( ) ( ) ( ) ( ) TNM046: Datorgrafik. Transformations. Linear Algebra. Linear Algebra. Sasan Gooran VT Transposition. Scalar (dot) product:

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1 TNM046: Datorgrafik Transformations Sasan Gooran VT 04 Linear Algebra ( ) ( ) =,, 3 =,, 3 Transposition t = 3 t = 3 Scalar (dot) product: Length (Norm): = t = = = Normaliation: ˆ = Linear Algebra = cos(θ) > 0 θ < 90 = 0 θ = 90 < 0 θ > 90 Eample: Find the angle between the two vectors below (answer:.68 radians or degrees) ( ) =,, ( ) =,, 3

2 Linear Algebra Cross (vector) product: = ( ) = ˆ ŷ ẑ 3 3 = sin(θ) ( ) =? ( ) =? 4 P = Basic Transformations (D) Translation P = O P T P T = ( t, t ) OP = OP + T à = + t = + t 5 Basic Transformations (D) Translation 6 3 A B What is the translation vector, from A to B? from B to A? 6

3 Basic Transformations (D) Rotation r θ ϕ P = r P = = r cosϕ = rsinϕ = r cos(ϕ +θ) = r cosϕ cosθ rsinϕ sinθ = rsin(ϕ +θ) = rsinϕ cosθ + r cosϕ sinθ = cosθ sinθ = sinθ + cosθ 7 Basic Transformations (D) Rotation Counter-clockwise rotation b angle θ about the coordinate origin P = R P à = cosθ sinθ sinθ cosθ 8 Basic Transformations (D) Rotation M = Rotation about an arbitrar point M M M r θ P = r P = = M + ( M )cosθ ( M )sinθ = M + ( M )sinθ + ( M )cosθ 9 3

4 Basic Transformations (D) Scaling Fied point: Coordinate origin = s = s à P = S P where s 0 S = 0 s s and s > 0 Scaling values less than reduces the sie of the object Scaling values greater than produce an enlargement s =s : Uniform scaling, which maintains the relative object properties 0 Basic Transformations (D) Scaling Arbitrar fied point: ( f, f ) = f + s ( f ) = f + s ( f ) 4 fied point: (0, 0) Scaled s=s=0.5 Object Scaled s=s= 3 fied point: (, ) Scaled s=s=0.5 Scaled s=s= Object 4 4 Matri representation Homogenous coordinates (translation) = + t t In order to represent this with a matri multiplication we add a third coordinate,. P = T P à = 0 t 0 t 0 0 4

5 Matri representation Homogenous coordinates (translation) T = 0 t 0 t 0 0 T is the translation matri. What is the inverse matri? T =? 3 Matri representation Homogenous coordinates (Rotation) = cosθ sinθ 0 sinθ cosθ P = R P R is the rotation matri. What is the inverse matri? R =? 4 Matri representation Homogenous coordinates (Scaling) s 0 0 = 0 s P = S P S is the scaling matri. What is the inverse matri? S =? 5 5

6 Composite Transformation If two successive translation vectors (t, t ) and (t, t ) are applied to a coordinate position P, the final transformed location P is calculated: P = T(t,t ) { T(t,t )* P } = { T(t,t ) T(t,t )}* P 0 t 0 t t 0 t 0 0 = 0 t + t 0 t + t Composite Transformation Two successive Rotations b θ and θ : P = R(θ ) { R(θ )* P } = { R(θ ) R(θ )}* P = R(θ +θ )* P Two successive Scaling-s: s s s s = s s s s General Pivot-Point Rotation Rotation about an arbitrar point ( M, M ): Step : Translate ( M, M ) to the coordinate origin (0,0) Step : Rotate about (0, 0) Step 3: Translate back to ( M, M ) 0 M = 0 M 0 0 cosθ sinθ 0 sinθ cosθ M 0 M 0 0 cosθ sinθ M ( cosθ)+ M sinθ = sinθ cosθ M ( cosθ) M sinθ

7 General Fied-Point Scaling Scaling about an arbitrar point ( f, f ): Step : Translate ( f, f ) to the coordinate origin (0,0) Step : Scale about (0, 0) Step 3: Translate back to ( f, f ) 0 f s f = 0 f 0 s 0 0 f s 0 f ( s ) = 0 s f ( s ) E: General Scaling Directions We can scale an object in other directions b rotating the object to align the desired scaling directions S θ Step : Rotate the object (b θ) to align Step : Scale in the direction of (, ) Step 3: Rotate back 0 S E: General Scaling Directions = cosθ sinθ 0 sinθ cosθ s s cosθ sinθ 0 sinθ cosθ s cos θ + s sin θ (s s )sinθ cosθ 0 = (s s )sinθ cosθ s sin θ + s cos θ What happens when s =s? 7

8 Reflection Reflection matri across =0 (-ais) Reflection matri across =0 (-ais) Reflection relative to the coordinate origin E: Reflection in = Step : Rotate counter clockwise (45 degrees) (=) à -ais Step : Reflect across the -ais Step 3: Rotate back = cosθ sinθ 0 sinθ cosθ θ = π / = cosθ sinθ 0 sinθ cosθ Shear = + sh = sh 0 = sh = 3 4 8

9 3D transformations, Translation 0 0 t 0 0 t = 0 0 t D Rotation About -ais Counter clockwise r θ r = a P = = a P = When rotating around -ais the -coordinate is unchanged 6 Rotation about -ais: Rotation about -ais: Wh changing the signs in front of sinθ? 3D Rotation = = cosθ sinθ 0 0 sinθ cosθ cosθ 0 sinθ sinθ 0 cosθ Rotation matri about -ais? 7 9

10 Rotation about an ais parallel to one of the coordinate aes Step : Translate so that the rotation ais coincides with the parallel coordinate ais Step : Perform the specified rotation about that ais Step 3: Translate back R(α) = T R co ais (α)*t 8 Eample: Counterclockwise rotation b 30 degrees about the ais passing through M:(,, 3) and M:(,, 3) (vector from M to M) The vector going from M to M is ( 3)-( 3)=(0 0), which is obviousl parallel to -ais 0 0 cos( π 6 ) 0 sin(π 6 ) 0 R( π 6 ) = sin( π ) 0 cos(π 6 ) Are there an other translations that work? 9 Rotation about an ais parallel to one of the coordinate planes Step : Translate the object so that the rotation ais passes through the coordinate origin. This vector is now on one of the coordinate planes. Step : Rotate the object so that the ais of rotation coincides with one of the coordinate aes (one rotation is needed) Step 3: Perform the specified rotation about that ais Step 4: Rotate back Step 5: Translate back R(α) = T R ais >co ais R co ais (α) R ais >co ais T 30 0

11 Eample: Counterclockwise rotation b 30 degrees about the ais passing through M:(,, 3) and M:(,, 4) (vector from M to M) The vector going from M to M is ( 4)-( 3)=(0 ), which is obviousl parallel to YZ-plane (The translation is as before) It is easil seen that if we rotate (0,, ) b 45 degrees and counterclockwise around X-ais, it will coincide with the Z-ais cos( π 4 ) sin(π 4 ) 0 R ais >co_ ais = 0 sin( π 4 ) cos(π 4 ) Eample: Continuation cos( π cos( π R = T 4 ) sin(π 4 ) 0 6 ) sin(π 6 ) sin( π sin( π 4 ) cos(π 4 ) 0 6 ) cos(π 6 ) cos( π 4 ) sin(π 4 ) 0 0 sin( π T ) cos(π 4 ) Where, T = Rotation about an arbitrar ais that is not parallel to an coordinate aes, or coordinate planes. Step : Translate the object so that the rotation ais passes through the coordinate origin Step : Rotate the object so that the ais of rotation coincides with one of the coordinate aes (two rotations are needed) Step 3: Perform the specified rotation about the coordinate ais Step 4: Rotate back Step 5: Translate back R(α) = T R ais >co ais R co ais (α) R ais >co ais T 33

12 M = m m m M = Ee m m m = U M M = ( m m, m m, m m ) U û = = (a, b,c) The vector is normalied U a + b + c =? 34 After translation (Step ) û = (a, b, c) 0 0 m 0 0 T = m 0 0 m Two rotations are needed to make u coincide with one of the coordinate aes. (step ) Here we find the rotations to coincide it with -ais The first rotation is about the -ais b φ to rotate u to the -plane (clockwise) The second rotation is about the -ais b θ to align that with û = (a, b, c) -ais (clockwise) θ ϕ v = (a, b, 0) 36

13 The first rotation: ˆ v = ˆ v set d = a + b cosϕ = a + b cosϕ ˆ v à cosϕ = a = (, 0, 0) (a, b, 0) = a d sinϕ = ± ( a d ) = ± b d if b > 0 --> 0 ϕ π --> sinϕ 0 if b < 0 --> π ϕ π --> sinϕ 0 Therefore: sinϕ = b d ϕ û = (a, b, c) v = (a, b, 0) 37 The second rotation: ẑ û = ẑ û cosθ = cosθ ẑ û = (0, 0,) (a, b, c) = c à cosθ = c Since 0 θ π then sinθ 0 sinθ = cos θ = c = d θ û = (a, b, c) Observe: a + b + c = and d = a + b 38 R ais >co ais = R(α) = T R ais >co ais R co ais (α) R ais >co ais T Notice: Clockwise rotations cosθ 0 sinθ sinθ 0 cosθ Step cosϕ sinϕ 0 0 sinϕ cosϕ R ais >co ais = c 0 d d 0 c a / d b / d 0 0 b / d a / d

14 Wh? R(α) = T R ais >co ais R co ais (α) R ais >co ais T Step 4 R ais >co ais Step 5 = a / d b / d 0 0 b / d a / d m 0 0 T = m 0 0 m c 0 d d 0 c R(α) = T R ais >co ais R co ais (α) R ais >co ais T Notice that in the approach presented here we coincided the ais of rotation with -ais. Therefore the coordinate ais is -ais. cos(α) sin(α) 0 0 sin(α) cos(α) 0 0 R co ais = Step 3 4 Eercises Eercise : Counterclockwise rotation b α about the ais passing through M:(,, 3) and M:(,, 5) (vector from M to M) Eercise : Counterclockwise rotation b α about the ais passing through M: (0, 0, 0) and M: (-, 3, -6) (vector from M to M) 4 4

15 Parallel Projection Just to make it consistent with the assignments in the labs, here we rearranged the, and -aes. The plane is the view plane and the -ais points towards the viewer. P = If the parallel projection on the view plane (-plane or =0) occurs along the -direction, then: = = P = = = Parallel Projection If the view plane is parallel to -plane, = v. P = If the parallel projection on the view plane (= v ) occurs along the -direction, then: P = = = = v = v Coordinate Sstem Transformation The aes of the two coordinate sstems are parallel. Assume that a point P has coordinates,, and coordinates u, v, n V N ( ) t in ( ) t in coordinate sstem. O = o o o U 0 0 o 0 0 T = o 0 0 o O T transforms (translates) sstem to sstem. 45 5

16 Coordinate Sstem Transformation The aes of the two coordinate sstems are parallel. u 0 0 o v 0 0 = o n 0 0 o o 0 0 = o 0 0 o u v n In order not to get confused think of the coordinates of the origin of the sstem (O ), which is (0,0,0) in, but ( ) t O = o, o, o in sstem. 46 Coordinate Sstem Transformation One of the aes are parallel (here and N) but UV plane is rotated b θ in relation to plane. V N θ O = ( o, o, o ) t U The view coordinate sstem is transformed to b a translation followed b a clockwise rotation about -ais. O 47 Coordinate Sstem Transformation The -aes are parallel but are rotated b θ. u v n = cosθ sinθ 0 0 sinθ cosθ o 0 0 o 0 0 o o 0 0 = o 0 0 o cosθ sinθ 0 0 sinθ cosθ u v n 48 6

17 Coordinate Sstem Transformation None of the aes are parallel First a translation is needed. Assume then that the unit vectors of the V coordinate aes of the sstem can be epressed in coordinates. U O = ( o, o, o ) t û = (u N,u,u ) ˆv = (v, v, v ) ˆn = (n, n, n ) O 49 Coordinate Sstem Transformation None of the aes are parallel u u u u o v v = v v o n n n n o o u v n = o u v n o u v n u v n 50 Coordinate Sstem Transformation Few points: In order to better understand the transformations in previous page, think of a point that has the coordinate (0,0,) in the sstem, then ou can see that it is first transformed to (n,n,n ) and then translated. What is the coordinates of a point with coordinates (,0,0) in the sstem? Make sure that ou have understood these transformations b tring some obvious coordinates. Notice that the inverse of a matri whose rows (or columns) make an orthonormal base is equal to its transpose. 5 7

18 Coordinate Sstem Transformation V N In practice ou usuall have the position of the camera (O ), the direction it is looking towards (N) and an up direction (V ) which is not necessaril perpendicular to N. If we now find an orthonormal base for the sstem then we have the transformation matri. O = ( o, o, o ) t O 5 Coordinate Sstem Transformation V V N ˆn = N = (n, n, n ) N In some literature this might be defined as ˆn = N Then U has to be perpendicular to the plane of N and V : V N û = V N = (u,u,u ) U N V has to be perpendicular to both U and N. O ˆv = ˆn û = (v, v, v ) 53 The transformations are now as shown on Page 50 8