Processing math: 8%
Map Projections
for a Spherical Planet
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Prof. David Bernstein
James Madison University
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Computer Science Department
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bernstdh@jmu.edu
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Motivation
- Some Observations:
- The Earth is not flat
- Maps usually are (especially on computer displays)
- The Implication:
- We need a way to project points on the surface of the
Earth onto a map
Thinking About Map Projections
- Using a Physical Model:
- Algebraically:
Physical Models
- The Idea:
- Shine a light onto or through a transparent Earth
and capture the shadows cast by the opaque features
- The "Parameters":
- The shape of the screen (called the projection surface)
- The position of the projection surface
- The location of the light source
Projection Surfaces
Positioning a Planar Projection Surface
Positioning a Cylindrical Projection Surface
Positioning a Conical Projection Surface
Light Sources
Polar Azimuthal Orthographic Projection
Notation
Polar Azimuthal Orthographic Projection (cont.)
- Derivation of the Projection:
- \cos \phi = d / R \Rightarrow d = R \cos \phi
- \cos \lambda = p_{2}/d \Rightarrow p_{2} = d \cos \lambda
= R\cos \phi \cos \lambda
- \sin \lambda = p_{1}/d \Rightarrow p_{1} = d \sin \lambda
= R \cos \phi \sin \lambda
- Derivation of the Inverse:
- d = ||\bs{p}||
- \cos \lambda = \frac{p_{2}}{||\bs{p}||} \Rightarrow
\lambda = \cos^{-1}(\frac{p_{2}}{||\bs{p}||})
- \cos \phi = d / R \Rightarrow
\phi = \cos^{-1}(\frac{||\bs{p}||}{R})
Polar Azimuthal Orthographic Projection (cont.)
The World
Cylindrical Stereographic Projection
Notation
Cylindrical Stereographic Projection (cont.)
- Derivation of the Projection:
- Since the triangles are "similar" \frac{b}{R+a} = \frac{p_{2}}{R+R}, hence p_{2} = (\frac{b}{a+R})2R
- \sin \phi = b/R \Rightarrow b = R \sin \phi
- \cos \phi = a/R \Rightarrow a = R \cos \phi
- So, p_{2} = \frac{R \sin \phi}{R+R\cos \phi}2R=\frac{R \sin \phi}{R(1+\cos \phi)}2R = \frac{\sin\phi}{1+\cos\phi}2R
- p_{1} = \lambda R since it is only determined by
where the cylinder is "cut"
- Derivation of the Inverse:
- \lambda = p_{1}/R
- \tan \phi = p_{2}/R \Rightarrow
\phi = \tan^{-1}(p_{2}/R) (but you have to be careful
about the domain of the \tan)
Desirable Properties of Projections
- The Most Common:
- Conformal (i.e., angles are preserved)
- Equal Area (i.e., areas are in constant proportion)
- Equidistant (i.e., distances are in constant proportion)
- An Important Mathematical Result:
- A single projection can not be both conformal
and equal area
Sinusoidal Projection
- Derivation of an Equal Area Projection:
- Uniformly space the parallels so p_{2} = \phi R
- To be equal area given these parallels, the length of
each parallel has to equal the "circumference" at that
latitude (which is R \cos \phi)
- The projected parallel must be proportional to this
circumference so it must have length k 2 \pi R \cos
\phi
- To be consistent at the equator (where the "circumference"
is 2 \pi R), we must have k = 0.5
- Assuming a constant horizontal scale, we must have
\frac{p_{1}}{\pi R \cos \phi} = \frac{\lambda}{\phi}
- So, p_{1} = \lambda R \cos \phi
- Note:
- We did not use a physical model
Sinusoidal Projection (cont.)
The World
Equatorial Cylindrical Equal Area Projection
- Parameters:
- \lambda_{0} is the standard longitude
(i.e., the horizontal center of the projection)
- Projection:
- p_{1} = R (\lambda - \lambda_{0})
- p_{2} = R \sin(\phi)
- Inverse:
- \lambda = \lambda_{0} + \frac{p_{1}}{R}
- \phi = \sin^{-1}(p_{2} / R)
Equatorial Cylindrical Equal Area Projection (cont.)
The World
Equatorial Cylindrical Conformal Projection (Mercator)
- Parameters:
- \lambda_{0} is the standard longitude
- Projection:
- p_{1} = R (\lambda - \lambda_{0})
- p_{2} = R \ln[\tan(\pi / 4 + \phi / 2)]
- Inverse:
- \lambda = \lambda_{0} + \frac{p_{1}}{R}
- \phi = 2 [\tan^{-1}(e^{\frac{p_{2}}{R}}) - \pi / 4]
Equatorial Cylindrical Conformal Projection (cont.)
The World
Conical Equal Area Projection (Albers)
- Parameters:
- \lambda_{0} and \phi_{0} are the longitude and latitude of the origin of the transformed coordinates
- \phi_{1} and \phi_{2} are
the two standard parallels (i.e., the parallels where the
cone intersects the sphere)
- Preliminary Calculations:
- n = 0.5 (\sin(\phi_{1}) + \sin(\phi_{2}))
- c = \cos^{2}(\phi_{1}) + 2 n \sin(\phi_{1})
- \rho_{0} = \frac{\sqrt(c - 2 n \sin(\phi_{0}))}{n}
- Projection:
- \rho = \frac{\sqrt{c - 2 n \sin(\phi)}}{n}
- \theta = n(\lambda - \lambda_{0})
- p_{1} = R \rho \sin(\theta)
- p_{2} = R (\rho_{0} - \rho \cos(\theta))
- Inverse:
- a = \sqrt{(p_{1}/R)^{2}+(\rho_{0}-p_{2}/R)^{2}}
- b = \tan^{-1}((p_{1}/R)/(\rho_{0} - p_{2}/R))
- \phi = \sin^{-1}(\frac{c- a^{2} n^{2}}{2n})
- \lambda = \lambda_{0} + b/n
Conical Equal Area Projection (cont.)
The World
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