1

Game MathematicsGame Mathematics

MatricesMatrices VectorsVectors Fixed-point Real NumbersFixed-point Real Numbers Triangle MathematicsTriangle Mathematics Intersection IssuesIntersection Issues Euler AnglesEuler Angles Angular DisplacementAngular Displacement QuaternionQuaternion Differential Equation BasicsDifferential Equation Basics

2

Essential Mathematics for Game DevelopmentEssential Mathematics for Game Development

Matrix basicsMatrix basics– DefinitionDefinition

– TransposeTranspose

– AdditionAddition

3

MatricesMatrices

A = (aij) =

a11 .. a1n

. . . .am1 .. amn

C = A T cij = aji

C = A + B cij = aij + bij

– Scalar-matrix multiplicationScalar-matrix multiplication

– Matrix-matrix multiplicationMatrix-matrix multiplication

4

C = A cij = aij

C = A B cij = aikbkj

k = 1

r

cij = row x column

Transformations in Transformations in MatrixMatrix form form– A point or a vector is a row matrix (de facto convention)A point or a vector is a row matrix (de facto convention)

5

V = [x y z]

– Using matrix notation, a point Using matrix notation, a point VV is transformed under is transformed under translation, scaling and rotation as :translation, scaling and rotation as :

V’ = V + DV’ = VSV’ = VR

where D is a translation vector andS and R are scaling and rotation matrices

6

– To make translation be a linear transformation, we To make translation be a linear transformation, we introduce the introduce the homogeneous coordinate systemhomogeneous coordinate system

V (x, y, z, w)

where w is always 1

– Translation TransformationTranslation Transformationx’ = x + Tx

y’ = y + Ty

z’ = z + Tz

V’ = VT

[x’ y’ z’ 1] = [x y z 1]

= [x y z 1] T

1 0 0 00 1 0 00 0 1 0Tx Ty Tz 1

7

– Scaling TransformationScaling Transformation

x’ = xSx

y’ = ySy

z’ = zSz

V’ = VS

[x’ y’ z’ 1] = [x y z 1]

= [x y z 1] S

Sx 0 0 00 Sy 0 00 0 Sz 00 0 0 1

Here Sx, Sy and Sz are scaling factors.

8

– Rotation TransformationsRotation Transformations

1 0 0 00 cos sin 00 -sin cos 00 0 0 1

Rx =

Ry =

Rz =

cos 0 -sin 0 0 1 0 0sin 0 cos 00 0 0 1

cos sin 0 0-sin cos 0 0 0 0 1 0 0 0 0 1

9

– Net Transformation matrixNet Transformation matrix

– Matrix multiplication are Matrix multiplication are not commutativenot 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

A vector is an entity that possesses A vector is an entity that possesses magnitudemagnitude an and d directiondirection..

A 3D vector is a triple :A 3D vector is a triple :– VV = (v = (v11, v, v22, v, v33)), where each component , where each component vvii is a scalar. is a scalar.

A ray (directed line segment), that possesses A ray (directed line segment), that possesses positionposition, , magnitudemagnitude and and directiondirection..

10

VectorsVectors

(x1,y1,z1)

(x2,y2,z2)

V = (x2-x1, y2-y1, z2-z1)

AddAddiition of vectorstion of vectors

Length of vectorsLength of vectors

11

X = V + W = (x1, y1, z1) = (v1 + w1, v2 + w2, v3 + w3)

V

W

V + W

V

W

V + W

|V| = (v12 + v2

2 + v32)1/2

U = V / |V|

Cross product of vectorsCross product of vectors– DefinitionDefinition

– ApplicationApplication» A normal vector to a polygon is calculated from 3 (non-collinear) A normal vector to a polygon is calculated from 3 (non-collinear)

vertices of the polygon.vertices of the polygon.

12

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)

NpV2

V1

polygon defined by 4 points

Np = V1 X V2

»Normal vector transformationNormal vector transformation

13

N(X) = detJ J-1T N(x)

where X = F(x) J the Jacobian matrix, Ji(x) =

F(x)xi

"Global and Local Deformations of Solid Primitives" Alan H. BarrComputer Graphics Volume 18, Number 3 July 1984

Normal vector transformation: Normal vector transformation: ExampleExample

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’ = zwhere r is the radius of the cylinder and r = 1/.

14

Dot product of vectorsDot product of vectors– DefinitionDefinition

– ApplicationApplication

15

|X| = V . W = v1w1 + v2w2 + v3w3

V

W

cos =V . W

|V||W|

Fixed Point Arithmetics : N bits (signed) IntegerFixed Point Arithmetics : N bits (signed) Integer– Example : N = 16 gives range –32768 Example : N = 16 gives range –32768 a aii 32767 32767

– We can use fixed scale to get the decimalsWe can use fixed scale to get the decimals

16

Fixed Point Arithmetics (1/2)Fixed Point Arithmetics (1/2)

a = ai / 28

1 1 18 integer bits

8 fractional bits

ai = 315, a = 1.2305

Multiplication then Requires RescalingMultiplication then Requires Rescaling

Addition just Like Normal Addition just Like Normal

17

Fixed Point Arithmetics (2/2)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

Compression for Floating-point Real Compression for Floating-point Real NumbersNumbers– 4 bytes reduced to 2 bytes4 bytes reduced to 2 bytes– Lost some accuracy but affordableLost some accuracy but affordable– Network data transferNetwork data transfer

Software 3D RenderingSoftware 3D Rendering

18

Fixed Point Arithmetics - ApplicationFixed Point Arithmetics - Application

19

h

ha

hb

hc

Aa

Ac

Ab

h = ha + hb + hc

where A = Aa + Ab + Ac

If (Aa < 0 || Ab < 0 || Ac < 0) thenthe point is outside the triangle

“Triangular Coordinates”

Aa Ab Ac

A A A

p

(xa,ya,za)

(xb,yb,zb)

(xc,yc,zc)

Triangular CoordinatesTriangular Coordinates

20

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)

Triangle Area – 2DTriangle Area – 2D

(xa,ya,za)

(xb,yb,zb)

(xc,yc,zc)

21

Area of a triangle in 3D

Triangle Area – 3DTriangle Area – 3D

(x0,y0,z0)

(x1,y1,z1)

(x2,y2,z2)

u

v

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!

p1

p0

p2

Terrain FollowingTerrain Following Hit TestHit Test Ray CastRay Cast Collision DetectionCollision Detection

22

Triangular Coordinate System - ApplicationTriangular Coordinate System - Application

Ray CastRay Cast Containment TestContainment Test

23

IntersectionIntersection

24

Ray Cast – The RayRay Cast – The Ray

x = x0 + (x1 – x0) ty = y0 + (y1 – y0) t, t = 0,z = z0 + (z1 – z0) t

{

Shoot a ray to calculate the intersection Shoot a ray to calculate the intersection between the ray and modelsbetween the ray and models

Use a parametric equation to represent a rayUse a parametric equation to represent a ray

8

The ray emits from (xThe ray emits from (x00,y,y00,z,z00)) Only the t Only the t 0 is the answer candidate 0 is the answer candidate The smallest positive t is the answerThe smallest positive t is the answer

25

Ray Cast – The PlaneRay Cast – The Plane

Each triangle in the Models has its plane Each triangle in the Models has its plane equationequation

UUse se ax + by + cz + d = 0ax + by + cz + d = 0 as the plane as the plane equationequation

((a, b, c)a, b, c) is the plane normal vector is the plane normal vector |d||d| is the distance of the plane to origin is the distance of the plane to origin Substitute the ray equation into the plane Substitute the ray equation into the plane

equationequation SSolve olve tt for finding the intersection for finding the intersection CCheck whether or not the intersect point heck whether or not the intersect point

insider the triangle insider the triangle

26

Intersection = 1, inside

Intersection = 2, outside

Intersection = 0, outside

Trick : Parametric equation for a ray which is parallel to the x-axis

x = x0 + t y = y0 , t = 0,

{ 8

(x0, y0)

2D Containment Test2D Containment Test

“ if the No. of intersection is odd, the point is inside, otherwise, it is outside”

27

3D Containment Test3D Containment Test

“ if the No. of intersection is odd, the point is inside, otherwise, is outside”

Same as the 2D containment testSame as the 2D containment test

Rotation vs OrientationRotation vs Orientation

28

Orientation: relative to a reference Orientation: relative to a reference alignmentalignment

Rotation: Rotation: change object from one orientation change object from one orientation

to anotherto another Represent an orientation as a Represent an orientation as a

rotation from the reference rotation from the reference alignmentalignment

29

A A rotation is described as a sequence of rotation is described as a sequence of rotations about three mutually orthogonal rotations about three mutually orthogonal coordinates axes fixed in space (e.g. coordinates axes fixed in space (e.g. world world coordinate systemcoordinate system))– X-roll, X-roll, Y-Y-roll, roll, Z-Z-rollroll

TThere are 6 possible ways to define a rotationhere are 6 possible ways to define a rotation– 3!3!

R(1, 2, ) represents an x-roll, followed by y-roll, followed by z-roll

R(1, 2, c2c3 c2s3 -s2 0 s1s2c3-c1s3 s1s2s3+c1c3 s1c2 0 c1s2c3+s1s3 c1s2s3-s1c3 c1c2 0 0 0 0 1 where si = sini and ci = cosi

Euler AnglesEuler Angles

Left hand system

30

Interpolation happening on each angleInterpolation happening on each angle Multiple routes for interpolationMultiple routes for interpolation MMore keys for constraintsore keys for constraints Can lead to gimbal lockCan lead to gimbal lock

z

x

y

Rz

x

y

R

Euler Angles & InterpolationEuler Angles & Interpolation

31

RR((, , nn), ), nn is the rotation axis is the rotation axis

n

r Rr

n

r

rv

rh

V

rv

V

Rrv

rh = (n.r)nrv = r - (n.r)n , rotate into position Rrv

V = nxrv = nxr

Rrv = (cos)rv + (sin)V-> Rr = Rrh + Rrv

= rh + (cos)rv + (sin)V = (n.r)n + (cos)r - (n.r)n) + (sin) nxr = (cos)r + (1-cos) n (n.r) + (sin) nxr

Angular DisplacementAngular Displacement

32

Sir William Hamilton (1843)Sir William Hamilton (1843) FFrom Complex numbers (a + rom Complex numbers (a + iib), b), i i 22 = -1 = -1 16,16,October, 1843, October, 1843, Broome BridgeBroome Bridge in in DublinDublin 1 1 realreal + 3 + 3 imaginaryimaginary = 1 = 1 quaternionquaternion qq = a + b = a + bii + c + cjj + d + dkk ii22 = = jj22 = = kk22 = -1 = -1 ijij = = kk & & jiji = - = -kk, cyclic permutation , cyclic permutation ii--jj--kk--ii qq = ( = (ss, , vv), where (), where (ss, , vv) = ) = ss + + vvxxii + + vvyyjj + + vvzzkk

QuaternionQuaternion

33

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

Quaternion AlgebraQuaternion Algebra

34

Take a pure quaternion p = (0, r)and a unit quaternion q = (s, v) where qq = 1and 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 = (cos, sinn), |n| = 1

Rq(p) = (0, (cos2- sin2)r + 2sin2 n(n.r) + 2cossin nxr) = (0, cos2r + (1 - cos2)n(n.r) + sin2 nxr)

Conclusion :The act of rotating a vector r by an angular displacement (, n) is the same as taking this displacement, ‘lifting’ it into quaternion space, by using a unit quaternion (cos(/2), sin(/2)n)

Quaternion VS Angular DisplacementQuaternion VS Angular Displacement

35

1-2y2-2z2 2xy-2wz 2xz+2wy 02xy+2wz 1-2x2-2z2 2yz-2wx 02xz-2wy 2yz+2wx 1-2x2-2y2 0 0 0 0 1

q = (w,x,y,z)

Conversion: Quaternion to MatrixConversion: Quaternion to Matrix

36

M0 M1 M2 0M3 M4 M5 0M6 M7 M8 0 0 0 0 1

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]; }

Conversion: Matrix to QuaternionConversion: Matrix to Quaternion

http://en.wikipedia.org/wiki/Conversion_between_quaternions_and_Euler_angles

37

Spherical linear interpolation, Spherical linear interpolation, slerpslerp

A

B

P

t

slerp(q1, q2, t) = q1 + q2

sin((1 - t))

sin sinsin(t)

Quaternion InterpolationQuaternion Interpolation

38

Initial value problemsInitial value problems OODEDE

– Ordinary differential equationOrdinary differential equation

NNumerical solutionsumerical solutions– EEuler’s methoduler’s method

– TThe midpoint methodhe midpoint method– Kunge –Kutta methodsKunge –Kutta methods

Differential Equation BasicsDifferential Equation Basics

39

An ODEAn ODE

VVector fieldector field SSolutionsolutions

– SSymbolic solutionymbolic solution– NNumerical solutionumerical solution

x = f (x, t)

where f is a known functionx is the state of the system, x is the x’s time derivative

x & x are vectorsx(t0) = x0, initial condition

.

Start here Follow the vectors …

Initial Value ProblemsInitial Value Problems

.

.

40

A numerical solutionA numerical solution– A simplification from A simplification from Tayler seriesTayler series

DDiscrete time steps starting with initial valueiscrete time steps starting with initial value Simple but not accurateSimple but not accurate

– Bigger steps, bigger errorsBigger steps, bigger errors– Step error: OStep error: O((h22) ) errorserrors– Total error: O(h)Total error: O(h)

Can be unstableCan be unstable Not efficientNot efficient

x(t + h) = x(t) + h f(x, t)

Euler’s MethodEuler’s Method

41

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 stepx = h f(x0 , t0)

b. Evaluate f at the midpointfmid = f ( x0+x/2, t0 +h/2 )

c. Take a step using the midpointx( t+ h) = x(t) + h fmid

. ..

a

b

c

Error term

The Midpoint MethodThe Midpoint Method

42

Midpoint = Midpoint = Runge-KuttaRunge-Kutta method of method of order 2order 2

Runge-KuttaRunge-Kutta method of order 4 method of order 4– Step error: OStep error: O(h(h55))– Total error: O(hTotal error: O(h44))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

The The Runge-KuttaRunge-Kutta Method Method

DynamicsDynamics– Particle systemParticle system

Game FX SystemGame FX System

43

Initial Value Problems - ApplicationInitial Value Problems - Application

http://upload.wikimedia.org/wikipedia/en/4/44/Strand_Emitter.jpg