevation of the Sun above Albemarle. (The right-hand legs of the triangles.)

3) The angles, particularly the angle of incidence of the solar rays (angle A) are the same in both triangles because the triangles are similar.

Make sure you understand this model before proceeding!

Ray

pat

h t

o A

lbem

arle

an

d d

ista

nce

to

Su

n

Stic

k 1

m h

igh

Shadow 1m long

Millions of meters

NOT TO SCALE!!!

B

A

*

Let’s take a minute to see what the setup is intended to model. Notice that there are two similar right triangles. Our model seeks to scale the characteristics of the small one we’ve made to the large one defined by the geometry of the Sun/Earth system, assuming a flat Earth. 1) The flat ground where we’ve driven the stick is intended to model the (hypothetically)

flat Earth. The length of the shadow cast by the stick will model the distance from Albemarle to Chippewa Falls. (The bottom legs of the triangles).

2) We have driven the stick vertically to model the fact that the sun is directly overhead at Albemarle. The stick’s length (1m) therefore becomes a model for the elevation of the Sun above Albemarle. (The right-hand legs of the triangles.)

3) The angles, particularly the angle of incidence of the solar rays (angle A) are the same in both triangles because the triangles are similar.

Make sure you understand this model before proceeding!

Ray

pat

h t

o A

lbem

arle

an

d d

ista

nce

to

Su

n

Stic

k 1

m h

igh

Shadow 1m long

Millions of meters

NOT TO SCALE!!!

B

A

*

Chippewa Falls

The small triangle that we can see and measure at Chippewa Falls is a model for the large triangle formed by the Sun, Albemarle, and Chippewa Falls. Notice that the stick and its shadow are the two legs of a right triangle, and the ray path is the hypotenuse. This group has made sure the stick is 1m long. It is a simple matter for them to measure the shadow, but they must measure very, very precisely! At Chippewa Falls the shadow would be exactly 1.00m long, the same length as the stick.

Shadow 1m long

Stic

k 1

m h

igh

B

A

Shadow 1m long

Stic

k 1

m h

igh

Chippewa Falls

Now remember basic trig: Tan(A) = length of opposite leg/length of adjacent leg = stick length/shadow length.

In this particular case the stick and the shadow are both 1m, so tan(A) = 1m/1m In other words, the tangent of A is 1. We can use the arctan (or tan-1) function of a calculator, or check a table in a book, to see which angle corresponds to a tangent of 1.

B

A

The page at left is copied from the CRC

Handbook of Standard Mathematical Tables

and shows a table of trig functions like you’d

find in any trig book. Notice first that the

numbers in the “Tan” column (or any of the

others) are all different. Each Tan value

corresponds to exactly one angle.

With Tan = 1, the corresponding angle is 45°,

which is the value for angle A in the diagram.

Ray

pat

h t

o A

lbem

arle

Remember that we are hypothesizing a flat Earth, so the drawing shows the hypothetically flat ground between Chippewa Falls and Albemarle. What we are about to do is true if and only if this hypothesis is correct! If the Earth is not flat, what we are about to do, and to conclude from it, is utter nonsense. Just be forewarned.

Stic

k 1

m h

igh

B

A

Ray

pat

h t

o A

lbem

arle

The ground distance from Albemarle to Chippewa Falls can be carefully measured from a map and found to be ~5000km. Remember tan(A) = opposite leg/adjacent leg, regardless of the scale of the triangle.

In this case (since we already know that A = 45°) [tan(45)=X/5000km], where X is the elevation of the sun above Albemarle.

~5000.0001km

Remember that tan(45) = 1

We can rewrite the equation as: 1=(X/5000km)

Rearranging we find X = ~5000km.

X (= ~5000km)

(The 1m shadow is insignificant in comparison to the 5000km [5,000,000m] between the locations, so we can safely ignore it

in our calculations).

B

B

A

WE NEED TO STOP AGAIN AND BE SURE WE UNDERSTAND WHAT WE HAVE JUST DONE

We have used our flat-Earth hypothesis to calculate a specific value for a specific physical distance -- the sun sits 5000km above the Earth’s surface! Our math is perfectly correct. If our assumptions are all correct then the sun has just been proven to be 5000km up. Forget that you “know” that this value is wrong (it’s really ~150 million km), we are interested in testing our hypothesis purely by its own internal logic. (And, after all, how do you “know” how far it is to the sun?) If our flat-Earth hypothesis fails, there will be no remaining reason to take the 5000km height of the sun seriously.

We can now treat this proven value as a new hypothesis or sub-hypothesis,

and propose a test for it.

If the sun sits 5000km above Albemarle when we view it from Chippewa falls,

it should also sit 5000km above Albemarle if we view it from anywhere else.

If that does not happen, what do we conclude?

New Iberia

Shadow 0.5774m long

Stic

k 1

m h

igh

When we set up our experiment in Louisiana we see something a little different from what we saw in Wisconsin. Careful measurement of the shadow reveals it to be 57.74cm (which we have to convert to 0.5774m), just a little over half as long as it was farther north. Because this shadow leg is shorter than at Chippewa Falls, the angle of solar incidence must be different. That is, angles A and B here are not the same size as in Wisconsin. We get the size of the angle A as we always have: tan(A) = 1m/0.5774m = 1.7319 and arctan (1.7319) = 60°. (Angle A is 60°)

B

A

The sun is exactly

that-a-way

New Iberia

Stic

k

Path

of

sola

r ra

ys

New Iberia Albemarle

Again we have two similar triangles – the small (stick/shadow/ray-path) one and the (Sun/Albemarle/New Iberia) one. The measured ground distance between Albemarle and New Iberia is ~3335km. Solving the tangent function again gives us a value for X, just as it did for the Chippewa Falls triangle. Tan(A) = (solar elevation/ground distance) or tan(60) = (X/3335km) or 1.7319 = (X/3335km) Rearranging, we get: X = (3335km*1.7319) = ~5775km

X

B

3335km

B

A

Both New Iberia and Chippewa Falls ?

5000km 3335km

Height of sun is not predictable from observations at Albemarle I. Height of sun is 5775km according to observations at New Iberia. Height of sun is 5000km according to observations at Chippewa Falls.

When we put all the places on one diagram, with their distances drawn to scale, we see clearly that there’s a problem. Our math indicates that the sun must be 5000km above the ground if we take one triangle seriously and 5775km if we take the other seriously. This is a 15.5% discrepancy, and this is substantial. It seems that two of our groups will be arguing over who has the height correct, and we need some clear thinking to help settle the matter.

It’s time to slow down again and figure out what we’ve done. The triangular relationships at Chippewa Falls proved that if the Earth is flat, the sun is ~5000km directly above Albemarle at noon on the equinox. However, the triangular relationships at New Iberia now prove that if the Earth is flat, the sun is ~5775km directly above Albemarle at noon on the equinox. There are 3 possible explanations, which we will examine individually:

1) There is some problem with our measurements or our mathematics.

2) How high the sun is depends on where you happen to be when you observe it.

3) There is some problem with our assumptions.

1) There is some problem with our measurements or our mathematics.

This is simply not the case. If you are appropriately skeptical, the measurements would be the first place to check. Get a map, measure the distances, use the map scale to convert them to kilometers, and you will find the “we” have “measured” them correctly. We are using the correct ground distance in our calculations to within about 1km, actually. (I didn’t get these off a map, I got them in a much more precise way). Our solutions are also mathematically correct. Take them to a mathematician if you want to check. The mathematics of triangles is so straightforward that any trig solution like this is as good as a direct observation. Any scientist would take our mathematical solution here as being as factually true as if we had directly observed the entire system at once, say by standing on Saturn. Mathematical proofs establish factual truth, so within the model we have constructed (with it’s included assumptions) the sun is both 5000 and 5775km above the ground at the same time!

2) How high the sun is depends on where you happen to be.

Neither group wins the argument about who’s right. The mathematical solutions of both groups are correct. Within our model, the sun is both 5000 and 5775km above the ground, at the same time. Anyone who needs an explanation for why the sun can’t be in two places at once, go and ask a first grader.

3) There is a problem with our assumptions.

As Sherlock Holmes tells Watson, probably at least once in every story and novel, “when you have eliminated all the possible explanations but one, that one, no matter how bizarre, must be correct”. In most cases scientists can’t be sure they have “all the possible explanations” in front of them, but in this case we do. Either the conclusion is true, or it’s not. If it’s not, either it is a problem in our solution (our math) or a problem in our model. The conclusion is not true and there’s no problem with our math, therefore there must be some problem with our model. Our model only has two assumptions in it: a flat earth, and a small, nearby sun, and the second is essentially irrelevant to our problem. That is, we will encounter the same logical conflict if the sun is close and nearby as we do if it is distant and large. What is responsible for our absurd conflicting conclusions?

Instead of a assuming that the sun’s rays approach a flat surface at different angles, let’s play with the idea that the Earth’s surface is tilted by differing amounts in different places. In this case we will assume a huge, distant sun because this allows the rays to approach Earth in parallel paths. In this model, the paths don’t change orientation from one part of the Earth to another, the ground does!

Ray path

Albemarle Is. – Ground not tilted.

(no shadow)

(A=60°)

New Iberia, LA - grounds tilts 30°. (.5775m shadow)

Ray path

Ray path

Chippewa Falls, WI - ground tilts 45°

(1m shadow)

(Ground tilts=45°)

The next slide simply arranges the three diagrams in a way that should make sense.

(A=90°)

(Ground tilts 30°)

(A=45°) (Ground tilts 0°)

Ray path

Albemarle Is. (no shadow) Ray

path

Chippewa Falls, WI (1m shadow)

Center of Earth

Equ

ato

r (0

°)

30°

45°

New Iberia, LA (.5775m shadow)

Ray path

30°

45°

0°

B

The larger in each pair of the similar “triangles” we’ve been solving aren’t really triangles. That’s why our distance calculations are screwy. A triangle is constructed by drawing straight lines between three points. A straight line distance between Albemarle and Chippewa Falls would go through the curved Earth. If we did this for these three points, the triangle would not be a right triangle and so the tangent function would not work anyway. It only works for right triangles. The small triangles were okay, the flaw in our model was to assume we could scale up because of a flat Earth.

Chippewa Falls

Albemarle Island

A

Be sure you understand that we cannot conclude from our experiment that

the Earth is spherical, we only have enough information to conclude that it

is not flat.

Another way of saying this is that the alternative hypothesis to the flat Earth

hypothesis is a not-flat Earth hypothesis.

Similar experiments to this one, arrayed across both longitude and latitude

in various ways would consistently be in agreement with a spherical Earth,

as the next slide suggests. So even though we haven’t proven it with our

experiment we could easily do so. Therefore we’ll be treating it as spherical

from now on.

Dawn Dusk

Noon Mid-afternoon Mid-morning

At different times of day we’d see different shadow lengths at different places, and could solve them as we have done for the shadows at noon. We’ve just

done the “noon” version of the experiment, and you are certainly familiar with the dawn/dusk version of the experiment if you’ve ever watched your own

shadow or the shadow of a stationary object through the day. The shady side of your house in the morning is sunny side in the afternoon and vice-versa.

Our conclusion would always be the

same: the best shape to explain the

shadows would be “circular”, just as

we saw a few slides back.

If a thing is circular in every direction

that’s the same thing as it being

spherical.

The spherical Earth model solves all our problems. At any point on Earth the trig functions give sensible patterns because the sun’s rays come in from the same direction everywhere and the curvature of the Earth ensures that the ground, and therefore the stick, orientations are correct. The shadow length depends on latitude, or distance in terms of degrees of arc along the surface rather than the solar position. No matter what the distance to the sun, as long as it is big enough that the rays strike Earth along parallel paths, the triangle solutions are sensible. A flat Earth is geometrically absurd. Therefore we reject the flat-Earth hypothesis because of our tests, and conclude that the EARTH IS NOT FLAT. Acceptance the spherical Earth is really a separate issue. However, other experiments similar to ours would easily suggest that no other shape would give consistently sensible trigonometric results.

Of course, there are other ways of knowing the Earth is spherical, as Neil

Armstrong can tell you.

Photographs of the Whole Earth from NASA

RETURN