The Mathematics of Maps Lecture 4. Dennis The The Mathematics of Maps Lecture 4 1/29
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1 The Mathematics of Maps Lecture 4 Dennis The The Mathematics of Maps Lecture 4 1/29
2 Mercator projection Dennis The The Mathematics of Maps Lecture 4 2/29
3 The Mercator projection (1569) Dennis The The Mathematics of Maps Lecture 4 3/29
4 The Mercator projection (1569) Importance: Navigation. Dennis The The Mathematics of Maps Lecture 4 3/29
5 The Mercator projection (1569) Importance: Navigation. Conformal map, i.e. angles are preserved. Dennis The The Mathematics of Maps Lecture 4 3/29
6 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
7 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
8 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
9 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 The last one makes the Mercator map useful for navigation. Dennis The The Mathematics of Maps Lecture 4 4/29
10 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
11 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
12 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
13 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) g(φ) = ln(sec φ + tan φ). Dennis The The Mathematics of Maps Lecture 4 5/29
14 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) g(φ) = ln(sec φ + tan φ). Ψ M : { X = θ, π θ π Y = ln(sec φ + tan φ), π 2 φ π 2. Dennis The The Mathematics of Maps Lecture 4 5/29
15 Distortion in the Mercator projection Dennis The The Mathematics of Maps Lecture 4 6/29
16 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
17 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
18 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 Need φ, θ as functions of X, Y, i.e. inverse Mercator map Ψ 1 M. Dennis The The Mathematics of Maps Lecture 4 7/29
19 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
20 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
21 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
22 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
23 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
24 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 Dennis The The Mathematics of Maps Lecture 4 9/29
25 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
26 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
27 Stereographic projection Dennis The The Mathematics of Maps Lecture 4 10/29
28 Stereographic projection (from north pole) Dennis The The Mathematics of Maps Lecture 4 11/29
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
30 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
31 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
32 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
33 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). Hence, Ψ N : X = 2x 1 z, Y = 2y 1 z. Dennis The The Mathematics of Maps Lecture 4 12/29
34 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
35 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 Dennis The The Mathematics of Maps Lecture 4 13/29
36 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
37 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 Towards north pole, lim R. (High distortion.) φ π 2 Towards south pole, lim R 0. φ π + 2 Dennis The The Mathematics of Maps Lecture 4 13/29
38 Stereographic projection is conformal Halley (1695): Stereographic projection is conformal. Dennis The The Mathematics of Maps Lecture 4 14/29
39 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
40 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 ) Conformality means λ p = λ m. For f (φ) = 2(sec φ + tan φ), we can verify this is true. (Note f (φ) > 0 on ( π 2, π 2 ).) Dennis The The Mathematics of Maps Lecture 4 14/29
41 Distortion in the stereographic projection Dennis The The Mathematics of Maps Lecture 4 15/29
42 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
43 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
44 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
45 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
46 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 ) S 2. Dennis The The Mathematics of Maps Lecture 4 17/29
47 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 ) S 2. Dennis The The Mathematics of Maps Lecture 4 17/29
48 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 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
49 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 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
50 Circles images Dennis The The Mathematics of Maps Lecture 4 18/29
51 From Mercator to stereographic projection Dennis The The Mathematics of Maps Lecture 4 19/29
52 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
53 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
54 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
55 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
56 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
57 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. Dennis The The Mathematics of Maps Lecture 4 21/29
58 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
59 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. The Mercator-to-stereographic map is conformal and decomposes: Dennis The The Mathematics of Maps Lecture 4 21/29
60 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
61 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
62 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
63 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
64 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
65 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
66 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. Let s cut open the cone and flatten it. How to specify the map? Dennis The The Mathematics of Maps Lecture 4 24/29
67 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
68 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
69 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
70 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. Require ρ( π 2 ) = 0 and ρ (φ) < 0. Dennis The The Mathematics of Maps Lecture 4 25/29
71 General conic conformal projections Under Ψ C, meridians and parallels are orthogonal. Dennis The The Mathematics of Maps Lecture 4 26/29
72 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 Dennis The The Mathematics of Maps Lecture 4 26/29
73 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
74 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
75 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
76 Conical maps Additional requirements: Dennis The The Mathematics of Maps Lecture 4 27/29
77 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
78 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
79 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
80 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. ρ(φ) = cot φ 0 ( sec φ0 + tan φ 0 sec φ + tan φ ) sin φ0. Dennis The The Mathematics of Maps Lecture 4 27/29
81 Limiting cases The conic maps Ψ C have the following limiting cases: Dennis The The Mathematics of Maps Lecture 4 28/29
82 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
83 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
84 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
85 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. (Recall: complex conjugation and east west direction was clockwise earlier.) Dennis The The Mathematics of Maps Lecture 4 28/29
86 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
87 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. Dennis The The Mathematics of Maps Lecture 4 29/29
88 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 ). Dennis The The Mathematics of Maps Lecture 4 29/29
89 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
90 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. Higher dim? Different structures? Differential geometry! Dennis The The Mathematics of Maps Lecture 4 29/29
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