Camera projections & ray casting

Drag the world point. Push it further away (larger Z) and its image-plane dot creeps back toward the center — the same real-world displacement covers less of the frame the farther away it happens. That shrinking-with-distance effect is perspective, and it's the whole reason a camera needs more than a ruler to describe.

A pinhole camera projects a 3D point onto the 2D image plane in two stages. First, extrinsics move the point from world coordinates into the camera's own frame — for a camera at center $C$ with no rotation, that's just a subtraction:

Then intrinsics turn the camera-space point into a pixel: divide by depth to get the perspective effect, then scale by focal length $f$ and shift by the principal point $(c_x, c_y)$:

• $P_{\text{world}}$, $C$ — the 3D point and the camera's center, both in world coordinates.
• $P_{\text{cam}} = (X, Y, Z)$ — the same point re-expressed relative to the camera.
• $Z$ — depth: distance in front of the camera along its viewing axis.
• $f$ — focal length, in pixels: how strongly the camera magnifies the scene.
• $(c_x, c_y)$ — the principal point: the pixel where the camera's forward axis itself lands.
• $(u, v)$ — the resulting pixel coordinates on the image plane.

1. Extrinsics: world to camera

   With the camera sitting at the world origin, $P_{\text{cam}} = P_{\text{world}}$ exactly — the world point $(1, 1, 5)$ is already in camera coordinates.

2. Perspective divide

   Divide by depth: $X/Z = 1/5 = 0.2$ and $Y/Z = 1/5 = 0.2$. This is the step that makes farther points shrink toward the center — the same $X$ produces a smaller ratio at larger $Z$.

3. Intrinsics: scale and shift

   With $f = 100$ and $(c_x, c_y) = (50, 50)$: $u = 100(0.2) + 50 = 70$ and $v = 100(0.2) + 50 = 70$. The point lands at pixel $(70, 70)$.

Drag the point and watch $Y/Z$ directly — it's the single number that decides $v$ once $f$ and $c_y$ are fixed. Push $Y$ far negative or $Z$ close to the camera and $Y/Z$ blows up, sending the point flying off the edge of the frame.

Two more points through the same camera ($f = 100$, $c_x = c_y = 50$):

1. P = (2, -1, 4), camera at the origin

   • Perspective divide: $X/Z = 2/4 = 0.5$, $Y/Z = -1/4 = -0.25$
   • Scale and shift: $u = 100(0.5) + 50 = 100$, $v = 100(-0.25) + 50 = 25$

   Pixel $(100, 25)$.

2. P = (3, 1, 5), camera moved to C = (1, 0, 0)

   1. Extrinsics: $P_{\text{cam}} = (3, 1, 5) - (1, 0, 0) = (2, 1, 5)$
   2. Perspective divide: $X/Z = 2/5 = 0.4$, $Y/Z = 1/5 = 0.2$
   3. Scale and shift: $u = 100(0.4) + 50 = 90$, $v = 100(0.2) + 50 = 70$

   Pixel $(90, 70)$ — moving the camera changed the pixel even though the world point never moved.