Frames of Reference in Space - Rotation Transforms for Computer Graphics - page 148

⎡
⎣

⎤
⎦

cos α sin α 00

− sin α cos α 00

R − 1

α,z =

0 0 1 0

0 0 0 1

⎡
⎣

⎤
⎦

cos α sin α 0 − t x cos α − t y sin α

− sin α cos α 0 − t y cos α + t x sin α

R − 1

α,z T − 1 t x ,t y , 0 =

.

0 0 1 0

0 0 0 1

Let's test this transform by making α = 90°, and t x = t y = 1, as shown in

Fig. 10.4 (b):

⎡
⎣

⎤
⎦

010 − 1

− 100 1

0010
0001

R − 1

90 ° ,z T − 1

1 , 1 , 0 =

which, if used on the unit cube shown in Fig. 10.4 (c), produces
⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

010 − 1 00001111

00110011
01010101
11111111

−

100 1
0010
0001

⎡
⎣

⎤
⎦

− − − −

100
11110000
01010101
11111111

1 100 1

=

as confirmed by Fig. 10.4 (d).
Let's explore what happens if we swap the rotation and translation transforms to
T − 1

t x ,t y , 0 R − 1

α,z . This now implies that the frame of reference is rotated α about the z -
axis, and then translated (t x ,t y , 0 ) in the rotated frame's space, which is not what we
had planned. Here are the three transforms for rotating a frame of reference about
an off-set axis:

⎡
⎣

⎤
⎦

1 0 0 0

0 cos α sin α − t y cos α − t z sin α

R − 1

α,x T − 1

0 ,t y ,t z =

0 − sin α cos α − t z cos α + t y sin α

0 0 0 1

⎡
⎣

⎤
⎦

cos α 0 − sin α − t x cos α + t z sin α

01 0 0

α,y T − 1

R − 1

t x , 0 ,t z =

− −

sin α 0 cos α t z cos α t x sin α

00 0 1

⎡
⎣

⎤
⎦

− −

cos α sin α 0 t x cos α t y sin α

− sin α cos α 0 − t y cos α + t x sin α

α,z T − 1

R − 1

t x ,t y , 0 =

.

0 0 1 0

0 0 0 1