Practice problem set

Try these by hand or in code. Solutions follow. Working through problems manually is the fastest way to spot which formulas you've actually internalized versus which ones you've just read.

Problems

P1. Convert 40° 26' 46.302" N, 79° 58' 56.903" W to decimal degrees.

P2. What's the distance (Haversine) between Paris (48.8566, 2.3522) and New York (40.7128, -74.0060)?

P3. What's the initial bearing from Paris to New York?

P4. At zoom level 8, which tile contains Tokyo (35.6762, 139.6503)?

P5. A drone flies 2 km on bearing 045° from (0, 0). Where does it end up?

P6. Compute a bounding box for a 10 km radius around Karachi (24.8607, 67.0011).

P7. What's the area of a polygon with vertices (in lon, lat): (0, 0), (1, 0), (1, 1), (0, 1)? (Use the spherical formula.)

P8. A point at zoom 10 has world pixel coordinates (550000, 380000). What's its lat/lon?

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Solutions

S1.

$$\text{lat} = 40 + 26/60 + 46.302/3600 = 40.4462°$$ $$\text{lon} = -(79 + 58/60 + 56.903/3600) = -79.9825°$$

S2. Haversine:

$$\Delta\varphi = -8.1438° = -0.14217 \text{ rad}, \quad \Delta\lambda = -1.33271 \text{ rad}$$ $$a = \sin^2(-0.07109) + \cos(0.85245)\cos(0.71081)\sin^2(-0.66635) \approx 0.1946$$ $$c = 2 \arctan2(\sqrt{0.1946}, \sqrt{0.8054}) \approx 0.9170$$ $$d \approx 6{,}371{,}008.8 \times 0.9170 \approx 5{,}840 \text{ km}$$

S3.

$$y = \sin(-1.33271)\cos(0.71081) \approx -0.7359$$ $$x = \cos(0.85245)\sin(0.71081) - \sin(0.85245)\cos(0.71081)\cos(-1.33271) \approx 0.2936$$ $$\theta = \arctan2(-0.7359, 0.2936) \approx -68.18°$$ $$\text{bearing} \approx 291.82° \quad \text{(WNW — NY is to the west and slightly north)}$$

The WNW bearing makes geographic sense: New York is almost due west of Paris, with a slight northward component visible from the non-trivial latitude difference.

S4.

$$n = 256$$ $$x = \lfloor 256 \times (139.6503 + 180)/360 \rfloor = 227$$ $$y = \lfloor 256 \times 0.3939 \rfloor = 100$$

Tile: (8, 227, 100)

S5.

$$\delta = 2000 / 6{,}371{,}008.8 = 0.000314$$ $$\theta = \pi/4$$ $$\varphi_2 = \arcsin(0.000222) \approx 0.01272°$$ $$\lambda_2 = \arctan2(0.000222, 1.0000) \approx 0.01272°$$

End: (0.0127°, 0.0127°)

Equal latitude and longitude offsets make sense — at the equator, going NE means equal N and E components.

S6.

$$\Delta\varphi = 10000/6{,}371{,}008.8 \approx 0.0899°$$ $$\Delta\lambda = 0.0899/\cos(24.86°) \approx 0.0992°$$

Box: lat $[24.7708, 24.9506]$, lon $[66.9019, 67.1003]$

S7. Apply the spherical polygon area formula at small scale:

$$\text{Area} \approx 12{,}180 \text{ km}^2$$

Sanity check: at the equator, 1° × 1° ≈ 111 × 111 ≈ 12,321 km². Close.