1. CS 325 Introduction to Computer Graphics 02 / 17 / 2010 Instructor: Michael Eckmann

2. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Today’s Topics Questions/comments? Clipping –Polygons 3d –right handed vs. left handed coordinate systems –3 dimensional vectors –dot product –cross product –parametric equation of a line in 3 dimensions –equation of a plane in 3 dimensions

3. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Line Clipping Why use Liang-Barsky (L-B) vs. Cohen-Sutherland (C-S)‏ line clipping?

4. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Line Clipping Liang-Barsky (L-B) vs. Cohen-Sutherland (C-S)‏ –L-B in general more efficient updates of u1 & u2 require only one divide window edge intersections are computed only once when u1 & u2 are final –C-S repeatedly calcs intersections along a line path even though line may be completely outside clipping window each intersection calc requires a divide and a multiply Both line clipping algorithms can be extended to 3-d A third line clipping algorithm, Nicholl-Lee-Nicholl (N-L-N), which we will not go into is faster than both L-B & C-S but it cannot be extended to 3-d.

5. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Polygon Clipping Polygon clipping –Polygons can be clipped against successive infinite extensions of the clipping window edges –Keep track of new vertices (intersections with edge of clipping window) at each stage. –Notice the number of vertices increased in the example below.

6. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Polygon Clipping

7. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Polygon Clipping The text describes two polygon clipping algorithms. The Sutherland-Hodgman algorithm and the Weiler-Atherton algorithm. A major difference between the two is that the Sutherland-Hodgman algorithm produces as output one polygon. Think about a concave polygon that when clipped, really should result in two polygons. Example on board.

8. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 3d math (Ch. 5, Appdx A, Sec. 3-15)‏ Coordinate systems Vectors Lines Plane Homogeneous Coordinates of a 3d point Transformations –Translation –Scaling –Rotation (about x, y, z)‏ –Shear (in xy)‏

9. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Right-handed 3d coordinate system

10. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Left-handed 3d coordinate system

11. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors A vector is a directed line segment that has magnitude (length) and direction. We can define a vector as the difference between two points. (Example on board.)‏ fig A-13 in text. In 2 dimensions: V = P 2 -P 1 = (v x, v y )‏ In 3 dimensions: V = P 2 -P 1 = (v x, v y, v z )‏ The v x, v y, and v z values are the projections of the line segment onto the x, y and z axes, respectively. Magnitude of a Vector is determined by the Pythagorean theorem: For 2d: |V| = sqrt(v x 2 + v y 2 ) and for 3d: |V| = sqrt(v x 2 + v y 2 + v z 2 )‏

12. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors For 2d vectors: the direction of a vector is often specified as an angle in relation to the horizontal. For 3d vectors: the direction of a vector is often specified as three angles in relation to the positive axes. (note: The alpha in the diagram would only be accurate if V was on the y-z plane.)‏ cos α = v x / |V| cos β = v y / |V| cos γ = v z / |V| cos 2 α + cos 2 β + cos 2 γ = 1

13. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors A vector has a direction and a magnitude. Does it have a position in the coordinate system?

14. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors A vector has a direction and a magnitude. Does it have a position in the coordinate system? –No –Example: –p1 = (5,5,5), p2 = (7,7,5), p3 = (2,7,3), p4 = (4,9,3)‏ –The vectors generated by [p2 – p1] and [p4 – p3] result in the same vector which is [2,2,0] what's its magnitude? what's its direction? what's its position?

15. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors A vector has a direction and a magnitude. Does it have a position in the coordinate system? –No –Example: –p1 = (5,5,5), p2 = (7,7,5), p3 = (2,7,3), p4 = (4,9,3)‏ –The vectors generated by [p2 – p1] and [p4 – p3] result in the same vector which is [2,2,0] what's its magnitude? sqrt (4+4) = sqrt(8) = 2*sqrt(2)‏ what's its direction? –cos -1 (sqrt(2)/2) = 45 degrees –cos -1 (0) = 90 degrees what's its position? –it has no position

16. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors Vector addition: V 1 + V 2 = (v 1x +v 2x, v 1y +v 2y, v 1z +v 2z )‏ Scalar multiplication: sV = (sv x, sv y, sv z )‏ Dot product (aka scalar product) of 2 vectors results in a scalar: V 1 ● V 2 = |V 1 | |V 2 |cos θ θ is the (smaller) angle between the two vectors alternatively: V 1 ● V 2 = v 1x v 2x +v 1y v 2y +v 1z v 2z What would be the dot product of two perpendicular vectors (those with the angle between them being 90 degrees)?

17. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors Cross product (aka vector product) of 2 vectors results in a vector: V 1 x V 2 = u |V 1 | |V 2 |sin θ θ is the angle between the two vectors u is a unit vector (length = 1) perpendicular to both V 1 and V 2 u's direction is determined by the right-hand rule Right-hand rule is: with your right hand, grasp the axis perpendicular to the plane of the two vectors and make sure that the direction of your fingers curve from v1 to v2. u's direction is the direction of your thumb. Alternatively: V 1 x V 2 = (v 1y v 2z – v 1z v 2y, v 1z v 2x – v 1x v 2z, v 1x v 2y – v 1y v 2x )‏ Cross product is not commutative, nor associative. V 1 x V 2 = - (V 2 x V 1 )‏

18. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors Cross product of two vectors is a vector that is perpendicular to the two vectors and has magnitude equal to the area of the parallelogram formed by the two vectors. (picture on board)‏

19. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 vectors Recap –a vector has magnitude and direction (but no position)‏ –addition of 2 vectors results in a vector –a scalar times a vector results in a vector –Cross product of two vectors results in a vector –but –dot product of two vectors results in a scalar

20. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 lines Parametric equation of a line in 3 dimensions –Given points P 1 and P 2 the equation of a line that contains these points is: x = x 1 + t(x 2 – x 1 )‏ y = y 1 + t(y 2 – y 1 )‏ z = z 1 + t(z 2 – z 1 )‏ –Given a point P 1 and vector V the equation of a line that contains the point and is in the direction of V is: x = x 1 + t(x v )‏ y = y 1 + t(y v )‏ z = z 1 + t(z v )‏ Line = P 1 + t (P 2 – P 1 )‏ Line = P 1 + Vt

21. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Planes (part of sec. 3.15)‏ The general plane equation: Ax + By + Cz + D = 0 A, B, C, D are constants and (x,y,z) are the coordinates of the points on the plane. A useful form of the plane equation is: A'x + B'y + C'z + D' = 0 where A'=A/d, B'=B/d, C'=C/d, D'=D/d, and d = sqrt(A 2 + B 2 + C 2 )‏ Because then it's easy to find the distance between a point (x 1, y 1, z 1 ) and the plane which is simply: A'x 1 + B'y 1 + C'z 1 + D' A normal vector to a plane is perpendicular to the plane. If the equation of the plane is Ax + By + Cz + D = 0, then a normal to the plane is the vector (A, B, C)‏ Pictures on board.

22. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 The plane equation From the drawing on the board –Equation of the plane: Ax + By + Cz + D = 0 –Given three non collinear points, P 1, P 2, P 3 these points uniquely determine the plane –The cross product of P 1 - P 2 and P 3 - P 2, gives us a normal vector N. –For an arbitrary point P = (x, y, z), P is on the plane if N ● [ P - P 2 ] = 0 –Why?

23. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 The plane equation From the drawing on the board –Equation of the plane: Ax + By + Cz + D = 0 –Given three non collinear points, P 1, P 2, P 3 these points uniquely determine the plane –The cross product of P 1 - P 2 and P 3 - P 2, gives us a normal vector N. –For an arbitrary point P = (x, y, z), P is on the plane if N ● [ P - P 2 ] = 0 –Why? Because if N is normal to the plane, and P-P 2 is a vector on the plane, then the angle between these is 90 degrees and so the dot product will be 0, since cos(90) = 0. –If P is not on the plane, therefore P-P 2 is not on the plane, so what happens?

24. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 The plane equation For an arbitrary point P = (x, y, z), P is on the plane if N ● [ P - P 2 ] = 0 –So, we can derive the plane equation like so: –N = (A, B, C) and is normal to the plane –P = (x, y, z) and represents any point on the plane –P 2 = (x 2, y 2, z 2 ) is some fixed point on the plane N ● [ P - P 2 ] = [A,B,C] ● [x- x 2,y-y 2, z- z 2 ]= A(x- x 2 ) + B(y-y 2 ) + C(z- z 2 ) = Ax + By + C z + A(- x 2 ) + B(-y 2 ) + C(- z 2 ) = 0 Notice that x, y and z are variables and the rest are constant values so this gives Ax + By + C z + D = 0 where D = A(- x 2 ) + B(-y 2 ) + C(- z 2 )‏

25. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Inverse Transforms and Identity The inverse of a Matrix is the matrix that when multiplied by the original results in the Identity matrix. The identity matrix is an nxn matrix with 1's along the main diagonal (from upperleft to lowerright) and 0's elsewhere. MM -1 = I = M -1 M

26. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 Reminder about Homogeneous Coordinates When we transform some homogeneous point (x,y,z,1) we may end up with something other than 1 in the homogeneous parameter slot i.e. (x 1,y 1,z 1,w), w!= 1. To figure out which point to plot in 3d, we need to divide each coordinate by w x' = x 1 / w y' = y 1 / w z' = z 1 / w 1 = w / w so we plot (x',y',z')‏

27. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 3d transformations Translation Scale Rotation Shear

28. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 3d Translation 3d Translation in homogeneous coordinates is a direct extension of translation in 2d. [ 1 0 0 t x ] [x] [x+t x ] [ 0 1 0 t y ] [y] = [y+t y ] [ 0 0 1 t z ] [z] [z+t z ] [ 0 0 0 1 ] [1] [ 1 ]

29. Michael Eckmann - Skidmore College - CS 325 - Spring 2010 3d Scale [ s x 0 0 0 ] [x] [s x x] [ 0 s y 0 0 ] [y] = [s y y] [ 0 0 s z 0 ] [z] [s z z] [ 0 0 0 1 ] [1] [ 1 ] 3d Scale has the same problem/feature that 2d scaling has, namely it also translates.