1. Game Mathematics

2. Fixed-point Real Numbers Triangle Mathematics Intersection Issues
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 Matrix basics Definition Transpose Addition a11 .. a1n . .
am1 .. amn
A = (aij) =
C = A T cij = aji
C = A + B cij = aij + bij

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

5. Transformations in Matrix form
A point or a vector is a row matrix (de facto convention)
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

6. Translation Transformation
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
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 Sy
Sz 0
[x’ y’ z’ 1] = [x y z 1]
= [x y z 1] S
Here Sx, Sy and Sz are scaling factors.

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

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
[x” y” z” 1] = [x y z 1] M3
M1 M2 = M2 M1

10. A vector is an entity that possesses magnitude and direction.
Vectors
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.
(x1,y1,z1)
V = (x2-x1, y2-y1, z2-z1)
(x2,y2,z2)

11. Addition of vectors Length of vectors 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|

12. Cross product of vectors
Definition
Application
A normal vector to a polygon is calculated from 3 (non-collinear) vertices of the polygon.
X = V X W
= (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 = V1 X V2 V1

13. Normal vector transformation
N(X) = detJ J-1T N(x)
where X = F(x)
J the 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

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

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

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

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 Coordinatesha
(xa,ya,za) Ac p hb Ab 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
= 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 p0 u p1 v p2 u = p1 – p0 v = p2 – p0
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 |u x v |
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
x = x0 + (x1 – x0) t
y = y0 + (y1 – y0) t, t = 0,
z = z0 + (z1 – z0) t {
Shoot a ray to calculate the intersection between the ray and models
Use a parametric equation to represent a ray 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 = y , t = 0, { 8
“if the No. of intersection is odd, the point is inside,
otherwise, it is outside”

27. 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. There are 6 possible ways to define a rotation
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) = c2c c2s s
s1s2c3-c1s3 s1s2s3+c1c3 s1c2 0
c1s2c3+s1s3 c1s2s3-s1c3 c1c2 0
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. R(q, n), n is the rotation axis
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 rv V 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 + 2sin2q n(n.r) + 2cosqsinq nxr)
= (0, cos2qr + (1 - cos2q)n(n.r) + sin2q nxr)
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 x2-2z2 2yz-2wx 0
2xz-2wy 2yz+2wx 1-2x2-2y2 0
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

37. Quaternion Interpolation
Spherical linear interpolation, slerp A P t B f
sin((1 - t)f)
sin(tf)
slerp(q1, q2, t) = q 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 f is 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 . .
Start here
Follow the vectors …

40. Discrete time steps starting with initial value
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
Dynamics
Particle system
Game FX System