Focal Length

Dear Group

I have a question concerning the simulation of varous cameras in Radiance.

Hi Marcus,

I am really sorry -- in my earlier (April) response, I was going from memory, which is something I should avoid these days... The formula I gave you was based on the assumption that the height of a 35mm frame is 35mm, which is just plain wrong. 35mm refers to the total width of the film, including the registration (sprocket) holes. The actual size of the imageable frame is 24x36mm. I found the following excellent page explaining focal lengths and how to use them, which explains everything:

  http://www.panoramafactory.com/equiv35/equiv35.html

The formula you fixed from the book:

  h35 = 114.6*atan(17/f)
  v35 = 114.6*atan(11.5/f)

is close to correct for a 35mm landscape frame. It should actually be:

  h35 = 2*180/PI*atan(36/2/f)
  v35 = 2*180/PI*atan(24/2/f)

The values of 17 and 11.5mm may have been derived from measurements of an actual slide frame, which were probably slightly smaller than the official size. The answer is pretty close either way.

To get the view angles for other focal lengths, simply substitute the target film dimensions for the 36 and 24mm values in the formmula above. You'll have to verify these sizes yourself. (I'm not sure for example if a 4x5 transparency has a 40x50mm imaging area or something smaller.)

I hope this helps this time.
-Greg