Computer Aided Engineering Design

TRANSFORMATIONS AND PROJECTIONS 51

OP* = OE + EP* = x*i + y*j + z*k

= –w + * = – + ( – *) + ( – *) + ( – *)

w

z

w

w

z

xx

w

z

yy

w

z

kPPk⎛ i jzz

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

k

Thus,x = OP · i =

w

z (x – x*),y* = OP* · j =

w

z (y – y*) and OP* · k =

w

z

( – *)zz yielding

x

wx

zw

y

wy

zw

* = z

+

, * =

+

, and * = 0

This suggests that the image of P as seen from E on the plane of projection (z = 0) is given by

P* =

wx

zw

wy

+ zw

,

+

⎡ , 0, 1

⎣⎢

⎤

⎦⎥

which can be expressed using the 4 × 4 matrix as

PPP* =

*

*

0

1

=

+

+

0

1

0

+ 1

= =

10 0 0

01 0 0

00 0 0

001 1

ers

x

y

wx

zw

wy

zw

x

y

z

w w

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

≡

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎢

⎢⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

x

y

z

1

(2.34)

We can develop similar perspective projection matrices for the human eye to be on the x- and y-
axis, respectively, using cyclic symmetry. For the view point Ex at x = – w on the x-axis, a line joining

Ex and P will intersect the y-z image plane at P

wy

xw

wz

yz xw

* 0

+ +

⎡ 1

⎣⎢

⎤

⎦⎥

. Similarly, if the view point

is shifted to Ey at y = –w on y-axis, the line joining Ey and P will intersect z-x image plane at

P wx

yw

wz

zx yw

*

+

0

+

⎡ 1

⎣⎢

⎤

⎦⎥

.

Example 2.9. A line P 1 P 2 has coordinates P 1 (4, 4, 10) and P 2 (8, 2, 4) and the observer’s eye Ez is
located at (0, 0, – 4). Find the perspective projection of the line on the x-y plane.
Any point P on a given line can be written in the parametric form P = (1 – u)P 1 + uP 2 , where
u∈ [0 1]. When u = 0, P = P 1 and when u = 1, P = P 2. The perspective image of P on the x-y image
plane as seen from Ez can be obtained as follows:

P = (1 – u)[4 4 10] + u[8 2 4] = [4(1 + u) 2(2 – u) 2(5 – 3u)]

Using the transformation in Eq. (2.34), the perspective image of P on x-y plane is

P* =

10 0 0

01 0 0

00 0 0

4(1 + )

2(2 – )

2(5 – 3 )

1

=

4(1 + )

2(2 – )

0

7 – 3

2

8(1 + )

(7 – 3

001

4

1

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

≡

u

u

u

u

u

u

u

u))

4(2 – )

(7 – 3 )

0

1

u

u

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥