Processing: Creative Coding and Computational Art

Figure 13-13. Brick Tower sketch
This is a pretty long example, but I assure you you’ve seen it all before. I reused the
Point3D and Cube classes, which really simplified things. If you build your classes well, you
can continue to reuse or extend them, which will increase your efficiency over time. This is
one of the great benefits of OOP. The only somewhat complicated parts of the sketch are
the code that controls the alternating brick layer rotation and the crenellation code. Since
I knew what angle of rotation I needed to generate the brick ring, I wrote the expression
angle = 360.0/bricksPerLayer*i/2;
to control the alternating seams per each brick layer. Remember the order of operations
(which does make a difference): when the operators are of the same precedence level, the
expression is solved from left to right. This was the case with this expression, as multiplication
and division have the same precedence level. The expression evaluates alternating
values per loop cycle between an even multiple of the angle and one-half of an even multiple,
which gives me the alternating brick seams. (The actual angle value doesn’t really
matter.) Try removing the /2 from the expression to see what happens when the seams
don’t alternate.
The crenellation code uses the modulus operator, which I’ve discussed in numerous chapters
throughout the book; remember that it evaluates to the remainder of a division. I
used the operator so that every other brick would be drawn.
// add crenellation
if (i==brickLayers-1){
if (j%2 == 0){
bricks[j].create();
}
}
I guess with crenellation like this (one brick high), you’d need very tall bricks (or very short
soldiers).
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