Parametric Curves CS 318 Interactive Computer Graphics John C. - PowerPoint PPT Presentation

Parametric Curves CS 318 Interactive Computer Graphics John C. Hart

Linear Interpolation p 1 =( x 1 , y 1 ) • Need to get from point p 0 to point p 1 • y Define a parametric function p ( t ) p (0) = p 0 , p (1) = p 1 • Separate into coordinate functions p ( t ) = ( x ( t ), y ( t )) p 0 =( x 0 , y 0 ) x (0) = x 0 , x (1) = x 1 y (0) = y 0 , y (1) = y 1 x • Interpolate p ( t ) = p 0 + t ( p 1 – p 0 ) = (1- t ) p 0 + t p 1 x ( t ) = x 0 + t ( x 1 – x 0 ) = (1- t ) x 0 + t x 1 x y y ( t ) = y 0 + t ( y 1 – y 0 ) = (1- t ) y 0 + t y 1 t t

Hermite Interpolation p 1 =( x 1 , y 1 ) • From point p 0 along p ’ 0 to point p 1 toward p ’ 1 • y Define a parametric function p ( t ) p (0) = p 0 , p (1) = p 1 p’ (0) = p ’ 0 , p’ (1) = p ’ 1 • Separate into coordinate functions x (0) = x 0 , x (1) = x 1 p ’ 0 =( x ’ 0 , y ’ 0 ) p 0 =( x 0 , y 0 ) x ’(0) = x ’ 0 , x ’(1) = x ’ 1 x x ’(0) = C = x ’ 0 x (0) = D = x 0 , • Need cubic function x (1) = A + B + C + D = x 1 x ( t ) = At 3 + Bt 2 + Ct + D A + B = x 1 – x 0 – x ’ 0 x ’( t ) = 3 At 2 + 2 Bt + C x ’(1) = 3 A + 2 B + C = x ’ 1 • Solve 3 A + 2 B = x ’ 1 – x ’ 0 A = 2 x 0 – 2 x 1 + x ’ 0 + x ’ 1 = x ’ 1 – x ’ 0 – 2 x 1 + 2 x 0 + 2 x ’ 0 A = x ’ 1 + x ’ 0 – 2 x 1 + 2 x 0 B = -3 x 0 + 3 x 1 – 2 x ’ 0 – x ’ 1 = x 1 – x 0 – x ’ 0 – x ’ 1 – x ’ 0 + 2 x 1 B C = x ’ 0 , D = x 0 – 2 x 0 = 3 x 1 – 3 x 0 – x ’ 1 – 2 x ’ 0

Hermite Interpolation p 1 =( x 1 , y 1 ) • From point p 0 along p ’ 0 to point p 1 toward p ’ 1 • y Define a parametric function p ( t ) p (0) = p 0 , p (1) = p 1 p’ (0) = p ’ 0 , p’ (1) = p ’ 1 • Separate into coordinate functions x (0) = x 0 , x (1) = x 1 p ’ 0 =( x ’ 0 , y ’ 0 ) p 0 =( x 0 , y 0 ) x ’(0) = x ’ 0 , x ’(1) = x ’ 1 x • Need cubic function x ( t ) = At 3 + Bt 2 + Ct + D x ’( t ) = 3 At 2 + 2 Bt + C x y • Solve A = 2 x 0 – 2 x 1 + x ’ 0 + x ’ 1 t t B = -3 x 0 + 3 x 1 – 2 x ’ 0 – x ’ 1 C = x ’ 0 , D = x 0

Hermite Matrix p 1 =( x 1 , y 1 ) y p ( t ) = (2 p 0 – 2 p 1 + p ’ 0 + p ’ 1 ) t 3 + (-3 p 0 + 3 p 1 – 2 p ’ 0 – p ’ 1 ) t 2 + p ’ 0 t + d 0 =( x ’ 0 , y ’ 0 ) p 0 =( x 0 , y 0 ) p 0 (1) x      1 1 2 2 1 1 p 0        3 3 2 1 p       p 0 p 1   3 2 1 p ( ) t t t t 1      0 0 1 0 p ' 0     0     0 1 0 0 0 p ' 1 t t p’ 0 p’ 1 (2 t 3 – 3 t 2 + 1) p (t) = p 0 + (-2 t 3 + 3 t 2 ) p 1 + 0 0 ( t 3 – 2 t 2 + 1) p ’ 0 + ( t 3 – t 2 ) p ’ 1 t t

Linear Interpolation p 1 =( x 1 , y 1 ) • Need to get from point p 0 to point p 1 • y Define a parametric function p ( t ) p (0) = p 0 , p (1) = p 1 • Separate into coordinate functions p ( t ) = ( x ( t ), y ( t )) p 0 =( x 0 , y 0 ) x (0) = x 0 , x (1) = x 1 y (0) = y 0 , y (1) = y 1 x • Interpolate p ( t ) = p 0 + t ( p 1 – p 0 ) = (1- t ) p 0 + t p 1 x ( t ) = x 0 + t ( x 1 – x 0 ) = (1- t ) x 0 + t x 1 x y y ( t ) = y 0 + t ( y 1 – y 0 ) = (1- t ) y 0 + t y 1 t t

Interpolating Interpolations n ( t ) = (1- t ) B i n -1 ( t ) + tB i n - 1 ( t ) B i - 1 = 2 ( t ) = (1-t) B 0 1 ( t ) B 0 + = 2 ( t ) = (1- t ) B 1 1 ( t ) + t B 0 1 ( t ) B 1 = 2 ( t ) = t B 1 1 ( t ) B 2

3 ( t ) 3 ( t ) B 0 B 3 Bernstein Polynomials 1 3 ( t ) B 2 3 ( t ) B 1 • Defined for any degree n ( t ) = ( n i ) t i (1- t ) n - i B i • n choose i 0 0 1/3 2/3 1 i ) = n !/( i !( n – i )!) = ( n ( n - 1 ) + ( n - 1 1 ) i i - • Partition of unity 3 ( t )+ bB 1 3 ( t )+ cB 2 3 ( t )+ dB 3 3 ( t ) x ( t ) =aB 0 – Sum to one for any t in [0,1] x 1 S i =0.. n B i n ( t ) = 1 x ( t ) • Higher degrees lerps of lower degrees x 3 n ( t ) = ( n i ) t i (1- t ) n - i B i x 0 = ( n - 1 ) t i (1- t ) n - i + ( n - 1 1 ) t i (1- t ) n - i i i - n -1 ( t ) + tB i n - 1 ( t ) = (1- t ) B i - 1 x 2

p 1 Cubic Bezier Curves p 2 • Bernstein basis applied to points p ( t ) = S i ( 3 i ) t i (1- t ) 3- i p i • Bezier curve specified by four control p 0 points including two endpoints p 3 • Affine invariance: – Let M be a 4x4 transformation B 0 ( t ) B 3 ( t ) 1 – Then M p ( t ) = S i B i ( t ) M p i B 1 ( t ) B 2 ( t ) • Curve entirely contained in the convex hull of the control points 0 0 1/3 2/3 1

p 1 Cubic Bezier Matrix p 2 p ( t ) = (1- t ) 3 p 0 + 3(1- t ) 2 t p 1 + 3(1- t ) t 2 p 2 + t 3 p 3 = (1 – 3 t + 3 t 2 – t 3 ) p 0 + (3 t – 6 t 2 + 3 t 3 ) p 1 + (3 t 2 – 3 t 3 ) p 2 + t 3 p 0 p 3 p 3       1 3 3 1 p 0      3 6 3 0 p B 0 ( t ) B 3 ( t )         3 2 1 p ( ) t t t t 1   1     3 3 0 0 p 2         1 0 0 0 p B 1 ( t ) B 2 ( t ) 3 0 0 1/3 2/3 1

p ’ 3 Bezier v. Hermite p 3 p 2 p 1 = p 0 + 3 p ’ 0 p 2 = p 3 – 3 p ’ 3 • Bezier       1 1 3 1 p 0      2 6 3 0 p         3 2 1 p ( ) t t t t 1       p ’ 0 3 3 0 0 p p 1 p 0 2         1 0 0 0 p 3 • Hermite      2 2 1 1 p 0        3 3 2 1 p         3 2 3 p ( ) t t t t 1      0 0 1 0 p ' 0         1 0 0 0 p ' 3

Building Bernsteins n ( t ) = (1- t ) B i n -1 ( t ) + tB i n - 1 ( t ) B i - 1 = 2 ( t ) = (1-t) B 0 1 ( t ) B 0 + = 2 ( t ) = (1- t ) B 1 1 ( t ) + t B 0 1 ( t ) B 1 = 2 ( t ) = t B 1 1 ( t ) B 2

de Casteljau p 1 p 12 Algorithm p 2 p 012 1- t • Cascading lerps p 123 p 0123 p 01 = (1- t ) p 0 + t p 1 p 01 p 12 = (1- t ) p 1 + t p 2 t p 23 p 23 = (1- t ) p 2 + t p 3 p 0 p 012 = (1- t ) p 01 + t p 12 p 3 p 123 = (1- t ) p 12 + t p 23 p 0123 = (1- t ) p 012 + t p 123 • Subdivides curve at p 0123 – p 0 p 01 p 012 p 0123 – p 0123 p 123 p 23 p 3 • Repeated subdivision converges to curve

p 1 B-Spline Segment p 2 t =0 p ( t ) = ( – 1/6 p 0 +1/2 p 1 – 1/2 p 2 +1/6 p 3 ) t 3 + t =1 ( 1/2 p 0 – p 1 +1/2 p 2 ) t 2 + ( – 1/2 p 0 + 1/2 p 2 ) t + p 0 1/6 p 0 +2/3 p 1 +1/6 p 2 p 3 but makes more sense as… p ( t ) = ( – 1/6 t 3 + 1/2 t 2 – 1/2 t + 1/6) p 0 + – t 2 + 2/3) p 1 + ( 1/2 t 3 ( – 1/2 t 3 + 1/2 t 2 + 1/2 t + 1/6) p 2 + ( 1/6 t 3 ) p 3

B-Spline Basis ( t ) B 1 ( t ) B 2 2/3 p ( t ) = ( – 1/6 t 3 + 1/2 t 2 – 1/2 t + 1/6) p 0 + – t 2 + 2/3) p 1 + ( 1/2 t 3 1/6 ( t ) B 0 B 3 ( t ) ( – 1/2 t 3 + 1/2 t 2 + 1/2 t + 1/6) p 2 + ( 1/6 t 3 ) p 3 0 t 1 ( t ) p 1 + B 2 ( t ) p 2 + B 3 ( t ) p 3 = B 0 ( t ) p 0 + B 1 • Piecewise cubic approximation of a Gaussian bump function ( t ) B 2 B 1 ( t ) • Progressively weights ( t ) B 3 B 0 ( t ) points along spline p 3 p 2 p 1 p 0 0 1 0 1 0 1 0 1

p 0 Uniform B-Splines p 1 t =0 t =1 p 2 • p 3 t =2 Notation p 4 – d = degree of polynomial t =3 – k = order of polynomal = d + 1 p 5 t =4 – E.g. cubic: d = 3, k = 4 p 6 t =5 Segment i  t < i +1 uses k = d + 1 control • points p i to p i + d p 7 p (t) = S j 3 = 0 B j ( t mod 1) p i + j n  •  Normalized basis function N i , d ( t ) p ( t ) N ( t ) p i , k i N i , d ( t ) = B floor( t - i ) ( t mod 1) if i  t < i + d +1 •  i 0 – Otherwise its zero p 1 p 2 p 3 p 4 p 5 p 6 • Knot vector – e.g. [0,1,2,3,4,5,6,7] – in general [ t 0 , t 1 ,…, t n + k ] t 0 1 2 3 4 5

Non-Uniform knot vector: [0, .5, 1.3, 3.4, 4, 5.1, 6, 7] N =3 curve segments B-Splines d =3 (cubic) p 0 • Can specify an arbitrary parameter t i at each p 1 control point p i • Let N = # of polynomial curve segments p 2 t =1.3 • Parameters contained in a knot vector t =3.4 p 3 – Length ( N +1) + 2 d – 2 p 4 t =4 – [ t 0 , t 1 , t 2 , t 3 ,…, t N +2 d -2 ] p 5 – Cubic: [ t 0 , t 1 , t 2 , t 3 ,…, t N +4 ] t =5.1 p 6 • Domain of resulting curve is [ t d -1 , t N + d -1 ] – Cubic: domain = [ t 2 , t N+2 ] (segments [ t 0 , t 1 ], [ t 1 , t 2 ], [ t N +2 , t N +3 ] and [ t N +3 , t N +4 ] aren’t p 7 plotted) – Need d –1 “extra” knots at the beginning and end of the knot vector

Knot Multiplicity • Knot multiplicity = # of times a given knot 2 nd derivative appears in the knot vector discontinuity • Continuity = d – multiplicity • Cubic example – All knots unique – 2 nd derivative continuity – Multiplicity two – 1 st derivative cont. 1 st derivative discontinuity – Multiplicity three – 0 th derivative cont. – Multiplicity four – discontinuous • Endpoint interpolation – Knots of multiplicity d +1 at beginning and (0 th derivative) end of knot vector discontinuity – e.g. [0, 0, 0, 0, 1, 2, 3, 3, 3, 3]