3D to 2D projection

Question

I'm trying to write a game in Java (Processing, actually) with some cool 3D effects.

I have a choice of two 3D renderers, but neither have the quality or flexibility of the default renderer. I was thinking that if I could get a function to proje

So say I have a set of coordinates (x, y, z) floating in 3D space. How would I get where on the 2D screen that point should be drawn (perspective)?

Just to clarify, I need only the bare minimum (not taking the position of the camera into account, I can get that effect just by offsetting the points) - I'm not re-writing OpenGL.

And yes, I see that there are many other questions on this - but none of them seem to really have a definitive answer.

Answer 1

Here is your bare minimum.

column = X*focal/Z + width/2
row = -Y*focal/Z  + height/2

The notation is:

X, Y, Z are 3D coordinates in distance units such as mm or meters;

focal is a focal length of the camera in pixels (focal= 500 for VGA resolution is a reasonable choice since it will generate a field of view about 60 deg; if you have a larger image size scale your focal length proportionally); note that physically focal~1cm << Z, which simplifies formulas presented in the previous answer.

height and width are the dimensions of the image or sensor in pixels;

row, column - are image pixels coordinates (note: row starts on top and goes down, while Y goes up). This is a standard set of coordinate systems centered on camera center (for X, Y, Z) and on the upper-left image corner (for row, column, see green lines).

You don't need to use OpenGL indeed since these formulas are easy to implement. But there will be some side-effects such as whenever your object has several surfaces they won't display correctly since you have no way to simulate occlusions. Thus you can add a depth buffer which is a simple 2D array with float Z values that keeps track which pixels are closer and which are further; if there is an attempt to write more than once at the same projected location, the closer pixel always wins. Good luck.

Answer 2

Look into Pin hole camera model

http://en.wikipedia.org/wiki/Pinhole_camera_model

ProjectedX = WorldX * D / ( D + worldZ )

ProjectedY = WorldY * D / ( D + worldZ )

where D is the distance between the projection plane and eye