Figure 2: Orthographic projection.

As you can see, the view volume is defined by six planes:

Because the view volume and the canonical view volume are both axis-aligned boxes, there is no correction for distance in this type of projection. The end result is, in fact, a lot like the result in Figure 1 where you just dropped the z-coordinate of every point. Objects of the same size in 3D space appear the same size in the projection, even if one is much further from the camera than the other. Lines that are parallel in 3D space remain parallel in the final image. Using this kind of projection would be out of the question for something like a first-person shooter—imagine trying to play one of those without being able to tell how far away anything is!—but it does have its uses. You might use it in a tile-based game, for instance, especially one where the camera is positioned at a fixed angle. Figure 3 shows a simple example.

Figure 3: A simple example of orthographic projection.

So without further ado, start to figure out how this is going to work. The easiest approach may be to consider each of your three axes separately, and compute how to map points along that axis from the original view volume into the canonical view volume. You begin with the x-coordinate. A point within your view volume will have an x-coordinate on the range [l, r], and you want to transform it to the range [–1, 1].

Now, in preparation to scale the range down to the size you want, you subtract l from all terms to produce a zero on the left-hand side. Another approach you could take here would be to translate the range so that it centers on zero, rather than having one of its endpoints at zero, but the algebra is a bit neater this way, so I'll do it like this for the sake of readability.

Now that one end of your range is positioned at zero, you can scale it down to the size you want. You want the range of x-values to be two units wide, from 1 to –1, so you multiply through by 2/(r – l). Note that r – l is the width of your view volume and is thus always a positive number, so you don't have to worry about the inequalities changing directions.

Next, you subtract one from all terms to produce your desired range of [–1, 1].

A bit of basic algebra allows you to write the center term as a single fraction:

Finally, you split the center term into two fractions so that it takes the form px + q; you need to group your terms this way so that the equations you derive can be easily translated into matrix form.

The center term of this inequality now gives you the equation you need to transform x into the canonical view volume.

The steps required to obtain a formula for y are exactly the same—just substitute y for x, t for r, and b for l—so rather than repeat them here, I'll just show the result:

Finally, you need to derive a formula for z. It's a little different in this case because you're mapping z to the range [0, 1] rather than [–1, 1], but this should look very familiar. Here's your starting condition, a z-coordinate on the range [n, f]:

You subtract n from all terms so the lower end of the range is positioned at zero:

And now, all that's left is to divide through by f – n to produce a final range of [0, 1]. As before, note that f – n indicates the depth of your viewing volume and thus will never be negative.

Finally, you split this into two fractions so it takes the form pz + q:

This gives you your formula for transforming z:

Now, you're ready to write your orthographic projection matrix. To recap your work thus far, here are the three projection equations you've derived:

If you write this in matrix form, you get:

That's it! Direct3D provides a function called D3DXMatrixOrthoOffCenterLH() (what a mouthful!) that constructs an orthographic projection matrix based on this same formula; you can find it in the DirectX documentation. The "LH" in that unwieldy function name refers to the fact that you're using a left-handed coordinate system. But, what exactly does "OffCenter" mean?

The answer to that question leads you to a simplified form of the orthographic projection matrix. Consider a few points: First, in eye space, your camera is positioned at the origin and looking directly down the z-axis. And second, you usually want your field of view to extend equally far to the left as it does to the right, and equally far above the z-axis as below. If that is the case, the z-axis passes directly through the center of your view volume, and so you have r = –l and t = –b. In other words, you can forget about r, l, t, and b altogether, and simply define your view volume in terms of a width w, and a height h, along with your other clipping planes f and n. If you make those substitutions into the orthographic projection matrix above, you get this rather simplified version:

This equation is implemented by the Direct3D function D3DXMatrixOrthoLH(). You can almost always use this matrix instead of the more general, "off center" version that you derived above, unless you're doing something strange with your projection.

One further point before you finish this section. It's instructive to note that this matrix can be represented as the concatenation of two simpler transforms: a translation followed by a scale. This should make sense to you if you think about it geometrically because all you're doing in an orthographic projection is shifting points from one axis-aligned box to another; the viewing volume doesn't change its shape, only its position and its size. Specifically, you have:

This product form of your projection is perhaps a bit more intuitive because it lets you more easily visualize what's happening. First, the viewing volume is translated along the z-axis so that its near plane coincides with the origin; then, a scale is applied to bring it down to the dimensions of the canonical view volume. That's easy enough to understand, right? The matrix for an off-center orthographic projection also can be represented as the product of a transformation and a scale, but it's similar enough to the result shown above that I won't list it here.

That about wraps it up for orthographic projections, so now you can move onto something a little more challenging.