Tag: "Computer Graphics"

The "Little Planet" Effect

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The "Little Planet" effect is the colloquial name often used to refer to the mathematical concept of a stereographic projection. The end result is quite impressive, especially considering the little amount of code that is actually required to create the image. All that is required is a panoramic image with a 360° view from side to side, or a photo-sphere, such as used with Google Earth to provide an immersive view of a location.

Golden Gate Bridge Stereographic Projection

Stereographic Projection

This is a mapping from a spherical position onto a plane. You will commonly see this type of projection in cartography; two examples are mapping the earth and planispheres (celestial charts). This projection is useful because it is not possible to map from a sphere to a plane without some type of distortion. The stereographic projection preserves angles and distorts areas. This trade-off is preferred for navigation, which is typically performed with angles.

The projection is typically performed from one of the poles. However, it can originate from any point on the sphere. For simplicity, unless otherwise stated I will refer to projections that originate at the North Pole of the sphere. For a unit-sphere located at the origin, this would be at \(\matrix{[0 & 0 & 1]}\)

The distortion of the image depends on the placement of the plane relative to the sphere. The upper hemisphere exhibits most of the distortion. The distortion becomes more extreme the closer the point in the sphere's surface approaches the origin of the projection. The projection extends to infinity as the projection's origin is undefined.

If the plane bisects the sphere at the equator, the lower hemisphere will be projected within an area the size of the circumference of the sphere, and the upper hemisphere is projection on the plane outside of the sphere.

When the plane is located at the surface of the sphere, opposite from the projection's origin, the lower hemisphere will project over an area that is twice that of the sphere's equator. The image below illustrates this configuration.

Stereographic Projection

We can reference any point on the spheres surface with two angles representing the latitude \(\phi\) and longitude \(\lambda\), where:

\(-\pi \lt \lambda \lt \pi, -\cfrac{\pi}{2} \lt \phi \lt \cfrac{\pi}{2} \)

The following image is a scaled down version of the panorama that I used to generate the stereographic projection of the Golden Gate Bridge at the beginning of the article. Normally we would index the pixels in this image with two variables, \(x\) (width) and \(y\) (height).

Golden Gate Bridge Panorama

We can simplify the math to map a full-view panorama to a sphere by normalizing the dimensions for both the sphere and our surface map; that is, to reduce the scale to the unit scale of one. This means we will perform the surface map operation on a unit-sphere, and the dimensions of our panorama will then span from: \(-1 \lt x \lt 1, -1 \lt y \lt 1\).

The following image shows how the coordinate system from the sphere maps to the image:

normalized coordinates

Project the sphere onto the Plane

We will create a ray, to calculate the projection of a single point from the sphere to the plane. This ray begins at the projective origin, passes through the sphere and points at any other point on the surface of the sphere. This ray continues and will intersect with the projective plane. This intersection is the point of projection. Alternatively, we can calculate the intersection point on the sphere's surface, given a ray that points between the projective origin and the projection plane. This demonstrates that the stereographic projection is a bijective operation; there is a one-to-one correspondence for the values on both surfaces.

The diagram below depicts the projection in two-dimensions:

Side Intersection

If \(\lambda_0\) as the central longitude and (\phi_1\) as the central latitude, this relationship can be stated mathematically as:

\( \eqalign{ u &= k \cos \phi \sin(\lambda - \lambda_0) \cr v &= k [ \cos \phi_1 \sin \phi - \sin \phi_1 \cos \phi \cos(\lambda - \lambda_0)]\cr & where \cr k &= \cfrac{2R}{1+ \sin \phi_1 \sin \phi + \cos \phi_1 \cos \phi \cos(\lambda - \lambda_0)} } \)

The term \(k\) determines where the projected plane is located.

Codez Plz

That is enough theory and explanation. Here is the code that I used to generate the stereographic projections of the Golden Gate Bridge and Las Vegas. The code presented below is adapted from a project I just completed that used the Computer Vision library OpenCV. The only important thing to note in the code below is that Mat objects are used to store images and pixels are represented with a type of std::vector. You should have no problem converting the pixel access operations from the code below to whatever image processing API that you are using.

If you do run into problems, leave a comment and I will help you work through porting the code.

There are three functions:

Main Projection

This function works by creating a ray between the projection origin and a pixel location on the projection plane. The intersection of the sphere's surface is calculated, which indicates the location to sample from the sphere's surface map. Because we are dealing with discrete, pixelated digital images, this sampling process creates visual artifacts. To help improve the smoothness of the image, we use a bilinear filter to average the values of four surrounding pixels of the sample location from the sphere.


void RenderProjection(Mat &pano, long len, Mat &output) {
  output.create(len, len, CV_16UC3)
  long half_len = len / 2;
  Size sz       = pano.size();  
  for (long indexX = 0; indexX < len; ++indexX) {
    for (long indexY = 0; indexY < len; ++indexY) {
      double sphereX = (indexX - half_len) * 10.0 / len;
      double sphereY = (indexY - half_len) * 10.0 / len;
      double Qx, Qy, Qz;
      if (GetIntersection(sphereX, sphereY, Qx, Qy, Qz))
        double theta  = std::acos(Qz);
        double phi    = std::atan2(Qy, Qx) + k_pi;
        theta          = theta * k_pi_inverse;
        phi            = phi   * (0.5 * k_pi_inverse);
        double Sx      = min(sz.width -2.0, sz.width  * phi);
        double Sy      = min(sz.height-2.0, sz.height * theta);
        output.at<Vec3s>(indexY, indexX) = BilinearSample(pano, Sx, Sy);

Calculate the Intersection

This calculation is an optimized reduction of the quadratic equation to calculate the intersection point on the surface of the sphere.


bool GetIntersection(double u, double v,
  double &x, double &y, double &z)
  double Nx    = 0.0;
  double Ny    = 0.0;
  double Nz    = 1.0;
  double dir_x = u - Nx;
  double dir_y = v - Ny;
  double dir_z = -1.0 - Nz;
  double a = (dir_x * dir_x) + (dir_y * dir_y) + (dir_z * dir_z);
  double b = (dir_x * Nx) + (dir_y * Ny) + (dir_z * Nz);
  b *= 2;
  double d = b*b;
  double q = -0.5 * (b - std::sqrt(d));
  double t = q / a;
  x = (dir_x * t) + Nx;
  y = (dir_y * t) + Ny;
  z = (dir_z * t) + Nz;
  return true;

Bilinear Filter

The bilinear filter calculates a weighted-sum of the four surrounding pixels for a digital image sample.


Vec3s BilinearSample(Mat &image, double x, double y) {
  Vec3s c00 = image.at<vec3s>(int(y), int(x));
  Vec3s c01 = image.at<vec3s>(int(u), int(x)+1);
  Vec3s c10 = image.at<vec3s>(int(y)+1, int(x));
  Vec3s c11 = image.at<vec3s>(int(y)+1, int(x)+1);
  double X0 = x - floor(x);
  double X1 = 1.0 - X0;
  double Y0 = y - floor(y);
  double Y1 = 1.0 - Y0;
  double w00 = X0 * Y0;
  double w01 = X1 * Y0;
  double w10 = X0 * Y1;
  double w11 = X1 * Y1;
  short r  = short(c00[2] * w00 + c01[2] * w01
                 + c10[2] * w10 + c11[2] * w11);
  short g  = short(c00[1] * w00 + c01[1] * w01
                 + c10[1] * w10 + c11[1] * w11);
  short b  = short(c00[0] * w00 + c01[0] * w01
                 + c10[0] * w10 + c11[0] * w11);
  return make_BGR(b, g, r);

...and a helper function:


Vec3s make_BGR(short blue, short green, short red)
  Vec3s result;
  result[0] = blue;
  result[1] = green;
  result[2] = red;
  return result;

Here is another sample of a stereographic projection and the panorama that I used to create it:

Las Vegas Stereographic Projection

Las Vegas Panorama


The stereographic projection has been known and used since the time of the ancient Greeks. It was heavily used in the Age of Exploration to create maps of the world where the distortion was applied to distances, and the relative angles between local points is preserved. When it is applied to full-view panoramas a neat effect is created called, "The Little Planet Effect." With just a little bit of theory and some concepts from computer graphics we were able to turn this concept into code with less than 100 lines of code.

3D Projection and Matrix Transforms

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My previous two entries have presented a mathematical foundation for the development and presentation of 3D computer graphics. Basic matrix operations were presented, which are used extensively with Linear Algebra. Understanding the mechanics and limitations of matrix multiplication is fundamental to the focus of this essay. This entry demonstrates three ways to manipulate the geometry defined for your scene, and also how to construct the projection transform that provides the three-dimensional effect for the final image.


There are three types of transformations that are generally used to manipulate a geometric model, translate, scale, and rotate. Your models and be oriented anywhere in your scene with these operations. Ultimately, a single square matrix is composed to perform an entire sequence of transformations as one matrix multiplication operation. We will get to that in a moment.

For the illustration of these transforms, we will use the two-dimensional unit-square shown below. However, I will demonstrate the math and algorithms for three dimensions.

\( \eqalign{ pt_{1} &= (\matrix{0 & 0})\\ pt_{2} &= (\matrix{1 & 0})\\ pt_{3} &= (\matrix{1 & 1})\\ pt_{4} &= (\matrix{0 & 1})\\ }\) Unit-square model


Translating a point in 3D space is as simple as adding the desired offset for the coordinate of each corresponding axes.

\( V_{translate}=V+D \)

For instance, to move our sample unit cube from the origin by \(x=2\) and \(y=3\), we simply add \(2\) to the x-component for every point in the square, and \(3\) to every y-component.

\( \eqalign{ pt_{1D} &= pt_1 + [\matrix{2 & 3}] \\ pt_{2D} &= pt_2 + [\matrix{2 & 3}] \\ pt_{3D} &= pt_3 + [\matrix{2 & 3}] \\ pt_{4D} &= pt_4 + [\matrix{2 & 3}] \\ }\) Translation

There is a better way, though.

That will have to wait until after I demonstrate the other two manipulation methods.

Identity Matrix

The first thing that we need to become familiar with is the identity matrix. The identity matrix has ones set along the diagonal with all of the other elements set to zero. Here is the identity matrix for a \(3 \times 3\) matrix.

\( \left\lbrack \matrix{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1} \right\rbrack\)

When any valid input matrix is multiplied with the identity matrix, the result will be the unchanged input matrix. Only square matrices can identity matrices. However, non-square matrices can be used for multiplication with the identity matrix if they have compatible dimensions.

For example:

\( \left\lbrack \matrix{a & b & c \\ d & e & f \\ g & h & i} \right\rbrack \left\lbrack \matrix{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1} \right\rbrack = \left\lbrack \matrix{a & b & c \\ d & e & f \\ g & h & i} \right\rbrack \)

\( \left\lbrack \matrix{7 & 6 & 5} \right\rbrack \left\lbrack \matrix{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1} \right\rbrack = \left\lbrack \matrix{7 & 6 & 5} \right\rbrack \)

\( \left\lbrack \matrix{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1} \right\rbrack \left\lbrack \matrix{7 \\ 6 \\ 5} \right\rbrack = \left\lbrack \matrix{7 \\ 6 \\ 5} \right\rbrack \)

We will use the identity matrix as the basis of our transformations.


It is possible to independently scale geometry along each of the standard axes. We first start with a \(2 \times 2\) identity matrix for our two-dimensional unit square:

\( \left\lbrack \matrix{1 & 0 \\ 0 & 1} \right\rbrack\)

As we demonstrated with the identity matrix, if we multiply any valid input matrix with the identity matrix, the output will be the same as the input matrix. This is similar to multiplying by \(1\) with scalar numbers.

\( V_{scale}=VS \)

Based upon the rules for matrix multiplication, notice how there is a one-to-one correspondence between each row/column in the identity matrix and for each component in our point..

\(\ \matrix{\phantom{x} & \matrix{x & y} \\ \phantom{x} & \phantom{x} \\ \matrix{x \\ y} & \left\lbrack \matrix{1 & 0 \\ 0 & 1} \right\rbrack } \)

We can create a linear transformation to scale point, by simply multiplying the \(1\) in the identity matrix, with the desired scale factor for each standard axis. For example, let's make our unit-square twice as large along the \(x-axis\) and half as large along the \(y-axis\):

\(\ S= \left\lbrack \matrix{2 & 0 \\ 0 & 0.5} \right\rbrack \)

This is the result, after we multiply each point in our square with this transformation matrix; I have also broken out the mathematic calculations for one of the points, \(pt_3(1, 1)\):

\( \eqalign{ pt_{3}S &= \\ &= [\matrix{1 & 1}] \left\lbrack\matrix{2 & 0 \\ 0 & 0.5} \right\rbrack\\ &= [\matrix{(1 \cdot 2 + 1 \cdot 0) & (1 \cdot 0 + 1 \cdot 0.5)}] \\ &= [\matrix{2 & 0.5}] }\) Scale

Here is the scale transformation matrix for three dimensions:

\(S = \left\lbrack \matrix{S_x & 0 & 0 \\ 0 & S_y & 0 \\ 0 & 0 & S_z }\right\rbrack\)


A rotation transformation can be created similarly to how we created a transformation for scale. However, this transformation is more complicated than the previous one. I am only going to present the formula, I am not going to explain the details of how it works. The matrix below will perform a counter-clockwise rotation about a pivot-point that is placed at the origin:

\(\ R= \left\lbrack \matrix{\phantom{-} \cos \Theta & \sin \Theta \\ -\sin \Theta & \cos \Theta} \right\rbrack \)


As I mentioned, the rotation transformation pivots the point about the origin. Therefore, if we attempt to rotate our unit square, the result will look like this:

The effects become even more dramatic if the object is not located at the origin:

In most cases, this may not be what we would like to have happen. A simpler way to reason about rotation is to place the pivot point in the center of the object, so it appears to rotate in place.

To rotate an object in place, we will need to first translate the object so the desired pivot point of the rotation is located at the origin, rotate the object, then undo the original translation to move it back to its original position.

Translate object's pivot point to the origin Rotate Translate the rotated object back to it's original position

While all of those steps must be taken in order to rotate an object about its center, there is a way to combine those three steps to create a single matrix. Then only one matrix multiplication operation will need to be performed on each point in our object. Before we get to that, let me show you how we can expand this rotation into three dimensions.

Imagine that our flat unit-square object existed in three dimensions. It is still defined to exist on the \(xy\) plane, but it has a third component added to each of its points to mark its location along the \(z-axis\). We simply need to adjust the definition of our points like this:

\( \eqalign{ pt_{1} &= (\matrix{0 & 0 & 0})\\ pt_{2} &= (\matrix{1 & 0 & 0})\\ pt_{3} &= (\matrix{1 & 1 & 0})\\ pt_{4} &= (\matrix{0 & 1 & 0})\\ }\) Unit-square in 3 dimensions

So, you would be correct to surmise that in order to perform this same \(xy\) rotation about the \(z-axis\) in three dimensions, we simply need to begin with a \(3 \times 3\) identity matrix before we add the \(xy\) rotation.

\( R_z = \left\lbrack \matrix{ \phantom{-}\cos \Theta & \sin \Theta & 0 \\ -\sin \Theta & \cos \Theta & 0 \\ 0 & 0 & 1 } \right\rbrack \) Rotate about z-axis

We can also adjust the rotation matrix to perform a rotation about the \(x\) and \(y\) axis, respectively:

R_x = \( \left\lbrack \matrix{ 1 & 0 & 0 \\ 0 & \phantom{-}\cos \Theta & \sin \Theta \\ 0 & -\sin \Theta & \cos \Theta } \right\rbrack \) Rotate about x-axis
\( R_y = \left\lbrack \matrix{ \cos \Theta & 0 & -\sin \Theta\\ 0 & 1 & 0 \\ \sin \Theta & 0 & \phantom{-}\cos \Theta } \right\rbrack \) Rotate about y-axis

Composite Transform Matrices

I mentioned that it is possible to combine a sequence of matrix transforms into a single matrix. This is highly desirable, considering that every point in an object model needs to be multiplied by the transform. What I demonstrated up to this point is how to individually create each of the transform types, translate, scale and rotate. Technically, the translate operation, \(V'=V+D\) is not a linear transformation because it is not composed within a matrix. Before we can continue, we must find a way to turn the translation operation from a matrix add to a matrix multiplication.

It turns out that we can create a translation operation that is a linear transformation by adding another row to our transformation matrix. However, a transformation matrix must be square because it starts with the identity matrix. So we must also add another column, which means we now have a \(4 \times 4\) matrix:

\(T = \left\lbrack \matrix{1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ t_x & t_y & t_z & 1 }\right\rbrack\)

A general transformation matrix is in the form:

\( \left\lbrack \matrix{ a_{11} & a_{12} & a_{13} & 0\\ a_{21} & a_{22} & a_{23} & 0\\ a_{31} & a_{32} & a_{33} & 0\\ t_x & t_y & t_z & 1 }\right\rbrack\)

Where \(t_x\),\(t_y\) and \(t_z\) are the final translation offsets, and the sub-matrix \(a\) is a combination of the scale and rotate operations.

However, this matrix is not compatible with our existing point definitions because they only have three components.

Homogenous Coordinates

The final piece that we need to finalize our geometry representation and coordinate management logic is called Homogenous Coordinates. What this basically means, is that we will add one more parameter to our vector definition. We refer to this fourth component as, \(w\), and we initially set it to \(1\).

\(V(x,y,z,w) = [\matrix{x & y & z & 1}]\)

Our vector is only valid when \(w=1\). After we transform a vector, \(w\) has often changed to some other scale factor. Therefore, we must rescale our vector back to its original basis by dividing each component by \(w\). The equations below show the three-dimensional Cartesian coordinate representation of a vector, where \(w \ne 0\).

\( \eqalign{ x&= X/w \\ y&= Y/w \\ z&= Z/w }\)

Creating a Composite Transform Matrix

We now possess all of the knowledge required to compose a linear transformation sequence that is stored in a single, compact transformation matrix. So let's create the rotation sequence that we discussed earlier. We will a compose a transform that translates an object to the origin, rotates about the \(z-axis\) then translates the object back to its original position.

Each individual transform matrix is shown below in the order that it must be multiplied into the final transform.

\( T_c R_z T_o = \left\lbrack \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -t_x & -t_y & -t_z & 1 \\ } \right\rbrack \left\lbrack \matrix{ \phantom{-}\cos \Theta & \sin \Theta & 0 & 0 \\ -\sin \Theta & \cos \Theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ } \right\rbrack \left\lbrack \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ t_x & t_y & t_z & 1 \\ } \right\rbrack \)

Here, values are selected and the result matrix is calculated:

\(t_x = 0.5, t_y = 0.5, t_z = 0.5, \Theta = 45°\)

\( \eqalign{ T_c R_z T_o &= \left\lbrack \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -0.5 & -0.5 & -0.5 & 1 \\ } \right\rbrack \left\lbrack \matrix{ 0.707 & 0.707 & 0 & 0 \\ -0.707 & 0.707 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ } \right\rbrack \left\lbrack \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0.5 & 0.5 & 0.5 & 1 \\ } \right\rbrack \\ \\ &=\left\lbrack \matrix{ \phantom{-}0.707 & 0.707 & 0 & 0 \\ -0.707 & 0.707 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0.5 & -0.207 & 0 & 1 \\ } \right\rbrack }\)

It is possible to derive the final composite matrix by using algebraic manipulation to multiply the transforms with the unevaluated variables. The matrix below is the simplified definition for \(T_c R_z T_o\) from above:

\( T_c R_z T_o = \left\lbrack \matrix{ \phantom{-}\cos \Theta & \sin \Theta & 0 & 0 \\ -\sin \Theta & \cos \Theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ (-t_x \cos \Theta + t_y \sin \Theta + t_x) & (-t_x \sin \Theta - t_y \cos \Theta + t_y) & 0 & 1 \\ } \right\rbrack\)

Projection Transformation

The last thing to do, is to convert our 3D model into an image. We have three-dimensional coordinates, that must be mapped to a two-dimensional surface. To do this, we will project a view of our world-space onto a flat two-dimensional screen. This is known as the "projection transformation" or "projection matrix".

Coordinate Systems

It may be already be apparent, but if not I will state it anyway; a linear transformation is a mapping between two coordinate systems. Whether we translate, scale or rotate we are simply changing the location and orientation of the origin. We have actually used the origin as our frame of reference, so our transformations appear to be moving our geometric objects. This is a valid statement, because the frame of reference determines many things.

We need a frame of reference that can be used as the eye or camera. We call this frame of reference the view space and the eye is located at the view-point. In fact, up to this point we have also used two other vector-spaces, local space (also called model space and world space. To be able to create our projection transformation, we need to introduce one last coordinate space called screen space. Screen space is a 2D coordinate system defined by the \(uv-plane\), and it maps directly to the display image or screen.

The diagram below depicts all four of these different coordinate spaces:

Coordinate Spaces

The important concept to understand, is that all of these coordinate systems are relatively positioned, scaled, and oriented to each other. We construct our transforms to move between these coordinate spaces. To render a view of the scene, we need to construct a transform to go from world space to view space.

View Space

The view space is the frame of reference for the eye. Basically, our view of the scene. This coordinate space allows us to transform the world coordinates to a position relative to our view point. This makes it possible for us to create a projection of this visible scene onto our view plane.

We need a point and two vectors to define the location and orientation of the view space. The method I demonstrate here is called the "eye, at, up" method. This is because we specify a point to be the location of the eye, a vector from the eye to the focal-point, at, and a vector that indicates which direction is up. I like this method because I think it is an intuitive model for visualizing where your view-point is located, how it is oriented, and what you are looking at.

View Spaces

Right or Left Handed?

The view space is a left-handed coordinate system, which differs from the right-handed coordinate systems that we have used up to this point. In math and physics, right-handed systems are the norm. This places the \(y-axis\) up, and the \(x-axis\) pointing to the right (same as the classic 2D Cartesian grid), and the resulting \(z-axis\) will point toward you. For a left-handed space, the positive \(z-axis\) points into the paper. This makes things very convenient in computer graphics, because \(z\) is typically used to measure and indicate depth.

You can tell which type of coordinate system that you have created by taking your open palm and pointing it perpendicularly to the \(x-axis\), and your palm facing the \(y-axis\). The positive \(z-axis\) will point in the direction of your thumb, based upon the direction your coordinate space is oriented.

Let's define a simple view space transformation.

\( \eqalign{ eye&= [\matrix{2 & 3 & 3}] \\ at&= [\matrix{0 & 0 & 0}] \\ up&= [\matrix{0 & 1 & 0}] }\)

Simple Indeed.

From this, we can create a view located at \([\matrix{2 & 3 & 3}]\) that is looking at the origin and is oriented vertically (no tilt to either side). We start by using the vector operations, which I demonstrated in my previous post, to define vectors for each of the three coordinate axes.

\( \eqalign{ z_{axis}&= \| at-eye \| \\ &= \| [\matrix{-2 & -3 & -3}] \| \\ &= [\matrix{-0.4264 & -0.6396 & -0.6396}] \\ \\ x_{axis}&= \| up \times z_{axis} \| \\ &=\| [\matrix{-3 & 0 & 2}] \|\\ &=[\matrix{-0.8321 & 0 & 0.5547}]\\ \\ y_{axis}&= \| z_{axis} \times x_{axis} \| \\ &= [\matrix{-0.3547 & -0.7687 & -0.5322}] }\)

We now take these three vectors, and plug them into this composite matrix that is designed to transform world space to a new coordinate space defined by axis-aligned unit vectors.

\( \eqalign { T_{view}&=\left\lbrack \matrix{ x_{x_{axis}} & x_{y_{axis}} & x_{z_{axis}} & 0 \\ y_{x_{axis}} & y_{y_{axis}} & y_{z_{axis}} & 0 \\ z_{x_{axis}} & z_{y_{axis}} & z_{z_{axis}} & 0 \\ -(x_{axis} \cdot eye) & -(y_{axis} \cdot eye) & -(z_{axis} \cdot eye) & 1 \\ } \right\rbrack \\ \\ &=\left\lbrack \matrix{ -0.8321 & -0.3547 & -0.4264 & 0 \\ 0 & -0.7687 & -0.6396 & 0 \\ -0.5547 & -0.5322 & -0.6396 & 0 \\ 3.3283 & 4.6121 & 4.6904 & 1 \\ } \right\rbrack }\)

Gaining Perspective

The final step is to project what we can see from the \(eye\) in our view space, onto the view-plane. As I mentioned previous, the view-plane is commonly referred to as the \(uv-plane\). The \(uv\) coordinate-system refers to the 2D mapping on the \(uv-plane\). The projection mapping that I demonstrate here places the \(uv-plane\) some distance \(d\) from the \(eye\), so that the view-space \(z-axis\) is parallel with the plane's normal.

The \(eye\) is commonly referred to by another name for the context of the projection transform, it is called the center of projection. That is because this final step, basically, projects the intersection of a projected vector onto the \(uv-plane\). The projected vector originates at the center of projection and points to each point in the view-space.

Perspective Intersection

The cues that let us perceive perspective with stereo-scopic vision is called, fore-shortening. This effect is caused by objects appearing much smaller as they are further from our vantage-point. Therefore, we can recreate the foreshortening in our image by scaling the \(xy\) values based on their \(z\)-coordinate. Basically, the farther away an object is placed from the center of projection, the smaller it will appear.

Scaling based on distance

The diagram above shows a side view of the \(y\) and \(z\) axes, and is parallel with the \(x-axis\). Notice the have similar triangles that are created by the view plane. The smaller triangle on the bottom is the distance \(d\), which is the distance from the center of projection to the view plane. The larger triangle on the top extends from the point \(Pv\) to the \(y-axis\), this distance is the \(z\)-component value for the point \(Pv\). Using the property of similar triangles we now have a geometric relationship that allows us to scale the location of the point for the view plane intersection.

\(\frac{x_s}d = \frac{x_v}{z_v}\)

\(\frac{y_s}d = \frac{y_v}{z_v}\)

\(x_s = \frac{x_v}{z_v/d}\)

\(y_s = \frac{y_v}{z_v/d}\)

With this, we can define a perspective-based projection transform:

\( T_{project}=\left\lbrack \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1/d \\ 0 & 0 & 0 & 0} \right\rbrack \)

This is now where the homogenizing component \(w\) returns. This transform will have the affect of scaling the \(w\) component to create:

\( \eqalign{ X&= x_v \\ Y&= y_v \\ Z&= z_v \\ w&= z_v/d }\)

Which is why we need to access these values as shown below after the following transform is used:

\( \eqalign{ [\matrix{X & Y & Z & w}] &= [\matrix{x_v & y_v & z_v & 1}] T_{project}\\ \\ x_v&= X/w \\ y_v&= Y/w \\ z_v&= Z/w }\)

Scale the Image Plane

Here is one last piece of information that I think you will appreciate. The projection transform above defines a \(uv\) coordinate system that is \(1 \times 1\). Therefore, if you make the following adjustment, you will scale the unit-plane to match the size of your final image in pixels:

\( T_{project}=\left\lbrack \matrix{ width & 0 & 0 & 0 \\ 0 & height & 0 & 0 \\ 0 & 0 & 1 & 1/d \\ 0 & 0 & 0 & 0} \right\rbrack \)


We made it! With a little bit of knowledge from both Linear Algebra and Trigonometry, we have been able to construct and manipulate geometric models in 3D space, concatenate transform matrices into compact and efficient transformation definitions, and finally create a projection transform to create a two-dimensional image from a three-dimensional definition.

As you can probably guess, this is only the beginning. There is so much more to explore in the creation of computer graphics. Some of the more interesting topics that generally follow after the 3D projection is created are:

  • Shading Models: Flat, Gouraud and Phong.
  • Shadows
  • Reflections
  • Transparency and Refraction
  • Textures
  • Bump-mapping ...

The list goes on and on. I imagine that I will continue to write on this topic and address polygon fill algorithms and the basic shading models. However, I think things will be more interesting if I have code to demonstrate and possibly an interactive viewer for the browser. I have a few demonstration programs written in C++ for Windows, but I would prefer to create something that is a bit more portable.

Once I decide how I want to proceed I will return to this topic. However, if enough people express interest, I wouldn't mind continuing forward demonstrating with C++ Windows programs. Let me know what you think.


Watt, Alan, "Three-dimensional geometry in computer graphics," in 3D Computer Graphics, Harlow, England: Addison-Wesley, 1993

Foley, James, D., et al., "Geometrical Transformations," in Computer Graphics: Principles and Practice, 2nd ed. in C, Reading: Addison-Wesley, 1996

Geometry in 3D Computer Graphics

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I previously introduced how to perform the basic matrix operations commonly used in 3D computer graphics. However, we haven’t yet reached anything that resembles graphics. I now introduce the concepts of geometry as it’s related to computer graphics. I intend to provide you with the remaining foundation necessary to be able to mathematically visualize the concepts at work. Unfortunately, we will still not be able to reach the "computer graphics" stage with this post; that must wait until the next entry.

This post covers a basic form of a polygonal mesh to represent models, points and Euclidean vectors. With the previous entry and this one, we will have all of the background necessary for me to cover matrix transforms in the next post. Specifically the 3D projection transform, which provides the illusion of perspective. This will be everything that we need to create a wireframe viewer. For now, let's focus on the topics of this post.

Modeling the Geometry

We will use models composed of a mesh of polygons. All of the polygons will be defined as triangles. It is important to consistently define the points of each triangle either clockwise or counter-clockwise. This is because the definition order affects the direction a surface faces. If a single model mixes the order of point definitions, your rendering will have missing polygons.

The order of point definition for the triangle can be ignored if a surface normal is provided for each polygon. However, to keep things simple, I will only be demonstrating with basic models that are built from an index array structure. This data structure contains two arrays. The first array is a list of all of the vertices (points) in the model. The second array contains a three-element structure with the index of the point that can be found in the first array. I define all of my triangles counter-clockwise while looking at the front of the surface.

Generally, your models should be defined at or near the origin. The rotation transformation that we will explore in the next post requires a pivot point, which is defined at the origin. Model geometries defined near the origin are also easier to work with when building models from smaller components and using instancing to create multiple copies of a single model.

Here is a definition for a single triangle starting at the origin and adjacent to the Y plane. The final corner decreases for Y as it moves diagonally along the XZ plane:

0: 0 0 0
1: 1 0 0
2: 0 1 0
A 0 1 2
Single Triangle Model

This definition represents a unit cube starting at the origin. A cube has 8 corners to hold six square-faces. It requires two triangles to represent each square. Therefore, there are 8 points used to define 12 triangles:

0: 0 0 0
1: 1 0 0
2: 1 1 0
3: 0 1 0
4: 0 1 1
5: 1 1 1
6: 1 0 1
7: 0 0 1
A 7 6 4
B 4 6 5
C 5 6 1
D 5 1 2
E 5 2 3
F 5 3 4
G 4 3 7
H 0 7 3
I 7 0 6
J 6 0 1
K 0 3 1
L 1 3 2
Cube Model (Front)
Cube Model (Rear)


Recall from my previous post that a vector is a special type of matrix, in which there is only a single row or column. You may be aware of a different definition for the term "vector"; something like

A quantity having direction as well as magnitude, especially determining the position of one point in space relative to another.

Oxford Dictionary

From this point forward, when I use the term "vector", I mean a Euclidean vector. We will even use single-row and single-column matrices to represent these vectors; we’re just going to refer to those structures as matrices to minimize the confusion.

Notational Convention

It is most common to see vectors represented as single-row matrices, as opposed to the standard mathematical notation which uses column matrices to represent vectors. It is important to remember this difference in notation based on what type of reference you are following.

The reason for this difference, is the row-form makes composing the transformation matrix more natural as the operations are performed left-to-right. I discuss the transformation matrix in the next entry. As you will see, properly constructing transformation matrices is crucial to achieve the expected results, and anything that we can do to simplify this sometimes complex topic, will improve our chances of success.


As the definition indicates, a vector has a direction and magnitude. There is a lot of freedom in that definition for representation, because it is like saying "I don’t know where I am, but I know which direction I am heading and how far I can go." Therefore, if we think of our vector as starting at the origin of a Cartesian grid, we could then represent a vector as a single point.

This is because starting from the origin, the direction of the vector is towards the point. The magnitude can be calculated by using Pythagoras' Theorem to calculate the distance between two points. All of this information can be encoded in a compact, single-row matrix. As the definition indicates, we derive a vector from two points, by subtracting one point from the other using inverse of matrix addition.

This will give us the direction to the destination point, relative to the starting point. The subtraction effectively acts as a translation of the vectors starting point to the origin.

For these first few vector examples, I am going to use 2D vectors. Because they are simpler to demonstrate and diagram, and the only difference to moving to 3D vectors is an extra field has been added.

Here is a simple example with two points (2, 1) and (5, 5). The first step is to encode them in a matrix:

\( {\bf S} = \left\lbrack \matrix{2 & 1} \right\rbrack, {\bf D} = \left\lbrack \matrix{5 & 5} \right\rbrack \)

Two points

Let’s create two vectors, one that determines how to get to point D starting from S, and one to get to point S starting from D. The first step is to subtract the starting point from the destination point. Notice that if we were to subtract the starting point from itself, the result is the origin, (0, 0).

\( \eqalign{ \overrightarrow{SD} &= [ \matrix{5 & 5} ] - [ \matrix{2 & 1} ] \cr &= [ \matrix{3 & 4} ] } \)

S-to-D translated Vector SD

\( \eqalign{ \overrightarrow{DS} &= [ \matrix{2 & 1} ] - [ \matrix{5 & 5} ] \cr &= [ \matrix{-3 & -4} ] } \)

D-to-S translated Vector DS

Notice that the only difference between the two vectors is the direction. The magnitude of each vector is the same, as expected. Here is how to calculate the magnitude of a vector:


\( \eqalign{ v &= \matrix{ [a_1 & a_2 & \ldots & a_n]} \cr |v| &= \sqrt{\rm (a_1^2 + a_2^2 + \ldots + a_n^2)} } \)


I covered scalar multiplication and matrix addition in my previous post. However, I wanted to briefly revisit the topic to demonstrate one way to think about vectors, which may help you develop a better intuition for the math. For the coordinate systems that we are working with, vectors are straight lines.

Two vectors

When you add two or more vectors together, you can think of them as moving the tail of each successive vector to the head of the previous vector. Then add the corresponding components.

\( \eqalign{ v &= A+B \cr |v| &= \sqrt{\rm 3^2 + 4^2} \cr &= 5 }\)

Visualization of adding vectors

Regardless of the orientation of the vector, it can be decomposed into its individual contributions per axis. In our model for the vector, the contribution for each axis is the raw value defined in our matrix.

3 vectors to add together Arrange vectors head-to-tail

\( v = A+B+C \)

The original vectors decomposed into axis aligned vectors The sum of the original 3 vectors

\( \eqalign{ |v| &= \sqrt{\rm (3+2-3)^2 + (1+3+2)^2} \cr &= \sqrt{\rm 2^2 + 6^2} \cr &= 2 \sqrt{\rm 10} }\)

Unit Vector

We can normalize a vector to create unit vector, which is a vector that has a length of 1. The concept of the unit vector is introduced to reduce a vector to simply indicate its direction. This type of vector can now be used as a unit of measure to which other vectors can be compared. This also simplifies the task of deriving other important information when necessary.

To normalize a vector, divide the vector by its magnitude. In terms of the matrix operations that we have discussed, this is simply scalar multiplication of the vector with the inverse of its magnitude.

\( u = \cfrac{v}{ |v| } \)

\(u\) now contains the information for the direction of \(v\). With algebraic manipulation we can also have:

\( v = u |v| \)

This basically says the vector \(v\) is equal to the unit vector that has the direction of \(u\) multiplied by the magnitude of \(v\).

Dot Product

The dot product is used for detecting visibility and shading of a surface. The dot product operation is shown below:

\( \eqalign{ w &= u \cdot v \cr &= u_1 v_1 + u_2 v_2 + \cdots + u_n v_n }\)

This formula may look familiar. That is because this is a special case in matrix multiplication called the inner product, which looks like this (remember, we are representing vectors as a row matrix):

\( \eqalign{ u \cdot v &= u v^T \cr &= \matrix {[u_1 & u_2 & \cdots & u_n]} \left\lbrack \matrix{v_1 \\ v_2 \\ \vdots \\ v_n } \right\rbrack }\)

We will use the dot product on two unit vectors to help us determine the angle between these vectors; that’s right, angle. We can do this because of the cosine rule from Trigonometry. The equation that we can use is as follows:

\( \cos \Theta = \cfrac{u \cdot v} {|u||v|} \)

To convert this to the raw angle, \(\Theta\), you can use the inverse function of cosine, arccos(\( \Theta \)). In C and C++ the name of the function is acos().

\( \cos {\arccos \Theta} = \Theta \)

So what is actually occurring when we calculate the dot product that allows us to get an angle?

Extract angle from dot product

We are essentially calculating the projection of u onto v. This gives us enough information to manipulate the Trigonometric properties to derive \( \Theta \).

Projection on unit vectors

I would like to make one final note regarding the dot product. Many times it is only necessary to know the sign of the angle between two vectors. The following table shows the relationship between the dot product of two vectors and the value of the angle, \( \Theta \), which is in the range \( 0° \text{ to } 180° \). If you like radians, that is \( 0 \text{ to } \pi \).

\( w = u \cdot v = f(u,w) \)

\( f(u,w) = \cases{ w \gt 0 & \text { if } \Theta \lt 90° \cr w = 0 & \text { if } \Theta = 90° \cr w \lt 0 & \text { if } \Theta \gt 90° }\)

Theta < 90 degrees, therefore w is positive Theta = 90 degrees, therefore w is zero Theta /> 90 degrees, therefore w is negative

Cross Product

The cross product is used to calculate a vector that is perpendicular to a surface. The name for this vector is the surface normal. The surface normal indicates which direction the surface is facing. Therefore, it is used quite extensively for shading and visibility as well.

We have reached the point where we need to start using examples and diagrams with three dimensions. We will use the three points that define our triangular surface to create two vectors. Remember, it is important to use consistent definitions when creating models.

The first step is to calculate our planar vectors. What does that mean? We want to create two vectors that are coplanar (on the same plane), so that we may use them to calculate the surface normal. All that we need to be able to do this are three points that are non-collinear, and we have this with each triangle definition in our polygonal mesh model.

If we have defined a triangle \(\triangle ABC\), we will define two vectors, \(u = \overrightarrow{AB}\) and \(v = \overrightarrow{AC}\).

A single triangle surface planar vectors derived from the surface definition

If we consistently defined the triangle surfaces in our model, we should now be able to take the cross product of \(u \times v\) to receive the surface normal, \(N_p\), of \(\triangle ABC\). Again, the normal vector is perpendicular to its surface.

Counter-clockwise cross product Clockwise cross product

In practice, this tells us which direction the surface is facing. We would combine this with the dot product to help us determine if are interested in this surface for the current view.

So how do you calculate the cross product?

Here is what you need to know in order to calculate the surface normal vector:

\(\eqalign { N_p &= u \times v \cr &= [\matrix{(u_2v_3 – u_3v_2) & (u_3v_1 – u_1v_3) & (u_1v_2 – u_2v_1)}] }\)

To help you remember how to perform the calculation, I have re-written it in parts, and used \(x\), \(y\), and \(z\) rather than the index values.

\(\eqalign { x &= u_yv_z – u_zv_y \cr y &= u_zv_x – u_xv_z \cr z &= u_xv_y – u_yv_x }\)

Let's work through an example to demonstrate the cross product. We will use surface A from our axis-aligned cube model that we constructed at the beginning of this post. Surface A is defined completely within the XY plane. How do we know this? Because all of the points have the same \(z\)-coordinate value of 1. Therefore, we should expect that our calculation for \(N_p\) of surface A will be an \(z\) axis-aligned vector pointing in the positive direction.

Step 1: Calculate our surface description vectors \(u\) and \(v\):

\( Surface_A = pt_7: (0,0,1), pt_6: (1,0,1), \text{ and } pt_4: (0,1,1) \)

\( \eqalign { u &= pt_6 - pt_7 \cr &= [\matrix{1 & 0 & 1}] - [\matrix{0 & 0 & 1}] \cr &=[\matrix{\color{red}{1} & \color{green}{0} & \color{blue}{0}}] }\)

\( \eqalign { v &= pt_4 - pt_7 \cr &= [\matrix{0 & 1 & 1}] - [\matrix{0 & 0 & 1}] \cr &=[\matrix{\color{red}{0} & \color{green}{1} & \color{blue}{0}}] }\)

Step 2: Calculate \(u \times v\):

\(N_p = [ \matrix{x & y & z} ]\text{, where:}\)

\( \eqalign { x &= \color{green}{u_y}\color{blue}{v_z} – \color{blue}{u_z}\color{green}{v_y} \cr &= \color{green}{0} \cdot \color{blue}{0} - \color{blue}{0} \cdot \color{green}{1} \cr &= 0 }\)

\( \eqalign { y &= \color{blue}{u_z}\color{red}{v_x} – \color{red}{u_x}\color{blue}{v_z} \cr &= \color{blue}{0} \cdot \color{red}{0} - \color{red}{1} \cdot \color{blue}{0} \cr &= 0 }\)

\( \eqalign { z &= \color{red}{u_x}\color{green}{v_y} – \color{green}{u_y}\color{red}{v_x} \cr &= \color{red}{1} \cdot \color{green}{1} - \color{green}{0} \cdot \color{red}{0} \cr &= 1 }\)

\(N_p = [ \matrix{0 & 0 & 1} ]\)

This is a \(z\) axis-aligned vector pointing in the positive direction, which is what we expected to receive.

Notation Formality

If you look at other resources to learn more about the cross product, it is actually defined as a sum in terms of three axis-aligned vectors \(i\), \(j\), and \(k\), which then must be multiplied by the vectors \(i=[\matrix{1&0&0}]\), \(j=[\matrix{0&1&0}]\), and \(k=[\matrix{0&0&1}]\) to get the final vector. The difference between what I presented and the form with the standard vectors, \(i\), \(j\), and \(k\), is I omitted a step by placing the result calculations directly in a new vector. I wanted you to be aware of this discrepancy between what I have demonstrated and what is typically taught in the classroom.

Apply What We Have Learned

We have discovered quite a few concepts between the two posts that I have written regarding the basic math and geometry involved with computer graphics. Let's apply this knowledge and solve a useful problem. Detecting if a surface is visible from a specific point-of-view is a fundamental problem that is used extensively in this domain. Since we have all of the tools necessary to solve this, let's run through two examples.

For the following examples we will specify the view-point as \(pt_{eye} = [\matrix{3 & 2 & 3}]:\)

View-point origin (eye)

Also, refer to this diagram for the surfaces that we will test for visibility from the view-point. The side surface is surface \(D\) from the cube model, and the hidden surface is the bottom-facing surface \(J\).

Surfaces to test for visibility

Detect a visible surface

Step 1: Calculate our surface description vectors \(u\) and \(v\):

\( \eqalign{ Surface_D: &pt_5: (1,1,1), \cr &pt_1: (1,0,0), \cr &pt_2: (1,1,0) } \)

\( \eqalign { u &= pt_1 - pt_5 \cr &= [\matrix{1 & 0 & 0}] - [\matrix{1 & 1 & 1}] \cr &=[\matrix{0 & -1 & -1}] }\)

\( \eqalign { v &= pt_2 - pt_5 \cr &= [\matrix{1 & 1 & 0}] - [\matrix{1 & 1 & 1}] \cr &=[\matrix{0 & 0 & -1}] }\)

Step 2: Calculate \(u \times v\):

\(N_{pD} = [ \matrix{x & y & z} ]\text{, where:}\)

\( \eqalign { x &= -1 \cdot (-1) - (-1) \cdot 0 \cr &= 1 }\)

\( \eqalign { y &= -1 \cdot 0 - (-1) \cdot 0 \cr &= 0 }\)

\( \eqalign { z &= 0 \cdot 0 - (-1) \cdot 0 \cr &= 0 }\)

\(N_pD = [ \matrix{1 & 0 & 0} ]\)

Step 3: Calculate vector to the eye:

To calculate a vector to the eye, we need to select a point on the target surface and create a vector to the eye. For this example, we will select \(pt_5\)

\( \eqalign{ \overrightarrow{view_D} &= pt_{eye} - pt_5\cr &= [ \matrix{3 & 2 & 3} ] - [ \matrix{1 & 1 & 1} ]\cr &= [ \matrix{2 & 1 & 2} ] }\)

Step 4: Normalize the view-vector:

Before we can use this vector in a dot product, we must normalize the vector:

\( \eqalign{ |eye_D| &= \sqrt{\rm 2^2 + 1^2 + 2^2} \cr &= \sqrt{\rm 4 + 1 + 4} \cr &= \sqrt{\rm 9} \cr &= 3 \cr \cr eye_u &= \cfrac{eye_D}{|eye_D|} \cr &= \cfrac{1}3 [\matrix{2 & 1 & 2}] \cr &= \left\lbrack\matrix{\cfrac{2}3 & \cfrac{1}3 & \cfrac{2}3} \right\rbrack }\)

Step 5: Calculate the dot product of the view-vector and surface normal:

\( \eqalign{ w &= eye_{u} \cdot N_{pD} \cr &= \cfrac{2}3 \cdot 1 + \cfrac{1}3 \cdot 0 + \cfrac{2}3 \cdot 0 \cr &= \cfrac{2}3 }\)

Step 6: Test for visibility:

\(w \gt 0\), therefore \(Surface_D\) is visible.

Side Surface D

Detect a surface that is not visible

Step 1: Calculate our surface description vectors \(u\) and \(v\):

\( \eqalign{ Surface_J: &pt_6: (1,1,1), \cr &pt_0: (0,0,0), \cr &pt_1: (1,0,0) } \)

\( \eqalign { u &= pt_0 - pt_6 \cr &= [\matrix{0 & 0 & 0}] - [\matrix{1 & 0 & 1}] \cr &=[\matrix{-1 & 0 & -1}] }\)

\( \eqalign { v &= pt_1 - pt_6 \cr &= [\matrix{1 & 0 & 0}] - [\matrix{1 & 0 & 1}] \cr &=[\matrix{0 & 0 & -1}] }\)

Step 2: Calculate \(u \times v\):

\(N_{pJ} = [ \matrix{x & y & z} ]\text{, where:}\)

\( \eqalign { x &= 0 \cdot (-1) - (-1) \cdot 0 \cr &= 0 }\)

\( \eqalign { y &= -1 \cdot 0 - (-1) \cdot (-1) \cr &= -1 }\)

\( \eqalign { z &= (-1) \cdot 0 - 0 \cdot 0 \cr &= 0 }\)

\(N_pJ = [ \matrix{0 & -1 & 0} ]\)

Step 3: Calculate vector to the eye:

For this example, we will select \(pt_6\)

\( \eqalign{ \overrightarrow{view_J} &= pt_{eye} - pt_6\cr &= [ \matrix{3 & 2 & 3} ] - [ \matrix{1 & 0 & 1} ]\cr &= [ \matrix{2 & 2 & 2} ] }\)

Step 4: Normalize the view-vector:

Before we can use this vector in a dot product, we must normalize the vector:

\( \eqalign{ |eye_J| &= \sqrt{\rm 2^2 + 2^2 + 2^2} \cr &= \sqrt{\rm 4 + 4 + 4} \cr &= \sqrt{\rm 12} \cr &= 2 \sqrt{\rm 3} \cr \cr eye_u &= \cfrac{eye_J}{|eye_J|} \cr &= \cfrac{1}{2 \sqrt 3} [\matrix{2 & 2 & 2}] \cr &= \left\lbrack\matrix{\cfrac{1}{\sqrt 3} & \cfrac{1}{\sqrt 3} & \cfrac{1}{\sqrt 3}} \right\rbrack }\)

Step 5: Calculate the dot product of the view-vector and surface normal:

\( \eqalign{ w &= eye_{u} \cdot N_{pJ} \cr &= \cfrac{1}{\sqrt 3} \cdot 0 + \cfrac{1}{\sqrt 3} \cdot (-1) + \cfrac{1}{\sqrt 3} \cdot 0 \cr &= -\cfrac{1}{\sqrt 3} }\)

Step 6: Test for visibility:

\(w \lt 0\), therefore \(Surface_J\) is not visible.

Bottom Surface J

What has this example demonstrated?

Surprisingly, this basic task demonstrates every concept that we have learned from the two posts up to this point:

  • Matrix Addition: Euclidean vector creation
  • Scalar Multiplication: Unit vector calculation, multiplying by \(1 / |v| \).
  • Matrix Multiplication: Dot Product calculation, square-matrix multiplication will be used in the next post.
  • Geometric Model Representation: The triangle surface representation provided the points for our calculations.
  • Euclidean Vector: Calculation of each surfaces planar vectors, as well as the view vector.
  • Cross Product: Calculation of the surface normal, \(N_p\).
  • Unit Vector: Preparation of vectors for dot product calculation.
  • Dot Product: Used the sign from this calculation to test for visibility of a surface.


Math can be difficult sometimes, especially with all of the foreign notation. However, with a bit of practice the notation will soon become second nature and many calculations can be performed in your head, similar to basic arithmetic with integers. While many graphics APIs hide these calculations from you, don't be fooled because they still exist and must be performed to display the three-dimensional images on your screen. By understanding the basic concepts that I have already demonstrated, you will be able to better troubleshoot your graphics programs when something wrong occurs. Even if you rely solely on a graphics library to abstract the complex mathematics.

There is only one more entry to go, and we will have enough basic knowledge of linear algebra, Euclidean geometry, and computer displays to be able to create, manipulate and display 3D models programmatically. The next entry discusses how to translate, scale and rotate geometric models. It also describes what must occur to create the illusion of a three-dimensional image on a two-dimensional display with the projection transform.


Watt, Alan, "Three-dimensional geometry in computer graphics," in 3D Computer Graphics, Harlow, England: Addison-Wesley, 1993

Introduction to the Math of Computer Graphics

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Three-dimensional computer graphics is an exciting aspect of computing because of the amazing visual effects that can be created for display. All of this is created from an enormous number of calculations that manipulate virtual models, which are constructed from some form of geometric definition. While the math involved for some aspects of computer graphics and animation can become quite complex, the fundamental mathematics that is required is very accessible. Beyond learning some new mathematic notation, a basic three-dimensional view can be constructed with algorithms that only require basic arithmetic and a little bit of trigonometry.

I demonstrate how to perform some basic operations with two key constructs for 3D graphics, without the rigorous mathematic introduction. Hopefully the level of detail that I use is enough for anyone that doesn't have a strong background in math, but would very much like to play with 3D APIs. I introduce the math that you must be familiar with in this entry, and in two future posts I demonstrate how to manipulate geometric models and apply the math towards the display of a 3D image.

Linear Algebra

The type of math that forms the basis of 3D graphics is called linear algebra. In general, linear algebra is quite useful for solving systems of linear equations. Rather than go into the details of linear algebra, and all of its capabilities, we are going to simply focus on two related constructs that are used heavily within the topic, the matrix and the vertex.


The Matrix provides an efficient mechanism to translate, scale, rotate, and convert between different coordinate systems. This is used extensively to manipulate the geometric models for environment calculations and display.

These are all examples of valid matrices:

\( \left\lbrack \matrix{a & b & c \cr d & e & f} \right\rbrack, \left\lbrack \matrix{10 & -27 \cr 0 & 13 \cr 7 & 17} \right\rbrack, \left\lbrack \matrix{x^2 & 2x \cr 0 & e^x} \right\rbrack\)

Square matrices are particularly important, that is a matrix with the same number of columns and rows. There are some operations that can only be performed on a square matrix, which I will introduce shortly. The notation for matrices, in general, uses capital letters as the variable names. Matrix dimensions can be specified as a shorthand notation, and to also identify an indexed position within the matrix. As far as I am aware, row-major indexing is always used for consistency, that is, the first index represents the row, and the second represents the column.

\( A= [a_{ij}]= \left\lbrack \matrix{a_{11} & a_{12} & \ldots & a_{1n} \cr a_{21} & a_{22} & \ldots & a_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr a_{m1} & a_{m2} & \ldots & a_{mn} } \right\rbrack \)


The vector is a special case of a matrix, where there is only a single row or a single column; also a common practice is to use lowercase letters to represent vectors:

\( u= \left\lbrack \matrix{u_1 & u_2 & u_3} \right\rbrack \), \( v= \left\lbrack \matrix{v_1\cr v_2\cr v_3} \right\rbrack\)


The mathematical shorthand notation for matrices is very elegant. It simplifies these equations that would be quite cumbersome, otherwise. There are only a few operations that we are concerned with to allow us to get started. Each operation has a basic algorithm to follow to perform a calculation, and the algorithm easily scales with the dimensions of a matrix.

Furthermore, the relationship between math and programming is not always clear. I have a number of colleagues that are excellent programmers and yet they do not consider their math skills very strong. I think that it could be helpful to many to see the conversion of the matrix operations that I just described from mathematical notation into code form. Once I demonstrate how to perform an operation with the mathematical notation and algorithmic steps, I will also show a basic C++ implementation of the concept to help you understand how these concepts map to actual code.

The operations that I implement below assume a Matrix class with the following interface:


class Matrix
  // Ctor and Dtor omitted
  // Calculate and return a reference to the specified element.
  double& element(size_t row, size_t column);
  // Resizes this Matrix to have the specified size.
  void resize(size_t row, size_t column);
  // Returns the number rows.
  size_t rows();
  // Returns the number of columns.
  size_t columns();
  std::vector<double>    data;


Transpose is a unary operation that is performed on a single matrix, and it is represented by adding a superscript T to target matrix.

For example, the transpose of a matrix A is represented with AT.

The transpose "flips" the orientation of the matrix so that each row becomes a column, and each original column is transformed into a row. I think that a few examples will make this concept more clear.

\( A= \left\lbrack \matrix{a & b & c \cr d & e & f \cr g & h & i} \right\rbrack \)

\(A^T= \left\lbrack \matrix{a & d & g \cr b & e & h \cr c & f & i} \right\rbrack \)

The resulting matrix will contain the same set of values as the original matrix, only their position in the matrix changes.

\( B= \left\lbrack \matrix{1 & 25 & 75 & 100\cr 0 & 5 & -50 & -25\cr 0 & 0 & 10 & 22} \right\rbrack \)

\(B^T= \left\lbrack \matrix{1 & 0 & 0 \cr 25 & 5 & 0 \cr 75 & -50 & 10 \cr 100 & -25 & 22} \right\rbrack \)

It is very common to transpose a matrix, including vertices, before performing other operations. The reason will become clear for matrix multiplication.

\( u= \left\lbrack \matrix{u_1 & u_2 & u_3} \right\rbrack \)

\(u^T= \left\lbrack \matrix{u_1 \cr u_2 \cr u_3} \right\rbrack \)

Matrix Addition

Addition can only be performed between two matrices that are the same size. By size, we mean that the number of rows and columns are the same for each matrix. Addition is performed by adding the values of the corresponding positions of both matrices. The sum of the values creates a new matrix that is the same size as the original two matrices provided to the add operation.


\( A= \left\lbrack \matrix{0 & -2 & 4 \cr -6 & 8 & 0 \cr 2 & -4 & 6} \right\rbrack, B= \left\lbrack \matrix{1 & 3 & 5 \cr 7 & 9 & 11 \cr 13 & 15 & 17} \right\rbrack \)


\(A+B=\left\lbrack \matrix{1 & 1 & 9 \cr 1 & 17 & 11 \cr 15 & 11 & 23} \right\rbrack \)

The size of these matrices do not match in their current form.

\(U= \left\lbrack \matrix{4 & -8 & 5}\right\rbrack, V= \left\lbrack \matrix{-1 \\ 5 \\ 4}\right\rbrack \)

However, if we take the transpose of one of them, their sizes will match and they can then be added together. The size of the result matrix depends upon which matrix we perform the transpose operation on from the original expression.:

\(U^T+V= \left\lbrack \matrix{3 \\ -3 \\ 9} \right\rbrack \)


\(U+V^T= \left\lbrack \matrix{3 & -3 & 9} \right\rbrack \)

Matrix addition has the same algebraic properties as with the addition of two scalar values:

Commutative Property:

\( A+B=B+A \)

Associative Property:

\( (U+V)+W=U+(V+W) \)

Identity Property:

\( A+0=A \)

Inverse Property:

\( A+(-A)=0 \)

The code required to implement matrix addition is relatively simple. Here is an example for the Matrix class definition that I presented earlier:


void Matrix::operator+=(const Matrix& rhs)
  if (rhs.data.size() == data.size())
    // We can simply add each corresponding element
    // in the matrix element data array.
    for (size_t index = 0; index < data.size(); ++index)
      data[index] += rhs.data[index];
Matrix operator+( const Matrix& lhs,
                  const Matrix& rhs)
  Matrix result(lhs);
  result += rhs;
  return result;

Scalar Multiplication

Scalar multiplication allows a single scalar value to be multiplied with every entry within a matrix. The result matrix is the same size as the matrix provided to the scalar multiplication expression:


\( A= \left\lbrack \matrix{3 & 6 & -9 \cr 12 & -15 & -18} \right\rbrack \)


\( \frac{1}3 A= \left\lbrack \matrix{1 & 2 & -3 \cr 4 & -5 & -6} \right\rbrack, 0A= \left\lbrack \matrix{0 & 0 & 0 \cr 0 & 0 & 0} \right\rbrack, -A= \left\lbrack \matrix{-3 & -6 & 9 \cr -12 & 15 & 18} \right\rbrack, \)

Scalar multiplication with a matrix exhibits these properties, where c and d are scalar values:

Distributive Property:

\( c(A+B)=cA+cB \)

Identity Property:

\( 1A=A \)

The implementation for scalar multiplication is even simpler than addition.
Note: this implementation only allows the scalar value to appear before the Matrix object in multiplication expressions, which is how the operation is represented in math notation.:


void Matrix::operator*=(const double lhs)
  for (size_t index = 0; index < data.size(); ++index)
    data[index] *= rhs;
Matrix operator*( const double scalar,
                  const Matrix& rhs)
  Matrix result(rhs);
  result *= scalar;
  return result;

Matrix Multiplication

Everything seems very simple with matrices, at least once you get used to the new structure. Then you are introduced to matrix multiplication. The algorithm for multiplication is not difficult, however, it is much more labor intensive compared to the other operations that I have introduced to you. There are also a few more restrictions on the parameters for multiplication to be valid. Finally, unlike the addition operator, the matrix multiplication operator does not have all of the same properties as the multiplication operator for scalar values; specifically, the order of parameters matters.

Input / Output

Let's first address what you need to be able to multiply matrices, and what type of matrix you can expect as output. Once we have addressed the structure, we will move on to the process.

Given an the following two matrices:

\( A= \left\lbrack \matrix{a_{11} & \ldots & a_{1n} \cr \vdots & \ddots & \vdots \cr a_{m1} & \ldots & a_{mn} } \right\rbrack, B= \left\lbrack \matrix{b_{11} & \ldots & b_{1v} \cr \vdots & \ddots & \vdots \cr b_{u1} & \ldots & b_{uv} } \right\rbrack \)

A valid product for \( AB=C \) is only possible if number of columns \( n \) in \( A \) is equal to the number of rows \( u \) in \( B \). The resulting matrix \( C \) will have the dimensions \( m \times v \).

\( AB=C= \left\lbrack \matrix{c_{11} & \ldots & c_{1v} \cr \vdots & \ddots & \vdots \cr c_{m1} & \ldots & c_{mv} } \right\rbrack, \)

Let's summarize this in a different way, hopefully this arrangement will make the concept more intuitive:

Matrix multiplication dimensions

One last form of the rules for matrix multiplication:

  • The number of columns, \(n\), in matrix \( A \) must be equal to the number of rows, \(u\), in matrix \( B \):
    \( n = u \)

  • The output matrix \( C \) will have the number of rows, \(m\), in \(A\), and the number of columns, \(v\), in \(B\):
    \( m \times v \)

  • \(m\) and \(v\) do not have to be equal. The only requirement is that they are both greater-than zero:
    \( m \gt 0,\)
    \(v \gt 0 \)

How to Multiply

To calculate a single entry in the output matrix, we must multiply the element from each column in the first matrix, with the element in the corresponding row in the second matrix, and add all of these products together. We use the same row in the first matrix, \(A\), for which we are calculating the row element in \(C\). Similarly, we use the column in the second matrix, \(B\) that corresponds with the calculating column element in \(C\).

More succinctly, we can say we are multiplying rows into columns.

For example:

\( A= \left\lbrack \matrix{a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23}} \right\rbrack, B= \left\lbrack \matrix{b_{11} & b_{12} & b_{13} & b_{14} \cr b_{21} & b_{22} & b_{23} & b_{24} \cr b_{31} & b_{32} & b_{33} & b_{34} } \right\rbrack \)

The number of columns in \(A\) is \(3\) and the number of rows in \(B\) is \(3\), therefore, we can perform this operation. The size of the output matrix will be \(2 \times 4\).

This is the formula to calculate the element \(c_{11}\) in \(C\) and the marked rows used from \(A\) and the columns from \(B\):

\( \left\lbrack \matrix{\color{#B11D0A}{a_{11}} & \color{#B11D0A}{a_{12}} & \color{#B11D0A}{a_{13}} \cr a_{21} & a_{22} & a_{23}} \right\rbrack \times \left\lbrack \matrix{\color{#B11D0A}{b_{11}} & b_{12} & b_{13} & b_{14} \cr \color{#B11D0A}{b_{21}} & b_{22} & b_{23} & b_{24} \cr \color{#B11D0A}{b_{31}} & b_{32} & b_{33} & b_{34} } \right\rbrack = \left\lbrack \matrix{\color{#B11D0A}{c_{11}} & c_{12} & c_{13} & c_{14}\cr c_{21} & c_{22} & c_{23} & c_{24} } \right\rbrack \)

\( c_{11}= (a_{11}\times b_{11}) + (a_{12}\times b_{21}) + (a_{13}\times b_{31}) \)

To complete the multiplication, we need to calculate these other seven values \( c_{12}, c_{13}, c_{14}, c_{21}, c_{22}, c_{23}, c_{24}\). Here is another example for the element \(c_{23}\):

\( \left\lbrack \matrix{ a_{11} & a_{12} & a_{13} \cr \color{#B11D0A}{a_{21}} & \color{#B11D0A}{a_{22}} & \color{#B11D0A}{a_{23}} } \right\rbrack \times \left\lbrack \matrix{b_{11} & b_{12} & \color{#B11D0A}{b_{13}} & b_{14} \cr b_{21} & b_{22} & \color{#B11D0A}{b_{23}} & b_{24} \cr b_{31} & b_{32} & \color{#B11D0A}{b_{33}} & b_{34} } \right\rbrack = \left\lbrack \matrix{c_{11} & c_{12} & c_{13} & c_{14}\cr c_{21} & c_{22} & \color{#B11D0A}{c_{23}} & c_{24} } \right\rbrack \)

\( c_{23}= (a_{21}\times b_{13}) + (a_{22}\times b_{23}) + (a_{23}\times b_{33}) \)

Notice how the size of the output matrix changes. Based on this and the size of the input matrices you can end up with some interesting results:

\( \left\lbrack \matrix{a_{11} \cr a_{21} \cr a_{31} } \right\rbrack \times \left\lbrack \matrix{ b_{11} & b_{12} & b_{13} } \right\rbrack = \left\lbrack \matrix{c_{11} & c_{12} & c_{13} \cr c_{21} & c_{22} & c_{23} \cr c_{31} & c_{32} & c_{33} } \right\rbrack \)

\( \left\lbrack \matrix{ a_{11} & a_{12} & a_{13} } \right\rbrack \times \left\lbrack \matrix{b_{11} \cr b_{21} \cr b_{31} } \right\rbrack = \left\lbrack \matrix{c_{11} } \right\rbrack \)


To help you keep track of which row to use from the first matrix and which column from the second matrix, create your result matrix of the proper size, then methodically calculate the value for each individual element.

The algebraic properties for the matrix multiplication operator do not match those of the scalar multiplication operator. These are the most notable:

Not Commutative:

The order of the factor matrices definitely matters.

\( AB \ne BA \) in general

I think it is very important to illustrate this fact. Here is a simple \(2 \times 2 \) multiplication performed two times with the order of the input matrices switched. I have highlighted the only two terms that the two resulting answers have in common:

\( \left\lbrack \matrix{a & b \cr c & d } \right\rbrack \times \left\lbrack \matrix{w & x\cr y & z } \right\rbrack = \left\lbrack \matrix{(\color{red}{aw}+by) & (ax+bz)\cr (cw+dy) & (cx+\color{red}{dz}) } \right\rbrack \)

\( \left\lbrack \matrix{w & x\cr y & z } \right\rbrack \times \left\lbrack \matrix{a & b \cr c & d } \right\rbrack = \left\lbrack \matrix{(\color{red}{aw}+cx) & (bw+dx)\cr (ay+cz) & (by+\color{red}{dz}) } \right\rbrack \)

Product of Zero:

\( AB=0 \) does not necessarily imply \( A=0 \) or \( B=0 \)

Scalar Multiplication is Commutative:

\( (kA)B = k(AB) = A(kB) \)

Associative Property:

Multiplication is associative, however, take note that the relative order for all of the matrices remains the same.

\( (AB)C=A(BC) \)

Transpose of a Product:

The transpose of a product is equivalent to the product of transposed factors multiplied in reverse order

\( (AB)^T=B^TA^T \)


We are going to use a two-step solution to create a general purpose matrix multiplication solution. The first step is to create a function that properly calculates a single element in the output matrix:


double Matrix::multiply_element(
  const Matrix& rhs,
  const Matrix& rhs,
  const size_t i,
  const size_t j
  double product = 0;
  // Multiply across the specified row, i, for the left matrix
  // and the specified column, j, for the right matrix.
  // Accumulate the total of the products
  // to return as the calculated result.
  for (size_t col_index = 1; col_index <= lhs.columns(); ++col_index)
    for (size_t row_index = 1; row_index <= rhs.rows(); ++row_index)
      product += lhs.element(i, col_index)
               * rhs.element(row_index, j);
  return product;

Now create the outer function that performs the multiplication to populate each field of the output matrix:


// Because we may end up creating a matrix with
// an entirely new size, it does not make sense
// to have a *= operator for this general-purpose solution.
Matrix& Matrix::operator*( const Matrix& lhs,
                           const Matrix& rhs)
  if (lhs.columns() == rhs.rows())
    // Resize the result matrix to the proper size.
    this->resize(lhs.row(), rhs.columns());
    // Calculate the value for each element
    // in the result matrix.
    for (size_t i = 1; i <= this->rows(); ++i)
      for (size_t j = 1; j <= this->columns(); ++j)
        element(i,j) = multiply_element(lhs, rhs, i, j);
  return *this;


3D computer graphics relies heavily on the concepts found in the branch of math called, Linear Algebra. I have introduced two basic constructs from Linear Algebra that we will need to move forward and perform the fundamental calculations for rendering a three-dimensional display. At this point I have only scratched the surface as to what is possible, and I have only demonstrated how. I will provide context and demonstrate the what and why, to a degree, on the path helping you begin to work with three-dimensional graphics libraries, even if math is not one of your strongest skills.


Kreyszig, Erwin; "Chapter 7" from Advanced Engineering Mathematics, 7th ed., New York: John Wiley & Sons, 1993

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