The Mathematics of Maps Lecture 4. Dennis The The Mathematics of Maps Lecture 4 1/29

The Mathematics of Maps Lecture 4 Dennis The The Mathematics of Maps Lecture 4 1/29

Mercator projection Dennis The The Mathematics of Maps Lecture 4 2/29

The Mercator projection (1569) Dennis The The Mathematics of Maps Lecture 4 3/29

The Mercator projection (1569) Importance: Navigation. Dennis The The Mathematics of Maps Lecture 4 3/29

The Mercator projection (1569) Importance: Navigation. Conformal map, i.e. angles are preserved. Dennis The The Mathematics of Maps Lecture 4 3/29

The Mercator projection (1569) Importance: Navigation. Conformal map, i.e. angles are preserved. Here s a fun little puzzle: Mercator Puzzle Dennis The The Mathematics of Maps Lecture 4 3/29

Rhumb lines Rhumb line = path of constant compass bearing on the sphere, i.e. have constant angle with the corresponding parallel & meridian. Dennis The The Mathematics of Maps Lecture 4 4/29

Rhumb lines Rhumb line = path of constant compass bearing on the sphere, i.e. have constant angle with the corresponding parallel & meridian. parallels of latitude horizontal lines meridians of longitude vertical lines rhumb lines straight lines Dennis The The Mathematics of Maps Lecture 4 4/29

Mercator map formula For a map Ψ : X = θ, Y = g(φ), we calculated scale factors: L(Ψ(A s )) L(Ψ(B t )) λ p = lim = sec(φ), λ m = lim = g (φ). s 0 L(A s ) t 0 L(B t ) Dennis The The Mathematics of Maps Lecture 4 5/29

Mercator map formula For a map Ψ : X = θ, Y = g(φ), we calculated scale factors: L(Ψ(A s )) L(Ψ(B t )) λ p = lim = sec(φ), λ m = lim = g (φ). s 0 L(A s ) t 0 L(B t ) Conformality means λ p = λ m. Dennis The The Mathematics of Maps Lecture 4 5/29

Mercator map formula For a map Ψ : X = θ, Y = g(φ), we calculated scale factors: L(Ψ(A s )) L(Ψ(B t )) λ p = lim = sec(φ), λ m = lim = g (φ). s 0 L(A s ) t 0 L(B t ) Conformality means λ p = λ m. Requiring g(0) = 0, g (φ) > 0, g (φ) = sec(φ) Dennis The The Mathematics of Maps Lecture 4 5/29

Distortion in the Mercator projection Dennis The The Mathematics of Maps Lecture 4 6/29

Distance along a rhumb line Q: How can we find the distance along a rhumb line? Rhumb line line segment X (t) = at + b, Y (t) = ct + d. Dennis The The Mathematics of Maps Lecture 4 7/29

Distance along a rhumb line Q: How can we find the distance along a rhumb line? Rhumb line line segment X (t) = at + b, Y (t) = ct + d. For a path (φ(t), θ(t)), 0 t 1 on S 2, we saw in Lecture 2 a 1 formula for its length L = (φ ) 2 + cos 2 (φ)(θ ) 2 dt. 0 Dennis The The Mathematics of Maps Lecture 4 7/29

Inverse of the Mercator map Given Ψ M : { X = θ Y = ln(sec(φ) + tan(φ)), what is Ψ 1 M? Dennis The The Mathematics of Maps Lecture 4 8/29

Inverse of the Mercator map Given Ψ M : { X = θ Y = ln(sec(φ) + tan(φ)), what is Ψ 1 M? Using some trig. trickery... e Y = sec(φ) + tan(φ) = 1+sin(φ) cos(φ) = 1 cos(φ+ π 2 ) sin(φ+ π 2 ) = 2 sin 2 ( φ 2 + π 4 ) 2 sin( φ 2 + π 4 ) cos( φ 2 + π 4 ) = tan( φ 2 + π 4 ). Dennis The The Mathematics of Maps Lecture 4 8/29

Inverse of the Mercator map Given Ψ M : { X = θ Y = ln(sec(φ) + tan(φ)), what is Ψ 1 M? Using some trig. trickery... e Y = sec(φ) + tan(φ) = 1+sin(φ) cos(φ) = 1 cos(φ+ π 2 ) sin(φ+ π 2 ) = 2 sin 2 ( φ 2 + π 4 ) = tan( φ 2 sin( φ 2 + π 4 ) cos( φ 2 + π 4 ) 2 + π 4 ). Thus, Ψ 1 M : { θ = X φ = 2 arctan(e Y ) π 2 Dennis The The Mathematics of Maps Lecture 4 8/29

Distance along a rhumb line - 2 Ψ 1 M : { θ = X φ = 2 arctan(e Y ) π 2, Rhumb line: { X (t) = at + b Y (t) = ct + d Dennis The The Mathematics of Maps Lecture 4 9/29

Distance along a rhumb line - 2 Ψ 1 M : { θ = X φ = 2 arctan(e Y ) π 2, Rhumb line: { X (t) = at + b Y (t) = ct + d φ = 2eY c 1 + e 2Y, θ = a, cos(φ) = 2eY 1 + e 2Y. Dennis The The Mathematics of Maps Lecture 4 9/29

Distance along a rhumb line - 2 Ψ 1 M : { θ = X φ = 2 arctan(e Y ) π 2, Rhumb line: { X (t) = at + b Y (t) = ct + d φ = 2eY c 1 + e 2Y, θ = a, cos(φ) = 2eY 1 + e 2Y. L = 1 0 (φ ) 2 + cos 2 (φ)(θ ) 2 dt = a 2 + c 2 1 =... = 2 a 2 + c 2 (arctan(e c+d ) arctan(e d )). c (When c 0, we recover L = 2ed = cos(φ).) 1+e 2d 0 2e Y dt 1 + e2y Dennis The The Mathematics of Maps Lecture 4 9/29

Distance along a rhumb line - 2 Ψ 1 M : { θ = X φ = 2 arctan(e Y ) π 2, Rhumb line: { X (t) = at + b Y (t) = ct + d φ = 2eY c 1 + e 2Y, θ = a, cos(φ) = 2eY 1 + e 2Y. L = 1 0 (φ ) 2 + cos 2 (φ)(θ ) 2 dt = a 2 + c 2 1 =... = 2 a 2 + c 2 (arctan(e c+d ) arctan(e d )). c 0 2e Y dt 1 + e2y (When c 0, we recover L = 2ed = cos(φ).) This formula is 1+e 2d valid on the unit sphere S 2. On a sphere of radius R, this length would be rescaled by R. Dennis The The Mathematics of Maps Lecture 4 9/29

Stereographic projection Dennis The The Mathematics of Maps Lecture 4 10/29

Stereographic projection (from north pole) Dennis The The Mathematics of Maps Lecture 4 11/29

Stereographic projection (from north pole) parallels of latitude circles centred at 0 meridians of longitude rays emanating from 0 Dennis The The Mathematics of Maps Lecture 4 11/29

Stereographic projection (from north pole) parallels of latitude circles centred at 0 meridians of longitude rays emanating from 0 (Note: Viewed from above, East West is a clockwise rotation in the map.) Dennis The The Mathematics of Maps Lecture 4 11/29

Geometric construction On S 2, put a light source at the north pole, and cast the shadow of P S 2 onto the plane tangent at the south pole. Dennis The The Mathematics of Maps Lecture 4 12/29

Geometric construction On S 2, put a light source at the north pole, and cast the shadow of P S 2 onto the plane tangent at the south pole. Line NP intersects z = 1 plane at Q: ( 2x 1 z, 2y 1 z, 1). Dennis The The Mathematics of Maps Lecture 4 12/29

Polar coordinate formula Use spherical polar coords (x, y, z) = (cos φ cos θ, cos φ sin θ, sin φ) to write Ψ N. Dennis The The Mathematics of Maps Lecture 4 13/29

Polar coordinate formula Use spherical polar coords (x, y, z) = (cos φ cos θ, cos φ sin θ, sin φ) 2 cos φ cos θ 2 cos φ sin θ to write Ψ N. Then X = 1 sin φ, Y = 1 sin φ, and R = X 2 + Y 2 = 2 cos φ 2 cos φ(1 + sin φ) = 1 sin φ 1 sin 2 φ Ψ N : R = 2(sec φ + tan φ), Θ = θ Dennis The The Mathematics of Maps Lecture 4 13/29

Stereographic projection is conformal Halley (1695): Stereographic projection is conformal. Dennis The The Mathematics of Maps Lecture 4 14/29

Stereographic projection is conformal Halley (1695): Stereographic projection is conformal. For a map Ψ : R = f (φ), Θ = θ, we calculated scale factors L(Ψ(A s )) L(Ψ(B t )) λ p = lim = f (φ) sec(φ), λ m = lim = f (φ). s 0 L(A s ) t 0 L(B t ) Dennis The The Mathematics of Maps Lecture 4 14/29

Distortion in the stereographic projection Dennis The The Mathematics of Maps Lecture 4 15/29

Circles are mapped to circles Definition A circle on S 2 is the intersection of a plane Π in R 3 with S 2. A circle in the plane is either an ordinary circle or a line. Dennis The The Mathematics of Maps Lecture 4 16/29

Circles are mapped to circles Definition A circle on S 2 is the intersection of a plane Π in R 3 with S 2. A circle in the plane is either an ordinary circle or a line. Theorem Any circle on S 2 is mapped by Ψ N to a circle in the plane. Dennis The The Mathematics of Maps Lecture 4 16/29

Stereographic projection is circle preserving IDEA: Invert Ψ N : X = 2x 1 z, Y = 2y 1 z Dennis The The Mathematics of Maps Lecture 4 17/29

Stereographic projection is circle preserving IDEA: Invert Ψ N : X = 2x 1 z, Y = 2y 1 z ( P = (x, y, z) = 4X X 2 + Y 2 + 4, to get 4Y X 2 + Y 2 + 4, X 2 + Y 2 4 X 2 + Y 2 + 4 ) S 2. Dennis The The Mathematics of Maps Lecture 4 17/29

Stereographic projection is circle preserving IDEA: Invert Ψ N : X = 2x 1 z, Y = 2y 1 z ( P = (x, y, z) = 4X X 2 + Y 2 + 4, If P Π S 2, then ax + by + cz = d. to get 4Y X 2 + Y 2 + 4, X 2 + Y 2 4 X 2 + Y 2 + 4 ) S 2. Dennis The The Mathematics of Maps Lecture 4 17/29

Stereographic projection is circle preserving IDEA: Invert Ψ N : X = 2x 1 z, Y = 2y 1 z ( P = (x, y, z) = 4X X 2 + Y 2 + 4, to get 4Y X 2 + Y 2 + 4, X 2 + Y 2 4 X 2 + Y 2 + 4 If P Π S 2, then ax + by + cz = d. Plug above in to get 4(aX + by ) + c(x 2 + Y 2 4) = d(x 2 + Y 2 + 4) ) S 2. Dennis The The Mathematics of Maps Lecture 4 17/29

Stereographic projection is circle preserving IDEA: Invert Ψ N : X = 2x 1 z, Y = 2y 1 z ( P = (x, y, z) = 4X X 2 + Y 2 + 4, to get 4Y X 2 + Y 2 + 4, X 2 + Y 2 4 X 2 + Y 2 + 4 If P Π S 2, then ax + by + cz = d. Plug above in to get 4(aX + by ) + c(x 2 + Y 2 4) = d(x 2 + Y 2 + 4) This is the eqn of a circle if c d and a line if c = d. ) S 2. Dennis The The Mathematics of Maps Lecture 4 17/29

Circles images Dennis The The Mathematics of Maps Lecture 4 18/29

From Mercator to stereographic projection Dennis The The Mathematics of Maps Lecture 4 19/29

A simple complex relationship Identify the plane with C, i.e. Z = X + iy = Re iθ and similarly for tilded variables. Dennis The The Mathematics of Maps Lecture 4 20/29

A simple complex relationship Identify the plane with C, i.e. Z = X + iy = Re iθ and similarly for tilded variables. The function F is described in real coordinates as R = 2e Y, Θ = X Dennis The The Mathematics of Maps Lecture 4 20/29

A simple complex relationship Identify the plane with C, i.e. Z = X + iy = Re iθ and similarly for tilded variables. The function F is described in real coordinates as R = 2e Y, Θ = X Using the complex exponential, we have Z = Re i Θ = 2e Y e ix = 2e i(x iy ) = 2e iz. Dennis The The Mathematics of Maps Lecture 4 20/29

Complex differentiable functions Given F : C C, define C-differentiability via the usual formula F F (a + h) F (a) (a) = lim. h 0 h Dennis The The Mathematics of Maps Lecture 4 21/29

Complex differentiable functions Given F : C C, define C-differentiability via the usual formula F F (a + h) F (a) (a) = lim. BUT, this is much stronger than h 0 h R-differentiability of F considered as a map R 2 R 2. FACT: When F exists and is nonzero, F is a conformal mapping. Dennis The The Mathematics of Maps Lecture 4 21/29

New conformal maps from old Given a conformal map Ψ from the sphere to the plane, we can use a C-differentiable F (with F 0) to produce a new map Ψ: Dennis The The Mathematics of Maps Lecture 4 22/29

New conformal maps from old Given a conformal map Ψ from the sphere to the plane, we can use a C-differentiable F (with F 0) to produce a new map Ψ: Möbius transformations F (z) = az+b cz+d form a wonderful class of C-differentiable (hence conformal) maps. Möbius transformations revealed Dennis The The Mathematics of Maps Lecture 4 22/29

A family of conformal maps Here s another way from Mercator to stereographic projection: The Mercator and stereographic projections, and many in between. Dennis The The Mathematics of Maps Lecture 4 23/29

Conic conformal projections Lambert (1772): Gave a family of conic conformal map projections with Mercator and stereographic projections as limiting cases. Dennis The The Mathematics of Maps Lecture 4 24/29

Conic conformal projections Lambert (1772): Gave a family of conic conformal map projections with Mercator and stereographic projections as limiting cases. Also allow φ 0 [ π 2, 0), i.e. cone vertex appears below the sphere. φ 0 = π 2 stereographic projection from the north pole. Dennis The The Mathematics of Maps Lecture 4 24/29

General conic projections Ψ C : { X = ρ(φ) sin(τθ) Y = ρ 0 ρ(φ) cos(τθ) Dennis The The Mathematics of Maps Lecture 4 25/29

General conic projections Ψ C : { X = ρ(φ) sin(τθ) Y = ρ 0 ρ(φ) cos(τθ) Meridians: uniformly spaced rays emanating from (0, ρ 0 ). Thus, θ = 0 meridian is along the Y -axis. Dennis The The Mathematics of Maps Lecture 4 25/29

General conic projections Ψ C : { X = ρ(φ) sin(τθ) Y = ρ 0 ρ(φ) cos(τθ) Meridians: uniformly spaced rays emanating from (0, ρ 0 ). Thus, θ = 0 meridian is along the Y -axis. Parallels: Circular arc centred at (0, ρ 0 ) of radius ρ(φ) (for latitude φ). Pick φ 0 such that ρ(φ 0 ) = ρ 0. Dennis The The Mathematics of Maps Lecture 4 25/29

General conic conformal projections Under Ψ C, meridians and parallels are orthogonal. Dennis The The Mathematics of Maps Lecture 4 26/29

General conic conformal projections Under Ψ C, meridians and parallels are orthogonal. Scale factors: 2π λ p = lim (2πτ) = ρ(φ)τ sec(φ) s cos(φ) λ m = lim ρ(φ + t) ρ(φ) t 0 t = ρ (φ) = ρ (φ) s 0 ρ(φ) s Conformality λ p = λ m yields the DE: ρ (φ) = ρ(φ)τ sec(φ), ρ(φ 0 ) = ρ 0. Dennis The The Mathematics of Maps Lecture 4 26/29

General conic conformal projections Under Ψ C, meridians and parallels are orthogonal. Scale factors: 2π λ p = lim (2πτ) = ρ(φ)τ sec(φ) s cos(φ) λ m = lim ρ(φ + t) ρ(φ) t 0 t = ρ (φ) = ρ (φ) s 0 ρ(φ) s Conformality λ p = λ m yields the DE: Solving yields ρ (φ) = ρ(φ)τ sec(φ), ρ(φ 0 ) = ρ 0. ( ) sec φ0 + tan φ τ 0 ρ(φ) = ρ 0. sec φ + tan φ Dennis The The Mathematics of Maps Lecture 4 26/29

General conic conformal projections Under Ψ C, meridians and parallels are orthogonal. Scale factors: 2π λ p = lim (2πτ) = ρ(φ)τ sec(φ) s cos(φ) λ m = lim ρ(φ + t) ρ(φ) t 0 t = ρ (φ) = ρ (φ) s 0 ρ(φ) s Conformality λ p = λ m yields the DE: Solving yields ρ (φ) = ρ(φ)τ sec(φ), ρ(φ 0 ) = ρ 0. ( ) sec φ0 + tan φ τ 0 ρ(φ) = ρ 0. sec φ + tan φ Still have several parameters to play with: ρ 0, τ, φ 0. Dennis The The Mathematics of Maps Lecture 4 26/29

Conical maps Additional requirements: Dennis The The Mathematics of Maps Lecture 4 27/29

Conical maps Additional requirements: Parallel at latitude φ 0 is a standard parallel, i.e. length is preserved. Dennis The The Mathematics of Maps Lecture 4 27/29

Conical maps Additional requirements: Parallel at latitude φ 0 is a standard parallel, i.e. length is preserved. 2π cos φ 0 = 2πτρ 0, so τ = cos φ 0 ρ 0. Dennis The The Mathematics of Maps Lecture 4 27/29

Conical maps Additional requirements: Parallel at latitude φ 0 is a standard parallel, i.e. length is preserved. 2π cos φ 0 = 2πτρ 0, so τ = cos φ 0 ρ 0. Cone tangential to sphere: ρ 0 = cot φ 0. Dennis The The Mathematics of Maps Lecture 4 27/29

Limiting cases The conic maps Ψ C have the following limiting cases: Dennis The The Mathematics of Maps Lecture 4 28/29

Limiting cases The conic maps Ψ C have the following limiting cases: φ 0 0: Mercator projection Ψ M. Dennis The The Mathematics of Maps Lecture 4 28/29

Limiting cases The conic maps Ψ C have the following limiting cases: φ 0 0: Mercator projection Ψ M. φ 0 π 2 2 : ρ(φ) sec φ+tan φ. Get Ψ S, rotated π 2 clockwise. Dennis The The Mathematics of Maps Lecture 4 28/29

Limiting cases The conic maps Ψ C have the following limiting cases: φ 0 0: Mercator projection Ψ M. φ 0 π 2 2 : ρ(φ) sec φ+tan φ. Get Ψ S, rotated π 2 clockwise. φ 0 π + 2 : ρ(φ) 2(sec φ + tan φ). Get Ψ N, rotated π 2 clockwise, but then reflected in the X -axis. Dennis The The Mathematics of Maps Lecture 4 28/29

Broader perspective on these lectures Q: Given two spaces with some structure, are they equivalent? Dennis The The Mathematics of Maps Lecture 4 29/29

Broader perspective on these lectures Q: Given two spaces with some structure, are they equivalent? The sphere and the plane are geodesically equivalent; symplectically equivalent (This refers to area in dim 2); conformally equivalent. They are not metrically equivalent. Have Gaussian curvature κ that is intrinsic and invariant under isometries ( Theorema Egregium ). A cylinder is metrically equivalent to the plane (also geodesically, symplectically, conformally), i.e. one has ideal maps for a cylinder. Dennis The The Mathematics of Maps Lecture 4 29/29