This post is a continuation of a series of posts about WebGL. The first started with fundamentals and the previous was about 3D perspective projection. If you haven't read those please view them first.
In the last post we had to move the F in front of the frustum because the m4.perspective
function expects it to sit at the origin (0, 0, 0) and that objects in the frustum are -zNear
to -zFar
in front of it.
Moving stuff in front of the view doesn't seem the right way to go does it? In the real world you usually move your camera to take a picture of a building.
You don't usually move the buildings to be in front of the camera.
But in our last post we came up with a projection that requires things to be in front of the origin on the -Z axis. To achieve this what we want to do is move the camera to the origin and move everything else the right amount so it's still in the same place relative to the camera.
We need to effectively move the world in front of the camera. The easiest way to do this is to use an "inverse" matrix. The math to compute an inverse matrix in the general case is complex but conceptually it's easy. The inverse is the value you'd use to negate some other value. For example, the inverse of a matrix that translates in X by 123 is a matrix that translates in X by -123. The inverse of a matrix that scales by 5 is a matrix that scales by 1/5th or 0.2. The inverse of a matrix that rotates 30° around the X axis would be one that rotates -30° around the X axis.
Up until this point we've used translation, rotation and scale to affect the position and orientation of our 'F'. After multiplying all the matrices together we have a single matrix that represents how to move the 'F' from the origin to the place, size and orientation we want it. We can do the same for a camera. Once we have the matrix that tells us how to move and rotate the camera from the origin to where we want it we can compute its inverse which will give us a matrix that tells us how to move and rotate everything else the opposite amount which will effectively make it so the camera is at (0, 0, 0) and we've moved everything in front of it.
Let's make a 3D scene with a circle of 'F's like the diagrams above.
Here's the code.
function drawScene() {
var numFs = 5;
var radius = 200;
...
// Compute the matrix
var aspect = gl.canvas.clientWidth / gl.canvas.clientHeight;
var zNear = 1;
var zFar = 2000;
var projectionMatrix = m4.perspective(fieldOfViewRadians, aspect, zNear, zFar);
var cameraMatrix = m4.yRotation(cameraAngleRadians);
cameraMatrix = m4.translate(cameraMatrix, 0, 0, radius * 1.5);
// Make a view matrix from the camera matrix.
var viewMatrix = m4.inverse(cameraMatrix);
// move the projection space to view space (the space in front of
// the camera)
var viewProjectionMatrix = m4.multiply(projectionMatrix, viewMatrix);
// Draw 'F's in a circle
for (var ii = 0; ii < numFs; ++ii) {
var angle = ii * Math.PI * 2 / numFs;
var x = Math.cos(angle) * radius;
var z = Math.sin(angle) * radius;
// add in the translation for this F
var matrix = m4.translate(viewProjectionMatrix, x, 0, z);
// Set the matrix.
gl.uniformMatrix4fv(matrixLocation, false, matrix);
// Draw the geometry.
var primitiveType = gl.TRIANGLES;
var offset = 0;
var count = 16 * 6;
gl.drawArrays(primitiveType, offset, count);
}
}
Just after we compute our projection matrix you can see we compute a camera that goes around the 'F's like in the diagram above.
// Compute the camera's matrix
var cameraMatrix = m4.yRotation(cameraAngleRadians);
cameraMatrix = m4.translate(cameraMatrix, 0, 0, radius * 1.5);
We then compute a "view matrix" from the camera matrix. A "view matrix" is the matrix that moves everything the opposite of the camera effectively making everything relative to the camera as though the camera was at the origin (0,0,0)
// Make a view matrix from the camera matrix.
var viewMatrix = m4.inverse(cameraMatrix);
We then combine (multiply) those to make a viewProjection matrix.
// create a viewProjection matrix. This will both apply perspective
// AND move the world so that the camera is effectively the origin
var viewProjectionMatrix = m4.multiply(projectionMatrix, viewMatrix);
Finally we use that space as the starting space to place each `F'
var x = Math.cos(angle) * radius;
var z = Math.sin(angle) * radius;
var matrix = m4.translate(viewProjectionMatrix, x, 0, z);
In other words the viewProjection is the same for each F
. Same perspective,
same camera.
And voila! A camera that goes around the circle of 'F's. Drag the cameraAngle
slider
to move the camera around.
That's all fine but using rotate and translate to move a camera where you want it and point toward what you want to see is not always easy. For example if we wanted the camera to always point at a specific one of the 'F's it would take some pretty crazy math to compute how to rotate the camera to point at that 'F' while it goes around the circle of 'F's.
Fortunately there's an easier way. We can just decide where we want the camera and what we want it to point at and then compute a matrix that will put the camera there. Based on how matrices work this is surprisingly easy.
First we need to know where we want the camera. We'll call this the
cameraPosition
. Then we need to know the position of the thing we want
to look at or aim at. We'll call it the target
. If we subtract the
cameraPosition
from the target
we'll have a vector that points in the
direction we'd need to go from the camera to get to the target. Let's
call it zAxis
. Since we know the camera points in the -Z direction we
can subtract the other way cameraPosition - target
. We normalize the
results and copy it directly into the z
part of a matrix.
+----+----+----+----+ | | | | | +----+----+----+----+ | | | | | +----+----+----+----+ | Zx | Zy | Zz | | +----+----+----+----+ | | | | | +----+----+----+----+
This part of a matrix represents the Z axis. In this case the Z-axis of the camera. Normalizing a vector means making it a vector that represents 1.0. If you go back to the 2D rotation article where we talked about unit circles and how those helped with 2D rotation. In 3D we need unit spheres and a normalized vector represents a point on a unit sphere.
That's not enough info though. Just a single vector gives us a point on a unit sphere but which orientation from that point to orient things? We need to fill out the other parts of the matrix. Specifically the X axis and Y axis parts. We know that in general these 3 parts are perpendicular to each other. We also know that "in general" we don't point the camera straight up. Given that, if we know which way is up, in this case (0,1,0), We can use that and something called a "cross product" to compute the X axis and Y axis for the matrix.
I have no idea what a cross product means in mathematical terms. What I do know is that if you have 2 unit vectors and you compute the cross product of them you'll get a vector that is perpendicular to those 2 vectors. In other words, if you have a vector pointing south east, and a vector pointing up, and you compute the cross product you'll get a vector pointing either south west or north east since those are the 2 vectors that are perpendicular to south east and up. Depending on which order you compute the cross product in, you'll get the opposite answer.
In any case if we compute the cross product of our zAxis
and
up
we'll get the xAxis for the camera.
And now that we have the xAxis
we can cross the zAxis
and the xAxis
which will give us the camera's yAxis
Now all we have to do is plug the 3 axes into a matrix. That gives us a
matrix that will orient something that points at the target
from the
cameraPosition
. We just need to add in the position
+----+----+----+----+ | Xx | Xy | Xz | 0 | <- x axis +----+----+----+----+ | Yx | Yy | Yz | 0 | <- y axis +----+----+----+----+ | Zx | Zy | Zz | 0 | <- z axis +----+----+----+----+ | Tx | Ty | Tz | 1 | <- camera position +----+----+----+----+
Here's the code to compute the cross product of 2 vectors.
function cross(a, b) {
return [a[1] * b[2] - a[2] * b[1],
a[2] * b[0] - a[0] * b[2],
a[0] * b[1] - a[1] * b[0]];
}
Here's the code to subtract two vectors.
function subtractVectors(a, b) {
return [a[0] - b[0], a[1] - b[1], a[2] - b[2]];
}
Here's the code to normalize a vector (make it into a unit vector).
function normalize(v) {
var length = Math.sqrt(v[0] * v[0] + v[1] * v[1] + v[2] * v[2]);
// make sure we don't divide by 0.
if (length > 0.00001) {
return [v[0] / length, v[1] / length, v[2] / length];
} else {
return [0, 0, 0];
}
}
Here's the code to compute a "lookAt" matrix.
var m4 = {
lookAt: function(cameraPosition, target, up) {
var zAxis = normalize(
subtractVectors(cameraPosition, target));
var xAxis = normalize(cross(up, zAxis));
var yAxis = normalize(cross(zAxis, xAxis));
return [
xAxis[0], xAxis[1], xAxis[2], 0,
yAxis[0], yAxis[1], yAxis[2], 0,
zAxis[0], zAxis[1], zAxis[2], 0,
cameraPosition[0],
cameraPosition[1],
cameraPosition[2],
1,
];
},
And here is how we might use it to make the camera point at a specific 'F' as we move it.
...
// Compute the position of the first F
var fPosition = [radius, 0, 0];
// Use matrix math to compute a position on the circle.
var cameraMatrix = m4.yRotation(cameraAngleRadians);
cameraMatrix = m4.translate(cameraMatrix, 0, 50, radius * 1.5);
// Get the camera's position from the matrix we computed
var cameraPosition = [
cameraMatrix[12],
cameraMatrix[13],
cameraMatrix[14],
];
var up = [0, 1, 0];
// Compute the camera's matrix using look at.
var cameraMatrix = m4.lookAt(cameraPosition, fPosition, up);
// Make a view matrix from the camera matrix.
var viewMatrix = m4.inverse(cameraMatrix);
...
And here's the result.
Drag the slider and notice how the camera tracks a single 'F'.
Note that you can use "lookAt" math for more than just cameras. Common uses are making a character's
head follow someone. Making a turret aim at a target. Making an object follow a path. You compute
where on the path the target is. Then you compute where on the path the target would be a few moments
in the future. Plug those 2 values into your lookAt
function and you'll get a matrix that makes
your object follow the path and orient toward the path as well.
Before you go on you might want to check out this short note on matrix naming.
Otherwise let's learn about animation next.