Map Projections

Deducing Braun's Stereographic Cylindrical Projection

Imaginary development of stereographic cylindrical map, most of it shown from the reverse side; the Eastern hemisphere on the left was unrolled.  Light rays pierce the semitransparent globe drawing the right edge of the map.

The stereographic cylindrical projection designed by Braun can be visualized geometrically, but it is slightly more complicated than the previous azimuthal orthographic example.  Instead of a flat projection plane directly yielding the projected map, here the projection surface is a cylindrical sheet tightly rolled against the Equator.  Every meridian is drawn on this tube by light rays emanating from an equatorial point on the meridian directly opposite.  The tube is then cut along an arbitrary meridian and unrolled. 

This is a general procedure for creating some cylindrical projections, although not all of them employ such a simple model. In fact, projections like the equidistant cylindrical are defined by arbitrary constraints, not a perspective process.

Transversal cut of the tube tangent at Equator. Due to projection constraints, the length of tube (therefore the map's height) is twice the sphere's diameter

Equations for direct mapping are straightforward.

Projected meridians are straight vertical lines, regularly spaced, thus x = R .

Parallel is projected as a straight horizontal line, whose y-coordinate can be derived from the diagram: by a simple proportion, h / (w + R) = y / (R + R). Thus,

• h = R sin 
• w = R cos 
• y = 2R sin  / (1 + cos )

Making = / 2 and using the trigonometric identities sin 2 = 2 sin  cos  and cos 2 = 1 - 2 sin ², y = 2R2 sin  cos  / (1 + 1 - 2 sin²) = 4R sin  cos /2 (1 - sin² ) = 2R sin  cos  / cos².  Therefore,

• x = R
• y = 2R tan ( /2)

www.progonos.com/furuti    January 22, 2004