I remember my pythagorean theorem A^2+b^2=C^2 and thought you should be able to use that to figure out how "wide" your lens will see at a given distance from your subject. For example, the Sigma 10-20mm has a an angle of view of 102 degrees. I figure you should be able to split that in two to form two right triangles and figure out 1/2 the width of the view you get when you use the lens at 1 foot, 5 feet, 10 feet etc. However, without knowing two of the three variables in the equation how can you (or is it even possible to) figure out how wide your view from that lens is given you are x feet from the subject? Am I making sense? Just overthinking my new toy I guess.
Neal, The problem with your hypothesis is that the glass bends the light and there are usually numerous glass elements built into each lens, so it's not a perfect triangle with straight lines. An exercise like this would require more elaborate mathematics and physics so the easiest way to find these values might be to do some actual tests and record what you find for a few of the shorter distances, and then see if you discover a linear relationship. Sean
This also explains why the Nikkor 10.5mm fisheye has a FOV of 180 degrees, while the Sigma 10-20mm is only 102 degrees at its widest. - Jay
The general geometrical idea is correct for a rectilinear lens (i.e. non-fisheye). However, there are two important caveats. Angular field of view as quoted by the manufacturer is given for focus at infinity. As you focus closer, the angular field of view decreases for a fixed focal length, as the image principal plane on the lens is moved away from the image plane. Conversely, most modern lenses (especially zooms) focus not by extension, but by reducing focal length somewhat at closer distances. This tends to increase the angular field of view. Thus correctly estimating field of view over a wide range of object distances requires a lot more knowledge about the characteristics of each particular lens over its entire range of focus.
After doing some research on Trigonometry, would it be safe to say that the width of your field of view could be approximated as follows: Field of view~(TAN(Angle of View/2)*Distance to subject)*2 I understand that the curvature of the glass would impact this some, as Gerry's comments about the decreasing angle of view at shorter focal distances. However, I'm only looking for an approximation in my own head I guess. If I'm using my Sigma 10-20mm at 10mm and the angle of view is 102.4, at 10 feet away from a wall, I would get something which is approximately 24 feet wide, give or take a few inches. does that hold up?
Yes, that expression will give you a reasonable approximation for field of view as long as you are not too close to the subject. The factors that I mentioned above are usually only significant near close focus. For a wide angle lens, 10 feet is likely close enough to infinity (hyperfocal distance for a 10 mm @ f/4 is about 1.5 m) that you can ignore magnification and focal length change when focusing. Note that lens manufacuturers usually quote angular FOV for the diagonal, not the frame width. Your original question mentioned 1 foot, 5 feet, etc. distances. Predictions at close distance may not be as accurate depending on lens focal length and magnification. For example, my 200 mm Micro provides 1:1 magnification at close focus of just under 0.5 m. This magnification factor by itself would cut the FOV about in half. However, focal length of the 200 Micro is actually about 104 mm at close focus, almost completely canceling the effect of magnification on the FOV. My 105 mm f/2.5 AIS focuses by extension so the focal length does not change. Since at close focus of about 1 m its magnification is about 1:7, the angular FOV is reduced about 14% from that at infinity.
