Earth curvature math

Here’s a clear, compact math synopsis of Earth’s curvature:

🌍 Basic Geometry

The Earth is (approximately) a sphere with mean radius
[
R = 6,371 \text{ km} = 6,371,000 \text{ m}.
]

The surface drops away from a tangent line due to curvature. For short distances, you can use the parabolic approximation:

[
\text{Drop} = \frac{d^2}{2R}
]

where ( d ) is the horizontal distance from the tangent point.

📏 Common Results

(neglecting atmospheric refraction)

🔭 Including Refraction

Refraction slightly bends light downward, making Earth appear flatter. Engineers often use an effective radius:
[
R' = \frac{7}{6} R
]

so the apparent drop is:
[
h' = \frac{d^2}{2R'}
]

This reduces the apparent curvature by about 15%.

🌐 For Precise Geometry (Great-Circle Chord)

If you want the exact sagitta (drop) for any distance:

[
h = R \left( 1 – \cos\frac{d}{R} \right)
]

where ( d ) is arc distance along the surface (radians).
For small ( d ), this reduces to ( \frac{d^2}{2R} ).

🧮 Example

For ( d = 10 \text{ km} = 10,000 \text{ m} ):

[
h = \frac{(10,000)^2}{2(6,371,000)} \approx 7.85 \text{ m}
]

So at 10 km, the Earth’s surface curves away about 8 meters from a straight line tangent at your eye level.

Would you like me to include observer horizon distance formulas next (i.e., how far you can see before curvature hides objects)?

Perfect — here’s the Earth curvature + horizon distance math summary, compact and complete:

🌍 Horizon Distance & Curvature Summary

1️⃣ Basic Geometry

For an observer at height ( h ) above the surface, the line of sight to the horizon is tangent to Earth’s surface.
Let ( R = 6,371,000 \text{ m} ) (mean Earth radius).

Using the Pythagorean relation:

[
(R + h)^2 = R^2 + d^2
]

where ( d ) is the distance to the horizon (straight line along the surface).

Solve for ( d ):

[
d = \sqrt{(R + h)^2 – R^2}
]

Simplify (since ( h \ll R )):

[
\boxed{d \approx \sqrt{2Rh}}
]

2️⃣ Horizon Distance Examples

3️⃣ Refraction Correction

Atmospheric refraction slightly bends light, extending visible distance.
Use an effective radius:

[
R' = \frac{7}{6}R
]

Then the apparent horizon becomes:

[
d' = \sqrt{2R'h} = \sqrt{\frac{7Rh}{3}}
]

So refraction increases visible distance by about 15%.

4️⃣ Object Hidden by Curvature

For two heights (observer ( h_1 ), target ( h_2 )):

[
\boxed{d_{visible} = \sqrt{2Rh_1} + \sqrt{2Rh_2}}
]

If their separation ( d ) exceeds ( d_{visible} ), part of the target is hidden below the horizon.

5️⃣ Quick Engineering Approximation

For practical use (metric units):

[
\boxed{d(\text{km}) \approx 3.57 \sqrt{h(\text{m})}}
]

and inversely:

[
\boxed{h(\text{m}) \approx 0.0786\,d^2(\text{km})}
]

Example

At eye height ( h = 1.8 \text{ m} ):

[
d = 3.57 \sqrt{1.8} \approx 4.8 \text{ km}
]

An observer standing 1.8 m tall can see about 4.8 km to the geometric horizon.

Would you like me to include the formula for how much of a distant object (like a ship or building) is hidden below the horizon next?