3. How to calculate projections

The object is given as points in R$^ 3$. For an object described by line segments and polygons, all you need is the vertices; the images of the vertices can be connected up with line segments just as the vertices were. (Again, this ignores the question of hidden lines.)

The image is in the $x,y$-plane, which really consists of points $(x,y,0)$ in R$^ 3$, but we'd like images as points in R$^ 2$ so that they are ready to plot on paper or on a screen.

(I) Orthographic

Easy: $(x,y,z) \mapsto (x,y)$.

(II) Oblique

Here the rays are parallel and slanted. To describe the direction of the rays, we use a vector V. It turns out to be easiest if V is multiplied by a scalar to make the third coordinate 1; this will still describe the same rays Thus, we write V$= (a,b,1)$.

A good method is to apply a viewing transformation to make the rays perpendicular to the viewplane while leaving the viewplane alone. After applying this transformation to the object, we are back in Case (I). See Figure .

Since the origin stays fixed and parallel lines stay parallel, the transformation is a linear map. Thus, the map will be given by a $3 \times 3$ matrix $A$ taking $(1,0,0) \mapsto (1,0,0)$, $(0,1,0) \mapsto (0,1,0)$ (since these are both in the viewplane, which stays fixed), and $(a,b,1) \mapsto (0,0,1)$ (to make V perpendicular to the $x,y$-plane).

Notice that it would be easier if the transformation did the opposite, since then we'd have images of standard basis vectors. So, let $B$ be a $3 \times 3$ matrix taking

$(1,0,0) \mapsto (1,0,0)$
$(0,1,0) \mapsto (0,1,0)$
$(0,0,1) \mapsto (a,b,1)$

Therefore $B = \left[\begin{array}{ccc}1&0&0\\ 0&1&0\\ a&b&1\end{array}\right]$. Since $B$ is the inverse of $A$, we have $A = \left[\begin{array}{rrr}1&0&0\\ 0&1&0\\ -a&-b&1\end{array}\right]$.
(Here we can use the special method of taking inverses of matrices that are like $I$ except in the off-diagonal entries of one row or one column.)

Now, to find the image of a point, just multiply by $A$ and then project orthographically:

$(x,y,z) \mapsto (x,y,z)\left[\begin{array}{rrr}1&0&0\\ 0&1&0\\ -a&-b&1\end{array}\right] = (x-az,y-bz,z) \mapsto (x-az,y-bz)$, or briefly,

(3.1 ) .

In practice, you can either apply this method with the matrix multiplication or else program the final formula $(x,y,z) \mapsto (x-az,y-bz)$. You can think of this formula as saying that the image of $(x,y,z)$ consists of $(x,y)$ but offset linearly somewhat, depending on $z$. See Figure .

Figure: Viewing transformation of oblique projection

.9book/05dir/fig3.eps

Example: (a) Find the viewing transformation for an oblique projection from the direction $(2,3,5)$. (b) Under this projection, find the image of the $2 \times 2 \times 2$ cube with vertices $(\pm 1, \pm 1 \pm 1)$.

Solution: (a) Scaling $(2,3,5)$ we get V$= (0.4,0.6,1)$, so the viewing matrix is
$\left[\begin{array}{ccc}1&0&0\\ 0&1&0\\ -0.4&-0.6&1\end{array}\right]$.

(b) The top face, with points $(\pm 1, \pm 1, +1)$, goes to points $(\pm 1 - 0.4, \pm 1 -0.6)$. The bottom face, with points $(\pm 1, \pm 1, -1)$ goes to points $(\pm 1 + 0.4, \pm 1 + 0.6)$. In R$^ 2$, these are $2 \times 2$ squares offset from one another. Connect them up and you have the traditional oblique picture of a cube, as shown in Figure .

(Notice that in this example the cube straddles the viewplane, but this has no effect on the formulas. Rather, there is one formula, and it works no matter whether $z$ is positive or negative.)

Figure: Oblique image of cube in plane

.4book/05dir/obl.ps

(III) Perspective

First consider the case where the viewpoint is the origin and the viewplane is the plane $z = 1$. The object should be entirely above the $z = 0$ level, so its points have $z > 0$. For each point $(x,y,z)$ in the object, we need to scale it to make the third coordinate equal to $1$. The result is $(x,y,z) \mapsto (\frac xz, \frac yz, 1)$. In terms of the viewplane coordinates $(x,y)$, this is

(3.2 )

If instead the viewpoint is $(0,0,H)$ for some $H$, we scale up by a factor of $H$ instead and get

(3.3 )

Now consider the case where the setup is the other way around: The viewplane is $z = 0$ and the viewpoint is $(0,0,H)$. In the earlier formula, we simply replace $z$ by $H - z$ to switch 0 and $H$ as $z$-values, and we get $(x,y,z) \mapsto \left (\frac {Hx}{H - z}, \frac {Hy}{H - z} \right )$, or equivalently,

(3.4 )

This is logical: $x$ and $y$ have been scaled by an amount that depends on $z$.

Example. Suppose we want a perspective picture of the cube $(\pm 1, \pm 1, \pm 1)$ on the $x,y$-plane from the viewpoint $(0,0,4)$. This is the case $H = 4$. For the vertices $(\pm 1, \pm 1, +1)$ on the top face we get
$(\pm 1, \pm 1, 1) \mapsto (\pm \frac 43, \pm \frac 43)$.

Similarly, for the vertices on the bottom face we get

$(\pm 1, \pm 1, -1) \mapsto (\pm \frac 45, \pm \frac 45)$. Joining up these eight points, you get the ``square inside a square'' perspective image of the cube. See Figure .

Figure: Perspective image of cube in plane

.4book/05dir/persp.ps