INSTITUTIONEN FÖR SYSTEMTEKNIK LULEÅ TEKNISKA UNIVERSITET Exam Review David Carr Virtual Environments, Fundamentals Spring 2004 1 L Mar-4-05 SMM009, Exam Review Overview • Working problems • Summary of the course 2 L Mar-4-05 SMM009, Exam Review INSTITUTIONEN FÖR SYSTEMTEKNIK LULEÅ TEKNISKA UNIVERSITET Summary 3 L Mar-4-05 SMM009, Exam Review 1
Elements of Image Formation • Objects • Viewer • Light source(s) • Attributes that govern how light interacts with the materials in the scene • Note the independence of the objects, viewer, and light source(s) 4 L Mar-4-05 SMM009, Exam Review Three-Color Theory • Human visual system has two types of sensors - Rods: monochromatic, night vision - Cones + Color sensitive + Three types of cone + Only three values (the tri stimulus values) are sent to the brain • Need only match these three values - Need only three primary colors 5 L Mar-4-05 SMM009, Exam Review Pinhole Camera & Synthetic Camera Model projector Use trigonometry to find projection of a point p x p = -x/z/d y p = -y/z/d z p = d image plane projection of p center of projection These are equations of simple perspective 6 L Mar-4-05 SMM009, Exam Review 2
Following the Pipeline: Transformations • Much of the work in the pipeline is in converting object representations from one coordinate system to another - World coordinates - Camera coordinates - Screen coordinates • Every change of coordinates is equivalent to a matrix transformation 7 L Mar-4-05 SMM009, Exam Review Clipping • Just as a real camera cannot “see” the whole world, the virtual camera can only see part of the world space - Objects that are not within this volume are said to be clipped out of the scene 8 L Mar-4-05 SMM009, Exam Review Projection • Must carry out the process that combines the 3D viewer with the 3D objects to produce the 2D image - Perspective projections: all projectors meet at the center of projection - Parallel projection: projectors are parallel, center of projection is replaced by a direction of projection 9 L Mar-4-05 SMM009, Exam Review 3
Rasterization • If an object is visible in the image, the appropriate pixels in the frame buffer must be assigned colors - Vertices assembled into objects - Effects of lights and materials must be determined - Polygons filled with interior colors/shades - Must have also determine which objects are in front (hidden surface removal) 10 L Mar-4-05 SMM009, Exam Review Geometry and Representation • Geometry - Relationships among objects in an n-dimensional space - In computer graphics, objects exist in 3 dimensions - Minimum set of primitives for building more sophisticated objects + Scalars, Vectors, Points • Representations - Dimension and basis - Coordinate systems for representing vectors spaces - Frames for representing affine spaces - Change of frames and basis - Homogeneous coordinates 11 L Mar-4-05 SMM009, Exam Review Affine Spaces • Point + a vector space • Operations - Vector-vector addition → vector - Scalar-vector multiplication → vector - Point-vector addition → point - Scalar-scalar operations → scalar • For any point P and the scalars 0,1 define - 1 • P = P - 0 • P = 0 (zero vector) 12 L Mar-4-05 SMM009, Exam Review 4
Convexity • An object is convex if and only if for any two points in the object all points on the line segment between these points are also in the object P P Q Q 13 L Mar-4-05 SMM009, Exam Review Planes • A plane be determined by a point and two vectors or by three points P( α , β )=R+ α u+ β v P( α , β )=R+ α (Q-R)+ β (P-Q) 14 L Mar-4-05 SMM009, Exam Review Normals • Every plane has a vector n normal (perpendicular, orthogonal) to it • From point-two vector form P( α , β ) = R + α u + β v, we know we can use the cross product to find n = u × v and the equivalent form (P( α )-P) × n = 0 v u P 15 L Mar-4-05 SMM009, Exam Review 5
Frames • Coordinate System is insufficient to present points • If we work in an affine space we can add a single point, the origin , to the basis vectors to form a frame • Frame determined by (P 0 , v 1 , v 2 , v 3 ) • Within this frame, every vector can be written as v 2 v = α 1 v 1 + α 2 v 2 + … + α n v n v 1 P 0 • Every point can be written as P = P 0 + β 1 v 1 + β 2 v 2 + … + β n v n v 3 16 L Mar-4-05 SMM009, Exam Review A Single Representation • If we define 0•P = 0 and 1•P = P then we can write v = α 1 v 1 + α 2 v 2 + α 3 v 3 = [ α 1 α 2 α 3 0] [ v 1 v 2 v 3 P 0 ] T P = P 0 + β 1 v 1 + β 2 v 2 + β 3 v 3 = [ β 1 β 2 β 3 1] [ v 1 v 2 v 3 P 0 ] T • Thus we obtain the four-dimensional homogeneous coordinate representation v = [ α 1 α 2 α 3 0] T p = [ β 1 β 2 β 3 1] T 17 L Mar-4-05 SMM009, Exam Review Homogeneous Coordinates • The general form of 4 dimensional homogeneous coordinates is p=[x y x w] T • We return to a three dimensional point (for w ≠ 0 ) by x ← x/w y ← y/w z ← z/w • If w=0 , the representation is that of a vector • Note that homogeneous coordinates replaces points in three dimensions by lines through the origin in four dimensions 18 L Mar-4-05 SMM009, Exam Review 6
Working with Representations Within the two frames any point or vector has a representation of the same form a = [ α 1 α 2 α 3 α 4 ] in the first frame b = [ β 1 β 2 β 3 β 4 ] in the second frame where α 4 = β 4 = 1 for points and α 4 = β 4 = 0 for vectors and a=M T b The matrix M is 4 x 4 and specifies an affine transformation in homogeneous coordinates 19 L Mar-4-05 SMM009, Exam Review Affine Transformations • Every linear transformation is equivalent to a change in frames • Every affine transformation preserves lines • However, an affine transformation has only 12 degrees of freedom because 4 of the elements in the matrix are fixed and are a subset of all possible 4 x 4 linear transformations 20 L Mar-4-05 SMM009, Exam Review Matrix Representation in 4D Homogeneous Coordinates • This form is better for implementation because: - All affine transformations can be expressed this way - Multiple transformations can be concatenated together • Translation 1 0 0 d � � x � � 0 1 0 d y � � 0 0 1 d � � z � � 0 0 0 1 � � 21 L Mar-4-05 SMM009, Exam Review 7
Rotation About the Z-Axis • Rotation about z axis in three dimensions leaves all points with the same z - Equivalent to rotation in 2 dimensions in planes of constant z x’ = x cos θ – y sin θ y’ = x sin θ + y cos θ z’ = z cos sin 0 0 � � � � � � � • x- and y-axes similar sin cos 0 0 � � � � R z ( θ ) = 0 0 1 0 � � � � 0 0 0 1 � � 22 L Mar-4-05 SMM009, Exam Review Scaling • Expand or contract along each axis (fixed point of origin) • Reflections are negative scaling x’=s x x y’=s y y z’=s z z s 0 0 0 � � x p ’= Sp � � 0 s 0 0 y � � S = S (s x , s y , s z ) = 0 0 s 0 � � z � � 0 0 0 1 � � 23 L Mar-4-05 SMM009, Exam Review Shear • Equivalent to pulling faces in opposite directions • On the x-axis 1 cot 0 0 � � � x’ = x + y cot θ � � 0 1 0 0 � � H ( θ ) = y’ = y 0 0 1 0 � � z’ = z � � 0 0 0 1 � � 24 L Mar-4-05 SMM009, Exam Review 8
Inverses • Could compute inverse matrices by general formulas, • But, simple geometric observations are easier - Translation: T -1 (d x , d y , d z ) = T (-d x , -d y , -d z ) - Rotation: R -1 ( θ ) = R (- θ ) +Holds for any rotation matrix +Note that since cos(- θ ) = cos( θ ) and sin(- θ )=-sin( θ ), R -1 ( θ ) = R T ( θ ) - Scaling: S -1 (s x , s y , s z ) = S (1/s x , 1/s y , 1/s z ) 25 L Mar-4-05 SMM009, Exam Review Concatenation • Form arbitrary affine transformation matrices by multiplying - Rotation, translation, and scaling matrices • Same transformation is applied to many vertices: - Cost of forming a matrix M=ABCD is not significant compared to the cost of computing Mp for many vertices p • Difficulty is forming a desired transformation • Order of Transformations - Note that matrix on the right is the first applied - Mathematically, the following are equivalent p ’ = ABCp = A ( B ( Cp )) - Note many references use column matrices to present points. In terms of column matrices p T ’ = p T C T B T A T 26 L Mar-4-05 SMM009, Exam Review Projection Perspective Parallel 27 L Mar-4-05 SMM009, Exam Review 9
Three Aspects of Computer Viewing • Positioning the camera - Setting the model-view matrix • Selecting a lens - Setting the projection matrix • Clipping - Setting the view volume • All are implemented in the pipeline • Default OpenGL projection is Orthogonal Projection orthogonal 28 L Mar-4-05 SMM009, Exam Review Projections and Normalization • The default projection in the eye (camera) frame is orthogonal • For points within the default view volume x p = x 1 0 0 0 � � � � 0 1 0 0 y p = y � � M = � 0 0 0 0 � z p = 0 � � 0 0 0 1 � � • Most graphics systems use view normalization - All other views are converted to the default view by transformations that determine the projection matrix - Allows use of the same pipeline for all views 29 L Mar-4-05 SMM009, Exam Review Simple Perspective • Center of projection at the origin • Projection plane z = d , d < 0 • Consider top and side views x x p = z / d y y p = z / d z p = d 30 L Mar-4-05 SMM009, Exam Review 10
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