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# perspective - X - FOV

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Hello, Using gluLookAt, I know about the Y field of view, near\far clips, etc., but is the exact x field of view I end up with simply computed by multiplying my aspect ratio (1027 x 768 in this case) which gives a 1.3333 aspect ratio ( so this window is 1.3 times wider than high) by whatever angle I set as my vertical field of view to? In other words, say I have a 26 degree vertical field of view(for example). To get my visible x field of view do I simply multiply the 1.33333 aspect ratio from a 1024x768 window by 26 degrees to get 33.8 degrees (which represents my x field of view)? Or can xFOV be solved for by the simple ratio: 26(degrees yFOV)/768(pixels vertically) = w( degrees xFOV) / 1024(pixels horizontally) which would give me 34.67 degrees of x view. Thanks in advance for any help. J [edited by - Jehsup on January 23, 2004 4:17:42 PM]

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The aspect ratio is a measure of the screen width divided by the screen height so this won''t work for angles.

You''ll have to solve using:

xFOV = atan(tan(yFOV) / aspectratio)

I''ll derive it here if you need

(1) aspectratio = width / height (by defn)(2) tan(yFOV)   = height / dist  (by defn)(3) tan(xFOV)   = width  / dist  (by defn)By (1), height = width * aspectratio  By (2), height = tan(yFOV) * dist(4) obtained by setting (1) == (2), width * aspectratio = tan(yFOV) * dist;and thereforewidth = tan(yFOV) * dist / aspectratio;By (3) and (4)tan(xFOV) = (tan(yFOV) * dist / aspectratio) / dist          = (tan(yFOV) / aspectratio)ThereforexFOV      = atan(tan(yFOV) / aspectratio)

Hope you understand.
FReY

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Eheh, he, silly me...

you can get the solution from the first 3 equations ... no need to go off on a tangent like I did..

(1) aspectratio = width / height (by defn)(2) tan(yFOV)   = height / dist  (by defn)(3) tan(xFOV)   = width  / dist  (by defn)aspectratio = tan(yFOV) * dist / tan(xFOV) * dist            = tan(yFOV) / tan(xFOV)

I feel silly.

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What are you using as the "dist"? The far clipping plane?

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dist can be any distance between point of view and the plane. as you can see it will be removed from the equation anyway. i prefer to have things happening on a plane with dist("z") of 1, though it seems more common to think of everything being projected on the near plane.

anyway, assuming dist=1, then
height = 2*tan(fov/2);
width = 2*tan(xfov/2) = height*aspect;
tan(xfov/2)=tan(fov/2)*aspect;
xfov=2*atan(tan(fov/2)*aspect);

so now im only slightly confused. we end up with similiar results that still are a little exclusive.
one thing that always bothered me: why do i read tan(fov) everywhere and when using that it simply wont work (ie, drawing a rect at distance 1 with those dimensions wont fit). at the same, using 2*tan(fov/2) seems to work perfectly, the rect matches the viewport and using it for culling seems to give exact results.

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Jashup, Trienco is absolutely right on both fronts. It doesn''t matter what the ''dist'' is, although you can certainly calculate ''dist'' if you wanted.

Trienco you are absolutely right - you should be using the half-angle within the expressions above, this is because the above expressions are trigonometrically attained and it is defined on right-angle triangles only.

As a result, Trienco''s expression is more correct than mine :D

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