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See, im a simple hardcore Programmer. In fact i wrote some 3D-Programs till now.

I had nothing to do this weekend, so i made a 3D-based filemanager. My desktop is multihead so the point-of-view of the 3d-screens must synct and coact to give an collaborated navigation.

My question is,

Given: I got one Point-of-view (the null-point) I got 5 displays with a distance and rotation from null-point. I got one 3d-world with trees, stones and so on.

Searched: The rotation and forcal-length of those 5 cameras.

Hope you Understand my problem.

enter image description here

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The camera is always the eye(s) of the user. Actually, I'd recommend that the user should be able to adjust the 5 POVs individually to compensate for different display positioning (and different seat positions) –  Hagen von Eitzen Oct 28 '12 at 8:34
    
@HagenvonEitzen, yes i guess focal-length describe the lenses. I took the name from the program blender, default value is like 35 millimeters. –  Peter Rader Oct 28 '12 at 8:52

1 Answer 1

The camera position is that of the eye (same for all displays). The azimuth and elevation correspond to the direction from eye perpendiculary to the display. If that line does not hit the display at its center, this offset must be taken into account and the image shifted. Depending on how the azimuth/elevation transformation is performed and how your displays are positioned, you may also need to tilt the image accordingly. The distance between eye and display determines the projection offset (is that what is called focal length in spite of the absence of any lenses?) and together with the display resolution it determines the view angle.

Ultimately, you will need and find for each display a $3\times 3$ matrix $B_i$ such that the procdure is as follows:

Depending on the main camera position, there is an affine isometric transformation $$v\mapsto A v + b$$ with the property that the camera position maps to $0$, the looking direction maps to the positive $z$ axis and the sagittal plane is transformed to the $xy$ plane. Then on display $i$, the 3d point $v$ corresponds to $(x/z, y/z)$ provided $z>0$, with $$ \left(\begin{matrix}x\\y\\z\end{matrix}\right)=B_i(A v + b).$$ Note that only $A$ and $v$ while navigating throuhg the 3d model, whereas the $B_i$ are fixed (determined by the display and user seat arrangement).

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i can premis that the display is exact 90° to the head. –  Peter Rader Oct 28 '12 at 8:55

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