Cylindrical Equal Area Projection


Lines of Latitude
Lines of Longitude
Prime Meridian
Facts on Earth
Zenithal Projections
Gnomonic Polar
Gnomonic Equatorial
Stereographic Polar
Stereographic Equatorial
Orthographic Polar
Orthographic Equatorial
Simple Conic Projection
Cylindrical Equal-area Projection





Figure A

Figure B







Cylindrical Projections is constructed by the projection of parallels and meridians onto the surface of a cylinder with the earth circumscribed within it. As a result, the surface of the cylinder is tangent to the earth at the equator. When unfolded, the meridians will be seen as vertical, parallel and equally spaced lines that intersect perpendicularly to the equator (refer to Figure C). This is primarly because the cylinder touches the equator and as such touches each of the intervals with a meridian. Since the meridians around the earth is equally spaced and rotated around the entire sphere, the projection will consist of equally spaced lines for the entire map. With the cylinder being tangential to the equator, the length of such a map would be equal to the circumference of the earth or circle (2 * 3.1515926654... * radius). Similarly, projected parallels are seen as straight lines that vary in distance as they move away from the equator. The corresponding projection of the latitude CD in Figure B is drawn from the point that the latitude intersects the circle (at point D) through the entire map. Thus producing the map as seen below. The distances away from the equator AB in figure B can be accurately calculated using the relationship as deduced before: radius of the circle * sin (angular distance of latitude). This relationship for any arbitrary latitude such as BC in Figure B came about using the following logic. Given that latitude AB has angular distance <AOB and that DC is parallel to AB, one knows that the vertical distance DC is away from AB is equal to a line segment between the two parallels and perpendicular to both. Therefore, by drawing a perpendicular line from point D to E, this implies that <DEO is a right angle and that sin (<DOE) = DE/OD. Because OD is equivalent to the radius of the circle and <DOE is the angular distance of the latitude, therefore DE = circle's radius * sin (angular distance of latitude).

Figure C