Rasterization: a Practical Implementation

Distributed under the terms of the CC BY-NC-ND 4.0 License.

  1. An Overview of the Rasterization Algorithm
  2. The Projection Stage
  3. The Rasterization Stage
  4. The Visibility Problem, the Depth Buffer Algorithm and Depth Interpolation
  5. Perspective Correct Interpolation and Vertex Attributes
  6. Rasterization: a Practical Implementation
  7. Source Code (external link GitHub)

The Projection Stage

Reading time: 13 mins.

Quick Review

In the previous chapter, we gave a high-level overview of the rasterization rendering technique. It can be decomposed into two main stages: first, the projection of the triangle's vertices onto the canvas, then the rasterization of the triangle itself. Rasterization means in this case, "breaking apart" the triangle's shape into pixels or raster element squares; this is what pixels used to be called in the past. In this chapter, we will review the first step. We have already described this method in the two previous lessons, thus we won't explain it here again. If you have any doubts about the principles behind perspective projection, check these lessons again. However, in this chapter, we will study a couple of new tricks related to projection that are going to be useful when we will get to the lesson on the perspective projection matrix. We will learn about a new method to remap the coordinates of the projected vertices from screen space to NDC space. We will also learn more about the role of the z-coordinate in the rasterization algorithm and how it should be handled at the projection stage.

Keep in mind as already mentioned in the previous chapter, that the goal of the rasterization rendering technique is to solve the visibility or hidden surface problem, which is to determine with parts of a 3D object are visible and which parts are hidden.

Projection: What Are We Trying to Solve?

What are we trying to solve here at that stage of the rasterization algorithm? As explained in the previous chapter, the principle of rasterization is to find if pixels in the image overlap triangles. To do so, we first need to project triangles onto the canvas and then convert their coordinates from screen space to raster space. Pixels and triangles are then defined in the same space, which means that it becomes possible to compare their respective coordinates (we can check the coordinates of a given pixel against the raster-space coordinates of a triangle's vertices).

The goal of this stage is thus to convert the vertices making up triangles from camera space to raster space.

Projecting Vertices: Mind the Z-Coordinate!

In the previous two lessons, we mentioned that when we compute the raster coordinates of a 3D point what we need in the end are its x- and y-coordinates (the position of the 3D point in the image). As a quick reminder, recall that these 2D coordinates are obtained by dividing the x and y coordinates of the 3D point in camera space, by the point's respective z-coordinate (what we called the perspective divide), and then remapping the resulting 2D coordinates from screen space to NDC space and then NDC space to raster space. Keep in mind that because the image plane is positioned at the near-clipping plane, we also need to multiply the x- and y-coordinate by the near-clipping plane. Again, we explained this process in great detail in the previous two lessons.

$$ \begin{array}{l} Pscreen.x = \dfrac{ near * Pcamera.x }{ -Pcamera.z }\\ Pscreen.y = \dfrac{ near * Pcamera.y }{ -Pcamera.z }\\ \end{array} $$

Note that so far, we have been considering points in screen space as essentially 2D points (we didn't need to use the points' z-coordinate after the perspective divide). From now on though, we will declare points in screen-space, as 3D points and set their z-coordinate to the camera-space points' z-coordinate as follow:

$$ \begin{array}{l} Pscreen.x = \dfrac{ near * Pcamera.x }{ -Pcamera.z }\\ Pscreen.y = \dfrac{ near * Pcamera.y }{ -Pcamera.z }\\ Pscreen.z = { -Pcamera.z }\\ \end{array} $$

It is best at this point to set the projected point z-coordinate to the inverse of the original point z-coordinate, which as you know by now, is negative. Dealing with positive z-coordinates will make everything simpler later on (but this is not mandatory).

Figure 1: when two vertices in camera space have the same 2D raster coordinates, we can use the original vertices z-coordinate to find out which one is in front of the other (and thus which one is visible).

Keeping track of the vertex z-coordinate in camera space is needed to solve the visibility problem. Understanding why is easier if you look at Figure 1. Imagine two vertices v1 and v2 which when projected onto the canvas, have the same raster coordinates (as shown in Figure 1). If we project v1 before v2 then v2 will be visible in the image when it should be v1 (v1 is clearly in front of v2). However, if we store the z-coordinate of the vertices along with their 2D raster coordinates, we can use these coordinates to define which point is closest to the camera independently of the order in which the vertices are projected (as shown in the code fragment below).

// project v2
Vec3f v2screen;
v2screen.x = near * v2camera.x / -v2camera.z;
v2screen.y = near * v2camera.y / -v2camera.z;
v2screen.z = -v2cam.z;

Vec3f v1screen;
v1screen.x = near * v1camera.x / -v1camera.z;
v1screen.y = near * v1camera.y / -v1camera.z;
v1screen.z = -v1camera.z;

// If the two vertices have the same coordinates in the image then compare their z-coordinate
if (v1screen.x == v2screen.x && v1screen.y == v2screen.y && v1screen.z < v2screen.z) {
    // if v1.z < v2.z then store v1 in frame-buffer
Figure 2: the points on the surface of triangles that a pixel overlaps can be computed by interpolating the vertices making up these triangles. See chapter 4 for more details.

What we want to render though are triangles, not vertices. So the question is, how does the method we just learned about apply to triangles? In short, we will use the triangle vertices coordinates to find the position of the point on the triangle that the pixel overlaps (and thus it's z-coordinate). This idea is illustrated in Figure 2. If a pixel overlaps two or more triangles, we should be able to compute the position of the points on the triangles that the pixel overlap, and use the z-coordinates of these points as we did with the vertices, to know which triangle is the closest to the camera. This method will be described in detail in chapter 4 (The Depth Buffer. Finding the Depth Value of a Sample by Interpolation).

Screen Space is Also Three-Dimensional

Figure 3: screen space is three-dimensional (middle image).

To summarize, to go from camera space to screen space (which is the process during which the perspective divide is happening), we need to:

Practically, this means that our projected point is not a 2D point anymore, but a 3D point. Or to say it differently, that screen space is not two- by three-dimensional. In his thesis Ed-Catmull writes:

Screen-space is also three-dimensional, but the objects have undergone a perspective distortion so that an orthogonal projection of the object onto the x-y plane, would result in the expected perspective image (Ed-Catmull's Thesis, 1974).

Figure 4: we can form an image of an object in screen space by projecting lines orthogonal (or perpendicular if you prefer) to the x-y image plane.

You should now be able to understand this quote. The process is also illustrated in Figure 3. First, the geometry vertices are defined in camera space (top image). Then, each vertex undergoes a perspective divide. That is, the vertex x- and y-coordinates are divided by their z-coordinate, but as mentioned before, we also set the resulting projected point z-coordinate to the inverse of the original vertex z-coordinate. This, by the way, infers a change of direction in the z-axis of the screen space coordinate system. As you can see, the z-axis is now pointing inward rather than outward (middle image in Figure 3). But the most important thing to notice is that the resulting object is a deformed version of the original object but a three-dimensional object. Furthermore what Ed-Catmull means when he writes "an orthogonal projection of the object onto the x-y plane, would result in the expected perspective image", is that once the object is in screen space, if we trace lines perpendicular to the x-y image plane from the object to the canvas, then we get a perspective representation of that object (as shown in Figure 4). This is an interesting observation because it means that the image creation process can be seen as a perspective projection followed by an orthographic projection. Don't worry if you don't understand clearly the difference between perspective and orthographic projection. It is the topic of the next lesson. However, try to remember this observation, as it will become handy later.

Remapping Screen Space Coordinates to NDC Space

In the previous two lessons, we explained that once in screen space, the x- and y-coordinates of the projected points need to be remapped to NDC space. In the previous lessons, we also explained that in NDC space, points on the canvas had their x- and y-coordinates contained in the range [0,1]. In the GPU world though, coordinates in NDC space are contained in the range [-1,1]. Sadly, this is one of these conventions again, that we need to deal with. We could have kept the convention [0,1] but because GPUs are the reference when it comes to rasterization, it is best to stick to the way the term is defined in the GPU world.

You may wonder why we didn't use the [-1,1] convention in the first place then. For several reasons. Once because in our opinion the term "normalize" should always suggest that the value that is being normalized is in the range [0,1]. Also because it is good to be aware that several rendering systems use different conventions with respect to the concept of NDC space. The RenderMan specifications for example define NDC space as a space defined over the range [0,1].

Thus once the points have been converted from camera space to screen space, the next step is to remap them from the range [l,r] and [b,t] for the x- and y-coordinate respectively, to the range [-1,1]. The term l, r, b, and t relate to the left, right, bottom, and top coordinates of the canvas. By re-arranging the terms, we can easily find an equation that performs the remapping we want:

$$l < x < r$$

Where x here is the x-coordinate of a 3D point in screen space (remember that from now on, we will assume that points in screen space are three-dimensional as explained above). If we remove the term l from the equation we get:

$$0 < x - l < r - l$$

By dividing all terms by (r-l) we get:

$$ \begin{array}{l} 0 < \dfrac {(x - l)}{(r - l)} < \dfrac {(r - l)}{(r - l)} \\ 0 < \dfrac {(x - l)}{(r -l)} < 1 \\ \end{array} $$

We can now develop the term in the middle of the equation:

$$0 < \dfrac {x}{(r -l)} - \dfrac {l}{(r -l)}< 1$$

We can now multiply all terms by 2:

$$0 < 2 * \dfrac {x}{(r -l)} - 2 * \dfrac {l}{(r -l)}< 2$$

We now remove 1 from all terms:

$$-1 < 2 * \dfrac {x}{(r -l)} - 2 * \dfrac {l}{(r-l)} - 1 < 1$$

If we develop the terms and regroup them, we finally get:

$$ \begin{array}{l} -1 < 2 * \dfrac {x}{(r -l)} - 2 * \dfrac {l}{(r-l)} - \dfrac{(r-l)}{(r-l)}< 1 \\ -1 < 2 * \dfrac {x}{(r -l)} + \dfrac {-2*l+l-r}{(r-l)} < 1 \\ -1 < 2 * \dfrac {x}{(r -l)} + \dfrac {-l-r}{(r-l)} < 1 \\ -1 < \color{red}{\dfrac {2x}{(r -l)}} \color{green}{- \dfrac {r + l}{(r-l)}} < 1\\ \end{array} $$

This is a very important equation because the red and green terms of the equation in the middle of the formula will become the coefficients of the perspective projection matrix. We will study this matrix in the next lesson. But for now, we will just apply this equation to remap the x-coordinate of a point in screen space to NDC space (any point that lies on the canvas has its coordinates contained in the range [-1.1] when defined in NDC space). If we apply the same reasoning to the y-coordinate we get:

$$-1 < \color{red}{\dfrac {2y}{(t - b)}} \color{green}{- \dfrac {t + b}{(t-b)}} < 1$$

Putting Things Together

At the end of this lesson, we now can perform the first stage of the rasterization algorithm which you can decompose into two steps:

From there, it is extremely simple to convert the coordinates to raster space. We just need to remap the x- and y-coordinates in NDC space to the range [0,1] and multiply the resulting number by the image width and height respectively (don't forget that in raster space the y-axis goes down while in NDC space it goes up. Thus we need to change y's direction during this remapping process). In code we get:

float nearClippingPlane = 0.1;
// point in camera space
Vec3f pCamera;
worldToCamera.multVecMatrix(pWorld, pCamera);
// convert to screen space
Vec2f pScreen;
pScreen.x = nearClippingPlane * pCamera.x / -pCamera.z;
pScreen.y = nearClippingPlane * pCamera.y / -pCamera.z;
// now convert point from screen space to NDC space (in range [-1,1])
Vec2f pNDC;
pNDC.x = 2 * pScreen.x / (r - l) - (r + l) / (r - l);
pNDC.y = 2 * pScreen.y / (t - b) - (t + b) / (t - b);
// convert to raster space and set point z-coordinate to -pCamera.z
Vec3f pRaster;
pRaster.x = (pNDC.x + 1) / 2 * imageWidth;
// in raster space y is down so invert direction
pRaster.y = (1 - pNDC.y) / 2 * imageHeight;
// store the point camera space z-coordinate (as a positive value)
pRaster.z = -pCamera.z;

Note that the coordinates of points or vertices in raster space are still defined as floating point numbers here and not integers (which is the case for pixel coordinates).

What's Next?

We now have projected the triangle onto the canvas and converted these projected vertices to raster space. Both the vertices of the triangle and the pixel live in the same coordinate system. We are now ready to loop over all pixels in the image and use a technique to find if they overlap a triangle. This is the topic of the next chapter.


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