Map Projections. Jim Emery. Revision 7/21/ The Earth Ellipsoid 2. 2 The Fundamental Quadratic Forms of a Surface 6

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1 Map Projections Jim Emery Revision 7/21/2004 Contents 1 The Earth Ellipsoid 2 2 The Fundamental Quadratic Forms of a Surface 6 3 Principal Directions On A Surface 9 4 Principal Normal Curvatures on the Ellipsoid 11 5 The Coefficients of the First Fundamental Form For the Geodetic Representation of the Ellipsoid 13 6 Conformal Maps of Surfaces 14 7 A Conformal Map of the Ellipsoid to the Sphere 18 8 A Conformal Map of the Ellipsoid to the Plane: The Polar Stereographic Projection 22 9 The Lambert Projection: A conformal Map From An Ellipsoid To A Cone Bibliography 23 1

2 1 The Earth Ellipsoid We shall derive the parametric representation of the ellipsoid in the geodetic coordinates. The geodetic latitude is measured by the surface normal rather than by the line from the center of the ellipsoid. Traditionally the earth latitude is measured by astronomically determining the angle between the earth surface normal and the pole star. The geodetic latitude is the complement of this angle. The geocentric latitude is not so easily measured. Hence maps are traditionally given in terms of geodetic coordinates. The ellipsoid is generated by rotating the cross sectional ellipse about the polar axis. The equatorial radius is called a and the polar radius is called b. The equatorial radius is greater than the polar radius. Let the generating ellipse be given parametrically as p(θ) =(x, z) =(a cos(θ), bsin(θ)). The angle θ is the geocentric latitude. A tangent vector at the point p is A normal vector at p is t =( a sin(θ),bcos(θ)). n =(b cos(θ),asin(θ)). The geodetic latitude is the angle of the normal and is given by φ =tan 1 ( a sin(θ) b cos(θ) ). Let q(s) be a parametric representation of a line through p in the direction of the normal, q(s) =p(θ)+sn(θ)) =(a cos(θ)+sbcos(θ),bsin(θ)+sa sin(θ)) =((a + sb)cos(θ), (b + sa)sin(θ)) This line meets the polar axis. We want to find the distance d from p to this intersection point. At the intersection point the x coordinate is zero, so we have s = a/b 2

3 P a b θ φ Q Figure 1: Coordinates on the earth ellipsoid. The geocentric latitude is θ, the geodetic latitude is φ, the equatorial radius is a, the polar radius is b, P is a point on the ellipsoid, distance PQ is equal to the principal radius of normal curvature in the direction perpendicular to the meridian. PQ is normal to the surface. 3

4 and Then So that z =(b a 2 /b)sin(θ) = b2 a 2 sin(θ). b d 2 = a 2 cos 2 (θ)+(bsin(θ) sin(θ) b2 a 2 ) 2. b d(θ) = a b b 2 cos 2 (θ)+a 2 sin 2 (θ). We can write this as a function of the geodetic latitude φ, andthenwe shall call this distance N We have N(φ) =d(θ) = a b b 2 cos 2 (θ)+a 2 sin 2 (θ) = a cos(θ) 1+(a 2 /b 2 )tan 2 (θ) = a cos(θ) 1+tan 2 (φ) = a cos(θ) cos(φ) = a cos(φ) = a cos(φ) 1 1+tan 2 (θ) 1 1+(b 2 /a 2 )tan 2 (φ) = a 2 a 2 cos 2 (φ)+b 2 sin 2 (φ) = a 2 a 2 (1 sin 2 (φ)) + b 2 sin 2 (φ) a = 1 a2 b 2 sin 2 (φ) a 2 a = 1 ɛ 2 sin 2 (φ), 4

5 where a2 b ɛ = 2 is the eccentricity of the ellipse. We can write the ellipse in terms of N(φ) andφ. Wehave a 2 x = d(θ)cos(φ) =N(φ)cos(φ). and z = b sin(θ) 1 = b 1 1+tan 2 (θ) = b (b 2 /a 2 )tan 2 (φ) = b (b2 /a 2 )tan 2 (φ) 1+(b 2 /a 2 )tan 2 (φ) = b 2 sin(φ)a 2 a 2 a 2 cos 2 (φ)+b 2 sin 2 (φ) = b2 a sin(φ)n(φ) 2 = N(φ)(1 ɛ 2 )sin(φ). Generating the ellipsoid by rotation of the ellipse, we have the parametric surface P (φ, λ) =(N(φ)cos(φ)cos(λ),N(φ)cos(φ)sin(λ),N(φ)(1 ɛ 2 )sin(φ)), where φ is the geodetic latitude and λ is the longitude. We shall see that N(φ) is the principal radius of curvature of the surface in a direction perpendicular to the meridian (meridians are constant longitude curves). 5

6 2 The Fundamental Quadratic Forms of a Surface Let X(u, v) be a 2d surface in R 3.Let α =(α 1 (t),α 2 (t)) = (u, v) be a curve in the two dimensional domain. Then X α is a curve on the surface. Then dx α = D 1 XDα 1 + D 2 XDα 2 dt is a tangent to the curve and is an element of the tangent space. The first fundamental bilinear form is the inner product of such tangent vectors. B 1 (T 1,T 2 )=T 1 T 2. The coordinate curves are (t, 0) and (0,t) which have tangents D 1 X and D 2 X respectively. The quadratic form is B 1 (T 1,T 1 )=T 1 T 1. For every tangent vector T there is a curve α so that the tangent vector is the tangent of the curve on the surface. So let us write then where T = D 1 XDα 1 + D 2 XDα 2 = D 1 Xdu + D 2 Xdv. ds 2 = B 1 (T,T)=Edu 2 +2Fdudv+ Gdv 2 E = D 1 X D 1 X F = D 1 X D 2 X G = D 2 X D 2 X. The Weingarten linear transformation is a mapping from the tangent space to the tangent space given by W (T )= T N, 6

7 where T N is the covariant derivative of the surface normal vector field N in the direction of T. The derivative is defined as T N = dn(α(t), dt where α is a curve whose derivative is the tangent vector T. Because N is a unit vector N N =1 so that dn dt N =0 which implies that dn is in the tangent space. Corresponding to this linear dt transformation is the bilinear form The corresponding quadratic form is B 2 (T 1,T 2 )=W (T 1 ) T 2 Q 2 (T )= B 2 (T,T) B 2 is symmetric and is known as the second fundamental form. In some books the second fundamental form is given as the negative, Q 2 (T )=B 2 (T,T). Notice that N T =0 so dn dt T = N dt dt. Suppose the curve α with derivative T is parameterized by arclength, then T = dα ds and so Q 2 (T )= W(T) T = dn ds T = N dt ds = N K, where K is the curvature vector for the curve α, which in turn is equal to the curvature κ times the curve normal vector. The curve normal is not 7

8 necessarily the same as the surface normal. So we have the component of the curvature vector in the direction of N. κ n = N K = Q 2 ( dα ds ). This is called the normal curvature of the surface in the direction T. Now suppose β is a curve not parameterized by arclength, Let the arclength be s(t) = t 0 dβ dτ dτ so ds dt = dβ dt. So let the curve β be reparameterized by arclength as γ(s) =β(t(s)) then dγ ds = dβ dt dt ds = dβ dt / dβ dt. Hence the normal curvature in the direction T for any T is Then where To calculate Q 2 suppose κ n = Q 2 (T/ T ) = Q 2(T ) T 2 = Q 2(T ) Q 1 (T ). T = D 1 XDα 1 + D 2 XDα 2 = D 1 Xdu + D 2 Xdv = X u du + X v dv. Q 2 (T )= W (T ) T = (W (X u )du + W (X v )dv) (X u du + X v dv) = ((W (X u ) X u )du 2 +2(W (X u ) X v )dudv)+(w (X v ) X v )dv 2 = edu 2 +2fdudv + gdv 2. e = (W (X u ) X u )= N u X u = N X uu f = (W (X u ) X v )= N u X v = N X uv 8

9 and g = (W (X v ) X v )= N v X v = N X vv. Nowwecanwriteaunitnormalas So we can write e, f, g as N = X u X v EG F 2 e = X u X v X uu EG F 2 f = X u X v X uv EG F 2 and g = X u X v X vv. EG F 2 3 Principal Directions On A Surface If T is a unit tangent vector, then the normal curvature in the direction of T is κ n = Q 2 (T )= W (T ) T. There is a maximum and minimum value for this curvature. The directions giving the maximum and minimum are called the principal directions. Suppose the maximum value of κ n is in the direction of the unit vector T 1.LetT 2 be a unit vector that is perpendicular to T 1. Then every unit tangent vector T can be written as for some angle θ. The curvatures are T (θ) =cos(θ)t 1 +sin(θ)t 2 κ n (θ) =Q 2 (T (θ)) = W (T (θ)) T (θ) where = [(cos(θ)w (T 1 )+sin(θ)w (T 2 )) (cos(θ)t 1 +sin(θ)t 2 )] =cos 2 (θ)c 11 +2cos(θ)sin(θ)C 12 +sin 2 (θ)c 22, C ij = W (T i ) T j. 9

10 By assumption the maximum occurs at θ = 0, so the derivative is zero there. We have dκ n dθ = 2cos(θ)sin(θ)C 11 +2(cos 2 (θ) sin 2 (θ))c 12 +2cos(θ)sin(θ)C 22 So a zero derivative at 0, forces C 12 = 0. Because W maps each T to the tangent space, we can write for some a and b. Then and Hence W (T 1 )=at 1 + bt 2 a = W (T 1 ) T 1 = C 11 b = W (T 1 ) T 2 = C 12 =0. W (T 1 )= C 11 T 1 = κ n (0)T 1 So T 1 is an eigenvector of W and the eigenvalue is the negative of the maximum normal curvature. Similarly T 2 is an eigenvector with eigenvalue C 22. We have that κ n (θ) =cos 2 (θ)c 11 +sin 2 (θ)c 22. Clearly the maximum curvature is and the minimum curvature is κ n (0) = C 11 κ n (π/2) = C 22. Hence the maximum and minumum normal curvatures are negative eigenvalues of W and the corresponding principal directions T 1 and T 2 are orthogonal. We have found a formula for the normal curvature in a specified direction. In the direction the normal curvature is This is called Euler s formula. T (θ) =cos(θ)t max +sin(θ)t min, κ n (θ) =cos 2 (θ)κ max +sin 2 (θ)κ min. 10

11 4 Principal Normal Curvatures on the Ellipsoid Consider the ellipsoid surface f(u, v) where u is the geocentric latitude and v is the longitude. So f takes the form f(u, v) =(a cos(u)cos(v),acos(u)sin(v),bsin(u)). The coordinate curve tangent vectors are D 1 f =( a sin(u)cos(v), a sin(u)sin(v),bcos(u)), and D 2 f =( acos(u)sin(v),acos(u)cos(v), 0). Every tangent vector is a linear combination of these two vectors. Also for future reference we have the second derivative D 22 f =( a cos(u)cos(v), a cos(u)sin(v), 0). For the metric coefficients we have E = D 1 f D 1 f = a 2 sin 2 (u)+b 2 cos 2 (u) F =0 G = D 1 f D 1 f = a 2 cos 2 (u) Recall that in classical notation the first fundamental form, that is the surface metric, is ds 2 = Edu 2 +2Fdudv+ Gdv 2. The second fundamental form is The coefficient g is given by edu 2 +2fdudv + gdv 2. g = D 22f D 1 f D 2 f EG F 2. We have EG F 2 = a 2 cos 2 (u)(a 2 sin 2 (u)+b 2 cos 2 (u)). 11

12 And D 22 f D 1 f D 2 f = Det a cos(u)cos(v) a cos(u)sin(v) 0 a sin(u)cos(v) a sin(u)sin(v) b cos(u) a cos(u)sin(v) a cos(u)cos(v) 0 = a 2 b cos 3 (u)cos 2 (v)+a 2 b cos 3 (u)sin 2 (v) So we get g = = a 2 b cos 3 (u). ab cos 2 (u) a 2 sin 2 (u)+b 2 cos 2 (u). Consider a curve ω(t) =(u(t),v(t)). The normal curvature in the direction of the curve is κ n = e(du/dt)2 +2fdu/dtdv/dt + g(dv/dt) 2 E(du/dt) 2 +2F du/dtdv/dt + G(dv/dt) 2 Consider a curve ω with du/dt =0. Theu coordinate curve is the meridian curve and clearly by symmetry of the surface must be one of the principal normal curvature directions. Then curve ω has a tangent perpendicular to the meridian and the normal curvature in this direction is the second principal normal curvature. We have κ n = g(dv/dt)2 G(dv/dt) 2 = g G = ab cos 2 (u) 1 a 2 sin 2 (u)+b 2 cos 2 (u) a 2 cos 2 (u) b = a a 2 sin 2 (u)+b 2 cos 2 (u). The reciprocal is the radius of curvature and is = a a 2 sin 2 (u)+b 2 cos 2 (u) b 12 = d(u),

13 where d(u) =N(φ) is the distance from a surface point to the polar axis in the direction of the surface normal. So as stated in a previous section N(φ), where φ is the geodetic latitude, is the normal curvature of the ellipsoid in a direction perpendicular to the meridian. Recall that this principal radius of curvature N(φ)can be written as a (1 ɛ 2 sin 2 (φ)). 1/2 We can show that the other principal radius of curvature in the direction of the meridian is a(1 ɛ 2 ) M(φ) = (1 ɛ 2 sin 2 (φ)). 3/2 5 The Coefficients of the First Fundamental Form For the Geodetic Representation of the Ellipsoid The geodetic representation of the ellipsoid surface is where f(φ, λ) =(N cos(φ)cos(λ),ncos(φ)sin(λ, N(1 ɛ 2 )sin(φ)), N(φ) = a (1 ɛ 2 sin 2 (φ)) 1/2, and the eccentricity is a2 b ɛ = 2. a 2 Also the other principle curvature is The derivative of N is M(φ) = a(1 ɛ 2 ) (1 ɛ 2 sin 2 (φ)) 3/2. N = dn/dφ = aɛ2 sin(φ)cos(φ) (1 ɛ 2 sin 2 (φ)) 3/2. = ɛ2 sin(φ)cos(φ) 1 ɛ 2 M. 13

14 We have And Then D 1 f =[N cos(φ)cos(λ) N sin(φ)cos(λ), N cos(φ)sin(λ) N sin(φ)sin(λ), N (1 ɛ 2 )sin(φ)+n(1 ɛ 2 )cos(φ)]. D 2 f =( Ncos(φ)sin(λ),Ncos(φ)cos(λ), 0). E = D 1 f D 1 f =(N cos(φ)cos(λ)) 2 2N N cos(φ)sin(φ)cos 2 (λ)+(n sin(φ)cos(λ)) 2 +(N cos(φ)sin(λ)) 2 2N N cos(φ)sin(φ)sin 2 (λ)+(n sin(φ)sin(λ)) 2 +(N (1 ɛ 2 )sin(φ)) 2 2N N(1 ɛ 2 ) 2 sin(φ)cos(φ)+(n(1 ɛ 2 )cos(φ)) 2 = N 2 cos 2 (φ) 2N N cos(φ)sin(φ)+n 2 sin 2 (φ) +(1 ɛ 2 ) 2 [N 2 sin 2 (φ)+2n N sin(φ)cos(φ)+n 2 cos 2 (φ)] =(N cos(φ) N sin(φ)) 2 +(1 ɛ 2 )(N cos(φ)+n sin(φ)) 2 And this is equal to M 2, which may be verified by plotting the difference of the two expressions in the Maple program, or by deriving the identity. We have G = D 2 f D 2 f = N 2 cos 2 (φ)(sin 2 (λ)+cos 2 (λ)) = N 2 cos 2 (φ). Thecoordinatecurvesforφ and λ are perpendicular so F =0. 6 Conformal Maps of Surfaces Let f be a surface in R 3, f(u 1,v 1 ) R 3, with parameters u 1,v 1. Let g be asurfaceinr 3, g(u 2,v 2 ) R 3, with parameters u 2,v 2.Letpbeamapping from R 2 to R 2 taking parameters u 1,v 1 to u 2,v 2. This induces a mapping from f to g, givenby h = g p. So h(u, v) is the image of the point f(u, v). Let D be the domain of f and h. f u = f u = D 1f 14

15 and f v = f v = D 2f are a bases of tangent vectors to the surface f. surface domain D, with u = α 1 (t) then v = α 2 (t) f α If α(t) isacurveinthe is a curve on the surface. We get a tangent vector on the surface then D(f α) =D 1 fdα 1 + D 2 fdα 2. Let β(t) is a second curve in the surface domain D, with u = β 1 (t) v = β 2 (t) f β is a curve on the surface. We get a tangent vector on the surface D(f β) =D 1 fdβ 1 + D 2 fdβ 2. The cosine of the angle between these two vectors is The numerator is cos(θ) = D(f α) D(f β) D(f α) D(f β) D(f α) D(f β) =D 1 f D 1 f(dα 1 Dβ 1 )+D 1 f D 2 f(dα 1 Dβ 2 +Dα 2 Dβ 1 )+D 2 f D 2 f(dα 2 Dβ 2 ) Let E f = D 1 f D 1 f F f = D 1 f D 2 f 15

16 Then the numerator is G f = D 2 f D 2 f. D(f α) D(f β) =E f (Dα 1 Dβ 1 )+F f (Dα 1 Dβ 2 + Dα 2 Dβ 1 )+G f (Dα 2 Dβ 2 ) Similarly D(f α) = E f (Dα 1 Dα 1 )+2F f (Dα 1 Dα 2 )+G f (Dα 2 Dα 2 ) and D(f β) = E f (Dβ 1 Dβ 1 )+2F f (Dβ 1 Dβ 2 )+G f (Dβ 2 Dβ 2 ) So using this notation we have cos(θ) = E f (Dα 1 Dβ 1 )+F f (Dα 1 Dβ 2 + Dα 2 Dβ 1 )+G f (Dα 2 Dβ 2 ) Ef (Dα 1 Dα 1 )+2F f (Dα 1 Dα 2 )+G f (Dα 2 Dα 2 ) E f (Dβ 1 Dβ 1 )+2F f (Dβ 1 Dβ 2 )+G f (Dβ 2 Dβ 2 ) In classical notation a term such as wouldhavebeenwrittenas Dα 1 Dα 1 du 2 That is because Dα 1 = du dt and so du = du dt dt and the common factordt can be factored from all of the terms. The difficulty with the classical notation is that a du for example could refer either to a differential of the curve α or of the curve β. This can get very confusing. Suppose our mapping is to be conformal, meaning that angles between corresponding curves on the two surfaces are the same. So if we write the expression for the cosine of the angle between the images of the two curves on surface h, it should give the same value as the previous expression, thus 16

17 cos(θ) = E h (Dα 1 Dβ 1 )+F h (Dα 1 Dβ 2 + Dα 2 Dβ 1 )+G h (Dα 2 Dβ 2 ) E h (Dα 1 Dα 1 )+2F h (Dα 1 Dα 2 )+G h (Dα 2 Dα 2 ) E h (Dβ 1 Dβ 1 )+2F h (Dβ 1 Dβ 2 )+G h (Dβ 2 Dβ 2 ) Clearly this will be true if E f E h = F f F h = G f G h = m 2 for some number m. Conversely, if the map is conformal and the angle is preserved for all curve pairs, then by choice of special pairs of curves α and β, we can prove that we must have E f E h = F f F h = G f G h = m 2 for some m. So this is the general condition for a mapping p between a surface f and a surface g to be conformal, where h = g p. Considering the first fundamental form, we see that m is the differential scale factor. So each small distance ds between points in surface f is mapped to a distance mds on the surface g. This conformal condition allows us to find the conformal mapping, that is to determine the function p between coordinates. Because f and g are given, we can compute the E f,f f,g f and E g,f g,g f. But we need the E h,f h,g h. So we shall use the chain rule to get these. We have D 1 h 1 D 2 h 1 Dh = D 1 h 2 D 2 h 2 D 1 h 3 D 2 h 3 = D 1 g 1 D 2 g 1 D 1 g 2 D 2 g 2 D 1 g 3 D 2 g 3 [ D1 p 1 D 2 p 1 D 1 p 2 D 2 p 2 ] Then we have D 1 h = D 1 h 1 D 1 h 2 D 1 h 3 = D 1 g 1 D 1 p 1 + D 2 g 1 D 1 p 2 D 1 g 2 D 1 p 1 + D 2 g 2 D 1 p 2 D 1 g 3 D 1 p 1 + D 2 g 3 D 1 p 2 17

18 = D 1 gd 1 p 1 + D 2 gd 1 p 2. In a similar way we find that D 2 h = D 1 gd 2 p 1 + D 2 gd 2 p 2. Then E h = D 1 h D 1 h =(D 1 g D 1 g)(d 1 p 1 ) 2 +2(D 1 g D 2 g)(d 1 p 1 D 1 p 2 )+(D 2 g D 2 g)(d 1 p 2 ) 2 = E g D 1 p F gd 1 p 1 D 1 Dp 2 + G g D 1 p 2 2 and G h = D 2 D 2 h = E g D 2 p F g D 2 p 1 D 2 Dp 2 + G g D 2 p 2 2. In the next section we shall apply these methods to finding a conformal map between an ellipsoid and a sphere. 7 A Conformal Map of the Ellipsoid to the Sphere The ellipsoid surface is where f(u 1,v 1 )=(N cos(u 1 )cos(v 1 ),Ncos(u 1 )sin(v 1 ),N(1 ɛ 2 )sin(u 1 )), and the eccentricity is N(u 1 )= M(u 1 )= We found previously that a (1 ɛ 2 sin 2 (u 1 )) 1/2, ɛ = a(1 ɛ 2 ) (1 ɛ 2 sin 2 (u 1 )) 3/2, a2 b 2 a 2. E f = N 2 cos 2 (φ), F f =0, G f = N 2 cos 2 (φ). 18

19 The spherical surface is g(u 2,v 2 )=(R cos(u 2 cos(v 2 ),Rcos(u 2 )sin(v 2 ),Rsin(v 2 )), where R is the sphere radius. We find that E g = R 2, F g =0, G g = R 2 cos 2 (u 2 ). We can use a general function p of the form (u 2,v 2 )=p(u 1,v 1 )=(p 1 (u 1 ),c 1 v 1 + c 2 ). But we want the latitudes v 1 and v 2 to agree at 0, so we set c 2 =0. Alsoit turns out that the c 1 can be used to manipulate the location where the scale factor m can be set to a given value. But here we shall take c 1 =1. Sowe will use the function p(u 1,v 1 )=(p 1 (u 1 ),v 1 ). and So the partial derivatives of p are D 1 p 1 = du 2 du 1,D 2 p 1 =0,D 1 p 2 =0,D 2 p 2 =1. So from the previous section, for h = g p we have The conformality condition leads to the differential equation E h = E g ( du 2 du 1 ) 2 =(R du 2 du 1 ) 2 F h =0 G h = G g =(R cos(u 2 )) 2. E f E h = G f G h = m 2 19

20 M N cos(u 1 ) du 1 = 1 cos(u 2 ) du 2. We have so the equation becomes M N = 1 ɛ 2 1 ɛ 2 sin 2 (u 1 ), 1 ɛ 2 (1 ɛ 2 sin 2 (u 1 )) cos(u 1 ) du 1 = 1 cos(u 2 ) du 2. A rational function of sin(u) andcos(u) can always be integrated with the substitution z = tan(u/2). This substitution leads to a rational function of z which can always be integrated with partial fractions. For example in thecaseof 1 cos(u) du, we have cos(u) =cos 2 (u/2) sin 2 (u/2) =2cos 2 (u/2) 1 = 2 sec 2 (u/2) 1 = 1 tan2 (u/2) 1+tan 2 (u/2) and Then Similarly, = 1 z2 1+z 2 sin(u) = tan(u) = 2z 1+z 2, 2z 1 z 2. 1 du =ln( 1+z cos(u) 1 z ) 20

21 =ln( 1+z 1 z ) =ln( 1+tan(u/2) 1 tan(u/2) ) =ln( tan(π/4+u/2), where we have used partial fractions to integrate the rational function of z. This integral also has the form ln( tan(u)+sec(u) ) because sec(u)+tan(u) = 1+z2 1 z + 2z 2 1 z = 1+z 2 1 z. The second interval is 1 ɛ 2 (1 ɛ 2 sin 2 (u)) cos(u) du = ln(tan(π/4+u/2)( 1 ɛ sin(u) 1+ɛsin(u) )ɛ/2. Here is a verification method if the reader does not want to carry out the integration: Use Maple and set the integrand to say A, and the integrated expression to say B, andgiveɛ a numerical value, then plot the difference between A and the derivative of B. plot(a-diff(b,u),u=0..1.5); So if finally we remove the logarithm we get for the solution of our problem tan(π/4+u 2 /2) = tan(π/4+u 1 /2)( 1 ɛ sin(u) 1+ɛ sin(u) )ɛ/2. This defines our mapping function u 2 = p 1 (u 1 ). That is, given the latitude and longitude (u 1,v 1 ) of a point on the ellipsoid, we calculate the latitude and longitude of the image point on the sphere (u 2,v 2 ). At this point the sphere radius R is arbitrary. It can be used to change the scale factor m. 21

22 8 A Conformal Map of the Ellipsoid to the Plane: The Polar Stereographic Projection The stereographic projection of a point on a sphere of radius R, withthe south pole as a projection point, and a plane tangent to the north pole is given as j(u 2,v 2 )=(x, y) =(ρ(u 2 )cos(v 2 ),ρ(u 2 )sin(v 2 )) where u 2 is the latitude of the sphere, v 2 is the longitude, and ρ is the radius on the projection plane given as ρ =2R tan(π/4 u 2 /2). This is a conformal map from the sphere to the plane. We may compose j with the mapping of the ellipse to the sphere given in the previous section to get a conformal map from the ellipsoid to the plane. Thus ρ =2R tan(π/4 u 2 /2) 2R = tan(π/4+u 2 /2) 1+ɛsin(u) =2Rtan(π/2 u 1 /2)( 1 ɛ sin(u) )ɛ/2. Note that in these expressions taking reciprocals involves only some change of signs. Now we may absorb the factor 2R into a constant K, whichwemay select in order to change the scale factor. So we have ρ = K tan(π/2 u 1 /2)( 1+ɛsin(u 1) 1 ɛ sin(u 1 ) )ɛ/2. We can show that if we take K = 2a2 b (1 ɛ 1+ɛ )ɛ/2 that the scale factor will be unity at the north pole. This is a standard form for the polar stereographic projection. But suppose we want a unity scale factor at some specified geodetic latitude such as say 60 degrees (u 1 = π/3). This is called the latitude of true scale. To compute an appropriate K, we can calculate the radius of the latitude circle, and adjust K so that the latitude circle matches the corresponding circle on the plane, where the later has radius ρ. 22

23 9 The Lambert Projection: A conformal Map From An Ellipsoid To A Cone The Lambert projection is constructed in a manner quite similar to the projection from the ellipse to the sphere. The difference being that the second surface is a cone. The resulting equations are similar. The cone may be specified to be tangent to the ellipse at a certain latitude circle, or the cone may pass through two parallel circles rather than be tangent. The flat map is constructed by laying the cone flat, so that in general the map is not rectangular, rather being a region bounded by two circles and two straight lines. In the case of the tangent cone, there are two limiting cases, namely, (1) when the tangent circle is at zero latitude, in which case the cone apex goes to infinity and the cone becomes a cylinder, and the mapping is called the Mercator projection, and (2) when the tangent circle is at latitude 90 degrees and the cone becomes a plane, and then we get the polar stereographic projection of the previous section. There are a couple of constants in the lambert projection, which allow a special point to be defined where the scale factor is unity. Thus if we wanted an accurately scaled map of England so that at London the scale factor is unity, we could accomplish this. So historically maps were constructed in order that graphical measurements and calculations could be done on the physical maps, so that conformality and unity scale factor in the measurement area were very important. With computers, this no longer is quite so important, but map standards and surveying standards are still specified with these special Lambert projections. Lambert derived this projection in the 18th century. See the book Map Projections by Richardus and Adler for the details of the Lambert projection. 10 Bibliography [1]Richardus, Adler, Map Projections. [2]Yang Qihe,Snyder John P, Tobler Waldo R, Map Projection Transformation Principles and Applications, Taylor and Francis, London, [3]Pearson Frederick II, Map Projection Methods, Sigma Scientific, Blacksburg Virginia,