Claim status
Global plane not supported
In the record
Geoid and horizon in modern surveying practice
Testable
Horizon distance for measured observer and target heights
Method
Tangent geometry on a sphere, without atmospheric refraction
CALCULATOR / CASE FILE

How Far Is the Horizon? Horizon Distance Calculator

From a standing eye height of about 2 metres the horizon is roughly 5 kilometres away, and the distance grows with the square root of height: about 11 km from 10 m and 36 km from 100 m. The calculator gives the horizon distance for any eye height and shows how far away a taller object, such as a mast, can still be seen.

From the default eye height the water horizon lies 5,048 m away, and an elevated target adds 11,288 m, giving a shared sightline of 16.34 km.

A calm sea looks level, and the horizon can look like a straight division between water and sky. Neither observation requires an infinite plane. The interesting test begins when the observer changes height: the visible boundary moves farther away, and an elevated target can remain visible beyond the observer’s own sea-level horizon.

Run the calculation
A lighthouse on a cliff over the sea with tangent lines from its lamp to the horizonA claim about the world should fit the whole world.
RUN THE NUMBERS

How far can these two heights see?

Enter eye height and target-top height above the same water surface. The calculation uses straight light rays over a smooth sphere and reports line-of-sight lengths.

Calculation inputs
m

Your height assumption. Raising the eye extends its horizon; at low heights, four times the height gives about twice the range.

m

Your target-height assumption. A taller target can remain visible farther beyond the observer’s water horizon. Set zero for a water-level point.

Maximum shared straight sightline16.34km

The two tangent lengths combined; refraction and intervening obstructions are absent from this geometry.

Observer’s water-horizon distance5,048 m

Straight distance from the eye to the tangent point on the water.

Range contributed by target height11,288 m

An elevated target contributes its own tangent distance.

Working tape
  1. Twice Earth’s radius6,371,008.4 × 2 = 12,742,016.8
  2. Earth diameter + eye height12,742,016.8 + 2 = 12,742,018.8
  3. Eye height × adjusted diameter2 × 12,742,018.8 = 25,484,037.6
  4. Observer tangent distance√(25,484,037.6) = 5,048.171709
  5. Earth diameter + target height12,742,016.8 + 10 = 12,742,026.8
  6. Target height × adjusted diameter10 × 12,742,026.8 = 127,420,268
  7. Target tangent distance√(127,420,268) = 11,288.058646
  8. Add the two tangent distances5,048.171709 + 11,288.058646 = 16,336.230355
  9. Convert the sightline to kilometres16,336.230355 ÷ 1,000 = 16.33623
RUN THE NUMBERS

Separate curvature drop from hidden height

Enter surface arc distance and eye height. Drop is measured below the observer’s tangent plane; hidden height is the minimum radial target height that clears the horizon. Refraction k is an assumed uniform ray-curvature fraction, not measured weather.

Calculation inputs

Unit changes preserve the same physical inputs. The working tape stays in SI units.

km

Longer distances increase curvature drop and, beyond the eye horizon, hidden height.

m

A higher observer reduces hidden height but does not change tangent-plane drop at that distance.

Zero means straight rays. Positive k bends rays toward the surface; this constant-gradient scenario does not model a mirage.

Hidden target height17.54m

Minimum radial height at the target needed to meet the limiting ray; zero before the observer’s horizon.

Geometric tangent-plane drop31.39 m

A different quantity from hidden height: vertical displacement below the observer’s local tangent plane.

Surface arc to observer’s horizon5.05 km

An arc length, unlike the straight tangent length in the first instrument.

Working tape
  1. 1 − refraction coefficient1 − 0 = 1
  2. Effective spherical radius R/(1 − k)6,371,008.4 ÷ 1 = 6,371,008.4
  3. Surface distance in metres20 × 1,000 = 20,000
  4. Distance ÷ effective radius20,000 ÷ 6,371,008.4 = 0.003139
  5. Effective radius + eye height6,371,008.4 + 2 = 6,371,010.4
  6. Radius ÷ raised eye radius6,371,008.4 ÷ 6,371,010.4 = 1
  7. Angle to the eye horizonacos(1) = 0.000792
  8. Target angle beyond the eye horizon0.003139 − 0.000792 = 0.002347
  9. Keep zero before the horizonmax(0, 0.002347) = 0.002347
  10. Cosine of remaining anglecos(0.002347) = 0.999997
  11. Reciprocal cosine1 ÷ 0.999997 = 1.000003
  12. Secant minus one1.000003 − 1 = 0.000003
  13. Minimum target height above surface6,371,008.4 × 0.000003 = 17.544924
  14. Surface distance ÷ physical Earth radius20,000 ÷ 6,371,008.4 = 0.003139
  15. Cosine of full surface anglecos(0.003139) = 0.999995
  16. One minus cosine1 − 0.999995 = 0.000005
  17. Drop below local tangent plane6,371,008.4 × 0.000005 = 31.392179
  18. Effective radius × dip angle6,371,008.4 × 0.000792 = 5,048.170652
  19. Observer surface-arc horizon5,048.170652 ÷ 1,000 = 5.048171
01 / THE CLAIM

Local level is extended across the world

The claim is that water seeks its level, so a large body of water must form a single flat surface. Its immediate evidence is familiar: the sea does not look like a mound from the beach. The step requiring examination is the extension of a local horizontal into a plane that stays horizontal everywhere.

Level depends on the direction of gravity where it is measured. A local tangent plane can be horizontal at one location and cease to be horizontal elsewhere on a curved surface. Keeping this distinction allows water to be locally level all along a surface whose direction changes gradually with position.

02 / THE CASE

The observer stands above the surface being viewed

Surveyors describe the gravity-related reference for global mean sea level as the geoid. It is curved and gently irregular because Earth’s shape and mass distribution are irregular. Remove tides and currents from the ocean and its equilibrium surface follows that gravity field. The geoid is also extended through land to provide a reference for elevations.

For the horizon calculation, replace that detailed surface with a sphere. The last unobstructed straight ray from an elevated eye just touches it. The touching point is the geometric horizon. An eye above the water therefore looks slightly downward toward that boundary, even though the nearby surface is locally level.

A distant target can also extend above its own horizon. Seeing the upper part of a lighthouse or a mast beyond the distance to the observer’s water horizon is not an exception to the geometry. Both endpoint heights matter. A claim based only on the camera height has omitted a measurable part of the sightline.

03 / THE COMPUTATION

Two heights determine the available sightline

A radius drawn to a tangent point meets the sightline at a right angle. For Earth radius R and observer height h, the right triangle gives the straight distance d = √[h(2R + h)]. The same calculation applies to the target’s height. At the limiting shared sightline, those two tangent distances add together.

The opening assumptions place the eye 2 metres above the water and the target’s top 10 metres above it. With JPL’s mean Earth radius, the observer’s tangent distance is about 5.05 kilometres. The target contributes about 11.29 kilometres. The maximum unobstructed straight sightline between those heights is therefore about 16.34 kilometres.

The larger combined result is not a claim that the observer can see water itself that far away. The target stands above the water. Set its height to zero to recover the distance to a point on the surface. Raise either endpoint and more distance becomes available; at heights small compared with Earth’s radius, quadrupling one height approximately doubles that endpoint’s horizon distance.

These are straight-ray distances over a smooth sphere. For an actual observation, use the height above the intervening water, including the local water level, and account for any terrain. Atmospheric refraction bends the light path and can change the visible range. A photograph taken through an unknown near-water temperature profile does not supply a measured straight ray.

04 / THE FINDING

The boundary is close because the observer is low

For an eye 2 metres above the water, the geometric water horizon is about 5.05 kilometres away. A target standing 10 metres above the same surface extends the available straight sightline to about 16.34 kilometres. Its height accounts for the extra range beyond the observer’s own water horizon.

The nearby horizon gives a low observer a small view of a large curved surface. Raising either endpoint extends that view while the water remains locally level under gravity. Comparing an observation with these distances requires both endpoint heights and the light path: refraction changes the path, while a taller target changes the geometry.

05 / FURTHER READING

Observe one target from several heights

Keep the target and distance fixed while changing camera height, and record which named parts remain visible. Choose a stable shore target or fixed structure with surveyed feature heights, and measure the camera centre above the same surface datum used for the target. Record the water level rather than the tripod’s height above the pavement. Photograph from at least three camera heights without changing the target, keeping full uncropped frames, focal length and exposure metadata, and identify a feature’s lowest visible point, not merely whether the object appears.

Run the calculator for each height at the same distance, first with straight rays, then with an explicitly entered refraction scenario. Do not tune a coefficient merely to force one image to match. Repeat at different times with temperature and weather notes: a height-dependent change that persists across conditions is different from a transient stretched or inverted image.

06 / WHY PEOPLE BELIEVE IT

Why people believe it

Standing on a beach, the sea looks flat, and a flat-looking surface is easy to extend in the mind to the whole world. The error is one of scale: a horizon from 2 m eye height lies about 5 km away, and over that distance the surface drops by only 2 m, too little to see. Camera zoom strengthens the impression, because it enlarges what is visible without revealing what a lower sight line has hidden behind the horizon.

Parallel-arc length compared with polar projection radius

Flat Earth Map

The flat earth map is a polar disc projection, not the shape of the Earth. Try two cities, or see a 7,076 km southern arc drawn as 23,580 km, a 3.33× stretch.

Similar angles and a full-circle ratio

Eratosthenes Experiment

The Eratosthenes experiment measures the Earth from shadow angles at two places. A 7.2° difference over 800 km gives 40,000 km. Change the angle and distance.

Local vertical component of Earth’s angular velocity

Foucault Pendulum

A Foucault pendulum shows the Earth rotating: its swing plane turns by a rate set by latitude, 7.52 °/h clockwise at 30° north. Set the latitude and the hours.

Distance record divided by the claimed dome height

Firmament

The firmament is the biblical sky, read as a solid dome that spaceflight has not found. Apollo 13 went 400.2× the height of a 1,000 km dome.