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How to represent a camera in a 3D engine by a 3x3 matrix and a 3-vector

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· 11 months ago

This is a document I wrote some time ago that describes how a camera in a 3D engine can be represented by a 3x3 matrix and a 3-vector. This is the method that is used in the Crystal Space engine but it is general enough so that it can be used in other engines as well. This document is almost the exact mail that was sent so it is to be read as if you just asked the question to me ("how do you represent a 3D camera using a matrix and vector?") and I am answering you with this text :-)

Hi,

Ok, I will first try to give some general theory and try to make it clear to you and then I will apply this theory to answer your specific questions below.

A way to look at the matrix/vector representation of a camera is by seeing the matrix as a 3 dimension arrow pointing in some direction (the direction the camera is looking at) and the vector as the starting point of that arrow.

What does this matrix do? In fact it performs a linear transformation from 3D to 3D. With a 3x3 matrix you can represent every linear transformation from 3D to 3D. The matrix that we use for a camera is just a linear transformation matrix that transforms coordinates represented in one base to another. Let's assume that everything in the world is defined by using 3D vertices with an x, y, and z component. So a vertex is defined by three numbers: x, y, and z. These three numbers only have meaning when used relative to some base. A base is defined by three axis (if it is a 3D base that is).

So our camera matrix transforms 3D vertices from world space to camera space. This means that a vertex with position x,y,z in the world (in world space) will get new coordinates x',y',z' in camera space. The only reason that we want to apply this transformation is to make things easier for the rest of the 3D engine. Because after this transformation we can program the rest of the engine as if every vertex is represented in camera space. In other words; a vertex with coordinates (0,0,5) will be a vertex that lies just in front of the camera at distance 5. A vertex with coordinates (1,3,-5) is behind the plane of the camera and can thus be easily discarded. The test Z < 0 is thus an easy test to see if a vertex will be visible or not.

How does this transformation work? In fact it is just matrix algebra. For example, if the camera matrix is represented by M and the camera position is represented by P then we can write the equation to transform from world space to camera space by using:

C = M \cdot (W - P)

W is a 3D vector describing the position of a vertex in world space coordinates. C is a 3D vector describing the position of a vertex in camera space coordinates.

What this formula does is: first it translates the world space position so that the camera position is at (0,0,0). This is done by W-P. As you can see a vertex that would be on the same world space coordinates as the camera would be translated to (0,0,0). The result of this calculation is another 3D vector.

This vector is then multiplied by the camera matrix M to transform it to camera space. You can visualize this by treating M as an arrow pointing in some direction and the vertex lying somewhere relative to that arrow. By transforming with M (multiplying) we move the arrow until it points just the way we want it (with the Z-axis in front and so on).

A matrix by vector multiplication is defined as follows:

\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} a \cdot x + b \cdot y + c \cdot z \\ d \cdot x + e \cdot y + f \cdot z \\ g \cdot x + h \cdot y + i \cdot z \\ \end{bmatrix}

So for example, let's apply this formula in the initial configuration, with the camera pointing forwards in world space. The camera matrix is then equal to:

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}

and the vector is equal to

\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}

Translation of a vertex in world space coordinates to camera space coordinates makes no changes since the camera is at the origin of the world. Transformation results in the previous formula being applied:

\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 \cdot x + 0 \cdot y + 0 \cdot z \\ 0 \cdot x + 1 \cdot y + 0 \cdot z \\ 0 \cdot x + 0 \cdot y + 1 \cdot z \\ \end{bmatrix}

So as you can see, this does not change the vertex as it should be.

This is more or less the theory. If you have specific questions about this, don't hesitate to ask me.

Now I will try to answer your other questions.

Hi,
When I have initialized the matrix (as you sayd), how do you change it if you move the position/target of the camera. Are they the new 3-axis for the camera?

Ok, all the different kind of movements that you can do are again performed as transformations.

For example, let's say that you want to move forward a bit. If you would represent this movement in camera space than you would say that the camera moves from (0,0,0) (since the camera is at the origin in camera space) to (0,0,dist) with dist the distance that you want to move. This is because we defined camera space so that Z is in front of you.

But we want to know the position of the camera in world space! In fact, what we want to do is to transform the camera space position (0,0,dist) to world space. This would then be the new position for the camera.

So we need the inverse transformation. To calculate the inverse transformation you need to calculate the inverse of the matrix M. Let's call this inverse M'. We know that M * M' = I (the identity matrix).

Calculation of the inverse of a matrix is a bit complicated. If you are interested I can tell you how I do it but I don't really understand it myself (I just got the formulas from somewhere :-)

Starting from the equation:

M \cdot (W - P)

(our transformation from world to camera space) we would now like to calculate the new equation for the inverse transformation. We already have M' (the inverse matrix). Ok, let us multiply both sides of the equation by M'.

This gives:

M' \cdot C = M' \cdot M \cdot (W - P)

Since M' * M = I, this results in:

M' \cdot C = W - P

So the equation we are looking for is:

W = M' \cdot C + P

So this is the equation that transforms camera space to world space. Now we can use this to transform (0,0,dist) to the new camera space coordinates.

Other movements (like moving to the right (dist,0,0), upwards (0,dist,0), down (0,-dist,0) and so on) are all equivalent to this.

If you want to turn right then your position will not change but you will have to change the transformation matrix. This works differently. For example, to turn right you would want to rotate a certain angle around the Y axis (since the Y axis points upwards).

This rotation can be represented by the following matrix:

\begin{bmatrix} \cos(a) & 0 & -\sin(a) \\ 0 & 1 & 0 \\ \sin(a) & 0 & \cos(a) \\ \end{bmatrix}

(to see why this works, just try multiplying it by some vectors in 3D and see where they will transform too)

How can we then use this matrix to turn our camera to the right? An important thing to realize is that transformations can be combined by multiplying the matrices of the transformations. For example, if we have our matrix M transforming from world to camera space and we would like to apply the Y-axis rotation on the camera then you can see this as a combination of first the transformation from world to camera space followed by the rotation along the Y-axis. So instead of:

C = M \cdot (W - P)

we would want to do:

C = R \cdot M \cdot (W - P)

with R the rotation matrix.

Note that multiplication of matrices is not commutative. R \cdot M is not (always) the same as M \cdot R .

R \cdot M \cdot (W - P)

means: first apply transformation M on (W-P) and after that apply transformation R on that result.

So, we can conclude from this that we just have to multiply the camera matrix by R to get the new camera matrix:

M' = R \cdot M

Matrix multiplication works as follows:

\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} \begin{bmatrix} A & B & C \\ D & E & F \\ G & H & I \\ \end{bmatrix} = \begin{bmatrix} aA + bD + cG & aB + bE + cH & aC + bF + cI \\ dA + eD + fG & dB + eE + fH & dC + eF + fI \\ gA + hD + iG & gB + hE + iH & gC + hF + iI \\ \end{bmatrix}

Rotation along the other axis works similar. The rotation along the X axis is represented by:

\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(a) & -\sin(a) \\ 0 & \sin(a) & \cos(a) \\ \end{bmatrix}

The rotation along the Z axis is represented by:

\begin{bmatrix} \cos(a) & -\sin(a) & 0 \\ \sin(a) & \cos(a) & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}

With these formulas you should be able to do any movement of the camera that you want.