[MF3D.FORUM:105] Re: Image Size

[Richard Rylander <rlrylander@xxxxxxx>]

Greg Erker wrote:

>      Format Size F S Adj. S Immersion Factor
>      5P 23 x 21 44 2.0896 1.9967 0.3178
>      7P 23 x 28 44 2.4174 2.1910 0.3487
>      MF 50 x 50 78 2.6838 2.6838 0.4271
>      3Disc 24 x 36 43 2.7753 2.2660 0.3606
>      MF-W 56 x 58 78 2.9964 2.9443 0.4686
>      6 x 7 60 x 70 78 3.2876 3.0437 0.4844
>
> Not much time to talk today but I don't think there
> is any 6x7 camera that is actually 60x70mm. More like
> 50-some by 60-some. So the last line is overly optimistic.
>
> Where did you get 43mm for the 3Discover FL. Isn't it
> more like 50mm?
>
> Lastly, I've mounted some of my fireworks slides as
> 52x52mm to increase the fill-your-view factor. What
> immersion factor does that give?
>
> Greg E.

The format sizes and focal lengths are those used by Paul Talbot
(MF3D.FORUM.51)
and Alan Lewis (MF3D.FORUM.53) in their tables.  I used the same numbers
for easier
comparison to those results.

A 52 x 52mm format with 78mm focal length viewing lens would produce a
solid angle of
2.7780 steradians (the "adjusted solid angle" is the same since the
aspect ratio is 1), and an
Immersion Factor of 0.4421

If the viewing lens is 80mm, s = 2.7172, I.F. = 0.4325

The equation lets you [easily] find solid angles for other cases.  If
you don't need the inter-
mediate angular width and height ("aw", and "ah" in my original
message), the solid angle
can be calculated directly from the format width "w", height "h", and
focal length "f", as:

  s = 2*pi - 4*arccos(sin(arctan(w/(2*f)))*sin(arctan(h/(2*f))))

If the format is rectangular, the adjustment I used was to divide "s" by
the square root of
the format aspect ratio.  Finally, normalize by dividing by 2*pi
steradians to get an
"Immersion Factor" that is the fraction of a hemisphere the format
subtends.

Richard Rylander