math,3d,camera,projection-matrix , 3D : keep horizon drawn on screen at the same height regardless camera's zoom


3D : keep horizon drawn on screen at the same height regardless camera's zoom

Question:

Tag: math,3d,camera,projection-matrix

I have this strange use case.

These variables are known:

Now, I want the horizon (horizon position = (0,0,Infinite)) to be drawn on screen at a specific 2D height "YY". What must be the Camera X-axis rotation so that the horizon is drawn at "YY"?

You might ask why may I need this: well, when I change the camera zoom, the horizon changes position on screen (in all cases except when X-axis rotation = 0). I need to be able to change the camera zoom AND keep the horizon unchanged (relatively to its 2D position). AFAIK, this can only be achieved by changing the X-axis rotation accordingly.


Answer:

In the case of a well-defined symmetric view frustum, the solution can be found with some simple geometry:

Problem Visualization

I assume YY is in normalized device coordinates (ranging from -1 to 1). If they are in pixel coordinates, you have to convert them.

I have chosen the projection plane to be 1 unit far away from the camera. But any other distance would work as well. Then, the distance y' is simply

y' = YY * H/2

H/2 is the half screen height and can be calculated with:

H/2 = tan (fovy/2)

where fovy is the camera's field of view in the vertical direction.

You want to find angle alpha for your rotation. This is simply:

tan alpha = y' / 1 = YY * tan(fovy / 2)
    alpha = atan(YY * tan(fovy / 2)

Be aware of the direction. Positive values specify downward rotations.


For arbitrary projections, this problem can be solved analytically:

Assuming we have projection matrix P and view matrix V, we want to solve for:

w-clip(P * V * (0 0 1 0)^T) = (... YY ...)

Since you only want to allow translations and rotations about the x-axis for your camera, V has the form:

    / 1  0           0          tx \
V = | 0  cos alpha  -sin alpha  ty |
    | 0  sin alpha   cos alpha  tz |
    \ 0  0           0          1  /

This yields the equation:

YY = (p23 * cos alpha - p22 * sin alpha) / (p43 * cos alpha - p42 * sin alpha)

where pij is the entry of P in the i-th row and j-th column.

Use your favorite symbolic solver to get a solution for alpha and you can re-calculate your view-matrix.


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