Measure Earth yourself

Eratosthenes' experiment

Over 2,200 years ago Eratosthenes measured Earth's circumference using only a stick, a shadow, and geometry. You can repeat his calculation below.

Flat-earth claim

We only know Earth's shape and size from space photographs, which cannot be trusted.

What the measurement shows

Earth's circumference was computed long before spaceflight. Two shadows measured at the same moment in different places are enough to get about 40,000 km from a simple proportion — with no photograph at all.

Eratosthenes knew that at Syene (today's Aswan) at solstice noon the Sun lit the bottom of a well — it stood at the zenith. At the same moment in Alexandria a vertical gnomon cast a shadow tilted by about 7.2°. Since that is 1/50 of a full circle, Earth's circumference must be 50 times the distance between the cities.

The key assumption is one the flat model fails: the Sun's rays arrive parallel, and the angle difference comes from the curvature of the surface. On a flat Earth with a nearby Sun the shadows would behave quite differently — and would not give a consistent circumference from many city pairs.

Enter the distance between two places on the same meridian and the measured shadow-angle difference. The calculator finds the planet's circumference and radius.

Calculator

Your shadow measurement

Ready-made examples
no shadow7.2°
Measurement result
Calculated Earth circumference40,000 km
Resulting radius
6,366 km
Difference from the true value
0.2%
Two shadows gave a circumference within 0.2% of the real one (40,075 km) — with no space photograph at all.

The method assumes parallel sunrays and places on a similar meridian. The reference value is the equatorial circumference of 40,075 km.

What this shows
  • Two shadows and one proportion give a circumference near 40,000 km — with no space technology.
  • The result assumes parallel sunrays and a curved surface, which the flat model cannot reproduce.
  • Adding a third city tests the model: on a sphere every pair yields a consistent circumference.

Eratosthenes' experiment — questions

He compared the gnomon shadow angle in Alexandria with the absence of a shadow in Syene at the same noon. The ~7.2° difference is 1/50 of a circle, so he multiplied the distance between the cities by 50, getting a circumference of about 40,000 km.

Yes. You need a vertical stick, a shadow-length measurement at local solar noon, and a second person in another city on a similar meridian measuring at the same moment. From the two angles and the distance you compute the circumference with the same formula.

The method assumes the Sun's rays are nearly parallel (the Sun is very far away) and that the angle difference is due to surface curvature. On a flat model with a nearby Sun, different city pairs would give contradictory “circumferences”, whereas here they all agree on a single sphere.

Depending on the assumed length of the stadion, his result was within a few to a dozen percent of today's value (about 40,075 km at the equator). For a stick measurement more than two millennia ago, that agreement is remarkable.