Zenithal Gnomonic Equatorial 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

Figure C



For the equatorial case of the Gnomonic projection, the source of light is situated at the centre of the globe, like in all other gnomonic projections. But, unlike that of the polar case, the plane of projection (dark purple in Figure A) touches the globe at the equator (green line). When projected, all great circles are again represented as straight lines in the projection plane. This is depicted in Figure A where the green line on the plane is the projected equator and the light purple lines represent straight and parallel meridians. Unlike the equator, other parallels are curves conves to the equator. Note the red line on the projection plane in Figure A is where the projected parallel intersects with the particular light puple meridian.

Refering to Figure B, the mathematical calculations for the intersections of parallels with the meridians can be logically deduced. With the equator running from AB and touching the earth tangentially at point A, the central meridian becomes projected at the same point from the light ray AC. Because the light ray goes through the centre and touches the tangential point, it satisfies both conditions which leads to a perpendicular intersection with the equator. As such, <CAB = 90°. Therefore, triangle ACB becomes a right angle triangle. Another right angle triangle exist because of the fact that all meridians are perpendicular to the equator. As a result, all light rays that produces the projection of the equator is also perpendicular to the meridians. This is the case with the light ray BC with meridian DB, therefore, <CBD = 90°. The distance of the projected parallel away from the equator would be DB. DB's distance could be calculated using the two triangles DCB and ABC. Given that AC is the radius of the circle and <ACB is the angular distance of the longitude, BC = circle's radius * sec (angular distance of longitude) since cos (angular distance of longitude) = CA / BC. With <DCB equal to the angular distance of the latitude and <CBD = 90°, the length of DB = tan (angular distance of latitude) * circle's radius * sec (angular distance of longitude). This is because tan (<DCB or latitude) = DB / BC.

It is clear from Figure C that the meridians are constructed by extending light rays at the desired rotation away from the central meridian (or the line that touches the tangential point). The intersections of such rays with the equator (or the horizontal line tangential from the circle) will be equivalent to the distance away from each other. The relationship of these distances will be the circle's radius * tan (angular distance of longitude). This fact can be proven using the triangle AOB in Figure C. With <BOA equal to the angular distance of longitude and OA equal to the radius of the circle, the desired distance AB is equal to circle's radius * tan (angular distance of longitude) since tan (<BOA) = AB/OA.

As such the map in Figure C is produced.