Earth Curvature Drop
Approximate drop of the horizon over distance d.
Formula
Variables
Example
Over 5 km the Earth curves away by about 5000²/(2·6.37×10⁶) ≈ 2 m.
How it works
This formula estimates how far the Earth's curved surface falls below a straight tangent line after traveling a horizontal distance d. It comes from the geometry of a circle: the sagitta of a circle segment is exactly R - \sqrt{R^2 - (d/2)^2}, and for small distances compared with Earth's radius this simplifies to h \approx d^2/(2R). It is a good approximation when d is much smaller than R, which is usually true for everyday surveying, line-of-sight, and horizon problems.
When to use it
Use this when you want a quick estimate of how much the Earth curves away over a given distance, such as in visibility, surveying, photography, or engineering line-of-sight calculations. It is most reliable for distances that are tiny compared with Earth’s radius, not for global-scale paths.
Worked example, step by step
Find the Earth's curvature drop over a distance of 20 km. Convert to metres: d = 20 km = 20,000 m Use Earth’s radius: R = 6.37 × 10^6 m Substitute into the formula: h ≈ d^2 / (2R) h ≈ (20,000)^2 / (2 × 6.37 × 10^6) h ≈ 400,000,000 / 12,740,000 h ≈ 31.4 m So over 20 km, the Earth curves away by about 31 m.
Common mistakes to avoid
- Forgetting to convert distance into metres before substituting into the formula
- Using the formula for very large distances where the small-angle approximation is no longer accurate
- Confusing curvature drop with atmospheric refraction, which can noticeably change observed horizon distances
Did You Know?
The Earth curves away about 8 cm per kilometer squared — why distant ships seem to sink below the horizon.
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