Curves - Foundation of Free-form Surfaces
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1 Crves - Fondation of Free-form Srfaces
2 Why Not Simply Use a Point Matrix to Represent a Crve? Storage isse and limited resoltion Comptation and transformation Difficlties in calclating the intersections or crves and physical properties of objects Difficlties in design (e.g. control shapes of an existing object) Poor srface finish of manfactred parts
3 Advantages of Analytical Representation for Geometric Entities A few parameters to store Designers know the effect of data points on crve behavior, control, continity, and crvatre Facilitate calclations of intersections, object properties, etc.
4 Analytic Crves vs. Synthetic Crves Analytic Crves are points, lines, arcs and circles, fillets and chamfers, and conics (ellipses, parabolas, and hyperbolas) Synthetic crves inclde varios types of splines (cbic spline, B-spline, Beta-spline) and Bezier crves.
5 Crved Srfaces In CAD, We want to find a math form for representing crved srfaces, that : (a) look nice (smooth contors) (b) is easy to maniplate and manfactre (c) follows prescribed shape (airfoil design) To stdy the crved srface, we need to start from crves.
6 Parametric Representation * a crve * a srface [ ] ( ) = x( ), y( ), z( ) T P(, v) x(, v), y(, v), z(, v) P z =const = T -- p(x,y,z) n Umax v=const Umin -- p() y x We can represent any fnctions of crve (crved srface) sing parametric eqation.
7 Parametric Representation of Lines How is a line eqation converted by the CAD/ CAM software into the line database? How are the mathematical eqation correlated to ser commands to generate a line? z P, =0 P P -P P, = P P P P = P + ( P - P) P - P = ( P - P) x y P = P + ( P - P), 0
8 Lines P, = P, =0 P P -P P P P 0 ) ( ) ( ) ( + = + = + = z z z z y y y y x x x x ( - ), 0 = + P P P P x z y
9 Circle Representation (Non-parametric) (a) x + y = x = y = poor and non-niform definition sqare root complicated to compte
10 Circle Representation (parametric) π/ 3π/8 π/4 (b) x= cos y= sin better definition than (a) bt still slow π/8 0
11 Circle P n+ Representation 3 (parametric) P n Recrsive approach x y n x n + x y n = = r r cos sin θ θ = r cos( θ + dθ ) = r cosθ cos dθ n+ n+ = = x y n n cos d cos d θ θ + y x n n -- P n sin sin d d θ θ r sin -- P n+ θ sin d θ Observation: crves are represented by a series of line-segments Similarly all conic sections can be represented.
12 Ellipse π θ θ θ 0 sin cos = + = + = o o o z z B y y A x x The compter ses the same method as in the Representation 3 of circle to redce the amont of calclation.
13 Example
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15 Parabola explicit x = cy parametric x y z = = = x y z o o o + + A A 0
16 Hyperbola implicit x a b y = parametric x y z = = = a cosh bsinh z o 0
17 Parametric Representation of Synthetic Crves Analytic crves are sally not sfficient to meet geometric design reqirements of mechanical parts. Many prodcts need free-form, or synthetic crved srfaces. Examples: car bodies, ship hlls, airplane fselage and wings, propeller blades, shoe insoles, and bottles The need for synthetic crves in design arises on occasions: when a crve is represented by a collection of measred data points and (generation) when a crve mst change to meet new design reqirements. (modification)
18 The Order of Continity The order of continity is a term sally sed to measre the degree of continos derivatives (C 0, C, C ). y y i= i= i=3 i= i= i=3 x x Simplest Case Linear Segment y i = ai0 + ai x High order polynomial may lead to ripples y = a + a x a i i0 i in x n
19 Splines Ideal Order Splines a mechanical beam with bending deflections, or a smooth crve nder mltiple constraints. () () (3) (4) y x y y '' ( x) = R( x) EI a 6 = ( x) M EI b x + b EI i 3 i ( x) = x + x + c x + d = a i i i i Cbic Spline
20 Drafting Spline
21 Hermite Cbic Splines ( ) ( ) ( ) ( ) [ ] T z y x P,, = ( ) ( ) ( ) = = = c c c c z c c c c y c c c c x z z z z y y y y x x x x p " ( ) = x ( ) y ( ) z ( ) " # $ T = Ci i i=0 3 % 0 & & ( ) = 3 " # $ C 3 C C C 0 " ' ' ' ' ' ' # $ ( ( ( ( ( ( = [U T ][ C] 3 4 = coefficients to be determined Cbic Spline
22 Two End Points = = 0 P Bondary Conditions: Location of the two end points and their slopes Hermite Cbic Splines 3 i = Ci = C i= 0 ", P = 3 C3 + C + C P0 = C0 ' P0 = C # P = C 3 + C + C + C 0 " ' $# P = 3 C 3 + C + C =0 P r 0 p C + C + C0 p0 -- Pr 0 ' 4 3 eqations from two control points = p P r -- p Pr -- '
23 Hermite Cbic Splines C 0 = P 0 C = P ' 0 C = 3( P P 0 ) P ' 0 P ' C 3 = ( P 0 P ) + P ' 0 + P ' nknowns and eqations " 3 # P = Ci i = C # C # + C # + C # 0 P( ) + ( 3 i=0 = ( 3 + ) P 3 ' 0 + ) P0 + ( 3 + ( ) P ' ) P All parameters can be determined Hermite Cbic crve in vector form
24 Hermite Cbic Splines Eqation: ' 3 ' ) ( ) ( ) 3 ( ) 3 ( ) ( P P P P P " = 3 " # $ % %3 3 % % " & & & & # $ ' ' ' ' P 0 P P 0 ' P ' " & & & & & & # $ ' ' ' ' ' ' [ ] 0 = V M U H T 0 ) (3 ) 4 (3 ) 6 6 ( ) 6 (6 ) ( ' ' 0 0 ' = P P P P P Based on: Location of the two end points and their slopes In matrix form: Hermite Cbic crve in vector form
25 Limitations with Hermite Crves Hard to gess behavior between defined points for arbitrary end point slopes Limited to 3 rd degree polynomial therefore the crve is qite stiff
26 Bezier Crve p r p0 0 p r p p r p p r 3 p3 p r p p r p0 0 p r p p r 3 p3 p n ( ) = p B ( ) i= 0 i i, n n segment(each polygon) n+ vertices (each polygon) and nmber of control points [ 0, ]
27 Bezier Crve P. Bezier of the French atomobile company of Renalt first introdced the Bezier crve. A system for designing sclptred srfaces of atomobile bodies (based on the Bezier crve) - passes p 0 and - has end point derivatives: p n, the two end points. ' ' p p 0 0 = ' = nn( ( p p 0 )); p n ' = pn( n = p n n ( p n n ) pn ) - ses a vector of control points, representing the n+ vertices of a characteristic polygon.
28 Bernstein Polynomial B n B i, n ( ) = i i n ( n i) ( ) n i i, ( ) is a fnction of the nmber of crve segments, n. n= i 0 n = = = i( n i) 0 0
29 Bernstein Polynomial In the mathematical field of nmerical analysis, a Bernstein polynomial, named after Sergei Natanovich Bernstein, is a polynomial in the Bernstein form, that is a linear combination of Bernstein basis polynomials. A nmerically stable way to evalate polynomials in Bernstein form is de Castelja's algorithm which redces the comptational demand cased by the factorials.
30 An Example: If n =, then n+ = 3 vertices -- p r 0 -- p p r -- p p r i 0 n i( n i) 0 = = = 0 p n i= 0 ( ) = p B ( ) i i, n B i, n ( ) = i i n ( n i) ( ) n i p() = (" ) p0 + ( ) p + p p'() = ( ) p 0 + ( ) p + p p(0) = p r r r 0 p() = p p'(0) = ( p p0) r r r p'() = ( p p)
31 The order of Bezier crve is a fnction of the nmber of control points. For control points (n=3) always prodce a cbic Bezier crve. p p p0 p3 p p0 p p3 p p p0 p3 p3 p0 p p3 p p p p0 p p p p p0 p3 p p pn,p3 p p pn,p3 p
32 Convex Hll Property
33 An Example The coordinates of for control points relative to a crrent WCS are given by P 0 = " 0 T # $, P = " 3 0 T # $, P = " T # $, P3 = " 3 0 T # $ Find the eqation of the reslting Bezier crve. Also find points on crve for = 0, 4,, 3 4, P P? P 0 P 3
34 B i, n ( ) Soltion ( ) = P B + P B + P B + P B 0 P 0 0, 3, 3, 3 3, 3 = i i n ( n i) ( ) n i ( ) ( ) ( ) P = P( ) + 3P + 3P + P
35 Sbstitting the vales into his eqation gives [ ] T P( 0) = P0 = P = P0 + P + P + P3 = [ ] T = 0, ¼, ½, ¾, 3 3 P = P0 + P + P + P3 = [.5.75 ] T P = P0 + P + P + P3 = ( ) = P = [ 3 ] T 3 P 0 [ ] T P(,3) =/ P(3,3) - control points, P, P, P 3, & P 4, - points on crve, P() =/4 =3/4 P0(,) P3(3,)
36 Improvements of Bezier Crve Over the Cbic Spline The shape of Bezier crve is controlled only by its defining points (control points). First derivatives are not sed in the crve development as in the cbic spline. The order or the degree of the Bezier crve is variable and is related to the nmber of points defining it; n+ points define a nth degree crve. This is not the case for cbic splines where the degree is always cbic for a spline segment. The Bezier crve is smoother than the cbic splines becase it has higher-order derivatives.
37 B-Spline A Generalization from Bezier Crve Better local control Degree of reslting crve is independent to the nmber of control points.
38 Math Representation n P = P N ( ) 0 ( ) i i, k max i= 0 (k-) degree of polynomial with (n+) control points P, 0, P,... Pn n+ control points. N i, k ( ) B-spline fnction (to be calclated in a recrsive form) N ik, N ( ) N ( ) ( ) = ( i) + ( ) ik, i+, k i+ k i+ k i i+ k i+
39 Parametric Knots : parametric knots (or knot vales), for an open crve B-spline: j N ik, N ( ) N ( ) ( ) = ( i) + ( ) j ik, i+, k i+ k i+ k i i+ k i+ 0 j< k = j k+ k j n n k+ j > n where, 0 j n+k, ths if a crve with (k-) degree and (n+) control points is to be developed, (n+k+) knots then are reqired with 0 max = n k +
40
41 k-
42 Knot Vale Calclation N ik, N ( ) N ( ) ( ) = ( i) + ( ) ik, i+, k i+ k i+ k i i+ k i+ n = 3; 4 control points k = 4; 4-=3 cbic polynomial 0 j n+k= 7 n increases wider base k increases wider & taller
43 Properties of B-Spline Nmber of control points independent of degree of polynomial Linear k= vertex Qadratic B - Spline k=3 Cbic B - Spline k=4 Forth Order B - Spline k=5 vertex vertex vertex n=3; 4 control points The higher the order of the B-Spline, the less the inflence the close control point
44 Properties of B-Spline B-spline allows better local control. Shape of the crve can be adjsted by moving the control points. Local control: a control point only inflences k segments
45 Properties of B-Spline Repeated vales of a control point can pll a B-spline crve forward to vertex. ( Interactive crve control ) Same order polynomial Add more repeated control points to pll the crve
46 An Example Find the eqation of a cbic B-spline crve defined by the same control points as in the last example. How does the crve compare with the Bezier crve?
47 Example Problem for Finding the Bezier Crve
48 Example Problem for Finding the Bezier Crve
49 Finding the B-Spline Crve for the Same Example Problem
50 Vales to be Calclated n = 3; 4 control points k = 4; 4-=3 cbic polynomial j: 0 j n+k= 7 n increases wider base k increases wider & taller
51 Calclating the Knots, j
52 Calclating N i, N i, 0 < i i+ = = max and i+ otherwise and i = 0 0 = 0
53 Calclating Ni,k
54 Calclating Ni,k
55 Reslt n + control points: 3+=4 k degree crve: 4-=3 4 control points cbic polynomial
56 Non-Uniform Rational B-Spline Crve (NURBS) Rational B-Spline n P = P R ( ) 0 ( ) i i, k max i= 0 R i,k = h i N i,k ( ) n i =0 h i N i,k ( ) (h i " scalar ) If h =, then R ( ) N ( ),it is the representation of a B-Spline crve. i i, k = i, k Indstry Standard Today
57 h adds a degree of freedom to the crve, allowing to give more or less weight to each control point then
58
59 Development of NURBS Boeing: Tiger System in 979 SDRC: Geomod in 993 University of Utah: Alpha- in 98 Indstry Standard: IGES, PHIGS, PDES, Pro/E, etc.
60 Advantages of NURBS Serve as a genine generalizations of non-rational B-spline forms as well as rational and non-rational Bezier crves and srfaces Offer a common mathematical form for representing both standard analytic shapes (conics, qadratics, srface of revoltion, etc) and free-from crves and srfaces precisely. B-splines can only approximate conic crves. Provide the flexibility to design a large variety of shapes by sing control points and weights. increasing the weights has the effect of drawing a crve toward the control point. Have a powerfl tool kit (knot insertion/refinement/removal, degree elevation, splitting, etc. Invariant nder scaling, rotation, translation, and projections. Reasonably fast and comptationally stable. Clear geometric interpretations
61 Interpolation Using Hermite Crves
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63
64
65 Clamped or Free Ends
66 Disadvantages of Cbic Splines The order of the crve is always constant regardless of the nmber of data points. In order to increase the flexibility of the crve, more points mst be provided, ths creating more spline segments which are still of cbic order. The control of the crve is throgh the change of the positions of data points or the end slope change. The global control characteristics is not intitive.
67
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