What each construction needs
Every construction starts with the radius of the reduced earth: R = earth’s radius ÷ the denominator of the R.F. After that:
- Conical with one standard parallel. BP = R cot φ is the standard parallel’s radius. Parallels are arcs CE = 2πR × interval ÷ 360 apart; meridians run to the apex, XY = CE × cos φ apart on the standard parallel.
- Cylindrical equal area. Equator 2πR, divided every 2πR × interval ÷ 360. Parallels sit R sin φ from the equator.
- Mercator’s. The same equator. Parallels sit R × ln tan(45° + φ/2) from it: the book’s table value times R.
- Polar zenithal (BA and state boards). Circles round the pole of radius 2πR × (90° − φ) ÷ 360 (equidistant) or 2R sin((90° − φ) ÷ 2) (equal area).
Example: the book’s conical projection
NCERT asks for a conical projection with one standard parallel for 10°N to 70°N and 10°E to 130°E at 1:250,000,000 with a 10° interval. R = 640,000,000 ÷ 250,000,000 = 2.56 cm. The standard parallel is 40°N and the central meridian 70°E. The book draws the rest; worked out, BP = 2.56 × cot 40° = 3.05 cm, CE = 2 × 22/7 × 2.56 × 10 ÷ 360 = 0.45 cm and XY = 0.45 × cos 40° = 0.34 cm. Open the book problems to try it and the activity problems.
Where books and teachers differ
NCERT uses an earth radius of 640,000,000 cm and π = 22/7, but its Mercator example uses 250,000,000 inches. It keeps R = 2.56 in one example and rounds 2.133 to 2.1 in another. Set yours under Conventions and they print on every sheet. Answers worked with π = 22/7, 3.14 or 3.1416, and with R rounded or not, all count.
Questions
How do you calculate the radius of the reduced earth?
Divide the earth’s radius by the denominator of the R.F., in the same unit. NCERT uses 640,000,000 cm, so at 1:250,000,000 the reduced earth has R = 640,000,000 ÷ 250,000,000 = 2.56 cm. Every other length in the construction comes from R.
How do you find the radius of the standard parallel in a conical projection?
It is R cot φ, where φ is the standard parallel. In the book’s figure it is the tangent BP from the globe to the axis. For R = 2.56 cm and 40°N: 2.56 × cot 40° = 2.56 × 1.1918 = 3.05 cm. Draw the standard parallel as an arc of that radius from the apex.
How do I choose the standard parallel and central meridian?
Take the middle parallel and the middle meridian of the area. For 10°N to 70°N and 10°E to 130°E that is 40°N and 70°E. If the middle meridian falls between two graticule lines, as for 40°E to 30°W, mark half of XY on each side of it.
Where does NCERT’s Mercator table come from?
Each value is ln tan(45° + φ/2), the distance of a parallel from the equator on a globe of radius 1. Multiply it by R. The book prints 15° 0.265, 30° 0.549, 45° 0.881, 60° 1.317 and 75° 2.027; the last is cut rather than rounded (2.0276).
Why are parallels R sin φ apart in the cylindrical equal area projection?
The parallels are projected straight across from a circle of radius R, so each sits R sin φ above or below the equator. They crowd together near the poles, which makes up for the parallels being stretched to the length of the equator, so areas stay true.
Should I use π = 22/7 or 3.14?
NCERT writes 22/7, and some teachers use 3.14 or 3.1416. The difference is tiny at these sizes, so answers worked with any of them count. The key uses 22/7 or 3.1416, as you pick under Conventions.
Why doesn’t the printed key match my drawing?
Print at 100% or Actual size, not Fit to page. The key has a 10 cm bar (4 inches on inch problems): measure it with a ruler first. If it is right, compare your R, the radius of the standard parallel and the spacings with the key.
Sources, read October 27, 2026: NCERT Practical Work in Geography Part I, Class 11, chapter 4 Map Projections (PDF, 2026-27): the three worked constructions, the Mercator table and the activity; University of Mumbai (IDOL), TYBA Geography Paper VI, Tools and techniques in geography for spatial analysis I (Marathi, 2022, PDF), unit 1.2: polar zenithal equal area and equidistant constructions.