Graphics Reference
In-Depth Information
We begin by declaring the following conditions:
l
=
i
m
=
j n
=
k
=
−
OD
=
−
OQ
=
−
OP
d
=
d
k q
p
and Q is the point on the projection plane intersected by
p
. Our task is to find the scalars
and .
From Fig. 9.1 we see that
+
−
DQ
q
=
d
but
−
DQ has local components:
−
DQ
=
l
+
m
=
i
+
j
Therefore,
q
=
d
+
i
+
j
or
i
+
j
=
q
−
d
(9.1)
We now define
q
in terms of
p
,sowelet
q
=
p
where is a scalar that requires defining.
As
q
and
p
have the same projection on
k
, we find that
k
·
q
=
k
·
p
and
k
·
q
d
z
P
=
p
=
k
·
Substituting in Eq. (9.1) gives
d
z
P
+
=
+
+
−
i
j
x
P
i
y
P
j
z
P
k
d
k
Equating the
i
and
j
components gives
d
y
P
z
P
which are readily recognised as the perspective projection plane coordinates.
d
x
P
=
z
P
=
9.2.1 Horizontally oblique projection plane
Now let's consider the case where the projection plane is rotated about the vertical
m
-axis, as
shown in Fig. 9.2.
