Question

Prepare graticule for a Cylindrical Equal Area Projection for the world when R.F. is 1:150,000,000 and the interval is 15º apart.

Answer 1

1. Radius of the Earth (R): \[\frac{\text{Actual Radius of Earth}}{\text{Denominator of R.F.}} = \frac{640,000,000}{150,000,000} \approx 4.27\text{ cm}\]
2. Length of Equator: \[2\pi R = 2 \times \frac{22}{7} \times 4.27 \approx 26.81\text{ cm}\]
3. Spacing between Meridians: \[\frac{\text{Length of Equator} \times \text{Interval}}{360^\circ} = \frac{26.81 \times 15^\circ}{360^\circ} \approx 1.12\text{ cm}\] (dividing the equator into 24 equal parts)
4. Spacing of Parallels: Calculated using R sin θ (where θ is the latitude):
   • 15°: 1.11 cm
   • 30°: 2.14 cm
   • 45°: 3.02 cm
   • 60°: 3.70 cm
   • 75°: 4.12 cm
   • 90°: 4.27 cm