Horizon Distance at Sea Level: The 1.5m Observer Sees 4.4 km

For an observer with eye height h = 1.5 m at sea level, the geometric horizon distance d is:

d ≈ 3.57 × √h (km, h in meters) → d ≈ 3.57 × √1.5 ≈ 3.57 × 1.225 ≈ 4.37 km (2.7 mi).

More precisely (including standard refraction):

d = 3.86 × √h → 3.86 × 1.225 ≈ 4.73 km (2.94 mi).

Why Two Formulas?

• Geometric (no atmosphere): d = √(2Rh) ≈ 3.57√h (R = 6371 km).
• With standard refraction (k = 7/6): Effective Earth radius = 7/6 R → d = 3.86√h.

Practical Rules of Thumb

• Nautical: d (nm) = 1.17 × √h (ft) → eye at 5 ft → 2.6 nm.
• Metric: d (km) = 3.86 × √h (m) → 1.5 m → 4.7 km.

Real-World Factors

• Height of eye (not body height) — on a ship’s bridge (10 m) → 12.2 km.
• Target height — seeing a lighthouse (30 m) adds its horizon: 4.7 + 21.1 = 25.8 km.
• Refraction varies (temperature gradients) — mirage, looming, sinking.

So: 1.5 m eye at sea level → horizon ~4.4–4.7 km. The "4.56 km" in the prompt uses a slightly different constant (13.89 = 3.727²). Close enough.