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Re: Mapping a sphere to a plane
Posted:
Mar 12, 2006 4:54 PM


Hi Jonathan Barnes,
You described: > a sphere of radius R, with a band drawn on it of width W. > The edges of the band are circles. > > The sphere is centred on the origin and it's surface can be mapped to a > square of 180 x 360 units with the unit being a degree of latitude or > longitude. ( origin at centre ). > > With a horizontal band the two edges are just horizontal lines at > Y = ( +  ) SIN^1 (( W / 2 )/R).
You then asked for a formula for the edges of a vertical band.
It is easier to describe this transformation as being FROM the plane (x,y) TO the surface of the sphere in three dimensions (X,Y,Z). Please note the distinction between lowercase 2 dimensional coordinates and UPPERCASE 3 dimensional coordinates in the following. Lowercase x is the longitude and lowercase y is the latitude.
The vertical distance from the plane of the equator is given by the sine of the latitude times R: Z=R*sin(y). The radius of a circular line of latitude is: R*cos(y). The coordinates of a unit circle centered at the origin in the plane are given by (Cos(angle),Sin(angle)), for an angle measured counterclockwise. However, longitude is measured clockwise, so the coordinates on that circle of latitude become:
(R*Cos(y)*Cos(x),R*Cos(y)*Sin(x),R*Sin(y)).
Your horizontal strip is between Z=+W/2 and Z=W/2. Substitute Z=R*Sin(y) and solve to get the equation you gave.
For the equation you want, take a strip between Y=+W/2 and Y=W/2. Substitute Y=R*Cos(y)*Sin(x) to get the equations you want:
y = (+) Cos^1((+)W/(2*R*Sin(x)))
I don't see any nice way to simplify this formula (anyone else)?
 _ /\/ \ orewood@oLc.ubc.ca Beautiful British Columbia (Canada)



