1
\$\begingroup\$

Take a look at the picture.

enter image description here

I've got a camera(brown) and its ray(red). As I know ray has a direction and unlimited length(if you don't set it yourself). But now I need to determine ray's vector that starts at camera point and ends when it crosses a floor. How can I do that?

\$\endgroup\$
  • \$\begingroup\$ there are many rays that cross a floor, which one do you need? \$\endgroup\$ – ratchet freak Dec 12 '14 at 9:47
0
\$\begingroup\$

I assume you know the ray direction, otherwise there are infinite rays that cross the floor. It crosses the floor when your camera pos (x,y,z) offset by a scaled ray direction k*(a,b,c) has zero height, so:

y + k*b = 0 => (we only care about the y component)
k = -y/b

so the vector that you need to add to the camera is

(-y/b)*(a,b,c)

Of course, it goes without saying that you need to check for special case when b == 0, where the ray is parallel to the xz plane, so it will never cross it, unless on it (y==0)

\$\endgroup\$
  • \$\begingroup\$ Is there a mistake? Why k was -y/b and then became b/y? \$\endgroup\$ – Tony Dec 12 '14 at 11:15
  • \$\begingroup\$ duh, thanks, my bad, fixed it. it should stay -y/b. \$\endgroup\$ – Babis Dec 12 '14 at 11:24
0
\$\begingroup\$

The equation of the ray is p = camerapos + t*ray which is 3 equations (after splitting it up the coordinates):

p.x = camerapos.x + t*ray.x
p.y = camerapos.y + t*ray.y
p.z = camerapos.z + t*ray.z

then we know we want the point with p.y = 0 so lets fill it in:

0 = camerapos.y + t*ray.y
t = -camerapos.y / ray.y

then just fill in the t in the first equation: p = camerapos + (-camerapos.y / ray.y)*ray

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.