[MF3D.FORUM:120] Image Size - OOPS

[Richard Rylander <rlrylander@xxxxxxx>]

Oops!

The equation for solid angles subtended by various rectangular formats I gave in a previous message is correct.  Due to a misplaced parenthesis when programming my calculator, however, the values in the table of solid angles of various viewer formats were incorrect (I effectively divided instead of multiplying one of the factors).  Maybe I can get a job with NASA navigating probes to Mars.

The equation (always was correct) is repeated below with a revised table showing the correct solid angles.

Consider a rectangle "w" units wide, "h" units high, viewed from a point
"f" ('focal length' of viewer lens) units away from the center of the
rectangle.  The "1D" angular width, "aw", and angular height, "ah", [the
easy calculations] are:

    aw = 2*arctan(w/(2f))         ah = 2*arctan(h/(2f))

The "2D" solid angle, "s" is a little trickier:

    s = 2*pi - 4*arccos(sin(aw/2)*sin(ah/2))

In the [corrected] table below, "F" is the viewer focal length, "S" is the solid angle (in steradians),
"Adj. S" is the adjusted solid angle ("S" divided by the square root of the format aspect
ratio), and the "Immersion Factor" is the fraction of a hemisphere the adjusted solid angle
represents.

Format    Size      F      S      Adj. S   Immersion Factor
5P      23 x 21    44    0.2349   0.2245        0.0357
7P      23 x 28    44    0.3070   0.2782        0.0443
MF      50 x 50    78    0.3732   0.3732        0.0594
3Disc   24 x 36    43    0.4159   0.3396        0.0540
MF-W    56 x 58    78    0.4721   0.4639        0.0738
6 x 7   60 x 70    78    0.5900   0.5462        0.0869

Sorry if this caused any confusion, but then it wasn't intuitively obvious what the solid angles should be other than the limiting cases of 0 and 2*pi steradians.  Thanks to John Bercovitz for double checking the numerical results and letting me know something was wrong.

Richard Rylander