University of Tampere Computer Graphics 2013 School of Information Sciences Exercise

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1 University of Tampere Computer Graphics 2013 School of Information Sciences Exercise We can get the minmum of parabola f(x) = ax 2 + bx + c,a > 0,, when f (x) = 0. In this case x has the value of x = b/2a. Determine the value of t for which l : P + tu is closest to point Q. Hint: Look at the intersection formulas below and calculate the minimum of f(t) = d 2 (P +tu,q). 2. Let s write the ray of light in the form P +tu, u = 1 and t 0, where u is the direction of ray and P it s starting point. Let s look at the reflection of light in two dimensions. Determine the point in y-axis where a ray hits, which starts from origin, goes in the direction (1,2) and reflects from the line ) 1 2 l : (X (3,3)) ( ( 1, 1) = 0.? l 3. Determine the ray of light, which we get, when a ray starting from (1,2) reflects from a ball S, which has a center point (4,5) and radius 2. S 4. Determine a transformation, which scales points in the direction u with a coeffient s. 1

2 5. Programming environment. In this course we use OpenGL 4.0 programming environment in order to create graphics on the computer screen. Unfortunately such students whose computer does not have a graphics card that supports DirectX10/OpenGL 3.3 cannot compile/run the programs. However, the exercises can be done without compiling. Read the preface from and install one of the listed C/C++ compiler to your computer if you do not already have one. Install also freeglut and glew libraries and try to compile an example program given in chapter 1. Answer shortly to the following questions. What is the purpose of freeglut and glew libraries? What means a Shader Program? What is the strength of buffer objects? 6. We can use 3 3-matrices for rotations, translations, scaling and skewing. These are all transformations which has an inverse transformation. However, there are some useful transformations which are not invertible. Let s look at how the vector projection on a plane and cross product can be presented in a matrix form. Show that the projections can be calculated with 3 3-matrices as follows: ) Proj u v : uut u 2v, Proj u (I v : uut v. u 2 When u is a unit vector, the fromulas can be given in a form uu T ja I uu T. Show further, that also a cross product can be calculated with 3 3-matrices: 0 u 3 u 2 u v : Skew(u)v = u 3 0 u 1 v. u 2 u UsetheresultsfrompreviousassigmentandformamatrixS u,s,whichpresents the transformation from assigment Define the equation for a plane which is defined by the points P = (1,0,0), Q = (0,1,0), R = (0,2,3) in the form (X S) n = 0. Hint: Cross product. 2

3 About vector projections Vectors v and u are perpendicular to each other and u spans a sub space of R 3 which is complement to v. Let s define the projection as follows: Proj u v = u v u 2u, Proj u v = v Proj uv. For the projections Proj u v u, Proj u v u and v = Proj u v +Proj u v divide a vector v in two parts, from which one is perpendicular to u and another is parallel. Look at the picture. Proj u v v Proj u v u Let us show that Proj u v u: ( (Proj u v) u = v v u ) u 2u u = v u v u u 2(u u) = v u v u = v u v u = 0. u 2 u 2 Interserction of line and plane Let swritealineinaforml : X = P+tuandaplaneinaform(X Q) n = 0. { (X Q) n = 0 X = P +tu (P +tu Q) n = 0 (P Q) n+t(u n) = 0 t = (P Q) n u n = (Q P) n u n When a line is parallel to a plane u n = 0. Then, if a point in a line P is in a plane (P Q) n, the intersection is whole line, otherwise, line and plane do not intersect. 3

4 Intersection of a ball and a line Let s write a line l in a form l : P + tu, u = 1 and a ball S in a form S : d 2 (X,Q) = r 2, where d 2 (X,Q) is a square of distance and r 2 is a square of the radius of a ball. Squaring is not necessary, but it eases the calculations. Nowtheequationcanbewritteninaform X Q 2 = (X Q) (X Q) = r 2. { d 2 (X,Q) = r 2 X = P +tu d 2 (P +tu,q) = P +tu Q 2 = r 2 (P +tu Q) (P +tu Q) = r 2 (P Q+tu) (P Q+tu) = r 2 (P Q) (P Q+tu)+tu (P Q+tu) = r 2 (P Q) (P Q)+(P Q) (tu)+(tu) (P Q)+(tu) (tu) = r 2 P Q 2 +2t(u (P Q))+ u 2 t 2 = r 2 P Q 2 +2t(u (P Q))+t 2 = r 2 t 2 +2t(u (P Q))+ P Q 2 r 2 = 0 t = 2(u (P Q))± 4(u (P Q)) 2 4( P Q 2 r 2 ) 2 t = (u (P Q))± (u (P Q)) 2 P Q 2 +r 2 If the expression under the square root < 0, a line does not intersect a ball. If u 1, our expression gets the form: t = (u (P Q))± (u (P Q)) 2 u 2 P Q 2 + u 2 r 2 u 2. Properties of vector products (DP1) u v = v u (CP1) u v = v u (DP2) u (v+w) = u v+u w (CP2) u (v+w) = u v+u w (DP3) (αu) v = α(u v) (CP3) (αu) v = α(u v) (DP4) u u = 0 u = Θ (CP4) u (v w) = (u v) w u v u v = 0 u v u v = Θ u v u,v 4

5 Let the vectors u and v have an angle α between them. In this case following formulas are valid: u v = u v cosα, u v = u v sinα. 5