2
\$\begingroup\$

I have screen with dimensions: 800x600. The object that I want to rotate around is located at (400,300, -50) - center of the screen. Orthogonal projection: Near plan - 0.1f, far plane - 1000.0f left - 0, bottom - 0, right - 800, top - 600

Camera: position at (0,0,0) look at (0,0,-50) up vector (0,1,0)

Now what I'm doing to rotate is calculate the X of "look at" using: 0 + width * cos(A). Y remains the same (0). Z is calculated using: -50 + width * sin(A).

Now the problem that I'm facing is that after some angle the object starts to disappear, and it doesn't rotate around the object. What can be the problem?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ According to your math, your camera will orbit around 0,0,-50 with a radius of width. But you are saying the object you wish to orbit is at 400,300,-50. Correct? \$\endgroup\$ Commented May 17, 2013 at 22:06
  • \$\begingroup\$ yes, so in order to fix it, I need to change the look at values? Let's try it \$\endgroup\$ Commented May 18, 2013 at 8:23

1 Answer 1

Reset to default
1
\$\begingroup\$

The image of a point A under a rotation around another point B (an affine rotation if B is not the origin of the space) is A', with

A' = B + R*(A-B)

where R is the matrix of the associated linear rotation.

For example, in dimension 2, say you want to rotate A = (1,0) around B = (1,1) by 90 degrees counter-clockwise. That will yield (2,1). Make a picture if needed. And, if we call R the matrix of the linear counter-clockwise rotation of angle 90 degrees, we have

R = (0 -1)
    (1  0)

and

(1,1) + R*( (1,0) - (1,1) ) = (1,1) + R*(0,-1)
                            = (1,1) + (1,0)
                            = (2,1)

Now, the solution to your problem actually depends on your implementation of world and camera geometry. You can either rotate your camera around point B or you can "rotate the world" around the camera by the inverse of the previous rotation (as usual in the "ProjectionMatrix * CameraMatrix * ModelMatrix * positionOfVertex" part of a vertex shader, if it's OpenGL or DirectX you're using).

\$\endgroup\$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.