13472J/1.128J/2158J/16940J COMPUTATIONAL GEOMETRY Lecture 3 Kwanghee Ko T Maekawa N.M. Patrikalakis Massachusetts Institute of Technology Cambridge, MA 02139-4307, USA Copyright 2003 Massachusetts Institute of Technology Contents 3 Differential geometry of surfaces 3.1 Definition of surfaces 3.2 Curves on a surface 3.3 First fundamental form(arc length 3.4 Tangent plane 3.5 Normal vector 2234668 3.6 Second fundamental form II (curvature) 3.7 Principal curvatures Bibliography Reading in the Textbook Chapter 3, pp. 49-pp 72
13.472J/1.128J/2.158J/16.940J COMPUTATIONAL GEOMETRY Lecture 3 Kwanghee Ko T. Maekawa N. M. Patrikalakis Massachusetts Institute of Technology Cambridge, MA 02139-4307, USA Copyright c 2003 Massachusetts Institute of Technology Contents 3 Differential geometry of surfaces 2 3.1 Definition of surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3.2 Curves on a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3.3 First fundamental form (arc length) . . . . . . . . . . . . . . . . . . . . . . . . 4 3.4 Tangent plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.5 Normal vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.6 Second fundamental form II (curvature) . . . . . . . . . . . . . . . . . . . . . . 8 3.7 Principal curvatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Bibliography 13 Reading in the Textbook • Chapter 3, pp.49 - pp.72 1
Lecture 3 Differential geometry of surfaces 3.1 Definition of surfaces Implicit surfaces F(r,,a)=0 Example: 22+6+2=1 Ellipsoid, see Figure 3.1 Figure 3.1: Ellipsoid · Explicit surfaces If the implicit equation F(, y, a)=0 can be solved for one of the variables as a function of the other two, we obtain an explicit surface, as shown in Figure 3.2. Example: 2= 2(ax+ By2) Parametric surfaces x=ru, v),y=y(u, u), t=z(u, v) Here functions z(u, u), y(u, u), x(u, u) have continuous partial derivatives of the r th order, and the parameters u and v are restricted to some intervals(i.e.,u1≤u≤u2,n≤t≤v2) leading to parametric surface patches. This rectangular domain D of u, v is called parametric space and it is frequently the unit square, see Figure 3.3. If derivatives of the surface are continuous up to the rth order, the surface is said to be of class r, denoted
Lecture 3 Differential geometry of surfaces 3.1 Definition of surfaces • Implicit surfaces F(x, y, z) = 0 Example: x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 Ellipsoid, see Figure 3.1. x y z Figure 3.1: Ellipsoid. • Explicit surfaces If the implicit equation F(x, y, z) = 0 can be solved for one of the variables as a function of the other two, we obtain an explicit surface, as shown in Figure 3.2. Example: z = 1 2 (αx2 + βy 2 ) • Parametric surfaces x = x(u, v), y = y(u, v), z = z(u, v) Here functions x(u, v), y(u, v), z(u, v) have continuous partial derivatives of the r th order, and the parameters u and v are restricted to some intervals (i.e., u1 ≤ u ≤ u2, v1 ≤ v ≤ v2) leading to parametric surface patches. This rectangular domain D of u, v is called parametric space and it is frequently the unit square, see Figure 3.3. If derivatives of the surface are continuous up to the r th order, the surface is said to be of class r, denoted C r . 2
Figure 3.2: Explicit quadratic surfaces z=2(ax+ By").(a)Left: Hyperbolic paraboloid (a=-3, B=1).(b) Right: Elliptic paraboloid(a=l, B=3) In vector notation r=r(u,U where r=(a, g, a),r(u,v)=((u,u),y(u, u), 2(u, u) x=+ 1 y=u-}→ eliminate,→2=5(x2+y2) paraboloid 3.2 Curves on a surface Let r=r(u, v) be the equation of a surface, defined on a domain D(i.e, uI <usu U1 <U<U2). Let B(t=(u(t), u(t)be a curve in the parameter plane. Then r=ru(t), v(t)) is a curve lying on the surface, see Figure 3. 3. a tangent vector of curve B(t)is given by B(t=(i(t),i(t)) a tangent vector of a curve on a surface is given by By using the chain rule dr(u(t), v(t)) ar du ar du rui(t)+ri(t
Figure 3.2: Explicit quadratic surfaces z = 1 2 (αx2 + βy 2 ). (a) Left: Hyperbolic paraboloid (α = −3, β = 1). (b) Right: Elliptic paraboloid (α = 1, β = 3). In vector notation: r = r(u, v) where r = (x, y, z), r(u, v) = (x(u, v), y(u, v), z(u, v)) Example: r = (u + v, u − v, u 2 + v 2 ) x = u + v y = u − v z = u 2 + v 2 ⇒ eliminate u, v ⇒ z = 1 2 (x 2 + y 2 ) paraboloid 3.2 Curves on a surface Let r = r(u, v) be the equation of a surface, defined on a domain D (i.e., u1 ≤ u ≤ u2, v1 ≤ v ≤ v2). Let β(t) = (u(t), v(t)) be a curve in the parameter plane. Then r = r(u(t), v(t)) is a curve lying on the surface, see Figure 3.3. A tangent vector of curve β(t) is given by β˙(t) = (u˙(t), v˙(t)) A tangent vector of a curve on a surface is given by: dr(u(t), v(t)) dt (3.1) By using the chain rule: dr(u(t), v(t)) dt = ∂r ∂u du dt + ∂r ∂v dv dt = ruu˙(t) + rvv˙(t) (3.2) 3
(u(t),v(t)) 阝(t)=(u(t),v(t)yx Figure 3.3: The mapping of a curve in 2D parametric space onto a 3D biparametric surface 3.3 First fundamental form(arc length) Consider a curve on a surface r=ru(t), v(t)). The arc length of the curve on a surface is dt dt=r t dt V(rui+rui).(rui+rui)dt +(ry ru) √Ea2+2Faud+Ga2 The first fundamental form is defined as I dr dr=(rudu + rudu).(rudu+rudu) Edu2+2Fdudv+Gd2 E, F, G are called first fundamental form coefficients Note that E= ru. ru>0 and G ryry>0if ru#0 and ru+0. The first fundamental form I is positive definite. That is I>0 and I=0 if and only if du =0 and du =0 since e du+F Edu2 and EG-F=ru xTu2>0 I depends only on the surface and not on the parametrization The area of the surface can be derived as follows:
u v x y z D r(u,v) r(u(t),v(t)) β(t)=(u(t),v(t)) Parametric Space D 3D Space Figure 3.3: The mapping of a curve in 2D parametric space onto a 3D biparametric surface . 3.3 First fundamental form (arc length) Consider a curve on a surface r = r(u(t), v(t)). The arc length of the curve on a surface is given by ds = | dr dt|dt = |ru du dt + rv dv dt |dt = q (ruu˙ + rvv˙) · (ruu˙ + rvv˙)dt = q (ru · ru)du2 + 2rurvdudv + (rv · rv)dv2 = p Edu2 + 2Fdudv + Gdv2 (3.3) where E = ru · ru, F = ru · rv, G = rv · rv (3.4) The first fundamental form is defined as I = dr · dr = (rudu + rvdv) · (rudu + rvdv) = Edu2 + 2Fdudv + Gdv2 (3.5) E, F, G are called first fundamental form coefficients Note that E = ru · ru > 0 and G = rv · rv > 0 if ru 6= 0 and rv 6= 0. The first fundamental form I is positive definite. That is I ≥ 0 and I = 0 if and only if du = 0 and dv = 0 since I = 1 E (E du + F dv) 2 + EG − F 2 E dv2 and EG − F 2 = |ru × rv| 2 > 0. I depends only on the surface and not on the parametrization. The area of the surface can be derived as follows: 4
了 (uo+8 式(uvo+6v)-(uo,vo) y r(uo+8u vo)-=(uo, vo r(uo, vo) Figure 3.4: Area of an infinitessimal surface patch r 6A=|ra6×r2ov=|ru×rl|66v rn×rl2=(rn×r2)·(r1×rn) Using the vector identity (a x b). (cx d)=(a c(. d)-(a d)(b c), we get uIu (36) SA=VEG-F2 dudu A eg-F2 dudu Example: For the hyperbolic paraboloid r(u, u)=(u, 0,u2-u2), let us derive an expression for the area of a region of its surface corresponding to a the circle u2+u2< l in the parametric domain d We begin by forming expressions for the derivatives of the position vector r and the first fundamental form coeffients r (0,1,-2) Using Equation (3.8), we find EG-F2=(1+42)(1+42)-16x22=1+42+42>0 √1+4a2+42dude
r(u0,v0+δv)−r(u0,v0) r(u0+δu,v0)−r(u0,v0) r(u0,v0+δv) r(u0+δu,v0) r(u0,v0) δA Figure 3.4: Area of an infinitessimal surface patch. r(u0, v0 + δv) − r(u0, v0) ' ∂r ∂v δv r(u0 + δu, v0) − r(u0, v0) ' ∂r ∂u δu δA = |ruδu × rvδv| = |ru × rv|δuδv |ru × rv| 2 = (ru × rv) · (ru × rv) Using the vector identity (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c), we get |ru × rv| 2 = (ru · ru)(rv · rv) − (ru · rv) 2 (3.6) = EG − F 2 (3.7) δA = p EG − F2 δuδv, A = Z Z p EG − F2 dudv (3.8) Example: For the hyperbolic paraboloid r(u, v) = (u, v, u 2−v 2 ), let us derive an expression for the area of a region of its surface corresponding to a the circle u 2 +v 2 ≤ 1 in the parametric domain D. We begin by forming expressions for the derivatives of the position vector r and the first fundamental form coeffients. ru = (1, 0, 2u) rv = (0, 1, −2v) E = ru · ru = 1 + 4u 2 F = ru · rv = −4uv G = rv · rv = 1 + 4v 2 Using Equation (3.8), we find EG − F 2 = (1 + 4u 2 )(1 + 4v 2 ) − 16u 2 v 2 = 1 + 4u 2 + 4v 2 > 0 A = Z Z D p 1 + 4u 2 + 4v 2dudv 5