1 / 43

Game Mathematics

Game Mathematics. Essential Mathematics for Game Development. Matrices Vectors Fixed-point Real Numbers Triangle Mathematics Intersection Issues Euler Angles Angular Displacement Quaternion Differential Equation Basics. Matrices. a 11 .. a 1n . . . . a m1 .. a mn.

lej
Download Presentation

Game Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Game Mathematics

  2. Essential Mathematics for Game Development Matrices Vectors Fixed-point Real Numbers Triangle Mathematics Intersection Issues Euler Angles Angular Displacement Quaternion Differential Equation Basics

  3. Matrices a11 .. a1n . . . . am1 .. amn A = (aij) = C = A T cij = aji C = A + B cij = aij + bij • Matrix basics • Definition • Transpose • Addition

  4. C = aA cij = aaij r C = A B cij = Saikbkj k = 1 cij = row x column • Scalar-matrix multiplication • Matrix-matrix multiplication

  5. V = [x y z] • Using matrix notation, a point V is transformed under translation, scaling and rotation as : V’ = V + D V’ = VS V’ = VR where D is a translation vector and S and R are scaling and rotation matrices • Transformations in Matrix form • A point or a vector is a row matrix (de facto convention)

  6. To make translation be a linear transformation, we introduce the homogeneous coordinate system V (x, y, z, w) where w is always 1 • Translation Transformation x’ = x + Tx y’ = y + Ty z’ = z + Tz V’ = VT 1 0 0 0 0 1 0 0 0 0 1 0 Tx Ty Tz 1 [x’ y’ z’ 1] = [x y z 1] = [x y z 1] T

  7. Scaling Transformation x’ = xSx y’ = ySy z’ = zSz V’ = VS Sx 0 0 0 0 Sy 0 0 0 0 Sz 0 0 0 0 1 [x’ y’ z’ 1] = [x y z 1] = [x y z 1] S Here Sx, Sy and Sz are scaling factors.

  8. Rotation Transformations 1 0 0 0 0 cosq sinq 0 0 -sinq cosq 0 0 0 0 1 Rx = Ry = Rz = cosq 0 -sinq 0 0 1 0 0 sinq 0 cosq 0 0 0 0 1 cosq sinq 0 0 -sinq cosq 0 0 0 0 1 0 0 0 0 1

  9. Net Transformation matrix • Matrix multiplication are not commutative [x’ y’ z’ 1] = [x y z 1] M1 and [x” y” z” 1] = [x’ y’ z’ 1] M2 then the transformation matrices can be concatenated M3 = M1 M2 and [x” y” z” 1] = [x y z 1] M3 M1 M2 = M2 M1

  10. Vectors V = (x2-x1, y2-y1, z2-z1) (x1,y1,z1) (x2,y2,z2) • A vector is an entity that possesses magnitude and direction. • A 3D vector is a triple : • V = (v1, v2, v3), where each component vi is a scalar. • A ray (directed line segment), that possesses position, magnitude and direction.

  11. X = V + W • = (x1, y1, z1) • = (v1 + w1, v2 + w2, v3 + w3) W V + W W V + W V V |V| = (v12 + v22 + v32)1/2 U = V / |V| Addition of vectors Length of vectors

  12. X = VXW • = (v2w3-v3w2)i + (v3w1-v1w3)j + (v1w2-v2w1)k • where i, j and k are standard unit vectors : • i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) Np polygon defined by 4 points V2 • Np = V1XV2 V1 • Cross product of vectors • Definition • Application • A normal vector to a polygon is calculated from 3 (non-collinear) vertices of the polygon.

  13. N(X) = detJJ-1TN(x) where X = F(x) Jthe Jacobian matrix,Ji(x) = dF(x) dxi "Global and Local Deformations of Solid Primitives" Alan H. Barr Computer Graphics Volume 18, Number 3 July 1984 • Normal vector transformation

  14. Normal vector transformation: Example Given plan: S = S(x, 0, z), x in [0, 1], z in [0, 1] Transform the plane to the curved surface of a half cylinder: x’ = r – r cos (x/r) y’ = r sin (x /r) z’ = z where r is the radius of the cylinder and r = 1/p. 14

  15. |X| = V.W • = v1w1 + v2w2+v3w3 V q W V.W cosq = |V||W| • Dot product of vectors • Definition • Application

  16. Fixed Point Arithmetics (1/2) a = ai / 28 8 integer bits 1 1 1 8 fractional bits ai = 315, a = 1.2305 • Fixed Point Arithmetics : N bits (signed) Integer • Example : N = 16 gives range –32768  ai 32767 • We can use fixed scale to get the decimals

  17. Fixed Point Arithmetics (2/2) e = a.c = ai / 28 . ci/ 28  ei = (ai. ci) / 28 e = a+c = ai / 28 +ci/ 28  ei = ai + ci Multiplication then Requires Rescaling Addition just Like Normal

  18. Fixed Point Arithmetics - Application • Compression for Floating-point Real Numbers • 4 bytes reduced to 2 bytes • Lost some accuracy but affordable • Network data transfer • Software 3D Rendering

  19. Triangular Coordinates ha (xa,ya,za) Ac p Ab hb h (xb,yb,zb) Aa hc (xc,yc,zc) Aa Ab Ac h = ha + hb + hc where A = Aa + Ab + Ac If (Aa < 0 || Ab < 0 || Ac < 0) then the point is outside the triangle “Triangular Coordinates” A A A

  20. Triangle Area – 2D Area of a triangle in 2D xa ya A = ½ xb yb xc yc xa ya = ½ (xa*yb + xb*yc + xc*ya – xb*ya – xc*yb – xa*yc) (xa,ya,za) (xb,yb,zb) (xc,yc,zc)

  21. Triangle Area – 3D Area of a triangle in 3D p0 (x0,y0,z0) u p1 (x1,y1,z1) v (x2,y2,z2) p2 u = p1 – p0 v = p2 – p0 area = ½ |uxv| How to use triangular coordinates to determine the location of a point? Area is always positive!

  22. Triangular Coordinate System - Application Terrain Following Hit Test Ray Cast Collision Detection

  23. Intersection Ray Cast Containment Test

  24. Ray Cast – The Ray • Shoot a ray to calculate the intersection between the ray and models • Use a parametric equation to represent a ray x = x0 + (x1 – x0) t y = y0 + (y1 – y0) t, t = 0, z = z0 + (z1 – z0) t { 8 • The ray emits from (x0,y0,z0) • Only the t  0 is the answer candidate • The smallest positive t is the answer

  25. Ray Cast – The Plane • Each triangle in the Models has its plane equation • Use ax + by + cz + d = 0 as the plane equation • (a, b, c) is the plane normal vector • |d| is the distance of the plane to origin • Substitute the ray equation into the plane equation • Solve t for finding the intersection • Check whether or not the intersect point insider the triangle

  26. 2D Containment Test Intersection = 1, inside Intersection = 2, outside (x0, y0) Intersection = 0, outside Trick : Parametric equation for a ray which is parallel to the x-axis x = x0 + t y = y0 , t = 0, { 8 “if the No. of intersection is odd, the point is inside, otherwise, it is outside”

  27. 3D Containment Test • Same as the 2D containment test “if the No. of intersection is odd, the point is inside, otherwise, is outside”

  28. Rotation vs Orientation • Orientation: relative to a reference alignment • Rotation: • change object from one orientation to another • Represent an orientation as a rotation from the reference alignment

  29. Euler Angles • A rotation is described as a sequence of rotations about three mutually orthogonal coordinates axes fixed in space (e.g. world coordinate system) • X-roll, Y-roll, Z-roll • There are 6 possible ways to define a rotation • 3! R(q1, q2, q3) represents an x-roll, followed by y-roll, followed by z-roll R(q1, q2, q3) = c2c3 c2s3 -s2 0 s1s2c3-c1s3 s1s2s3+c1c3 s1c2 0 c1s2c3+s1s3 c1s2s3-s1c3 c1c2 0 0 0 0 1 where si = sinqi and ci = cosqi Left hand system

  30. Euler Angles & Interpolation • Interpolation happening on each angle • Multiple routes for interpolation • More keys for constraints • Can lead to gimbal lock R z z y y x x R

  31. Angular Displacement • R(q, n), n is the rotation axis rh = (n.r)n rv = r - (n.r)n , rotate into position Rrv V = nxrv = nxr Rrv = (cosq)rv + (sinq)V -> Rr = Rrh + Rrv = rh + (cosq)rv + (sinq)V = (n.r)n + (cosq)(r - (n.r)n) + (sinq) nxr = (cosq)r + (1-cosq) n(n.r) + (sinq) nxr V rv q r r Rr rh n n V rv q Rrv

  32. Quaternion • Sir William Hamilton (1843) • From Complex numbers (a + ib), i 2 = -1 • 16,October, 1843, Broome Bridge in Dublin • 1 real + 3 imaginary = 1 quaternion • q = a + bi + cj + dk • i2 = j2 = k2 = -1 • ij = k & ji = -k, cyclic permutation i-j-k-i • q = (s, v), where (s, v) = s + vxi + vyj + vzk

  33. Quaternion Algebra q1=(s1, v1)and q2=(s2, v2) q3= q1q2 = (s1s2 - v1.v2 , s1v2 + s2v1 + v1xv2) Conjugate of q = (s, v), q = (s, -v) qq = s2 + |v|2 = |q|2 A unit quaternion q = (s, v), where qq = 1 A pure quaternion p = (0, v) Noncommutative

  34. Quaternion VS Angular Displacement Take a pure quaternion p = (0, r) and a unit quaternion q = (s, v) where qq = 1 and define Rq(p) = qpq-1 where q-1 = q for a unit quaternion Rq(p) = (0, (s2 - v.v)r + 2v(v.r) + 2svxr) Let q = (cosq, sinq n), |n| = 1 Rq(p) = (0, (cos2q - sin2q)r + 2sin2qn(n.r) + 2cosqsinqnxr) = (0, cos2qr + (1 - cos2q)n(n.r) + sin2qnxr) Conclusion : The act of rotating a vector r by an angular displacement (q, n) is the same as taking this displacement, ‘lifting’ it into quaternion space, by using a unit quaternion (cos(q/2), sin(q/2)n)

  35. Conversion: Quaternion to Matrix 1-2y2-2z2 2xy-2wz 2xz+2wy 0 2xy+2wz 1-2x2-2z2 2yz-2wx 0 2xz-2wy 2yz+2wx 1-2x2-2y2 0 0 0 0 1 q =(w,x,y,z)

  36. Conversion: Matrix to Quaternion float tr, s; tr = m[0] + m[4] + m[8]; if (tr > 0.0f) { s = (float) sqrt(tr + 1.0f); q->w = s/2.0f; s = 0.5f/s; q->x = (m[7] - m[5])*s; q->y = (m[2] - m[6])*s; q->z = (m[3] - m[1])*s; } else { float qq[4]; int i, j, k; int nxt[3] = {1, 2, 0}; i = 0; if (m[4] > m[0]) i = 1; if (m[8] > m[i*3+i]) i = 2; j = nxt[i]; k = nxt[j]; s = (float) sqrt((m[i*3+i] - (m[j*3+j] + m[k*3+k])) + 1.0f); qq[i] = s*0.5f; if (s != 0.0f) s = 0.5f/s; qq[3] = (m[j+k*3] - m[k+j*3])*s; qq[j] = (m[i+j*3] + m[j+i*3])*s; qq[k] = (m[i+k*3] + m[k+i*3])*s; q->w = qq[3]; q->x = qq[0]; q->y = qq[1]; q->z = qq[2]; } M0 M1 M2 0 M3 M4 M5 0 M6 M7 M8 0 0 0 0 1 http://en.wikipedia.org/wiki/Conversion_between_quaternions_and_Euler_angles

  37. Quaternion Interpolation • Spherical linear interpolation, slerp A P B t f sin((1 - t)f) sin(tf) slerp(q1, q2, t) = q1 + q2 sinf sinf

  38. Differential Equation Basics • Initial value problems • ODE • Ordinary differential equation • Numerical solutions • Euler’s method • The midpoint method • Kunge –Kutta methods

  39. Initial Value Problems • An ODE • Vector field • Solutions • Symbolic solution • Numerical solution . x = f (x, t) where fis a known function x is the state of the system, x is the x’s time derivative x & x are vectors x(t0) = x0, initial condition . . Follow the vectors … Start here

  40. Euler’s Method x(t + h) = x(t) + h f(x, t) • A numerical solution • A simplification from Tayler series • Discrete time steps starting with initial value • Simple but not accurate • Bigger steps, bigger errors • Step error: O(h2) errors • Total error: O(h) • Can be unstable • Not efficient

  41. The Midpoint Method Error term . .. Concept : x(t0 + h) = x(t0) + h x(t0) + h2/2 x(t0) + O(h3) Result : x(t0 + h) = x(t0) + h[ f (x0 + h/2 f(x0 ,t0), t0 +h/2) ] Method : a. Compute an Euler step Dx = h f(x0 ,t0) b. Evaluate f at the midpoint fmid = f ( x0+Dx/2, t0 +h/2 ) c. Take a step using the midpoint x( t+ h) = x(t) + h fmid c b a

  42. The Runge-Kutta Method • Midpoint = Runge-Kutta method of order 2 • Runge-Kutta method of order 4 • Step error: O(h5) • Total error: O(h4) k1 = h f(x0, t0) k2 = h f(x0 + k1/2, t0 + h/2) k3 = h f(x0 + k2/2, t0 + h/2) k4 = h f(x0 + k3, t0 + h) x(t0+h) = x0 + 1/6 k1 + 1/3 k2 + 1/3 k3 + 1/6 k4

  43. Initial Value Problems - Application http://upload.wikimedia.org/wikipedia/en/4/44/Strand_Emitter.jpg • Dynamics • Particle system • Game FX System

More Related