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 Importance: Navigation. Conformal map, i.e. angles are preserved. Here’s a fun little puzzle: Mercator Puzzle
The Mercator projection (1569)
Dennis The The Mathematics of Maps – Lecture 4 3/29 Conformal map, i.e. angles are preserved. Here’s a fun little puzzle: Mercator Puzzle
The Mercator projection (1569)
Importance: Navigation.
Dennis The The Mathematics of Maps – Lecture 4 3/29 Here’s a fun little puzzle: Mercator Puzzle
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 7→
parallels of latitude 7→ horizontal lines meridians of longitude 7→ vertical lines rhumb lines 7→ straight lines
The last one makes the Mercator map useful for navigation.
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 The last one makes the Mercator map useful for navigation.
Rhumb lines
Rhumb line = path of constant compass bearing on the sphere, i.e. have constant angle with the corresponding parallel & meridian.
7→
parallels of latitude 7→ horizontal lines meridians of longitude 7→ vertical lines rhumb lines 7→ straight lines
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.
7→
parallels of latitude 7→ horizontal lines meridians of longitude 7→ vertical lines rhumb lines 7→ straight lines
The last one makes the Mercator map useful for navigation.
Dennis The The Mathematics of Maps – Lecture 4 4/29 0 Conformality means λp = λm . Requiring g(0) = 0, g (φ) > 0,
g 0(φ) = sec(φ) ⇒ g(φ) = ln(sec φ + tan φ).
( X = θ, −π ≤ θ ≤ π ΨM : π π . Y = ln(sec φ + tan φ), − 2 ≤ φ ≤ 2
Mercator map formula
For a map Ψ : X = θ, Y = g(φ) , we calculated scale factors:
L(Ψ(As )) L(Ψ(Bt )) 0 λp = lim = sec(φ), λm = lim = |g (φ)|. s→0 L(As ) t→0 L(Bt )
Dennis The The Mathematics of Maps – Lecture 4 5/29 Requiring g(0) = 0, g 0(φ) > 0,
g 0(φ) = sec(φ) ⇒ g(φ) = ln(sec φ + tan φ).
( X = θ, −π ≤ θ ≤ π ΨM : π π . Y = ln(sec φ + tan φ), − 2 ≤ φ ≤ 2
Mercator map formula
For a map Ψ : X = θ, Y = g(φ) , we calculated scale factors:
L(Ψ(As )) L(Ψ(Bt )) 0 λp = lim = sec(φ), λm = lim = |g (φ)|. s→0 L(As ) t→0 L(Bt )
Conformality means λp = λm .
Dennis The The Mathematics of Maps – Lecture 4 5/29 ⇒ g(φ) = ln(sec φ + tan φ).
( X = θ, −π ≤ θ ≤ π ΨM : π π . Y = ln(sec φ + tan φ), − 2 ≤ φ ≤ 2
Mercator map formula
For a map Ψ : X = θ, Y = g(φ) , we calculated scale factors:
L(Ψ(As )) L(Ψ(Bt )) 0 λp = lim = sec(φ), λm = lim = |g (φ)|. s→0 L(As ) t→0 L(Bt ) 0 Conformality means λp = λm . Requiring g(0) = 0, g (φ) > 0,
g 0(φ) = sec(φ)
Dennis The The Mathematics of Maps – Lecture 4 5/29 ( X = θ, −π ≤ θ ≤ π ΨM : π π . Y = ln(sec φ + tan φ), − 2 ≤ φ ≤ 2
Mercator map formula
For a map Ψ : X = θ, Y = g(φ) , we calculated scale factors:
L(Ψ(As )) L(Ψ(Bt )) 0 λp = lim = sec(φ), λm = lim = |g (φ)|. s→0 L(As ) t→0 L(Bt ) 0 Conformality means λp = λm . Requiring g(0) = 0, g (φ) > 0,
g 0(φ) = sec(φ) ⇒ g(φ) = ln(sec φ + tan φ).
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(As )) L(Ψ(Bt )) 0 λp = lim = sec(φ), λm = lim = |g (φ)|. s→0 L(As ) t→0 L(Bt ) 0 Conformality means λp = λm . Requiring g(0) = 0, g (φ) > 0,
g 0(φ) = sec(φ) ⇒ g(φ) = ln(sec φ + tan φ).
( X = θ, −π ≤ θ ≤ π ΨM : π π . Y = ln(sec φ + tan φ), − 2 ≤ φ ≤ 2
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 For a path (φ(t), θ(t)), 0 ≤ t ≤ 1 on S2, we saw in Lecture 2 a Z 1 q formula for its length L = (φ0)2 + cos2(φ)(θ0)2dt . 0 −1 Need φ, θ as functions of X , Y , i.e. inverse Mercator map ΨM .
Distance along a rhumb line
Q: How can we find the distance along a rhumb line?
Rhumb line 7→ line segment X (t) = at + b, Y (t) = ct + d .
Dennis The The Mathematics of Maps – Lecture 4 7/29 −1 Need φ, θ as functions of X , Y , i.e. inverse Mercator map ΨM .
Distance along a rhumb line
Q: How can we find the distance along a rhumb line?
Rhumb line 7→ line segment X (t) = at + b, Y (t) = ct + d .
For a path (φ(t), θ(t)), 0 ≤ t ≤ 1 on S2, we saw in Lecture 2 a Z 1 q formula for its length L = (φ0)2 + cos2(φ)(θ0)2dt . 0
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 7→ line segment X (t) = at + b, Y (t) = ct + d .
For a path (φ(t), θ(t)), 0 ≤ t ≤ 1 on S2, we saw in Lecture 2 a Z 1 q formula for its length L = (φ0)2 + cos2(φ)(θ0)2dt . 0 −1 Need φ, θ as functions of X , Y , i.e. inverse Mercator map ΨM . Dennis The The Mathematics of Maps – Lecture 4 7/29 Y 1+sin(φ) Using some trig. trickery... e = sec(φ) + tan(φ) = cos(φ) = π 2 φ π 1−cos(φ+ 2 ) 2 sin ( 2 + 4 ) φ π sin(φ+ π ) = φ π φ π = tan( 2 + 4 ). Thus, 2 2 sin( 2 + 4 ) cos( 2 + 4 ) ( θ = X Ψ−1 : M Y π φ = 2 arctan(e ) − 2
Inverse of the Mercator map
( X = θ −1 Given ΨM : , what is Ψ ? Y = ln(sec(φ) + tan(φ)) M
Dennis The The Mathematics of Maps – Lecture 4 8/29 Thus,
( θ = X Ψ−1 : M Y π φ = 2 arctan(e ) − 2
Inverse of the Mercator map
( X = θ −1 Given ΨM : , what is Ψ ? Y = ln(sec(φ) + tan(φ)) M
Y 1+sin(φ) Using some trig. trickery... e = sec(φ) + tan(φ) = cos(φ) = π 2 φ π 1−cos(φ+ 2 ) 2 sin ( 2 + 4 ) φ π sin(φ+ π ) = φ π φ π = tan( 2 + 4 ). 2 2 sin( 2 + 4 ) cos( 2 + 4 )
Dennis The The Mathematics of Maps – Lecture 4 8/29 Inverse of the Mercator map
( X = θ −1 Given ΨM : , what is Ψ ? Y = ln(sec(φ) + tan(φ)) M
Y 1+sin(φ) Using some trig. trickery... e = sec(φ) + tan(φ) = cos(φ) = π 2 φ π 1−cos(φ+ 2 ) 2 sin ( 2 + 4 ) φ π sin(φ+ π ) = φ π φ π = tan( 2 + 4 ). Thus, 2 2 sin( 2 + 4 ) cos( 2 + 4 ) ( θ = X Ψ−1 : M Y π φ = 2 arctan(e ) − 2
Dennis The The Mathematics of Maps – Lecture 4 8/29 2eY c 2eY ⇒ φ0 = , θ0 = a, cos(φ) = . 1 + e2Y 1 + e2Y
Z 1 q Z 1 Y 0 2 2 0 2 p 2 2 2e L = (φ ) + cos (φ)(θ ) dt = a + c 2Y dt 0 0 1 + e √ 2 a2 + c2 = ... = (arctan(ec+d ) − arctan(ed )) . c
2ed (When c → 0, we recover L = 1+e2d = cos(φ).) This formula is valid on the unit sphere S2. On a sphere of radius R, this length would be rescaled by R.
Distance along a rhumb line - 2
( ( θ = X Rhumb X (t) = at + b Ψ−1 : , M Y π line: φ = 2 arctan(e ) − 2 Y (t) = ct + d
Dennis The The Mathematics of Maps – Lecture 4 9/29 Z 1 q Z 1 Y 0 2 2 0 2 p 2 2 2e L = (φ ) + cos (φ)(θ ) dt = a + c 2Y dt 0 0 1 + e √ 2 a2 + c2 = ... = (arctan(ec+d ) − arctan(ed )) . c
2ed (When c → 0, we recover L = 1+e2d = cos(φ).) This formula is valid on the unit sphere S2. On a sphere of radius R, this length would be rescaled by R.
Distance along a rhumb line - 2
( ( θ = X Rhumb X (t) = at + b Ψ−1 : , M Y π line: φ = 2 arctan(e ) − 2 Y (t) = ct + d
2eY c 2eY ⇒ φ0 = , θ0 = a, cos(φ) = . 1 + e2Y 1 + e2Y
Dennis The The Mathematics of Maps – Lecture 4 9/29 2ed (When c → 0, we recover L = 1+e2d = cos(φ).) This formula is valid on the unit sphere S2. On a sphere of radius R, this length would be rescaled by R.
Distance along a rhumb line - 2
( ( θ = X Rhumb X (t) = at + b Ψ−1 : , M Y π line: φ = 2 arctan(e ) − 2 Y (t) = ct + d
2eY c 2eY ⇒ φ0 = , θ0 = a, cos(φ) = . 1 + e2Y 1 + e2Y
Z 1 q Z 1 Y 0 2 2 0 2 p 2 2 2e L = (φ ) + cos (φ)(θ ) dt = a + c 2Y dt 0 0 1 + e √ 2 a2 + c2 = ... = (arctan(ec+d ) − arctan(ed )) . c
Dennis The The Mathematics of Maps – Lecture 4 9/29 This formula is valid on the unit sphere S2. On a sphere of radius R, this length would be rescaled by R.
Distance along a rhumb line - 2
( ( θ = X Rhumb X (t) = at + b Ψ−1 : , M Y π line: φ = 2 arctan(e ) − 2 Y (t) = ct + d
2eY c 2eY ⇒ φ0 = , θ0 = a, cos(φ) = . 1 + e2Y 1 + e2Y
Z 1 q Z 1 Y 0 2 2 0 2 p 2 2 2e L = (φ ) + cos (φ)(θ ) dt = a + c 2Y dt 0 0 1 + e √ 2 a2 + c2 = ... = (arctan(ec+d ) − arctan(ed )) . c
2ed (When c → 0, we recover L = 1+e2d = cos(φ).)
Dennis The The Mathematics of Maps – Lecture 4 9/29 Distance along a rhumb line - 2
( ( θ = X Rhumb X (t) = at + b Ψ−1 : , M Y π line: φ = 2 arctan(e ) − 2 Y (t) = ct + d
2eY c 2eY ⇒ φ0 = , θ0 = a, cos(φ) = . 1 + e2Y 1 + e2Y
Z 1 q Z 1 Y 0 2 2 0 2 p 2 2 2e L = (φ ) + cos (φ)(θ ) dt = a + c 2Y dt 0 0 1 + e √ 2 a2 + c2 = ... = (arctan(ec+d ) − arctan(ed )) . c
2ed (When c → 0, we recover L = 1+e2d = cos(φ).) This formula is valid on the unit sphere S2. 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 parallels of latitude 7→ circles centred at 0 meridians of longitude 7→ rays emanating from 0
(Note: Viewed from above, East West is a clockwise rotation in the map.)
Stereographic projection (from north pole)
Dennis The The Mathematics of Maps – Lecture 4 11/29 (Note: Viewed from above, East West is a clockwise rotation in the map.)
Stereographic projection (from north pole)
parallels of latitude 7→ circles centred at 0 meridians of longitude 7→ rays emanating from 0
Dennis The The Mathematics of Maps – Lecture 4 11/29 Stereographic projection (from north pole)
parallels of latitude 7→ circles centred at 0 meridians of longitude 7→ 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 ~ ~ 2x 2y Line NP intersects z = −1 plane at Q:( 1−z , 1−z , −1). Hence, 2x 2y Ψ : X = , Y = . N 1 − z 1 − z
Geometric construction
On S2, put a light source at the north pole, and cast the shadow of P~ ∈ S2 onto the plane tangent at the south pole.
Dennis The The Mathematics of Maps – Lecture 4 12/29 Hence,
2x 2y Ψ : X = , Y = . N 1 − z 1 − z
Geometric construction
On S2, put a light source at the north pole, and cast the shadow of P~ ∈ S2 onto the plane tangent at the south pole.
~ ~ 2x 2y Line NP intersects z = −1 plane at Q:( 1−z , 1−z , −1).
Dennis The The Mathematics of Maps – Lecture 4 12/29 Geometric construction
On S2, put a light source at the north pole, and cast the shadow of P~ ∈ S2 onto the plane tangent at the south pole.
~ ~ 2x 2y Line NP intersects z = −1 plane at Q:( 1−z , 1−z , −1). Hence, 2x 2y Ψ : X = , Y = . N 1 − z 1 − z
Dennis The The Mathematics of Maps – Lecture 4 12/29 2 cos φ cos θ 2 cos φ sin θ Then X = 1−sin φ , Y = 1−sin φ , and
p 2 cos φ 2 cos φ(1 + sin φ) ⇒ R = X 2 + Y 2 = = 1 − sin φ 1 − sin2 φ
ΨN : R = 2(sec φ + tan φ), Θ = θ Towards north pole, lim R → ∞. (High distortion.) π − φ→ 2 Towards south pole, lim R → 0. π + φ→− 2
Polar coordinate formula
Use spherical polar coords (x, y, z) = (cos φ cos θ, cos φ sin θ, sin φ) to write ΨN .
Dennis The The Mathematics of Maps – Lecture 4 13/29 ΨN : R = 2(sec φ + tan φ), Θ = θ Towards north pole, lim R → ∞. (High distortion.) π − φ→ 2 Towards south pole, lim R → 0. π + φ→− 2
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 ΨN . Then X = 1−sin φ , Y = 1−sin φ , and
p 2 cos φ 2 cos φ(1 + sin φ) ⇒ R = X 2 + Y 2 = = 1 − sin φ 1 − sin2 φ
Dennis The The Mathematics of Maps – Lecture 4 13/29 Towards north pole, lim R → ∞. (High distortion.) π − φ→ 2 Towards south pole, lim R → 0. π + φ→− 2
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 ΨN . Then X = 1−sin φ , Y = 1−sin φ , and
p 2 cos φ 2 cos φ(1 + sin φ) ⇒ R = X 2 + Y 2 = = 1 − sin φ 1 − sin2 φ
ΨN : R = 2(sec φ + tan φ), Θ = θ
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 ΨN . Then X = 1−sin φ , Y = 1−sin φ , and
p 2 cos φ 2 cos φ(1 + sin φ) ⇒ R = X 2 + Y 2 = = 1 − sin φ 1 − sin2 φ
ΨN : R = 2(sec φ + tan φ), Θ = θ Towards north pole, lim R → ∞. (High distortion.) π − φ→ 2 Towards south pole, lim R → 0. π + φ→− 2
Dennis The The Mathematics of Maps – Lecture 4 13/29 For a map Ψ : R = f (φ), Θ = θ , we calculated scale factors
L(Ψ(As )) L(Ψ(Bt )) 0 λp = lim = f (φ) sec(φ), λm = lim = |f (φ)|. s→0 L(As ) t→0 L(Bt )
Conformality means λp = λm . For f (φ) = 2(sec φ + tan φ), we 0 π π can verify this is true. (Note f (φ) > 0 on (− 2 , 2 ).)
Stereographic projection is conformal
Halley (1695): Stereographic projection is conformal.
Dennis The The Mathematics of Maps – Lecture 4 14/29 Conformality means λp = λm . For f (φ) = 2(sec φ + tan φ), we 0 π π can verify this is true. (Note f (φ) > 0 on (− 2 , 2 ).)
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(As )) L(Ψ(Bt )) 0 λp = lim = f (φ) sec(φ), λm = lim = |f (φ)|. s→0 L(As ) t→0 L(Bt )
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(As )) L(Ψ(Bt )) 0 λp = lim = f (φ) sec(φ), λm = lim = |f (φ)|. s→0 L(As ) t→0 L(Bt )
Conformality means λp = λm . For f (φ) = 2(sec φ + tan φ), we 0 π π can verify this is true. (Note f (φ) > 0 on (− 2 , 2 ).)
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 Theorem 2 Any circle on S is mapped by ΨN to a “circle” in the plane.
Circles are mapped to “circles”
Definition 2 3 2 A circle on S is the intersection of a plane Π in R with S . 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 2 3 2 A circle on S is the intersection of a plane Π in R with S . A “circle” in the plane is either an ordinary circle or a line.
Theorem 2 Any circle on S is mapped by ΨN to a “circle” in the plane.
Dennis The The Mathematics of Maps – Lecture 4 16/29 Circles are mapped to “circles”
Definition 2 3 2 A circle on S is the intersection of a plane Π in R with S . A “circle” in the plane is either an ordinary circle or a line.
Theorem 2 Any circle on S is mapped by ΨN to a “circle” in the plane.
Dennis The The Mathematics of Maps – Lecture 4 16/29 to get
4X 4Y X 2 + Y 2 − 4 P = (x, y, z) = , , ∈ S2. X 2 + Y 2 + 4 X 2 + Y 2 + 4 X 2 + Y 2 + 4
If P ∈ Π ∩ S2, 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 6= d and a line if c = d.
Stereographic projection is circle preserving
2x 2y IDEA: Invert ΨN : X = 1−z , Y = 1−z
Dennis The The Mathematics of Maps – Lecture 4 17/29 If P ∈ Π ∩ S2, 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 6= d and a line if c = d.
Stereographic projection is circle preserving
2x 2y IDEA: Invert ΨN : X = 1−z , Y = 1−z to get
4X 4Y X 2 + Y 2 − 4 P = (x, y, z) = , , ∈ S2. X 2 + Y 2 + 4 X 2 + Y 2 + 4 X 2 + Y 2 + 4
Dennis The The Mathematics of Maps – Lecture 4 17/29 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 6= d and a line if c = d.
Stereographic projection is circle preserving
2x 2y IDEA: Invert ΨN : X = 1−z , Y = 1−z to get
4X 4Y X 2 + Y 2 − 4 P = (x, y, z) = , , ∈ S2. X 2 + Y 2 + 4 X 2 + Y 2 + 4 X 2 + Y 2 + 4
If P ∈ Π ∩ S2, then ax + by + cz = d.
Dennis The The Mathematics of Maps – Lecture 4 17/29 This is the eqn of a circle if c 6= d and a line if c = d.
Stereographic projection is circle preserving
2x 2y IDEA: Invert ΨN : X = 1−z , Y = 1−z to get
4X 4Y X 2 + Y 2 − 4 P = (x, y, z) = , , ∈ S2. X 2 + Y 2 + 4 X 2 + Y 2 + 4 X 2 + Y 2 + 4
If P ∈ Π ∩ S2, 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)
Dennis The The Mathematics of Maps – Lecture 4 17/29 Stereographic projection is circle preserving
2x 2y IDEA: Invert ΨN : X = 1−z , Y = 1−z to get
4X 4Y X 2 + Y 2 − 4 P = (x, y, z) = , , ∈ S2. X 2 + Y 2 + 4 X 2 + Y 2 + 4 X 2 + Y 2 + 4
If P ∈ Π ∩ S2, 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 6= d and a line if c = d.
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 The function F is described in real coordinates as
R˜ = 2eY , Θ˜ = X
Using the complex exponential, we have
˜ Z˜ = Re˜ iΘ = 2eY eiX = 2ei(X −iY ) = 2eiZ .
A simple complex relationship
iΘ Identify the plane with C, i.e. Z = X + iY = Re and similarly for tilded variables.
Dennis The The Mathematics of Maps – Lecture 4 20/29 The function F is described in real coordinates as
R˜ = 2eY , Θ˜ = X
Using the complex exponential, we have
˜ Z˜ = Re˜ iΘ = 2eY eiX = 2ei(X −iY ) = 2eiZ .
A simple complex relationship
iΘ Identify the plane with C, i.e. Z = X + iY = Re and similarly for tilded variables.
Dennis The The Mathematics of Maps – Lecture 4 20/29 Using the complex exponential, we have
˜ Z˜ = Re˜ iΘ = 2eY eiX = 2ei(X −iY ) = 2eiZ .
A simple complex relationship
iΘ Identify the plane with C, i.e. Z = X + iY = Re and similarly for tilded variables.
The function F is described in real coordinates as
R˜ = 2eY , Θ˜ = X
Dennis The The Mathematics of Maps – Lecture 4 20/29 A simple complex relationship
iΘ Identify the plane with C, i.e. Z = X + iY = Re and similarly for tilded variables.
The function F is described in real coordinates as
R˜ = 2eY , Θ˜ = X
Using the complex exponential, we have
˜ Z˜ = Re˜ iΘ = 2eY eiX = 2ei(X −iY ) = 2eiZ .
Dennis The The Mathematics of Maps – Lecture 4 20/29 BUT, this is much stronger than 2 2 R-differentiability of F considered as a map R → R . FACT: When F 0 exists and is nonzero, F is a conformal mapping. The Mercator-to-stereographic map is conformal and decomposes:
Complex differentiable functions
Given F : C → C, define C-differentiability via the usual formula F (a + h) − F (a) F 0(a) = lim . h→0 h
Dennis The The Mathematics of Maps – Lecture 4 21/29 FACT: When F 0 exists and is nonzero, F is a conformal mapping. The Mercator-to-stereographic map is conformal and decomposes:
Complex differentiable functions
Given F : C → C, define C-differentiability via the usual formula F (a + h) − F (a) F 0(a) = lim . BUT, this is much stronger than h→0 h 2 2 R-differentiability of F considered as a map R → R .
Dennis The The Mathematics of Maps – Lecture 4 21/29 The Mercator-to-stereographic map is conformal and decomposes:
Complex differentiable functions
Given F : C → C, define C-differentiability via the usual formula F (a + h) − F (a) F 0(a) = lim . BUT, this is much stronger than h→0 h 2 2 R-differentiability of F considered as a map R → R . FACT: When F 0 exists and is nonzero, F is a conformal mapping.
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 (a + h) − F (a) F 0(a) = lim . BUT, this is much stronger than h→0 h 2 2 R-differentiability of F considered as a map R → R . FACT: When F 0 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 az+b M¨obiustransformations F (z) = cz+d form a wonderful class of C-differentiable (hence conformal) maps. M¨obiustransformations revealed
New conformal maps from old
Given a conformal map Ψ from the sphere to the plane, we can use 0 a C-differentiable F (with F 6= 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 0 a C-differentiable F (with F 6= 0) to produce a new map Ψ:˜
az+b M¨obiustransformations F (z) = cz+d form a wonderful class of C-differentiable (hence conformal) maps. M¨obiustransformations 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 π Also allow φ0 ∈ [− 2 , 0), i.e. cone vertex appears below the sphere. π φ0 = − 2 stereographic projection from the north pole. Let’s cut open the cone and flatten it. How to specify the map?
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 π Also allow φ0 ∈ [− 2 , 0), i.e. cone vertex appears below the sphere. π φ0 = − 2 stereographic projection from the north pole. Let’s cut open the cone and flatten it. How to specify the map?
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 Let’s cut open the cone and flatten it. How to specify the map?
Conic conformal projections
Lambert (1772): Gave a family of conic conformal map projections with Mercator and stereographic projections as limiting cases.
π Also allow φ0 ∈ [− 2 , 0), i.e. cone vertex appears below the sphere. π φ0 = − 2 stereographic projection from the north pole.
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.
π Also allow φ0 ∈ [− 2 , 0), i.e. cone vertex appears below the sphere. π φ0 = − 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 Meridians: uniformly spaced rays emanating from (0, ρ0). Thus, θ = 0 meridian is along the Y -axis.
Parallels: Circular arc centred at (0, ρ0) of radius ρ(φ) (for latitude φ). Pick φ0 such that ρ(φ0) = ρ0. π 0 Require ρ( 2 ) = 0 and ρ (φ) < 0.
General conic projections
( X = ρ(φ) sin(τθ) ΨC : Y = ρ0 − ρ(φ) cos(τθ)
Dennis The The Mathematics of Maps – Lecture 4 25/29 Parallels: Circular arc centred at (0, ρ0) of radius ρ(φ) (for latitude φ). Pick φ0 such that ρ(φ0) = ρ0. π 0 Require ρ( 2 ) = 0 and ρ (φ) < 0.
General conic projections
( X = ρ(φ) sin(τθ) ΨC : Y = ρ0 − ρ(φ) cos(τθ)
Meridians: uniformly spaced rays emanating from (0, ρ0). Thus, θ = 0 meridian is along the Y -axis.
Dennis The The Mathematics of Maps – Lecture 4 25/29 π 0 Require ρ( 2 ) = 0 and ρ (φ) < 0.
General conic projections
( X = ρ(φ) sin(τθ) ΨC : Y = ρ0 − ρ(φ) cos(τθ)
Meridians: uniformly spaced rays emanating from (0, ρ0). Thus, θ = 0 meridian is along the Y -axis.
Parallels: Circular arc centred at (0, ρ0) of radius ρ(φ) (for latitude φ). Pick φ0 such that ρ(φ0) = ρ0.
Dennis The The Mathematics of Maps – Lecture 4 25/29 General conic projections
( X = ρ(φ) sin(τθ) ΨC : Y = ρ0 − ρ(φ) cos(τθ)
Meridians: uniformly spaced rays emanating from (0, ρ0). Thus, θ = 0 meridian is along the Y -axis.
Parallels: Circular arc centred at (0, ρ0) of radius ρ(φ) (for latitude φ). Pick φ0 such that ρ(φ0) = ρ0. π 0 Require ρ( 2 ) = 0 and ρ (φ) < 0.
Dennis The The Mathematics of Maps – Lecture 4 25/29 Scale factors:
s ρ(φ) 2π (2πτ) λp = lim = ρ(φ)τ sec(φ) s→0 s cos(φ)
ρ(φ + t) − ρ(φ) 0 0 λm = lim = |ρ (φ)| = −ρ (φ) t→0 t
Conformality λp = λm yields the DE:
0 ρ (φ) = −ρ(φ)τ sec(φ), ρ(φ0) = ρ0.
Solving yields
sec φ + tan φ τ ρ(φ) = ρ 0 0 . 0 sec φ + tan φ
Still have several parameters to play with: ρ0, τ, φ0 .
General conic conformal projections
Under ΨC , meridians and parallels are orthogonal.
Dennis The The Mathematics of Maps – Lecture 4 26/29 Conformality λp = λm yields the DE:
0 ρ (φ) = −ρ(φ)τ sec(φ), ρ(φ0) = ρ0.
Solving yields
sec φ + tan φ τ ρ(φ) = ρ 0 0 . 0 sec φ + tan φ
Still have several parameters to play with: ρ0, τ, φ0 .
General conic conformal projections
Under ΨC , meridians and parallels are orthogonal. Scale factors:
s ρ(φ) 2π (2πτ) λp = lim = ρ(φ)τ sec(φ) s→0 s cos(φ)
ρ(φ + t) − ρ(φ) 0 0 λm = lim = |ρ (φ)| = −ρ (φ) t→0 t
Dennis The The Mathematics of Maps – Lecture 4 26/29 Solving yields
sec φ + tan φ τ ρ(φ) = ρ 0 0 . 0 sec φ + tan φ
Still have several parameters to play with: ρ0, τ, φ0 .
General conic conformal projections
Under ΨC , meridians and parallels are orthogonal. Scale factors:
s ρ(φ) 2π (2πτ) λp = lim = ρ(φ)τ sec(φ) s→0 s cos(φ)
ρ(φ + t) − ρ(φ) 0 0 λm = lim = |ρ (φ)| = −ρ (φ) t→0 t
Conformality λp = λm yields the DE:
0 ρ (φ) = −ρ(φ)τ sec(φ), ρ(φ0) = ρ0.
Dennis The The Mathematics of Maps – Lecture 4 26/29 Still have several parameters to play with: ρ0, τ, φ0 .
General conic conformal projections
Under ΨC , meridians and parallels are orthogonal. Scale factors:
s ρ(φ) 2π (2πτ) λp = lim = ρ(φ)τ sec(φ) s→0 s cos(φ)
ρ(φ + t) − ρ(φ) 0 0 λm = lim = |ρ (φ)| = −ρ (φ) t→0 t
Conformality λp = λm yields the DE:
0 ρ (φ) = −ρ(φ)τ sec(φ), ρ(φ0) = ρ0.
Solving yields
sec φ + tan φ τ ρ(φ) = ρ 0 0 . 0 sec φ + tan φ
Dennis The The Mathematics of Maps – Lecture 4 26/29 General conic conformal projections
Under ΨC , meridians and parallels are orthogonal. Scale factors:
s ρ(φ) 2π (2πτ) λp = lim = ρ(φ)τ sec(φ) s→0 s cos(φ)
ρ(φ + t) − ρ(φ) 0 0 λm = lim = |ρ (φ)| = −ρ (φ) t→0 t
Conformality λp = λm yields the DE:
0 ρ (φ) = −ρ(φ)τ sec(φ), ρ(φ0) = ρ0.
Solving yields
sec φ + tan φ τ ρ(φ) = ρ 0 0 . 0 sec φ + tan φ
Still have several parameters to play with: ρ0, τ, φ0 .
Dennis The The Mathematics of Maps – Lecture 4 26/29 Parallel at latitude φ0 is a standard parallel, i.e. length is preserved. 2π cos φ = 2πτρ , so τ = cos φ0 . 0 0 ρ0
Cone tangential to sphere: ρ0 = cot φ0 . sec φ + tan φ sin φ0 ⇒ ρ(φ) = cot φ 0 0 . 0 sec φ + tan φ
Conical maps
Additional requirements:
Dennis The The Mathematics of Maps – Lecture 4 27/29 2π cos φ = 2πτρ , so τ = cos φ0 . 0 0 ρ0
Cone tangential to sphere: ρ0 = cot φ0 . sec φ + tan φ sin φ0 ⇒ ρ(φ) = cot φ 0 0 . 0 sec φ + tan φ
Conical maps
Additional requirements: Parallel at latitude φ0 is a standard parallel, i.e. length is preserved.
Dennis The The Mathematics of Maps – Lecture 4 27/29 Cone tangential to sphere: ρ0 = cot φ0 . sec φ + tan φ sin φ0 ⇒ ρ(φ) = cot φ 0 0 . 0 sec φ + tan φ
Conical maps
Additional requirements: Parallel at latitude φ0 is a standard parallel, i.e. length is preserved. 2π cos φ = 2πτρ , so τ = cos φ0 . 0 0 ρ0
Dennis The The Mathematics of Maps – Lecture 4 27/29 sec φ + tan φ sin φ0 ⇒ ρ(φ) = cot φ 0 0 . 0 sec φ + tan φ
Conical maps
Additional requirements: Parallel at latitude φ0 is a standard parallel, i.e. length is preserved. 2π cos φ = 2πτρ , so τ = cos φ0 . 0 0 ρ0
Cone tangential to sphere: ρ0 = cot φ0 .
Dennis The The Mathematics of Maps – Lecture 4 27/29 Conical maps
Additional requirements: Parallel at latitude φ0 is a standard parallel, i.e. length is preserved. 2π cos φ = 2πτρ , so τ = cos φ0 . 0 0 ρ0
Cone tangential to sphere: ρ0 = cot φ0 . sec φ + tan φ sin φ0 ⇒ ρ(φ) = cot φ 0 0 . 0 sec φ + tan φ
Dennis The The Mathematics of Maps – Lecture 4 27/29 φ0 → 0: Mercator projection ΨM . π − 2 π φ0 → 2 : ρ(φ) → sec φ+tan φ . Get ΨS , rotated 2 clockwise. π + π φ0 → − 2 : ρ(φ) → −2(sec φ + tan φ). Get ΨN , rotated 2 clockwise, but then reflected in the X -axis. (Recall: complex conjugation and east west direction was clockwise earlier.)
Limiting cases
The conic maps ΨC have the following limiting cases:
Dennis The The Mathematics of Maps – Lecture 4 28/29 π − 2 π φ0 → 2 : ρ(φ) → sec φ+tan φ . Get ΨS , rotated 2 clockwise. π + π φ0 → − 2 : ρ(φ) → −2(sec φ + tan φ). Get ΨN , rotated 2 clockwise, but then reflected in the X -axis. (Recall: complex conjugation and east west direction was clockwise earlier.)
Limiting cases
The conic maps ΨC have the following limiting cases: φ0 → 0: Mercator projection ΨM .
Dennis The The Mathematics of Maps – Lecture 4 28/29 π + π φ0 → − 2 : ρ(φ) → −2(sec φ + tan φ). Get ΨN , rotated 2 clockwise, but then reflected in the X -axis. (Recall: complex conjugation and east west direction was clockwise earlier.)
Limiting cases
The conic maps ΨC have the following limiting cases: φ0 → 0: Mercator projection ΨM . π − 2 π φ0 → 2 : ρ(φ) → sec φ+tan φ . Get ΨS , rotated 2 clockwise.
Dennis The The Mathematics of Maps – Lecture 4 28/29 (Recall: complex conjugation and east west direction was clockwise earlier.)
Limiting cases
The conic maps ΨC have the following limiting cases: φ0 → 0: Mercator projection ΨM . π − 2 π φ0 → 2 : ρ(φ) → sec φ+tan φ . Get ΨS , rotated 2 clockwise. π + π φ0 → − 2 : ρ(φ) → −2(sec φ + tan φ). Get ΨN , rotated 2 clockwise, but then reflected in the X -axis.
Dennis The The Mathematics of Maps – Lecture 4 28/29 Limiting cases
The conic maps ΨC have the following limiting cases: φ0 → 0: Mercator projection ΨM . π − 2 π φ0 → 2 : ρ(φ) → sec φ+tan φ . Get ΨS , rotated 2 clockwise. π + π φ0 → − 2 : ρ(φ) → −2(sec φ + tan φ). Get ΨN , rotated 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 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!
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 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!
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 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!
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 Higher dim? Different structures? Differential geometry!
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 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