Computer Graphics Zhen Jiang West Chester University Topics

Computer Graphics Zhen Jiang West Chester University

Topics • • • Curves 2 D and 3 D Clipping Projection and shadow Animation

Curves • Parametric cubic polynomial curves • Interpolation (interpolating polynomials) • Bezier Curve

Curves Where is the position for the next point? K=4 K=3 K=2 K=1

Curves Where is the position for the next point? K=4 K=3 K=2 K=1

Parametric cubic polynomial curves • P(u)=c 0+c 1 u+c 2 u 2+c 3 u 3 = u. Tc – Where u. T = [1, u, u 2, u 3] and c. T= [c 0, c 1, c 2, c 3 ]

Interpolation Where is the position for any point on this curve? K=4 K=3 K=2 K=1

Interpolation • P 0=p(0) = c 0 P 1=p(1/3) P 2=p(2/3) P 3=p(1) = c 0+c 1+c 2+c 3 • p. T=[p 0, p 1, p 2, p 3], A = [ • M=A-1= [ 1, 0, 0, 0 1, 1/3, (1/3)2, (1/3)3 1, 2/3, (2/3)2, (2/3)3 1, 1, 1, 1 1, 0, 0, 0 -5. 5, 9, -4. 5, 1 9, -22. 5, 18, -4. 5, 13. 5, -13. 5, 4. 5 ] • c=Mp • P(u)=u. TMp=b(u)p – where b(u) = [b 0(u), b 1(u), b 2(u), b 3(u), ] ]

Interpolation • • B 0(u)=-9/2(u-1/3)(u-2/3)(u-1) B 1(u)=27/2(u-2/3)u(u-1) B 2(u)=-27/2(u-1/3)u(u-1) B 3(u)=9/2(u-1/3)(u-2/3)u

Interpolation • Example – p. T=[p 0, p 1, p 2, p 3], where p 0= (0, 0), p 1=(1, 1), p 2, =(2, 3), p 3 =(3, 6) – (X, Y) when u=0? • B 0(0)=1, B 1(0)=0, B 2(0)=0, B 3(0)=0 • Since P(u)=b(u)p, X=b(u)Xp, Y=b(u)Yp, where Xp. T= [0, 1, 2, 3] and Yp. T= [0, 1, 3, 6]. 0 • X=[1, 0, 0, 0] [0, 1, 2, 3] T=[1, 0, 0, 0][ 1 ]=0 and Y=0; 2 3

Interpolation • Example – (X, Y) when u=1? • B 0(0)=0, B 1(0)=0, B 2(0)=0, B 3(0)=1 • Since P(u)=b(u)p, X=b(u)Xp, Y=b(u)Yp, where Xp. T= [0, 1, 2, 3] and Yp. T= [0, 1, 3, 6]. 0 0 0, 0, 0, 1 3 0, 0, 0, 1 • X=[ ][ ]= and Y= [ ][ 1 ]= 6 ; 1 2 3 3 6

Interpolation • Example – (X, Y) when u=3? • B 0(0)=-56, B 1(0)=189, B 2(0)=-216, B 3(0)=84 • Since P(u)=b(u)p, X=b(u)Xp, Y=b(u)Yp, where Xp. T= [0, 1, 2, 3] and Yp. T= [0, 1, 3, 6]. • X=[ • Y= [ -56, 189, -216, 84 ][ 0 1 2 3 0 1 3 6 ]= 9 ]=45

Bezier Curve • Two endpoints – p 0(x 0, y 0) origin endpoint, source – p 3(x 3, y 3) destination endpoint • Two control points – p 1(x 1, y 1) – p 2(x 2, y 2) • t form 0 to 1 • New point p(x(t), y(t)) • p(t)=c 0+c 1 t+c 2 t 2+c 3 t 3 – Where c 0=p 0, c 1=3(p 1 -p 0), c 2=3(p 2 -p 1)-c 1, c 3=p 3 -p 0 -c 1 -c 2.

/* Three control point Bezier interpolation mu ranges from 0 to 1, start to end of the curve */ XYZ Bezier 3(XYZ p 1, XYZ p 2, XYZ p 3, double mu) { double mum 1, mum 12, mu 2; XYZ p; mu 2 = mu * mu; mum 1 = 1 - mu; mum 12 = mum 1 * mum 1; p. x = p 1. x * mum 12 + 2 * p 2. x * mum 1 * mu + p 3. x * mu 2; p. y = p 1. y * mum 12 + 2 * p 2. y * mum 1 * mu + p 3. y * mu 2; p. z = p 1. z * mum 12 + 2 * p 2. z * mum 1 * mu + p 3. z * mu 2; return(p); }

/* Four control point Bezier interpolation mu ranges from 0 to 1, start to end of curve */ XYZ Bezier 4(XYZ p 1, XYZ p 2, XYZ p 3, XYZ p 4, double mu) { double mum 1, mum 13, mu 3; XYZ p; mum 1 = 1 - mu; mum 13 = mum 1 * mum 1; mu 3 = mu * mu; p. x = mum 13*p 1. x + 3*mu*mum 1*p 2. x + 3*mu*mu*mum 1*p 3. x + mu 3*p 4. x; p. y = mum 13*p 1. y + 3*mu*mum 1*p 2. y + 3*mu*mu*mum 1*p 3. y + mu 3*p 4. y; p. z = mum 13*p 1. z + 3*mu*mum 1*p 2. z + 3*mu*mu*mum 1*p 3. z + mu 3*p 4. z; return(p); }

/* General Bezier curve Number of control points is n+1 0 <= mu < 1 IMPORTANT, the last point is not computed */ XYZ Bezier(XYZ *p, int n, double mu) { int k, kn, nkn; double blend, muk, munk; XYZ b = {0. 0, 0. 0}; muk = 1; munk = pow(1 -mu, (double)n); nn--; if (kn > 1) { blend /= (double)kn; kn--; } if (nkn > 1) { blend /= (double)nkn; nkn--; } } /*end of while*/ b. x += p[k]. x * blend; b. y += p[k]. y * blend; b. z += p[k]. z * blend; } /*end of for*/ for (k=0; k<=n; k++) { return(b); nn = n; } kn = k; nkn = n - k; blend = muk * munk; muk *= mu; munk /= (1 -mu); while (nn >= 1) { blend *= nn;

2 D and 3 D (Transformation) • Coordinate system • Translation (p’=p+d) – (0, 0) by d(1, 2) – (1, 2) by d(2, 4) – (3, 6) by d(-3, -6) • Rotation cos Sin -sin cos p’ = [ -(3, 0) by =90˚ ]p=[ P (x’, y’) cos Sin -sin cos x y ][ ] • Scaling p’= r p (r > 1, big one; r < 1, small one) P (x, y)

2 D and 3 D (Transformation) • Mixed d For example, P’= r( P+d), where =[ cos -sin Sin cos ] ! What will it be if P’= r P+d, r (P+d), (r. P+d), r(P+d), or r. P+d? ! Undo if P’= r( P+d), P’’=P= ’(1/r. P’-d), where ’=[ cos sin ] -sin P r cos If P’’’= ’(1/r)P’-d, is P’’’=P? How about ’(1/r)(P’-d), (1/r) ’P’-d, , …, ?

2 D and 3 D (Transformation) • Translation (p’=p+d) – (0, 0, 0) by d(1, 2, 3) – (1, 2, 3) by d(2, 4, 6) – (3, 6, 9) by d(-3, -6, -9) • Rotation p’ = R p ( Rz cos 0 1 Ry =[ 0 -sin 0 • Scaling cos -sin 0 =[ sin cos 0 ], 0 0 1 sin 0 ], cos Rx =[ 1 0 0 2 0 cos -sin], 3 0 sin cos p’= r p (r > 1, big one; r < 1, small one) • Mixed

2 D and 3 D (Transformation) Y Y P Rz( ) P Rx( ) x x P Ry( ) Z Z

2 D and 3 D (Transformation) • 45 degree about the line through origin (0, 0, 0) and the point(1, 2, 3) with a fixed point of (4, 6, 5): gl. Matrix. Mode (GL_MODELVIEW); gl. Load. Identity(); gl. Translatef(4. 0, 6. 0, 5. 0); gl. Rotatef(45. 0, 1. 0, 2. 0, 3. 0); gl. Translatef(-4. 0, -6. 0, -5. 0);

2 D and 3 D (Transformation) Y’ Y P’ P 1 P X Z Z’ X P 1= P+d Z X’

2 D and 3 D (Transformation) P’ 1 Y’ Y’ P’ Y (1, 2, 3) X’ (0, 0, 0) Y P 2 P 1 Z’ X P 2= Rz P 1= Rz(P+d) Z Z X’

2 D and 3 D (Transformation) P’ 2 P’ 1 Y’ Y’ X Y P 2 Z’ X’ Y Z’ X P 3= Rx P 2= Rx. Rz(P+d) Z P 3 Z X’

2 D and 3 D (Transformation) P’ 3 Y Y’ X Y P 3 X Y X’ X P’ 2=? P’ 3 P’ 1=? P’ 2 Z’ Z P’ 3= Ry(45) P 3= Ry(45)Rx. Rz(P+d) P’=? P’ 1 Z Z Need solution? Check “undo” in 2 Ds.

Clipping

Clipping

Clipping (6, 18) (2, 10) (22, 18) (19, 9) (6, 6) (14, 4) (2, 2) (22, 6) (19, 2)

Clipping • Step 1: Start from a vertex – If it is a point inside window, save it as a vertex in the new polygon; status=1; – Otherwise, status=0. • Step 2: Pick the edge starting from this vertex until there is no new vertex of this polygon – If it has an intersection [0, 1) with any edge of window • status=!status, save the intersection as a vertex for the new polygon; • If status==1, save the endpoint of this segment as a vertex for new polygon; • Repeat step 2, start from the endpoint of this segment. – If it has no intersection with any edge of window • If status ==1, save the endpoint of this segment as a vertex for new polygon; • Repeat step 2, start from the endpoint of this segment. • Step 3: Connect all the vertices of the new polygon

Clipping • Does a segment from A(x 1, y 1) to B(x 2, y 2) have an intersection with an edge from P(x 3, y 3) to Q(x 4, y 4)? y up x 3=x 4 I=(1 - )A+ B, 0 ≤ ≤ 1 ylo = (x 3 -x 1)/(x 2 -x 1) y= (1 - )y 1+ y 2 0 ≤ ≤ 1 && ylo ≤ y ≤yup = (y 3 -y 1)/(y 2 -y 1) x= (1 - )x 1+ x 2 xmin y 3=y 4 xmax 0 ≤ ≤ 1 && xmin ≤ x ≤xmax