OK.

My name is Henry Schaefer and I will give the lecture today.

You talked about few transformations lastly and there you have the modeling transform

that transforms an object into world space.

So for example you have a sphere somewhere and the sphere is defined in a local coordinate

system and you want to transform it into a global

coordinate system you apply a modeling transform.

So every vertex is multiplied with a transformation matrix that transforms the sphere vertices.

Now you want to have a camera somewhere so in world coordinates you might have a camera

here and it's looking in this direction and you have a local coordinate system here.

So you apply an additional transform to transform the sphere vertices into the camera space

and then you want to have something like perspective projection.

So last time you talked about autographic projections now you want perspective projection

where you have vanishing points so you want to project these points onto the image plane

here where you have your pixels.

So the strategy is simple.

We project the objects directly towards the eye like here and we draw the objects where

they meet a view plane.

So we would draw the sphere here.

So this was done in the middle ages and the principle is simple and there we use the ratio

of similar triangles.

So you have the camera positioned at E, the eye, and you have a gazing direction G and

an image plane at a distance n and you want to know what is the y value in distance z

at your image plane, at your view plane.

And the ratio of similar triangles says that the ratio between your y on the view plane

to y is equal to n to z.

So you can compute your magnitude of y by n divided by z times y.

So the question is how to find a transformation matrix that transforms points like this.

So we have a few firsts for the perspective projection and we want to find a matrix that

shrinks objects at a distance.

So in perspective objects that are farther away seem to shrink and so you want to step

from there to this volume.

I made an image here too.

And we want, finally we want to apply a matrix to transform this volume into the canonical

view volume.

So we do not want to change the near and far values but we want these points to map to

the volume and we want the division by z.

The matrix that solves this problem is given here and we have used the similarity ratio

two and for example if you have a point we transform it to homogeneous coordinates so

we have an additional dimension for w and we transform this point using this matrix.

We get this point and then we want to remove the additional coordinate so we divide by

n divided by z and we get this 3D point.

Now let's see if the bounds are correct.

For example if we want to evaluate this transformation at the near plane and we plug n for z we get

what we expected.

The point on the near plane stays where it is.

The point on the far plane, the upper bound maps to the far plane in the z direction.

For this matrix we have an equivalence relation so we can scale it.

This means we can transform this matrix into a more appropriate form by multiplying it

with n and this removes the division here.