Understanding how glm::ortho()'s arguments affect vertex location after projection

Question

After searching many pages, glm documentation, tutorials...etc, I'm still confused on some things.

I'm trying to understand why I need to apply the following transformations to get my 800x600 (fullscreen square, assume the screen of the user is 800x600 for this minimal example) image to draw over everything. Assume I'm only drawing CCW triangles. Everything renders fine in my code, but I have to do the following:

// Vertex data (x/y/z), using EBOs
  0.0f, 600.0f, 1.0f,
800.0f,   0.0f, 1.0f,
  0.0f,   0.0f, 1.0f,
800.0f, 600.0f, 1.0f

// Later on...
glm::mat4 m, v, p;
m = scale(m, glm::vec3(-1.0, 1.0, 1.0));
v = rotate(v, glm::radians(180.0f), glm::vec3(0.0f, 1.0f, 0.0f));
p = glm::ortho(0.0f, 800.0f, 600.0f, 0.0f, 0.5f, 1.5f);

(Note that just since I used the variable names m, v, and p doesn't mean they're actually the proper transformation for that name, the above just does what I want it to)

I'm confused on the following:

• Where is the orthographic bounds? I assume it's pointing down the negative z-axis, but where do the left/right bounds come in? Does that mean [-400, 400] on the x-axis maps to [-1.0, 1.0] NDC, or that [0, 800] maps to it? (I assume whatever answer here applies to the y-axis). Then documentation just says Creates a matrix for an orthographic parallel viewing volume.

• What happens if you flip the following third and fourth arguments (I ask because I see people doing this and I don't know if it's a mistake/typo or it works by a fluke... or if it properly works regardless):

Args three and four here:

                              _____________
                             |  These two  |
p1 = glm::ortho(0.0f, 800.0f, 600.0f, 0.0f, 0.5f, 1.5f);
p2 = glm::ortho(0.0f, 800.0f, 0.0f, 600.0f, 0.5f, 1.5f);

• Now I assume this third question will be answered with the above two, but I'm trying to figure out if this is why my first piece of code requires me flipping everything on the x-axis to work... which I will admit I was just messing around with it and it happened to work. I figure I need a 180 degree rotation to turn my plane around so it's on the -z side however... so that just leaves me with figuring out the -1.0, 1.0, 1.0 scaling.

The code provided in this example (minus the variable names) is the only stuff I use and the rendering works perfectly... it's just my lack of knowledge as to why it works that I'm unhappy with.

EDIT: Was trying to understand it from here by using the images and descriptions on the site as a single example of reference. I may have missed the point.

EDIT2: As a random question, since I always draw my plane at z = 1.0, should I restrict my orthographic projection near/far plane to be as close to 1.0 as possible (ex: 0.99, 1.01) for any reason? Assume nothing else is drawn or will be drawn.

Answer 1

You can assume the visible area in a orthographic projection to be a cube given in view space. This cube is then mapped to the [-1,1] cube in NDC coordinates, such that everything inside the cube is visible and everything outside will be clipped away. Generally, the viewer looks along the negative Z-axis, while +x is right and +Y is up.

How are the orthographic bounds mapped to NDC space?

The side length of the cube are given by the parameters passed to glOrtho. In the first example, parameters for left and right are [0, 800], thus the space from 0 to 800 along the X axis is mapped to [-1, 1] along the NDC X axis. Similar logic happens along the other two axes (top/bottom along y, near/far along -z).

What happens when the top and bottom parameters are exchanged?

Interchanging, for example, top and bottom is equivalent to mirroring the scene along this axis. If you look at second diagonal element of a orthographic matrix, this is defined as 2 / (top - bottom). By exchanging top and bottom only the sign of this element changes. The same also works for exchanging left with right or near with far. Sometimes this is used when the screen-space origin should be the lower left corner instead of upper left.

Why do you have to rotate the quad by 180° and mirror it?

As described above, near and far values are along the negative Z-axis. Values of [0.5, 1.5] along -Z mean [-0.5, -1.5] in world space coordinates. Since the plane is defined a z=1.0 this is outside the visible area. By rotating it around the origin by 180 degrees moves it to z=-1.0, but now you are looking at it from the back, which means back-face culling strikes. By mirroring it along X, the winding order is changed and thus back and front side are changed.

Since I always draw my plane at Z = 1.0, should I restrict my orthographic projection near/far plane to be as close to 1.0 as possible?

As long as you don't draw anything else, you can basically choose whatever you want. When multiple objects are drawn, then the range between near and far defines how precise differences in depth can be stored.