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Basic mathematics for geometric modeling

Basic mathematics for geometric modeling. Coordinate Reference Frames. y. Cartesian Coordinate (2D) Polar coordinate. (x, y). x. r. . Y. P. P. y. r. r. y. . . x. x. x. Relationship : polar & cartesian. Use trigonometric, polar  cartesian x = r cos  , y = r sin 

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Basic mathematics for geometric modeling

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  1. Basic mathematics for geometric modeling

  2. Coordinate Reference Frames y • Cartesian Coordinate (2D) • Polar coordinate (x, y) x r 

  3. Y P P y r r y   x x x Relationship : polar & cartesian Use trigonometric, polar  cartesian x = r cos  , y = r sin  Cartesian  polar r =  x2 + y2,  = tan-1 (y/x)

  4. 3D cartesian coordinates y y x x z z Right-handed 3D coordinate system

  5. POINT • The simplest of geometric object. • No length, width or thickness. • Location in space • Defined by a set of numbers (coordinates) e.g P = (x, y) or P = (x, y, z) • Vertex of 2D/ 3D figure

  6. VECTOR • distance and direction • Does not have a fixed location in space • Sometime called “displacement”.

  7. VECTOR • Can define a vector as the difference between two point positions. y y2 V Q y1 P x x2 x1 V = Q – P = (x2 – x1, y2 – y1) = (Vx, Vy) Also can be expressed as V = Vxi + Vyj Component form

  8. VECTOR : magnitude & direction • Calculate magnitude using the Pythagoras theorem  distance • |V| = Vx2 + Vy2 • Direction •  = tan-1 (Vy/Vx)

  9. VECTOR : magnitude & direction V Q • Example 1 • If P(3, 6) and Q(6, 10). Write vector V in component form. • Answer • V = [6 - 3, 10 – 6] = [3, 4]

  10. VECTOR : magnitude & direction • Example 1 (cont) • Compute the magnitude and direction of vector V • Answer • Magnitud |V| = 32 + 42 • = 25 = 5 • Direction  = tan-1 (4/3) = 53.13

  11. Unit Vector • As any vector whose magnitude is equal to one • V = V |V| • The unit vector of V in example 1 is = [Vx/|V| , Vy/|V|] = [3/5, 4/5]

  12. VECTOR : 3D y • Vector Component • (Vx, Vy, Vz) • Magnitude • |V| = Vx2 + Vy2 + Vz2 • Direction •  = cos-1(Vx/|V|),  = cos-1(Vy/|V|),  =cos-1(Vz/|V|) • Unit vector • V = V = [Vx/|V|, Vy/|V|, Vz/|V|] |V| Vy V x Vz Vx z

  13. Scalar Multiplication • kV = [kVx, kVy, kVz] • If k = +ve  V and kV are in the same direction • If k = -ve  V and kV are in the opposite direction • Magnitude |kV| = k|V|

  14. Scalar Multiplication • Base on Example 1 • If k = 2, find kV and the magnitudes • Answer • kV = 2[3, 4] = [6, 8] • Magnitude |kV|= 62 + 82 = 100 = 10 • = k|V| = 2(5) = 10

  15. Vector Addition y y • Sum of two vectors is obtained by adding corresponding components • U = [Ux, Uy, Uz], V = [Vx, Vy, Vz] • U + V = [Ux + Vx, Uy + Vy, Uz + Vz] V V U + V U U x x

  16. P Q Vector Addition Q • Example • If vector P=[1, 5, 0], vector Q=[4, 2, 0]. Compute P + Q • answer • P + Q = [1+4, 5+2, 0+0] = [5, 7, 0] P

  17. Vector Addition & scalar multiplication properties • U + V = V + U • T + (U + V) = (T + U) + V • k(lV) = klV • (k + l)V = kV + lV • k(U + V) = kU + kV

  18. Scalar Product • Also referred as dot product or inner product • Produce a number. • Multiply corresponding components of the two vectors and add the result. • If vector U = [Ux, Uy, Uz], vector V = [Vx, Vy, Vz] • U . V = UxVx + UyVy + UzVz

  19. Scalar Product. • Example • If vector P=[1, 5, 0], vector Q=[4, 2, 0]. Compute P . Q • answer • P . Q = 1(4) + 5(2) + 0(0) • = 14

  20. U V Scalar Product properties • U.V = |U||V|cos  • angle between two vectors •  =cos –1(U.V) • |U||V| • Example • Find the angle between vector b=(3, 2) and vector c = (-2, 3) 

  21. Scalar Product properties Solution • b.c = (3, 2). (-2, 3) • 3(-2) + 2(3) = 0 • |b| = 32 + 22 = 13 = 3.61 • |c| = (-2)2 + 32 = 13 = 3.61 •  =cos –1 ( 0/(3.61((3.61)) •  = cos –1 ( 0 ) = 90

  22. Scalar Product properties • If U is perpendicular to V, U.V = 0 • U.U = |U|2 • U.V = V.U • U.(V+W) = U.V + U.W • (kU).V = U.(kV)

  23. Vector Product • Also called the cross product • Defined only for 3 D vectors • Produce a vector which is perpendicular to both of the given vectors. y c = a x b c x b a z

  24. Vector Product • To find the direction of vector C, use righ-hand rules C x B x B A z C A z

  25. B x A A x B C x x z B A z C B A Vector Product • To find the direction of vector C, use righ-hand rules

  26. exercise • Find the direction of vector C, (keluar skrin atau kedalam skrin) A x B Q B P x Q P A O M x N L x O M L N

  27. Vector Product • If vektor A = [Ax, Ay, Az], vektor B = [Bx, By, Bz] • A x B = i j k i j • Ax Ay Az Ax Ay • Bx By Bz Bx By = [ (AyBz-AzBy), (AzBx-AxBz), (AxBy-AyBx)]

  28. Vector Product P • Example • If P=[1, 5, 0], Q=[4, 2, 0]. Compute P x Q • Solution • P x Q = i j k i j • 1 5 0 1 5 • 4 2 0 4 2 • = [ (5.(0)-0.(5)), (0.(4)-1.(0)), (1.(2)-5.(4))] • = [ 0, 0, -18] Q

  29. Vector Product • Properties • U x V = |U||V|n sin  where n = unit vector perpendicular to both U and V • U x V = -V x U • U x (V + W) = U x V+ U x W • If U is parallel to V, U x V = 0 • U x U = 0 • kU x V = U x kV

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