# Lecture11_MatrixTransformsx - Ssugames.org

Software and s/w Development

Dec 2, 2013 (4 years and 6 months ago)

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Matrix Transformations

Reference:

“3D Math Primer for Graphics and Game Development, 2
nd

ed.”

Chapter 5

Chapter 6.x

http://www.teamten.com/lawrence/graphics/homogeneous/

Transform Matrices (4.2 + 5.1)

Take this vector:

Recall, if we think of this as a point, it means:

Start at the origin.

Move 0.5 units in the opposite direction of the
world

x
-
axis.

Move 3 units in the direction of the
world

y
-
axis

Move 9 units in the direction of the
world

z
-
axis.

Said mathematically:

9
3
5
.
0
v

1
0
0
*
9
0
1
0
*
3
0
0
1
*
5
.
0
v

9
3
5
.
0
*
1
0
0
0
1
0
0
0
1
Expressing this as a matrix*vector expression:

(assuming +x = [1,0,0], +y = [0,1,0], +z = [0,0,1])

Transform Matrices, cont.

Some notation:

Let
𝑤𝑜 𝑙

,
𝑤𝑜 𝑙


, and
𝑤𝑜 𝑙


be the
basis vectors
of a matrix M.

M is a
transformation matrix
(because
we’re using it to “transform” a vector)

Although technically in this case, the matrix
doesn’t actually change the original vector.

The first
b.v
., , represents the world +x
direction.

The second
b.v
., , represents the
world +y direction
.

The third
b.v
., , represents the world +z
direction.

v
M
v
v
M

*
9
3
5
.
0
1
0
0
0
1
0
0
0
1

0
0
1

0
1
0

1
0
0
Transform Matrices, cont.

Big deal, right?

Yes…but…what if we express that point relative to a “non
-
standard” coordinate system?

In other words, use a “non
-
standard” set of basis vectors.

Example:

Let’s assume +y and +z are [0,1,0] and [0,0,1], respectively
(as before)

But…let’s make +x be [
-
1,0,0]

Now the matrix is:

1
0
0
0
1
0
0
0
1
MysteryMat
Transform Matrices, cont.

Let’s watch what happens as we “transform” a few
points with this matrix…

1
0
0
0
1
0
0
0
1
MysteryMat

2
1
5
1
p

3
4
1
2
p

0
6
3
3
p

2
1
5
*
1
'
1
p
MysteryMat
p

3
4
1
*
2
'
2
p
MysteryMat
p

0
6
3
*
3
'
3
p
MysteryMat
p

p3

P3’

p2

P2’

p1

P1’

MysteryMat

does a x
-
axis

“Mirror”.

Transform Matrices, cont.

Let’s try to apply this idea of basis vectors and
local coordinate systems to a new matrix (4.2
in the book)

What does this matrix do?

Method1: Try to plot a few points…OK, but not the
best solution (
we might not pick points that show the transform
in action. Also, it takes a lot of time!
)

Method2: Look at the local coordinate system
defined by M…(see the next slide)

1
0
0
0
2
1
0
1
2
M
Transform Matrices, cont.

1
0
0
0
2
1
0
1
2
M

0
1
2
0
b

0
2
1
1
b

1
0
0
2
b

worldX

worldY

b0 =
localX

b1 =
localY

The
localX

and
localY

axes are rotated (~27 degrees)
and

scaled

from their
worldX

and
worldY

counter
-
parts.
localZ

=
worldZ
, so

this transformation leaves the z
-
values alone.

Imagine “deforming”
worldX

into b0,
worldY

into b1,
worldZ

into b2.

This is the same “deformation” the matrix will apply to all points

that we transform using this matrix.

Transform Matrices, cont.

1
0
0
0
2
1
0
1
2
M
worldX

worldY

b0 =
localX

b1 =
localY

worldX

worldY

Let’s say these are defined in
world space.

0
6
.
0
0
,
0
1
25
.
0
,
0
6
.
0
5
.
0
,
0
0
5
.
0
,
0
0
0
4
3
2
1
0
p
p
p
p
p

0
2
.
1
6
.
0
'
,
0
25
.
2
5
.
0
'
,
0
7
.
1
4
.
0
'
,
0
5
.
0
1
'
,
0
0
0
'
4
3
2
1
0
p
p
p
p
p

When we multiply (transform) them by M (just as we did on the last slide), these points are
transformed
in the same way the world axes are transformed into the basis vectors of M
.

Rotation Transform (5.1)

Now the problem of defining new transforms
just boils down to defining the (transformed)
basis vectors.

Let’s apply this to the problem of creating a
matrix which rotates
θ

degrees around the z
axis.

Rotation Transform (5.1), cont.

Put yourself in the direction of +z (looking at
the origin):

worldX

worldY

θ

θ

Now, let’s define the basis vectors…

Sin(
θ
)

Cos(
θ
)

0
)
sin(
)
cos(

localX
-
Sin(
θ
)

Cos(
θ
)

0
)
cos(
)
sin(

localY

1
0
0
localZ
Rotation Transformation (5.1), cont.

So, if we collect the 3 local vectors (basis
vectors into a matrix), we have…

We can similarly come up with matrices for x
-
axis and y
-
axis rotations…

1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(

Z
Rot

)
cos(
)
sin(
0
)
sin(
)
cos(
0
0
0
1

X
Rot

)
cos(
0
)
sin(
0
1
0
)
sin(
0
)
cos(

Y
Rot
Note: These matrices are the transpose of the matrices in the book (since the

author uses a Left
-
handed system)

Rotations

I'll show you a simpler (
imo
) version of 5.1.3
(rotation about an arbitrary axis) with matrix
concatenations.

Similarly for 5.2.2 (scaling about an arbitrary
axis)

Scale Transform (5.2)

The scale transform matrix is even simpler:

worldX

worldY

worldZ

Let’s say we want to scale:

2x in the x
-
direction

1x in the y
-
direction

½ in the z
-
direction

localX

localY

localZ

5
.
0
0
0
0
1
0
0
0
2
S

sz
sy
sx
S
0
0
0
0
0
0
In general:

sx

in the x
-
direction

sy

in the y
-
direction

sz

in the z
-
direction

If
sx

=
sy

=
sz
, the operation

is called a
universal scale.

Mirror (Reflection) and Shear (5.4 +
5.5)

[Do it on the board]

Translation (6.4)

Let’s say we want to develop a matrix for
translation (let’s say by amounts
tx
,
ty
, and
tz
).

Seems simple, right?

Wrong!

Try it

Translation (6.4)

Try it another way:

Try to come up with values for a
-
i

such that
all
points
transformed by the matrix are moved
by +4(x), +5(y), +6(z):

You can’t do it!

i
h
g
f
e
d
c
b
a
T

3
2
1
*
3
7
5
T

7
9
2
*
13
14
2
T
and

Translation

Up
to this point we've been looking at
linear
transformations

(you can do these with a 3x3 (for
3d) matrix
).

These
only

transform points
.

Translation is an example of an
affine
transformation

(a linear transform followed by a
linear transform).

You need to look at the problem in a higher
dimension…

…enter
homogeneous space
!

Translation, cont.

But, by using so
-
called
homogeneous
coordinates
, we can.

To visualize this, we'll look at a (x
-
)shear
matrix in 2d.

1
ℎ
0
1

Suppose
shearX

is 0.5:

Translations, cont.

Represent 2d points in 3d space, with w as 1.0

x

y

x

y

w

2D

3D

Translations, cont.

Now transform by a 3d shear in the x direction
by 2.0 units

x

y

x

y

w

2D

3D

1
0
2
0
1
0
0
0
1

Tanslations
, cont.

So…a shear in 3d, produces a
translation

in
2d.

We can do a translation in 3d by shearing in
4d.

We can't really visualize it

you just have to trust
the math

1
0
0
0
1
0
0
0
1
0
0
0
1
tz
ty
tx
T
Translation, cont.

Note, if you put a w of 0, the translation doesn't
affect the vector

Useful for vectors.

W should be:

1

if this is a point.

0

if this is a vector (direction + magnitude).

We'll see another application of this in
perspective projections.

In this case we'll end up with w values
after the
transformation

that are not 0 nor 1.

In this case we'll have to do a w
-
divide (homogeneous
divide)

Translation, cont.

Now, we can do translation, by using a 4x4
translation matrix like this:

Take our problem cases from before:

1
0
0
0
1
0
0
0
1
0
0
0
1
tz
ty
tx
T

1
3
2
1
*
1
0
0
0
6
1
0
0
5
0
1
0
4
0
0
1
1
3
7
5

1
7
9
2
*
1
0
0
0
6
1
0
0
5
0
1
0
4
0
0
1
1
13
14
2
Because it’s a point

It’s still a point

Translation, cont.

An extra bonus:

0
0
0
4
*
1
0
0
0
6
1
0
0
5
0
1
0
4
0
0
1
0
0
0
4
This one’s a vector

It’s still a vector

Note how the vector is unchanged.
Remember that you can re
-
draw a
vector anywhere and it still has the
same length and magnitude

here’s
proof!

4x4 Transformation matrices

OK, so now that we’re using 4x4 matrices for
translation, what about all the other matrices?

You can use 4x4 matrix, with the “old” 3x3
(linear) matrix embedded in the upper
-
left.

Or…think of it this affine transformation as a linear
transform (the 3x3 upper
-
left part) and a 0
-
vector
translation afterwards.

Example:
RotZ

now becomes:

1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
0
)
sin(
)
cos(

Z
Rot
Matrix Inverse (6.2)

The inverse of the matrix ( )has the property:

If M is a transform matrix, its inverse is the
opposite transform.

In general, the inverse is non
-
trivial to compute.

But in graphics…the "building block" matrices we've
seen so far do have an easy inverse.

I
M
M

1
*
1

M
Matrix Inverse, cont.

1
0
0
0
1
0
0
0
1
0
0
0
1
1
tz
ty
tx
T

1
0
0
0
0
/
1
0
0
0
0
/
1
0
0
0
0
/
1
1
sz
sy
sx
S
Rotation matrices are an example of an
orthonormal

matrix
. The basis vectors of these

matrices have unit length and are perpendicular to each other.

Q: How can you determine if three axes: p, q, and r are perpendicular?

A: p
-
dot
-
q, p
-
dot
-
r, and q
-
dot
-
r are all 0.

orthonormal

matrices is:

T
M
M

1
Transform Matrix Concatenation (5.6)

Suppose you want to apply the following
transforms to an object (in this order):

Translate (10,0,0)

Scales (2, 2, 2)

Rotate 15 degrees around the z
-
axis.

Transform Matrix Concatenation (8.7)

Numerically:

After Multiplying by the translation (10,0,0) matrix:

After multiplying by the scale (2,2,2) matrix:

After multiplying by the rotation of 15 degrees around the
z axis:

1
0
5
.
3
2
,
1
0
6
0
,
1
0
5
.
3
2
,
1
0
0
2
,
1
0
0
2
4
3
2
1
0
p
p
p
p
p

1
0
5
.
3
8
'
,
1
0
6
10
'
,
1
0
5
.
3
12
'
,
1
0
0
12
'
,
1
0
0
8
'
4
3
2
1
0
p
p
p
p
p

1
0
0
0
0
1
0
0
0
0
1
0
10
0
0
1
T

1
0
7
16
'
'
,
1
0
12
20
'
'
,
1
0
7
24
'
'
,
1
0
0
24
'
'
,
1
0
0
16
'
'
4
3
2
1
0
p
p
p
p
p

1
0
9
.
10
643
.
13
'
'
'
,
1
0
767
.
16
213
.
16
'
'
'
,
1
0
973
.
12
37
.
21
'
'
'
,
1
0
212
.
6
182
.
23
'
'
'
,
1
0
14
.
4
455
.
15
'
'
'
4
3
2
1
0
p
p
p
p
p

1
0
0
0
0
2
0
0
0
0
2
0
0
0
0
2
S

1
0
0
0
0
1
0
0
0
0
)
15
cos(
)
15
sin(
0
0
)
15
sin(
)
15
cos(
R
Transform Matrix Concatenation, cont.

You can get the same
net

effect by
concatenating these 3 transforms.

1
0
0
0
0
1
0
0
0
0
1
0
10
0
0
1
*
1
0
0
0
0
2
0
0
0
0
2
0
0
0
0
2
*
1
0
0
0
0
1
0
0
0
0
)
15
cos(
)
15
sin(
0
0
)
15
sin(
)
15
cos(
*
*
T
S
R
C

1
0
0
0
0
2
0
0
0
0
2
0
20
0
0
2
*
1
0
0
0
0
1
0
0
0
0
)
15
cos(
)
15
sin(
0
0
)
15
sin(
)
15
cos(

1
0
0
0
0
2
0
0
18
.
5
0
93
.
1
52
.
0
32
.
19
0
52
.
0
93
.
1
1
0
0
0
0
2
0
0
)
15
sin(
*
20
0
)
15
cos(
*
2
)
15
sin(
*
2
)
15
cos(
*
20
0
)
15
sin(
*
2
)
15
cos(
*
2
VERY IMPORTANT: In a right
-
handed system, multiply in the
reverse

order you want to

apply the transform (In a left
-
handed system, you do it in the forward order).

You
won’t

get the same composite matrix if you don’t follow this rule.

Transform Matrix Concatenation, cont.

Let’s test it.

1
0
5
.
3
2
,
1
0
6
0
,
1
0
5
.
3
2
,
1
0
0
2
,
1
0
0
2
4
3
2
1
0
p
p
p
p
p

1
0
0
0
0
2
0
0
18
.
5
0
93
.
1
52
.
0
32
.
19
0
52
.
0
93
.
1
C
Transform each of the points by C, and we get…

1
0
9
.
10
643
.
13
'
,
1
0
767
.
16
213
.
16
'
,
1
0
973
.
12
37
.
21
'
,
1
0
212
.
6
182
.
23
'
,
1
0
14
.
4
455
.
15
'
4
3
2
1
0
p
p
p
p
p

Which is the same points we got by transforming by the individual matrices!

Big Deal, right?

No!
To see why, we’ll need to analyze the costs…(see next slide)

Matrix multiplication costs

Matrix4x4 * Matrix4x4 uses 16 dot
-
products of Vector4’s.

Matrix4x4 * Vector4 uses 4 dot
-
products of Vector4’s.

Each Vector4
-
dot
-
Vector4 operation involves:

4 multiplies

So…

Matrix4x4 * Matrix4 = 64 multiplies, 48 adds

Matrix4x4 * Vector4 = 16 multiplies, 12 adds

Note: The GPU has specialized circuitry for doing this

often much more quickly than the CPU, but it still takes
time.

Analyzing the costs

Let’s say we have 3 transform matrices we want to
apply: M1, M2, M3 to n points.

MethodI
: Apply the transforms individually.

for
i

in range(n):

p1 = M1 * points[
i
]

p2 = M2 * p1

p3 = M3 * p2
# The transformed point

MethodII
: Create a combined transform. Then use it to
transform all n points.

C = M3 * M2 * M1

for
i

in range(n):

p3 = C * points[
i
] # The transformed point

Analyzing the costs, cont

What are the costs of the two methods?

MethodI
:

N * 3 * (Matrix4x4 times Vector4 cost)

Which is:

N * 3 * 16 multiplies

N * 3 * 12 adds

MethodII
:

2 * (Matrix4x4 times Matrix4x4 cost) + N * (Matrix4x4
times Vector4 cost)

Which is:

2 * 64 + N * 16 multiplies

2 * 48 + N * 12 adds

Which is better?

Analyzing the costs, cont

MethodI
:

N * 3 * 16 multiplies

N * 3 * 12 adds

MethodII
:

2 * 64 + N * 16 multiplies

2 * 48 + N * 12 adds

Comparison of Method1
vs

Method2 (using 3 transforms)

Method1

Method2

Verdict?

How much better?

N

mult

mult

1

48

36

144

108

Method1

33.3%

1/3 of Method2's mult's

2

96

72

160

120

Method1

66.7%

2/3 of Method2's mult's

4

192

144

192

144

Draw

8

384

288

256

192

Method2

66.7%

2/3 of Method1's mult's

16

768

576

384

288

Method2

50.0%

1/2 of Method1's mult's

32

1536

1152

640

480

Method2

41.7%

64

3072

2304

1152

864

Method2

37.5%

128

6144

4608

2176

1632

Method2

35.4%

256

12288

9216

4224

3168

Method2

34.4%

512

24576

18432

8320

6240

Method2

33.9%

1024

49152

36864

16512

12384

Method2

33.6%

2048

98304

73728

32896

24672

Method2

33.5%

4096

196608

147456

65664

49248

Method2

33.4%

8192

393216

294912

131200

98400

Method2

33.4%

16384

786432

589824

262272

196704

Method2

33.3%

32768

1572864

1179648

524416

393312

Method2

33.3%

65536

3145728

2359296

1048704

786528

Method2

33.3%

131072

6291456

4718592

2097280

1572960

Method2

33.3%

262144

12582912

9437184

4194432

3145824

Method2

33.3%

524288

25165824

18874368

8388736

6291552

Method2

33.3%

1048576

50331648

37748736

16777344

12583008

Method2

33.3%

2097152

100663296

75497472

33554560

25165920

Method2

33.3%

4194304

201326592

150994944

67108992

50331744

Method2

33.3%

1/3 of Method1's
mults

Note: the 3 above and in the
1/3 below is because we’re
doing 3 transforms.

If concatenating more than p
matrices, we would have 1/p
of the cost of
MethodI
.

Verdict: Matrix concatenation, except
for very small meshes, is
always

faster
than individual transforms.

Rotation Transform, not around the origin

A good example of matrix concatenation.

Normally, a z
-
axis rotation of 45 degrees on
this:

Produces this:

45

Rotation Transform, not around the origin, cont.

world
origin
.

But what if we want this:

To transform to this (rotate about a
local
origin
)?

45

Rotation Transform, not around the
origin, cont.

We can’t change how rotation works…

Hint: there are 3 (basic) transforms we need
to apply.

1) Translate the center of the object to the world
origin.

2) Rotate around the z
-
axis.

3) Translate the object by the
inverse

of Step1.

Rotation Transform, not around the
origin, cont.

1) Translate the center of the object to the world origin.

2) Rotate around the z
-
axis.

3) Translate the object by the
inverse

of Step1.

1
0
0
0
0
1
0
0
5
.
2
0
1
0
6
0
0
1
T

1
0
0
0
0
1
0
0
0
0
)
45
cos(
)
45
sin(
0
0
)
45
sin(
)
45
cos(
R

1
0
0
0
0
1
0
0
5
.
2
0
1
0
6
0
0
1
1
T
Rotation Transform, not around the origin, cont.

Now, we can combine these transforms into one composite
matrix.

Remember to multiply in the reverse order.

1
0
0
0
0
1
0
0
5
.
2
0
1
0
6
0
0
1
T

1
0
0
0
0
1
0
0
0
0
)
45
cos(
)
45
sin(
0
0
)
45
sin(
)
45
cos(
R

1
0
0
0
0
1
0
0
5
.
2
0
1
0
6
0
0
1
1
T
First step.

Second step.

Third step.

T
R
T
RotLocalZ
*
*
1

1
0
0
0
0
1
0
0
*
)
cos(
*
)
sin(
0
)
cos(
)
sin(
*
)
sin(
*
)
cos(
0
)
sin(
)
cos(
ty
ty
tx
tx
ty
tx
RotLocalZ

So…to rotate
θ

tx,ty,tz
), use this matrix…

1
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
0
)
sin(
)
cos(
1
0
0
0
1
0
0
0
1
0
0
0
1
tz
ty
tx
tz
ty
tx

Let’s say you want to rotate
θ

around a non
-
standard axis.

Let’s say the axis is the vector

0
7
.
0
4
.
0
Rotation about an arbitrary axis, cont.

Three steps:

Align the purple axis with (any) of the 3 (world)
axes

this is a rotation.

Rotate
θ

Undo the first rotation.

Rotation about an arbitrary axis, cont.

R1

= Align the purple axis with (any) of the 3
(world) axes

this is a rotation.

R2

= Rotate
θ

R3

= Undo the first rotation.

R
axis

= R3 * R2 * R1

R2 is easy

assume we’re rotating around the x
-
axis (but it
could be any of the 3 world axes)

R1 and R3
seem

hard unless you think in terms of
basis vectors…

Rotation about an arbitrary axis, cont.

Let’s take
R3 (
Undo the first rotation) first.

A rotation.

Rotates the
worldX

axis to the axis we’re rotating

So…
localX

= given axis.

The other local (basis) vectors aren’t shown, but
they need to be perpendicular to
localX
.

It actually doesn’t matter what they are

just that
they’re perpendicular to
localX
.

Have we ever done anything like this before?

YES! With the camera’s local coordinate system in
the
raytracer
!

Rotation about an arbitrary axis, cont.

1.
LocalX

= b0=given axis.

(make sure it’s normalized)

2.
LocalY

= b1=
LocalX

x

(normalize it afterwards)

3.
LocalZ

= b2=
LocalX

x
LocalY

One problem: In Step2, if
LocalX

is
[0,1,0]T
,

the result of cross
-
product’ing

it
will be the zero vector (which means we won’t have a valid rotation
matrix). Because of this, you often use this rule instead:

0
1
0

1
0
0
0
1
0
LocalXx
LocalXx
LocalY
if

if

0
1
0
LocalX

0
1
0
LocalX
where

9
.
0

2
1
0
3
b
b
b
R

Rotation about an arbitrary axis, cont.

Now, for R1.

Recall: This is the opposite of R3.

In other words, the
inverse
.

Recall also: The inverse of an
orthonormal

matrix
is simply its transpose!

All rotation matrices are
orthonormal

(test it if you’re
not sure)

T
R
R
3
1