Trigonometry. Field of view is 45 degrees = 800 pixels. Half the screen is 22.5 degrees = 400 pixels. By trig, 1000 tan 22.5 = 414.21 metres, which we know is 400 pixels, so each pixel is 1.04 metres in width. If your aspect ratio is correct, the height will be the same.

So: `(distanceToPlane * tan(fov / 2)) / (horizontalResolution / 2) = visible width of pixel in metres.`

I think your solution works. I realize now that i needed some more information to extimate the LOD geometry to use.

The problem is how to compute the distance in pixels between two points (wherever they are) when projected in the view plane. Suposing the info related to fov and the view plane distance is the same as in post above. How can we solve this ?

Im thinking in calculating the he distances between two extreme points in the bounding box of an object. Then based on the distance between the two in pixels i would choose what LOD geometry to use.

For instance if an object is going to be projected in a rectangular region of 10x10 pixels on the screen i would probably choose a LOD geometry with at most 10 vertices wide and 10 vertices equaly distributed.

Or i would use a flat image. Which probably would be a much better solution if the object is far enough and considering it is only 10x10 pixels. Just what are the known problems on using flats to replace geometry ?