Claim status
Curved-surface geometry matches the measurement
In the record
Eratosthenes, third century BCE
Testable
Meridian distance and solar zenith-angle difference
Method
Similar angles and a full-circle ratio
FLAT EARTH / CASE FILE

Eratosthenes experiment: 40,000 km around Earth

An 800 km meridian separation and a 7.2° difference in solar zenith angle imply a circumference of 40,000 km.

Two vertical sticks need not cast matching shadows. Measure their separation and the difference in their noon angles, and the mismatch becomes a ruler for a much larger circle.

Run the calculation
An engraved circular map with red horizon measurement linesA claim about the world should fit the whole world.
01 / THE CLAIM

One pair of shadows, two possible stories

Different shadow angles can be described by a curved surface under nearly parallel sunlight, or by a nearby light over a plane. A pair of sticks alone does not settle every model: the distance to the light and additional locations matter. The calculation here asks what circumference follows from the parallel-ray spherical model.

02 / THE CASE

The separation must match the measured angle

Eratosthenes used the difference in the Sun’s direction between two locations to estimate Earth’s circumference. The geometric idea survives without choosing a disputed modern length for the ancient stadion.

For a repeatable experiment, choose locations near one meridian, make the sticks vertical and measure at local solar noon on the same date. Civil noon is not necessarily solar noon. Use the north–south surface distance, not a road journey or a diagonal distance between arbitrary cities.

The inputs below are an illustrative 800 km and 7.2°, not a claimed reconstruction of the original survey. The angle is the difference between the two solar zenith angles; observations on opposite sides of the subsolar point require the signed angles rather than subtracting two unsigned lengths.

03 / THE COMPUTATION

Scale the measured arc to a complete circle

If an arc of length s subtends angle θ, it occupies θ/360 of the complete circumference. Therefore C = 360s/θ. Holding the angle fixed, doubling the measured distance doubles C; holding distance fixed, doubling the angle halves it.

The relative angle-error control applies an interval around the entered angle while holding distance fixed. A larger angle gives the lower circumference and a smaller angle gives the upper one. These are sensitivity bounds, not a statistical confidence interval and not a correction for an incorrectly measured baseline.

RUN THE NUMBERS

Turn two shadow angles into a circumference

Enter the north–south baseline and the positive difference in solar zenith angles. The rays are parallel and the surface is spherical.

Calculation inputs
km

A longer measured arc gives a proportionally larger circumference at the same angle.

°

Use the signed-angle difference as a positive magnitude. Smaller differences produce larger circumferences.

%

An illustrative interval around the angle; 2% of 7.2° is 0.144°. Distance is held fixed.

Inferred circumference40,000km

The circumference implied by the entered arc and angle.

Lower sensitivity bound39,216 km

The larger angle produces the smaller circumference.

Upper sensitivity bound40,816 km

The smaller angle produces the larger circumference.

Working tape
  1. Baseline × full-circle angle800 × 360 = 288,000
  2. Scale the observed arc to 360°288,000 ÷ 7.2 = 40,000
  3. Convert angle error to a fraction2 ÷ 100 = 0.02
  4. Larger-angle factor1 + 0.02 = 1.02
  5. Smaller-angle factor1 − 0.02 = 0.98
  6. Circumference at the larger angle40,000 ÷ 1.02 = 39,215.686275
  7. Circumference at the smaller angle40,000 ÷ 0.98 = 40,816.326531
04 / THE FINDING

A short baseline reaches around the planet

The opening 7.2° angle is one fiftieth of a circle. Fifty copies of an 800 km arc therefore make 40,000 km. A 2% angular interval changes that result asymmetrically: division by a smaller angle increases the answer more than division by the equally larger angle reduces it.

Repeat the measurement with a different baseline and compare the inferred circumference. Agreement across more than two sites tests a model more strongly than choosing a lamp height that fits only one pair of shadows.

All Flat Earth
Tangent geometry on a sphere, without atmospheric refraction

Horizon distance

The water horizon sits 5,048 m away and an elevated target adds 11,288 m, giving 16.34 km of shared sightline. Set both heights and watch it move.

Parallel-arc length compared with polar projection radius

Flat Earth map distortion

A southern parallel of 7,076 km is drawn 23,580 km long on a north-polar disk, a 3.33× stretch. Move the latitude and watch the error grow with it.

Orthographic projection of an ideal circular disk

Flat disk eclipse shadow

A tilted disk throws an oval shadow 100 cm by 50 cm, an axis ratio of 0.5. A sphere casts a circle at every angle. Set the tilt and compare the edges.

Local vertical component of Earth’s angular velocity

Foucault pendulum

At 30° north the ideal swing plane turns 7.52 °/h clockwise. Latitude and elapsed hours control the drift; the equator gives no rotational precession.