In 240 BCE a librarian in Alexandria compared two noon shadows and got a size for the planet. Sailors watched hulls disappear before masts. Travelers found that Polaris sank by a degree for every sixty-nine miles they went south. Anyone with a camera can measure the curve of the Earth’s shadow on an eclipsed Moon. Ships came home from the direction they left. Today satellites and ground stations measure their distance apart to the centimeter. Those six methods share no instruments, and in most cases no century. Each one returns a radius. They land within a few percent of each other.
| Eratosthenes, two noon shadows, 240 BCE using the short stadion — see below | 6,267 km | −1.6% |
| Horizon distance from a 2 m eye height | 6,250 km | −1.9% |
| Polaris falling 1° per 111 km | 6,360 km | −0.2% |
| The Earth’s shadow on an eclipsed Moon | 6,427 km | +0.9% |
| Circumnavigation, 40,075 km round | 6,378 km | +0.1% |
| Satellite geodesy (WGS-84) | 6,371 km | — |
Full spread: 6,250 to 6,427 km. That is 2.8% of the accepted value, across methods separated by more than two thousand years. Two of them need nothing but a stick or a protractor.
The Eratosthenes row carries an extra uncertainty the others do not: nobody knows which stadion he used. Worked with the long stadion instead, his answer becomes 7,361 km. That is an ambiguity about an ancient unit rather than about his geometry, and section 01 gives both figures.
At noon on the summer solstice the Sun stood directly over Syene, in southern Egypt, and threw no shadow down a well there. On the same day in Alexandria, about 800 km north, a vertical stick threw a shadow of 7.2°. On a flat Earth under a distant Sun both sticks would agree. They did not.
Eratosthenes took the difference to be a slice of the whole circle. 7.2° is a fiftieth of 360°, so the planet is fifty times the distance between the two cities. That gives a circumference near 39,400 km and a radius of about 6,267 km, low by 1.6%.
Schools repeat the measurement twice a year, at each equinox, through several independent programs on several continents. Around 450 Greek schools took part in one such round in 2017; students in Ghana, Canada, Chile, the United States, China, India, Australia and across Europe have paired up online to exchange measurements and work it out together. A school with no partner at its own longitude is told to use a virtual one on the equator, with a shadow of zero. That is Syene, reconstructed as a default value.
In June 2025 the science YouTuber SciManDan ran a crowd-sourced version on the solstice. Participants measured the shadow of a vertical object at their own local noon and sent in their latitude, the object’s height and the shadow’s length. He received 1,014 separate results from 43 countries. Three points near a common longitude reproduced Eratosthenes’ method to within about 4 percent.
The thousand points together did something he could not. Plotted as solar elevation against latitude, they trace a straight line, which is what a distant Sun over a sphere requires. A nearby Sun over a flat plane predicts a curve that flattens toward the edges and never reaches zero. Those are different shapes, not different numbers, and a thousand measurements decide between them without any argument about ancient units at all.
Stand on a beach with your eyes 2 m above the water and the horizon is about 5 km away. That distance is not set by haze or by the limits of your eye. Climb a 10 m dune and it moves out to about 11 km, and it keeps growing as the square root of your height, which is what a curved surface requires and a flat one does not offer at all.
Run it backwards and the horizon distance gives a radius of roughly 6,250 km. The same geometry says a ship 20 km out has its lowest 17 m already below the line, and that is the part worth watching, because it is visible from any shoreline: the hull goes first, then the deck, then the mast. Perspective shrinks a whole object evenly. It never eats the bottom while leaving the top.
Any argument about the horizon reaches “eight inches per mile squared” within a few minutes. The rule is correct. It is worth knowing exactly what it computes, because that is where it gets misused.
Draw a straight line out from your feet, level where you stand. It touches the sphere at that one point and continues straight; the surface curves away beneath it. The further out you go, the wider the gap between the line and the surface, and the gap grows as the square of the distance.
That is worth unpacking, because it is the part people skip. The gap does not open at a steady rate. It opens faster the further out you go. Right at your feet the surface and the line are heading the same way, so the gap opens at no rate at all. A mile out, the surface has curved and is heading slightly downward relative to the line, so the gap is opening. Two miles out it is heading downward twice as steeply, so the gap opens twice as fast.
The rate of opening grows steadily with distance, and the total gap is that rate averaged over the trip, multiplied by the distance traveled. Both of those grow with distance, so the gap grows with distance times distance. It is the same reason a dropped stone falls four times as far in two seconds as in one: its speed grows steadily with time, so the distance grows with time squared.
The geometry gives:
drop = d² ÷ 2R
drop — how far the surface sits below the straight line
d — how far out along that line you have gone
R — the radius of the Earth, 3,959 miles
At one mile, d² is one square mile. Dividing by 2R, which is 7,918 miles, gives 0.00012629 miles. The answer arrives in miles because square miles divided by miles leaves miles. To express it in inches, multiply by the number of inches in a mile: 0.00012629 × 63,360 = 8.00 inches.
That constant is not arbitrary. It is 1 ÷ 2R converted into inches, so it is fixed by the radius and nothing else. On Mars the same rule would read 15 inches per mile squared; on the Moon, 29. The 8 is the size of the Earth, written in miles and inches.
| Distance | Drop below the tangent line |
|---|---|
| 1 mi | 8 in |
| 2 mi | 32 in |
| 4 mi | 128 in |
| 8 mi | 512 in |
The tangent line touches the sphere at a single point: where you are standing, at ground level. The formula answers one question — how far below that line is the surface, d miles out.
A different question is how much of a distant object you cannot see. That depends on your eye height, because your eye is above the ground. With your eye at height h, your line of sight meets the surface at a distance √(2Rh) away, and that point is your horizon. Nothing is hidden before it. Beyond it, the amount hidden is measured from the horizon point:
hidden = (d − √(2Rh))² ÷ 2R
hidden — how much of the distant object is below your line of sight
h — your eye height above the ground
√(2Rh) — the distance to your own horizon
Take an object 10 miles away, with your eye 6 feet above the ground. First convert the eye height to miles, so it matches d and R: 6 ÷ 5,280 = 0.0011364 miles.
The horizon. 2Rh is 2 × 3,959 × 0.0011364 = 8.9977. Its square root is 3.00 miles. Everything nearer than that is fully visible.
What is hidden. The object lies 7 miles beyond the horizon. Squaring gives 49, and dividing by 2R gives 49 ÷ 7,918 = 0.0061891 miles. Multiplied by 5,280 feet in a mile, that is 32.7 feet.
The tangent drop, for comparison. 8 × 10² = 800 inches, or 66.7 feet.
The two differ because they measure different things. The tangent drop assumes an eye height of zero, and at zero eye height the two agree exactly.
People sometimes say the drop formula is wrong and you should use the haversine formula instead. The two are not alternatives. They answer different questions.
d² ÷ 2R answers: how far below the straight line is the ground? You give it how far out you are looking. It gives you back a drop.
The haversine formula answers: how far apart are two places? You give it the latitude and longitude of each. It gives you back a distance along the surface. It says nothing about curvature drop, horizons, or what you can see from where.
So one feeds the other. The haversine formula works out the distance, and the drop formula uses that distance.
Both of them, on a real case. Willis Tower in Chicago is at 41.8789° N, 87.6359° W. Grand Mere State Park, across the lake on the Michigan shore, is at 42.0286° N, 86.5464° W. Put those four numbers into the haversine formula and it returns 56.9 miles.
Now the second formula, with your eye 6 feet above the ground. Your horizon is 3.00 miles away, so the tower stands 53.9 miles past it. Square that: 2,909. Divide by 2R: 2,909 ÷ 7,918 = 0.367 miles. Multiply by 5,280 feet in a mile and 1,940 feet of the view is below your line of sight. Willis Tower is 1,451 feet tall, so the whole building sits below the line. Climb a 20-foot dune and the hidden height only falls to 1,766 feet, which is still taller than the tower.
Polaris sits almost exactly above the north pole of the Earth’s axis. Measure its angle above your horizon and you have your latitude, to within a degree, with a protractor and a weight on a string. Hang the weight beside the protractor so the string marks straight down, sight along the flat edge until Polaris is centered, and read the angle between the string and your sightline. Entry 175 walks through it as its ninth test, including the objection you will get. Drive 111 km south and that angle drops by exactly one degree. Keep going and it reaches the horizon at the equator and disappears. Keep going further and a second pole rises behind you, turning the other way.
One degree per 111 km, over 360 degrees, gives a circumference near 40,000 km and a radius of 6,360 km. This is the cheapest accurate measurement on the page, and it was the working basis of navigation for centuries before anyone could measure longitude at all.
During a lunar eclipse the Earth passes between the Sun and the Moon and drops its shadow across the lunar surface. The edge of that shadow is always a circular arc. Always, at every eclipse, at every orientation, whatever time of year and whichever part of the Earth happens to be facing the Moon.
A disc casts a circular shadow only when it is face-on. Turn it and the shadow becomes an ellipse, then a line. Only a sphere throws a round shadow from every angle. And the arc can be measured rather than merely observed: the shadow runs about 2.6 Moon-widths across, which after allowing for the Sun’s angular size gives an Earth about 3.7 lunar diameters wide, or a radius near 6,427 km.
Set off in any direction, keep going, and you arrive back where you started after about 40,075 km. That is a radius of 6,378 km, and it has been done by ship, by aircraft and by bicycle.
Getting back to your starting point does not by itself settle the shape. You can walk a circle around the middle of a dinner plate and return to where you began.
What settles it is doing that at different latitudes and comparing how far you traveled. An east–west loop is longest at the equator and shrinks as you move away from it, in both directions, north and south. On a disc with the north pole at the center, southern loops have to get longer, because the circles drawn on that map get bigger the further out you go. A disc has no way to shrink them without inventing distances nobody can fly.
Ground stations time laser pulses bounced off satellites. Radio telescopes on different continents observe the same distant quasar and compare when its signal arrives at each. GPS receivers solve for their own position from the timing of signals from several satellites at once. All three are cross-checked against each other continuously.
The resulting model, WGS-84, gives a mean radius of 6,371 km and an equatorial bulge of about 21 km, and it is the model inside every navigation receiver in the world.
Any single line above can be attacked, and each has been. Refraction complicates the horizon. Eratosthenes’ unit is uncertain. The eclipse measurement is coarse. Circumnavigation alone is ambiguous.
None of that touches the structure of the case. Six methods that share no physics, no instruments and in most cases no century return the same number to within 2.8%. Knock out any one and the radius is still measured five other ways. To break the case you would have to explain why shadows, a horizon, a star’s altitude, an eclipse, a closed loop and a satellite constellation all conspire to return 6,371 km — and the conspiracy would have to have been running since 240 BCE.
That is what convergence means, and it is why science does not need proof. It needs independent measurements that agree, and an explanation for why they should.
The demand that opens most of these conversations is “prove the Earth is a sphere.” It is fair to ask what the alternative has produced.
The flat model has never returned a radius. It has never predicted a horizon distance, a rate for Polaris, the width of the Earth’s shadow on the Moon, or the length of an east–west route in the southern hemisphere. Not a wrong number in any of those cases — no number at all.
That is the difference between a model and a position. A model produces figures you can go and check, and risks being wrong. Everything on this page can be falsified by a measurement, and each section names how.
Three of the six are within reach of anybody, tonight or this weekend.
Polaris. A protractor, a straw and a weight on a string. Measure the angle, compare it to your latitude. Then do it again 100 km north or south.
The vanishing ship. A shoreline, binoculars and a calm day. Watch the order in which a departing vessel disappears.
Eratosthenes. A stick, a friend several hundred kilometers away, and one agreed date. Both of you measure the shadow at local solar noon.
What accuracy to expect from Polaris. A homemade sight reads to about half a degree. Over a 5-degree baseline — roughly 555 miles of north-south travel — that is a 10% error, which puts the radius somewhere near 6,371 ± 600 km. That is poor by survey standards and completely decisive against a flat Earth, which predicts no such relationship at all.
How far apart you and your friend need to be for Eratosthenes. The shadow angles differ by the same number of degrees as your separation in latitude:
| Separation | Difference in shadow angle |
|---|---|
| 200 km | 1.8° |
| 500 km | 4.5° |
| 800 km | 7.2° — Eratosthenes’ own baseline |
Below about 200 km the difference is inside what a stick and a protractor can resolve, and the test stops discriminating.
The main reference has the full workups, the sources, and the comeback you will get for each one, with the answer to it.