Computer Game Basic 1 Computer Graphics and Its

Computer Game : Basic 1. Computer Graphics and Its application 2. 2 D Coordinate Geometry 3. 1 3. 2 3. 3 3. 4 3. 5 3. 6 2 D geometry -- points and polygons Object transformation and Coordinates System transformation 2 D Transformation -- Scaling, Translation, Rotation. Homogeneous Coordinates Composite 2 D Transformation Other Transformation -- Reflection (Mirror) and Shearing (Skip) 4. 2 D Viewing Operations 4. 1 Window-to-Viewport Mapping 4. 2 2 D Clipping 5. 3 D Coordinate System 5. 1 Coordinates System -- Right-handed and Left-Hand Coordinate system (a) 5. 2 3 D gemoetry -- points and polyhedra (a) 5. 3 3 D Transformation -- Scaling, Translation and Rotation. 5. 4 Composite 3 D Transformation -- Rotation about arbitrary Axis 5. 5 Other Transformation -- Reflection and Shearing 6. 3 D Viewing Operations 6. 1 Projections -- Orthgraphic Projections and Perspective Projections(a 6. 2 View Transformation 6. 3 View Volume and Z-clipping 7. Visual Realism (Skip) 7. 1 7. 2 7. 3 7. 4 7. 5 7. 6 CG Rendering techniques Hidden line removal algorithm z-buffer algorithm illumination and shading Ray-tracing algorithm Color and light Page: 1 東吳資訊管理 江清水

Computer Graphics vs. Pattern Recognition 3 D arrows with 3 D box Pattern Recognition CG Page: 2 Computer Graphics 東吳資訊管理 江清水

Computer Graphics Application Some of today's most popular applications in computer graphics are : • Computer Game • Presentation Graphics • Electronic design and circuit layout • Mechanical design and drafting • Process Control • Data Analysis – 2 D contour; 3 D continuous mesh surface plot – air flow and fluid flow – 4 D image; structural analysis of any properties • • • CG Simulation Chemical and molecular analysis User Interface Animation Art Geometric and solid modeling Page: 3 東吳資訊管理 江清水

2 D Geometry - points and polygons A point in 2 D is represented by two real numbers (X, Y) Y X A line segment is represented by its two end points (X 1, Y 1) (X 2, Y 2), or a 2 x 2 matrix [ ] X 1 Y 1 X 2 Y X CG Page: 4 東吳資訊管理 江清水

A polygon is represented by an list of points (X 1, Y 1), (X 2, Y 2), . . . (Xn, Yn) or a [ ] X 1 Y 1 n x 2 matrix : : Xn Yn For example, a triangle is represent by X 1 Y 1 X 2 Y 2 X 3 Y 3 [ ] Y X CG Page: 5 東吳資訊管理 江清水

Object transformation and Coordinates System transformation Object transformation is different from the coordinates system transformation Y (2, 4) 2 1 (2, 2) 1 2 CG Page: 6 (6, 2) 8 X 東吳資訊管理 江清水

Object Transformation Y Scale (1/2) about the origin X Y Translation (2, 1) X Rotation (p/2) counter clock wise about the origin Y X CG Page: 7 東吳資訊管理 江清水

Coordinates system transformation Y* Scale (1/2) about the origin 2 1 Translation (2, 1) X* 3 Y X Rotation (p/2) counter clock wise about the origin Y X If not described explicitly, transformation always means object transformation CG Page: 8 東吳資訊管理 江清水

2 D Translations There are 3 "basic" transformations: (1) Translation (2) Scaling (3) Rotation A series of transformations can be combined (concatenated) into one. (1) Translation : T(Tx, Ty) Y (x', y') Ty (x, y) Tx X CG Page: 9 東吳資訊管理 江清水

(2) Scaling (about the origin) : S(Sx, Sy) Y Y (x', y') (x, y) X X • What about "Mirror Images"? S(-2, -1) • How do we avoid distortion? • What happens when the scale factor equal to zero? CG Page: 10 東吳資訊管理 江清水

(3) Rotation (about the origin counterclockwise) : Y (x', y') (x, y) q a x y x' y' = = X R cos a R sin a R cos(a + q) = R(cos a cos q - sin a sin q) R sin(a + q) = R(sin a cos q + cos a sinq) So, x ' = xcosq - ysinq y ' = xsinq + ycosq CG Page: 11 東吳資訊管理 江清水

Concatenation -- A series of transformations can be combined (concatenated) into one. Example: scaling about arbitrary point. Y (x, y) (a, b) X 1. Translate so that point (a, b) becomes the temporary origin: x 1 = x - a y 1 = y - b 2. Scale to the correct size: x 2 = Sx*x 1 y 2 = Sy*y 1 3. Translate again to restore the coordinates of (a, b): x 3 = x 2 + a y 3 = y 2 + b CG Page: 12 東吳資訊管理 江清水

Algebraic representations for transformations are very limited: (1) A special procedure is need for each basic transformation and other known concatenated forms. (2) It is difficult to provide general transformation capabilities in algebraic form in the computer. (3) How can one find the inverse of a concatenated equation (to restore the original position, for example)? CG Page: 13 東吳資訊管理 江清水

Let the point (x, y) be a row vector [x y]: x' = ax + by and y' = cx + dy can be expressed in the matrix equation: [x' y'] = [x y] * [ ] 2 x 2 Let P' = [x' y'] and P = [x y], this becomes the matrix equation: P' = P * T where P' =[x' y'] , a c P = [x y] , T = [ b d ] Consider three basic transformations, can we find a "T" for each? • What about translations? X'=X+Tx, Y'=Y+Ty No, P' =[x' y']=[x y] [ ]+[T 1 0 0 1 x Ty]=P*T+Q • What about scaling? X'=Sx*X, Y'=Sy*Y Yes, P' =[x' y']=[x y] [ S 0 ] =P*T x y • What about rotation? x'=xcosq-ysinq, y'=xsinq+ycosq Yes, P' =[x' y']=[x y] CG Page: 14 [ cosq -sinq cosq ] 東吳資訊管理 江清水

Homogeneous Coordinates Let the point (x, y) be a row vector [x y 1]: x' = ax + by + e and y' = cx + dy + f can be expressed in the matrix equation: [x' y' 1] = [x y 1] [ ] 3 x 3 This becomes the matrix equation P' = P * T where P'=[x' y' 1], P= [x y 1], T= [ ] Now, can we find a "T" for each transformation? • What about translations? • What about scaling? Yes, T= [ [ • What about rotation? Yes, T= CG Page: 15 [ ] ] ] 東吳資訊管理 江清水

To check, what should the matrix be if we: (1) translate by (0, 0) ? P'=P* [ ]=P*[ ]=P (2) scale by (1, 1) ? P'=P* [ ]=P*[ ]=P (3) rotate by 0 o ? P'=P* [ ]=P*[ ]=P (4) rotate by 90 o ? (x', y') = (0, 1) P'=[x' y' 1]=[x y 1] (x, y) = (1, 0) =[1 0 1]* [ [ ] ]=[0 1 1] Draw the pictures and find the new vertices of the triangle for the following transformations (See the figure below) : Y A(2, 2) C(4, 1) B (2, 1) CG Page: 16 X 東吳資訊管理 江清水

(5) translate A by (2, 1). Y A (2, 2) C (4, 1) X B (2, 1) (6) translate A by (-2, 1). Y A (2, 2) C (4, 1) X B (2, 1) (7) scale A by (2, 1). Y A (2, 2) C (4, 1) X B (2, 1) (8) scale A by (-1, 1). Y A (2, 2) B (2, 1) CG Page: 17 C (4, 1) X 東吳資訊管理 江清水

(9) scale A by (1, 0). Y A (2, 2) B (2, 1) C (4, 1) X (10) scale A by (0, 0). Y A (2, 2) B (2, 1) C (4, 1) X (11) rotate A by 0 o. Y A (2, 2) B (2, 1) C (4, 1) X o. (12) rotate A by 90 Y A (2, 2) B (2, 1) CG Page: 18 C (4, 1) X 東吳資訊管理 江清水

Composite 2 D Transformation Concatenation is easy now. Consider our scaling example: Example: We want to reduce the size or the box, without changing the position of point (a, b) Y (x, y) (a, b) X Functional form and matrix form for each transformations: (1) translate: T(-a, -b) (2) scale : S(Sx, Sy) (3) translate: T(a, b) CG Page: 19 東吳資訊管理 江清水

Combining the matrix equations: P 3 =P 2* T (-a, -b)=(P 1*S(Sx, Sy) )*T(a, b) =P*T(-a, -b) *S(Sx, Sy)*T(a, b) Multiply the matrices : Now, the matrix equation is : or, in algebraic form : Thus, a single 3 x 3 matrix can represent any combination of basic transformations in a simple form. CG Page: 20 東吳資訊管理 江清水

Window-to-viewport Mapping We can define a window as a rectangular region of the world coordinate space, and the viewport as a rectangular region of the device coordinate system. (1, 1) (0, 1/2) VIEWPORT WINDOW (Normalize) (0, 0) (-1, 1) (3/4, 1) (0, 1/2) (Distortion) WINDOW (-1, 1) CG (1/4, 0) Page: 21 東吳資訊管理 江清水

The problem is to "map" coordinates from the window units into the viewport units. sxv sxw (xw, yw) (xv, yv) syw (cxw, cyw) (cxv, cyv) window viewport syv Derive the mapping from (xw, yw) , a coordinate in the window, to (xv, yv), the corresponding point in the viewport: * Remove window coordinate system x 1 = y 1 = sxw (x 1, y 1) syw (cxw, cyw) window CG Page: 22 東吳資訊管理 江清水

sxv * Convert units x 2 = y 2 = (x 2, y 2) syv (cxv, cyv) viewport * Remove to viewport center xv = yv = so xv = yv = (1) What happens if the ratios sxv/sxw and syv/syw are not equal? (2) If you want a set of data in the window, how do you decide the size of the window? CG Page: 23 東吳資訊管理 江清水

(3) Can the window-to-viewport mappintg represented in matrix form? CG Page: 24 東吳資訊管理 江清水

u Homework: Find the transformation matrix that will map points contained in a window whose lower left corner is at (2, 2) and upper right corner is at (6, 5) onto a viewport that has a lower left corner at (1/2, 1/2) and upper right corner at (1, 1). CG Page: 25 東吳資訊管理 江清水

(4) Derive the window to viewport mapping for viewports defined on devices which have the origin at the upper left of the screen, with the horizontal coordinate increasing right, and the vertical coordinate increasing down. y (100, 100) h (80, 80) * * (0, 0) CG window Page: 26 x v viewport (100, 100) 東吳資訊管理 江清水

2 D Clipping The process of clipping decides which part, if any, of a primitive lies inside the window. Clipping The algorithms used for line clipping can be divided into two parts: (1) Check all the line segments and separate those that intersect the window boundary. (The curve is represented by a sequence of line segment) (2) Clip the line segments obtained from step 1 by calculating their intersections with the window boundaries. CG Page: 27 東吳資訊管理 江清水

E F D G Ymax A B C Ymin Xmax (1) How can we sure that a line segment is totally inside the window (such as A) ? Both endpoints of the segment fall within the boundaries. (2) How can we sure that a line segment is totally outside the window (such as E, F, G)? Let two endpoints are (x 1, y 1), (x 2, y 2) If ( max{x 1, x 2} < xmin or min{x 1, x 2} > xmax or max{y 1, y 2} < ymin or min{y 1, y 2} > ymax) then the line segment is totally outside the window. We introduce an efficient procedure, the Cohen-Sutherland algorithm, to check the intersection property and further clip the line segment if necessary. CG Page: 28 東吳資訊管理 江清水

Cohen-Sutherland Algorithm The 2 D plane is divided into 9 regions. Each region is designated by a 4 -bit (base 2) integer code. Each bit of the code is set to 1 (true) or 0 (false), starting with the left most one. The bits are assigned by the follow rule: bit 1 = 1 if the region is above the window. bit 2 = 1 if the region is below the window. bit 3 = 1 if the region is on the right of the window. bit 4 = 1 if the region is on the left of the window. (1001) (1000) (1010) Ymax (0001) (0000) (0010) Ymin (0001) (0100) (0110) Xmin CG Page: 29 Xmax 東吳資訊管理 江清水

(1) Given a point (x, y), we would like to assign a "region code" to it by using Cohen-Suterlandalgorithm. In C this can be coded: code = 0 if (x < xmin) code = 1 if (x > xmax) code = 2 if (y < ymin) code += 4 if (y > ymax) code += 8 (2) According to the C code in (1), what code will be assigned if the point is on the boundary of the window? (0000) * 0 (3) How about the point on the line Xmax and not on the boundary of the window? (0100) or (1000) depending on y CG Page: 30 東吳資訊管理 江清水

(4) Draw precise picture for the partition of 2 D plane according to the C code in (1). (Using solid line and dotted line) Let's consider the line clipping now. At first, the visibility of the line segment is determined as follows: (1) Visible -> Both endoint codes are (0000). CG Page: 31 東吳資訊管理 江清水

(2) Invisible -> Bitwise logical AND of the endpoint codes is not (0000). (3) Indeterminate -> Bitwise logical AND of the endpoint codes is (0000), but at least one of the endpoint does not have a (0000) code. On the third situation, the line endpoints are "pushed" towards the boundary line until the first or the second situation occurs. Ymax (0001) (0010) (0000) Xmin CG Page: 32 Ymin Xmax 東吳資訊管理 江清水

The algorithm steps: 1. Calculate IC 1 and IC 2, the code of the two endpoints. 2. If (IC 1 + IC 2. eq. 0) then "Visible". 3. If ( (IC 1. AND. IC 2). NE. 0) then "Invisible". 4. If ( IC 1. EQ. 0) then { Swap(IC 1, IC 2); Swap(X 1, X 2); Swap(Y 1, Y 2); } // At lease one point is outside the // viewport. Insure that point #1 is // one of the outside point. 5. Push endpoint #1 towards Xmin, Xmax, Ymin, or Ymax, depending on which region endpoint #1 occupies (via IC 1). bit 1 = 1 toward Ymax bit 2 = 1 toward Ymin bit 3 = 1 toward Xmax bit 4 = 1 toward Xmin If endpoint #1 towards Ymax , then { Ymax = Y = X 1 + t(Y 2 - Y 1), find t and calculate X = X 1 + t(X 2 - X 1) } 6. Go To Step #1 above. CG Page: 33 東吳資訊管理 江清水

Example: Let Xmin = 2, Xmax = 8, Ymin = 2 and Ymax = 8, check the visibility of the following segments using the Cohen-Sutherland Algorithm and, if necessary, clip them against the appropriate window boundaries. Line AB: A(3, 10) B(6, 12) Line CD: C(4, 1) D(10, 6) B(6, 12) A(3, 10) Ymax =8 D(10, 6) Ymin=2 C(4, 1) Xmin=2 CG Page: 34 Xmax=8 東吳資訊管理 江清水

Example: Let Xmin = 2, Xmax = 8, Ymin = 2 and Ymax = 8, check the visibility of the following segments using the Cohen-Sutherland Algorithm. Line AB: A(3, 10) B(6, 12) Line CD: C(4, 1) D(10, 6) CG Page: 35 東吳資訊管理 江清水

How will the algorithm handle these examples? CG (A) (B) (C) (D) Page: 37 東吳資訊管理 江清水

3 D Coordinates System Right-handed Coordinates System X Y Y X Left-handed Coordinates System X Y Y X X Y We will use vight-handed Coordinates System in our class. CG Page: 38 東吳資訊管理 江清水

5. 2 3 D Geometry We can use the way to represent 3 D points and polyhedras as the way to represent 2 D points and polygons. For example: Y (0, 2, 0) (0, 0, 1) (3, 0, 0) X Z 0 3 0 0 2 0 0 1 [ ] represents a tetrahedron In homogenlous coordinates system represents the same tetrahedron CG Page: 39 東吳資訊管理 江清水

3 D Transformations Let's discuss 3 D transformations, going directly to matrix and functional forms: (1) Translation: [x' y' z' 1] = [x y z 1] * [ ] (2) Scaling: [x' y' z' 1] = [x y z 1] * (3) Rotation: • • CG Rotating about X(Y or Z) does not change X's. The value of Y' and Z' does not depend on X. Row 4 and column 4 are always the same. The problem is to locate sin, -sin and cos. Page: 40 東吳資訊管理 江清水

(a) Rotate about Z: Y X Z Eye [y' z' 1] = [y z 1] * [ cosq sinq 0 -sinq cosq 0 0 0 1 [x' y' z' 1] = [x y z 1] * [ cosq -sinq 0 0 ] sinq 0 cosq 0 0 1 0 0 0 1 ] (b) Rotate about Y: X Y X Z Z CG Page: 41 東吳資訊管理 江清水

[ [x' z' 1] = [x z 1] *[ cosq sinq -sinq cosq 0 0 [z' x' 1] = [z x 1] * 0 0 1 cosq -sinq 0 sinq cosq 0 0 0 1 [x' y' z' 1] = [x y z 1] * [ ] ] cosq 0 -sinq 0 0 1 0 0 sinq 0 cosq 0 0 1 ] (c) Rotate about X: Y X Z [y' z' 1] = [y z 1] * [ cosq sinq 0 -sinq cosq 0 0 0 1 [x' y' z' 1] = [x y z 1] * CG Page: 42 [ ] 1 0 0 cosq sinq 0 0 -sinq cosq 0 0 1 ] 東吳資訊管理 江清水

Homework: Rotate the block counterwisely p/2 degree about X, Y, Z axis respectively. Find new coordinate for each vertex. Y Rotate about X Y (0, 3, 0) (1, 0, 0) X X (0, 0, 2) Rotate about Y Z Z Rotate about Z Y X X Z Z CG Y Page: 43 東吳資訊管理 江清水

Projection Different eye position has different view. CG Page: 44 東吳資訊管理 江清水

After the projection, the distance between your eyes and the object vertices is unknown. With the help of hidden-line removal algorithm, you can know which vertex is closer to you. CG Page: 45 東吳資訊管理 江清水

Orthgraphic Parallel Projections vs. Perspective Projections Orthgraphic Parallel Projections: The center of projection is located at infinity and the projectors are prependicular to the projection plane. Perspective projections: The center of projection is located at a finite distance from the projection plane. CG Page: 46 東吳資訊管理 江清水

Orthgraphic parallel Projection : Y [x, y, z] [x, y, 0] X Z Projection Plane : XY-Plane (Z=0) Projector : parallel to Z axis The Problem for finding projection point are simple: project to [ x y 0 ] [ x y z ] will CG Page: 47 東吳資訊管理 江清水

Perspective Projection : How to get a picture? (1) Freeze the scene. (2) Freeze you head (eyes). Perspective: CG Page: 48 東吳資訊管理 江清水

What information we need to project a picture into a projection plane via perspective projection? (1) eye position (2) object position (3) picture plane Anything else? (4) Up direction CG Page: 49 東吳資訊管理 江清水

Consider the eye coordinate system: Ye [x y z] [xp yp 0] Xe Ze [0 0 d] = Eye Position Projection plane: XY-plane (Z=0) Eye position: [0 0 d] Projector: the line from [x y z] to a given eye position. The problem for finding projection point becomes: Find the intersection point for projcetion plane and projector. CG Page: 51 東吳資訊管理 江清水

(1) Look alone the -Xe direction: Ye Ye [x y z] [xp yp 0] Ze Xe [0 0 d] = Eye Position yp Ze (x, y, z) y -z d yp : y = d : d-z yp= (d/(d-z)) * y (2) Look alone the -Ye direction: x Y (x, y, z) -z xp [x y z] e [xp yp 0] Ze Xe [0 0 d] = Eye Position Xe d Ze xp : x = d : d-z xp= (d/(d-z)) * x CG Page: 52 東吳資訊管理 江清水

(3) Represented in matrix form : xp = (d/(d-z)) * x yp= (d/(d-z)) * y [xp yp 0 1]=[x y z 1] [ d/(d-z) 0 0 0 0 1 ] In homogeneous coordinates system, the points [ x y z 1] and [wx wy wz w] (w*0) represents the same points. So, the point [xp yp 0 1] = [dx/(d-z) dy/(d-z) 0 1] = [x y 0 1 -(z/d)] =[x y z 1] CG Page: 53 [ 1 0 0 ] 0 0 0 -1/d 0 1 東吳資訊管理 江清水

(4) Questions : (a) How to draw a 3 D line into the picture plane ? Y [x't y't 0] [xt yt zt] . [x f yf zf] [x'f y'f 0] X Z (b) What happens when Z (c) What about d CG Page: 54 東吳資訊管理 江清水

Exercise: u (1) Draw The rectangle whose vertices are P=(10, -10), Q=(-10, -10) R=(-10, 10), S=(10, 10) and XY plane is projection plane, (0, 0, 15) is the Y eye position. Y 2 R . Z CG 1 Q . . 3 P S (0, 0, 5) 4 X X (0, 0, 15) Page: 55 東吳資訊管理 江清水

(2) u Move your eye from (0, 0, 15) to ((0, 0, 15), what you will see? Y Y 2 R . Z CG 1 Q . . 3 (0, 0, 5) P S 4 X X (0, 0, 15) Page: 56 東吳資訊管理 江清水

(d) How about z > d (i. e. , point is behind the eye) ? Y X Z (e) How about z d (i. e. , point is near the eye) ? Y Y X Z CG X Z Page: 57 東吳資訊管理 江清水

View Transformation In general, we may want to observe object from some other position (in stead of (0, 0, d)) and direction (in stead of - Ze), say from point (xf , yf , zf ) to point (xt , yt , zt ). Yw (Xf , Yf , Zf ) (Xt , Yt , Zt ) view vector Xw world data representing object Zw Notice that the world data and view vector now are given in world coordinates, not eye coordinates. CG Page: 58 東吳資訊管理 江清水

We must transform the object data from world coordinates into eye coordinates before we can perform orthographic parallel or perspective projection. This transformation called viewing transformation. Where is the eye coordinate system? Yw Ye Xe Ze Xw world data Zw make the arbitrary Y (Up We e direction) which perpedicular to Ze CG Page: 59 東吳資訊管理 江清水

We want a "neat" viewing transformation, V, such that [xe ye ze 1] = [xw yw zw 1] * V i. e. , multiply all world points by V and it will be just as if the eye was at (0, 0, d) or (0, 0, ), depend on perspeative or orthographic parallel projection, looking at (0, 0, 0). Visualize the process as transforming the Ze and Ye axes to Zw and Yw axes. This is accomplished using four basic transformations: 1. Translate the vector so that (xt , yt , zt ) = (0, 0, 0). 2. Rotate the result so that the transformed (xf , yf , zf )' swings into the Yw-Zw plane. 3. Rotate the result so that (xf , yf , zf )'' swings down onto the Zw axis. 4. Rotate this result so that Ye swings down onto the Yw axis. CG Page: 60 東吳資訊管理 江清水

Step 1. The first step is just a simple translation. Yw Ye Xe Ze Xw Zw T 1 = = CG Page: 61 [ ] 東吳資訊管理 江清水

Step 2. Rotate a degrees about Yw to swing eye' into the Yw-Zw plane. tan(2 p- a)=X'/Z'=tan(- a) a = Check the sign (check rotation direction) Step 3. Swing eye'' down by b degrees to (0, 0, d) by rotating eye'' about Xw. b = Yw eye'' eye' b a Xw Zw CG Page: 62 東吳資訊管理 江清水

Now, the eye coordinate system and world coordinate system are shown in the following picture. Yw Ye Xe q q Xw Ze Zw Step 4. Rotate Xe about Zw q degrees. (asswne Xe = (X 3, Y 3, 0)) q = -tan-1 Y 3'/X 3' CG Page: 63 東吳資訊管理 江清水

Our transformation equation becomes: [xe ye ze 1] = [xw yw zw 1] T 1 R 2 R 3 R 4 To check your transformatios, try: eye = (0, 1, 0) Yw aim = (0, 0, 0) (0, 1, 0) Ye = (1, 0, 0) w Xw (1, 0, 0)w (0, 0, 1)w Zw Ze Ye Xe CG Page: 64 東吳資訊管理 江清水

Homework: Try again with Ye = (0, 0, 1) and other values are the same. Yw (0, 1, 0)w (0, 0, 1)e (0, 0, 0)w (0, 0, 0)e Xw (1, 0, 0)w (0, 0, 1)w (-1, 0, 0)e (0, 1, 0)e Zw CG Page: 65 東吳資訊管理 江清水

View Volume We saw that as z -> d, severe distortion occures, like a "fisheye" lens. Also, data far from the observer (z << 0) becomes dots! To eliminate these effects, we'll introduce two clipping planes, parallel to the Xe-Ye plane: (1) A front clipping plane at z=zfront (2) A back clipping plane at z=zback Ye Back Clipping Plane Xe Ze Projection Plane Front Clipping Plane This creates a viewing volume, which is a pyramid, and is now a 6 -sided frustum. CG Page: 66 東吳資訊管理 江清水

Z-Clipping Now the full clipping procedure becomes: (1) Clip on front and back planes. (2) Perspective projection to 2 D window. (3) Clip on the 2 D window. To determine the visibility of a point P(x, y, z), just test its z value: If ((z > zfront) or (z<zback)) then "Invisible", else "Visible". (1) Under what conditions that a line segment will not be clipped? P 1 = (X 1, Y 1, Z 1) Zback Z 1, P 2 = (X 2, Y 2, Z 2) Z 2 Zfront (2) How to clip a line segment using Z-clipping? (1) Z = Z 1 + (Z 2 -Z 1)t (2)replace Z = Zfront or Zend , find t (3)find (X Y Z) = (X 1 Y 1 Z 1) + t(X 2 -X 1, Y 2 -Y 1, Z 2 -Z 1) CG Page: 67 東吳資訊管理 江清水

Three types of perspective projections: (1) one-point (2) two-point (3) three-point CG Page: 68 東吳資訊管理 江清水