# Perspective projection

I am working on an application that uses perspective projection using:

glMatrixMode(GL_PROJECTION);
gluPerspective(fFOV, fAspect , fNearPlane, fFarPlane);
with the fFOV = 45 degrees. fNearPlane = 0.1 and fFarPlane = 500.

I can place an object (rectangle) and see it rendered on screen. What I am trying to do is place it somewhere along the Z-axis so that all the four corners fit the four corners of the screen.
What is the math required for this ? I know the size of the viewport as well.

gluPerspective fFOV is the vertical fov.
So you have fFOV/2 above and below the view center.
Lets put the rectangle at distance zdist from the camera.
Then its height should be :

tan(fFOV/2)*zdist = h/2
h = tan(fFOV/2)*zdist/2

and width should be :
w = h*viewportw/viewporth

It works beuatifully. Thank you.

Height = tan((fFOV * M_PI/ 180)/2.0) * (mTranslateZ) * 2 ;
Width = Height * viewport[2] / viewport[3] ;

Viewport is obtained by:

GLint viewport[4];
glGetIntegerv(GL_VIEWPORT,viewport);

Hello,

I have a similar problem where I wish to draw a one by one rectangle in the top left corner of the screen at a given depth.

Computing the width / height at a given depth gives the top-left max visible point of that plane.

y = Height / 2.0f; // positive
x = Width / 2.0f; // negative since 0, 0 is the center of the screen.

However drawing from that point doesn’t show anything on the screen.

I assume this point is visible since it would be on the border of the clipping planes and I draw toward the center of the screen, am I correct? The resulting drawing should be inside the clipping volume.

Any ideas? (I am using glRect to draw)

a) be sure that your model view matrix is an identity
b) your point should be these
xCorner =-width * .5f;
yCorner = height * .5f;
xRect = xCorner + delta;
yRect = yCorner - delta;

And if you really want to use glRect (that is deprecated) you should call it like this in order to have the normal facing the camera.
glRect(xCorner, yRect, xRect, yCorner);

Hello,

Two things:
a) I am not sure to understand why my matrix should be an identity. You were right it wasn’t. I translated the modelview matrix 20 unity forward.

I computed the height and width when the translation was done. Shouldn’t it work?

I mean the volume doesn’t change if I am 20 unit forward in the volume, right? I assumed when the drawing is over that I look from the identity matrix.

Maybe I am using the modelview matrix incorrectly. I thought that once the volume is built thanks to the projection matrix, I could position my object inside it and always view them.

The idea behind my example was to draw a rectangle of a fixed size, let say 5 x 10 and position it at a certain depth at the top left corner. Depending on the depth level I could then add others rectangles so I could fill the screen, but always using the initial value of the rectangle 5 x 10.
i.e. at a translation of n units forward z I could draw 10 rectangles on the same row and at a translation of 2n units forward z I could draw 20 rectangles on the same row.

b) I am using a book (OpenGL SuperBible), and it is using glRect (in the first chapters for now), what is the new alternative?