2Surfaces. Design with Bézier Surfaces

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1 You don t see something until you have the right metaphor to let you perceive it. James Gleick Surfaces Design with Bézier Surfaces S : r(u, v) = Bézier surfaces represent an elegant way to build a surface, initially used to design automobile bodies. Nowadays are largely used in computer graphics and computer-aided design (CAD), to describe font characters or paths in 3D animation. The main idea is to use some control points, in order to generate any shape you can imagine. A general parametric equation, for (n+1) (m+1) control points, is given by the formula: n m B i,n (u)b j,m (v)p ij i= j= where B i,n (u) = C i nu i (1 u) n i are the Bernstein polynomials, P ij the control points, and u, v [, 1]. In computer graphics the most used Bézier surfaces are the bicubic Bézier surfaces (m=n=3), mostly because they are a nice balance between simplicity and complexity, providing freedom for the artist with a minimal amount of complexity for the programmer and renderer. 1

2 Surfaces, Examples: Regular surfaces: A set S R 3 is a regular surface if each of its points p has a neighborhood V and there exists an open set D R and a mapping r : D V S: r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, (u, v) D with the following properties: i) r is bijective and continuous with a continuous inverse ii) every component x(u, v), y(u, v), z(u, v) is continuously differentiable iii) in every point of D: r u r v (,, ) The mapping r will be called a parametrization (surface patch) of V. the above definition means that locally a surface can be deformed into a plane from now on by a surface we will understand a regular surface. a parametric surface is a surface that can be described by a single parametrization r : D S Graphs of Functions If f : D R R is continuously differentiable then: r(x, y) = x i + y j + f(x, y) k is a global surface patch (parametrization) of the graph of f, because: ( r x r y = f ) x, f y, 1 (,, ).

3 the graph is a surface given in the explicit equation z = f(x, y) and every point on the graph has coordinates (x, y, f(x, y)) in particular for f(x, y) = x y one gets the surface: you can use this link to generate surfaces as graphs of functions. Surfaces of Revolution c 1 (u) if we rotate a regular 3D curve c(u) = c (u), u (a, b), whose image c 3 (u) does not intersect the Oz-axis, around the Oz-axis we get a parametric surface with the following parametrization: cos v sin v c 1 (u) r(u, v) = sin v cos v c (u), 1 c 3 (u) u (a, b), v (, π). cos v sin v the matrix sin v cos v acting, in the above formula, on the 1 curve coordinates is called the rotation matrix around the Oz-axis. with the other rotation matrices we can rotate around any axis of an Oxyz frame. Torus: a torus can be obtained rotating a circle, lying in the Oxz-plane with origin at (a,, ) and radius < R < a, around the Oz-axis. 3

4 such a surface will be parametrized as r : [, π] [, π] R 3 : cos v sin v a + R cos u r(u, v) = sin v cos v 1 a + R sin u = (a + R cos u)cos v i + (a + R cos u)sin v j + (a + R sin u) k Sphere: the sphere with center O(x, y, z ) and radius R is a surface but it doesn t admit a single global parametrization, since it can be covered only by at least two surface patches: the northern and the southern hemisphere. the northern hemisphere is a parametric surface with the parametrization r : [, π ] [, π] R 3 : r(u, v) = (x + Rsin u cos v) i + (y + Rsin u sin v) j + (z + Rcos u) k. the sphere can be seen as a surface of revolution. 4

5 Straight Circular Cylinder: a parametrization of a cylinder with radius R is: r(u, v) = R cos u i + R sin u j + v k where u π, a v b Flat Approximations of Surfaces: in the sequel we ll consider only parametric surfaces in order to simplify some tedious computations Coordinate curves: If r is a parametrization of a parametric surface S, then for fixed values u, v the mappings: u r(u, v ), v r(u, v) are called coordinate curves on S. meridians and parallels are coordinate curves on a sphere 5

6 Tangent plane : The tangent plane at a point M(u, v ) to a surface S is the plane that contains the point M and the tangent vectors r u (u, v ) and r v (u, v ) to the coordinate curves. It has the normal direction given by r u (u, v ) r v (u, v ). α : x x M y y M z z M x u(u, v ) y u(u, v ) z u(u, v ) = x v(u, v ) y v(u, v ) z v(u, v ) Affine Approximation of a Function the graph of a function f : D R R has the parametrization: r(x, y) = x i + y j + f(x, y) k thus the tangent plane at M(a, b, f(a, b)) is: x a y b z f(a, b) α : f 1 x (a, b) = f 1 y (a, b) which implies: α : z = f(a, b) + f f (a, b)(x a) + (a, b)(y b) x y g(x,y) the tangent plane is the graph of the function: g(x, y) = f(a, b) + f f (a, b)(x a) + (a, b)(y b) x y g contains the first terms of the Taylor expansion of f in (a, b), thus in a neighborhood of (a, b) we have: f(x, y) f(a, b) + f f (a, b)(x a) + (a, b)(y b) x y 6

7 the function g is not linear, because of the term f(a, b), but it is affine and we call it the affine approximation of f around (a, b). = the tangent plane at the graph of f (surface) can be seen as a geometric visualization of the Taylor expansion of order 1! Analysis on a Surface: we call the expressions: E(u, v) = r u (u, v), r u (u, v) F (u, v) = r u (u, v), r v (u, v) G(u, v) = r v (u, v), r v (u, v) the coefficients of the first fundamental form of S. a regular plane curve c : { u = u(t) v = v(t) in D generates a regular space curve: r c : r(u(t), v(t)) = x(u(t), v(t)) i + y(u(t), v(t)) j + z(u(t), v(t)) k on a parametric surface with parametrization r : D S. Length of a curve: The length of the arc between A(t = a) and B(t = b), determined on the curve r c, is given by: l AB = b a [ u(t)] E(u(t), v(t)) + u(t) v(t)f (u(t), v(t)) + [ v(t)] G(u(t), v(t)) dt for a closed curve like in the last figure one has to specify which of the two possible arcs between A and B are considered. (or the curve needs an orientation) Integration on parametric surfaces: For a parametric surface r : D S we define the scalar surface integral of a function f : S R as: f(x, y, z)ds = f(r(u, v)) r u r v dudv = f(r(u, v)) EG F dudv S D and we call the formal expression ds = r u r v dudv the surface area element. D 7

8 a scalar surface integral reduces to a double integral over D! the area of a surface patch r : D V can be obtained via the formula: Area(V ) = 1 ds = r u r v dudv V D Solved Problems Problem 1. Compute the mass of a cylindrical surface parametrized by: r(u, v) = a cos ui + a sin uj + vk, where u π, v H. The density in every point (x, y, z) is given by μ(x, y, z) = z (x + y ). Solution: The mass of this cylindrical surface is given by the scalar surface integral: m = μ(x, y, z)ds, S where S is the surface (side surface of the cylinder in this figure) and ds is the surface area element. The surface area element has the formula ds = r u r v dudv and one can reduce the surface integral to the double integral: m = μ(x, y, z)ds = μ(r(u, v)) r u r v dudv S D 8

9 In our case the set D is the set of (u, v), thus D = [, π] [, H] (have a look at the cylinder parametrization). First of all: r u = a sin ui + a cos uj + K. and: implies: r v = i + j + 1k i j k r u r v = a sin u a cos u = a cos ui + a sin uj 1 = r u r v = (a cos u) + (a sin u) = a. Hence ds = a dudv and by its very definition: m = μ(x, y, z)ds = μ(r(u, v))a dudv S Formulas of μ and r = F ubini = = π π [,π] [,H] [,π] [,H] v ((a cos u) + (a sin u) )adudv ( ) H v ((a cos u) + (a sin u) )adv du ( ) H v a a dv du = π a 3 H3 H3 du = πa3 3 3 Bonus: Let us compute the area of the cylindrical surface. By its very definition: Area(S) = 1 ds = 1 r u r v dudv = 1 a dudv F ubini = S π D ( ) H adv du = π [,π] [,H] Ha du = πah This formula is the famous πrg. Here R=a is the base circle radius and G = H the length of the cylinder s generatrix. Problem. Compute the length of the curve obtained for: { u = t [, t, π ] v = t on the surface x + y + z = R. Solution: The given equation is the implicit equation of a sphere with center O(,, ) and radius R. The general implicit equation of a sphere is: (x x O ) + (y y O ) + (z z O ) = R 9

10 We consider now the spherical coordinates on the sphere. The given curve is located in the northern hemisphere because u = t [, π ]. The northern hemisphere is a parametric surface with the parametrization given by the spherical coordinates: r(u, v) = Rsin u cos v i + Rsin u sin v j + Rcos u k. Remember: the whole sphere is not a parametric surface! The 3D curve obtained for u = v = t is: x = R sin t cos t [ c : y = R sin t sin t, t, π ] z = R cos t The curve starts at A(,, R) = A(t = ) and ends at B(, R, ) = B(t = π ). In order to find its length one needs to compute the coefficients of the first fundamental form: l AB = E(u, v) = r u r u = R (cos u cos v + sin v sin u + sin u) = R F (u, v) = r u r v = R sin u( cos u cos v sin v + cos u sin v cos v) = G(u, v) = r v r v = R sin u(sin v + cos v) = R sin u Now the length of c is: = π π = R π [ u(t)] E(u(t), v(t)) + u(t) v(t)f (u(t), v(t)) + [ v(t)] G(u(t), v(t)) dt [t ] R + (t )(t ) + [t ] R sin t dt 1 + sin t dt and so forth... :)) Proposed problems Problem 1. Consider the surface: x = sin u cos v S : y = sin u sin v z = cos u, (u, v) (, π) (, π) i) Find the coefficients of the first fundamental form ii) Compute the curvature and the torsion of the coordinate curve c : v = π 4 iii) Compute the measure of the angle between the tangent line at A(u = ) to the curve c and the line: x 1 = y = z

11 Problem. Consider the surface: S : r(u, v) = e u cos vi + e u cos vj + e u k, (u, v) R (, π) Find: i) The equation of the tangent plane and tangent line at M(u =, v = π ) ii) The equation of the binormal line and the osculating plane at A(u = ) belonging to the curve c : v = π 4 lying on the surface S. iii) The distance from the abovementioned binormal line to the line: x 1 = y + 1 = z Problem 3. Consider the triangle whose sides are the curves: u =, v =, u + v = 1 lying on the surface: S : x = u cos v y = u sin v z = sin v, (u, v) R (, π) Computer the perimeter and the measures of the angles corresponding to this triangle. Problem 4. Consider the surface: S : r(u, v) = ui + vj + sin(u + v)k i) Find the equation of the normal line and the tangent plane at O(,, ) and compute the angle between the normal line and the unit vector k. ii) Compute the area element of this surface and the area of the surface patch obtained for (u, v) [, π ] [, π ]. 11

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