Processing: Creative Coding and Computational Art

Beyond box() and sphere()
While there are many common 3D primitives (including cylinders, pyramids, cones, and
toroids), Processing’s built-in primitive-creation functions are limited to the measly box()
and sphere(). I’ll show you how to construct a few of these—which is mostly a matter of
creating algorithms. What I mean by this is that you need a set of logical instructions to
create these forms (unless you’re a math genius, which I’m not). Sometimes forms seem
complex until you see the steps to generate them. Before switching shapes, I’ll develop an
algorithm to construct a brick tower (shown in Figure 13-13) that includes a crenellated
roof, using the existing Cube class.
The tower will be constructed of rings of bricks with a rotational offset between the brick
rows, allowing the vertically adjacent brick seams to alternate, like in a regular brick wall.
The first part of the algorithm generates a single ring of bricks; next, the graphics context
(the object that controls the actual drawing) is translated the distance equal to the height
of a brick and rotated one-half the ring rotation angle so that the vertical seams alternate.
Then the first step is repeated until some specified height is reached. Finally, the last step
involves creating the crenellation, which uses the same basic routine as step 1 (generating
a brick ring), but with every other brick removed (actually never added).
To generate the initial brick ring, I used the following trig expressions:
z = cos(radians(angle))*radius;
x = sin(radians(angle))*radius;
These expressions, which were discussed earlier in the chapter, can be used to generate a
regular polygon that is flipped to face the top and bottom of the display window. In addition
to placing the bricks in a ring in the xz plane, I also need to rotate the bricks so that
they are aligned with the ring path. I accomplished this by rotating each brick the same
number of degrees I displaced it to form the ring. To help understand this, imagine that
you have a brick floating in front of your eyes. The brick’s wide side is facing you, as if you
were looking straight at a building. Now displace the brick 90 degrees around your head
(but don’t rotate it), so that it’s next to your left ear. The brick’s end is now facing your ear.
To make the wide side of the brick again face your head (actually, your ear), you need to
rotate the brick around its center point the same number of degrees it was displaced (i.e.,
90 degrees).
Following is the code for the example, which requires the Cube and Point3D classes:
// Brick Tower
// Point3D and Cube classes required.
float bricksPerLayer = 16.0; // value must be even
float brickLayers = 25.0;
Cube[]bricks = new Cube[int(bricksPerLayer*brickLayers)];
float brickWidth = 60, brickHeight = 25, brickDepth = 25;
float radius = 150.0;
float angle = 0;
3D
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