So early this morning I couldn't sleep and I ended up writing an article on flickr in one of the many "what lens should I buy" discussions that goes on there. I have noticed, in my days on flickr, that certain pieces of equipment and certain techniques have a following, and often get recommended simply because of the following rather than because the equipment/technique is actually suited to the purpose of the person asking. In an effort to supply a counterbalancing opinion, I found myself in need of trigonometry.
The Argument
One such "cult" item is the "nifty fifty" (the EF 50mm f/1.8) lens made for Canon EOS cameras. It is very sharp, very fast, and very cheap ($80). If you are on a tight budget (or even if you aren't) it makes sense to have one for your EOS camera unless you have a better 50mm prime, or don't need a 50mm prime.
While I will not argue that it is probably one of the best value-for-money lenses, it is not versatile at all, and yet it seems to get hailed as a magic-bullet lens. I regularly see people making claims like "it never comes off my camera". And after having used it myself, I can only conclude that these people shoot one type of thing and one type of thing only, or it never comes off because they don't own any other lenses.
There is a certain love affair with the 50mm focal length because it was the standard focal length for 35mm film for decades. But in the age of digital SLR cameras, things are different for the less expensive consumer DSLRs. These DSLR's tend to use an image sensor that is smaller than 35mm film. The APS-C style sensor, or crop sensor, does not render the entire image cast by a standard lens, but only a smaller piece in the center. This results in an apparent magnification factor of 1.6. Hence if you put a 50mm lens on a crop-sensor camera, it's like working with an 80mm lens (50 x 1.6 = 80). The end result is a smaller-than-expected "field of view" (FOV).
On an old Canon 35mm film camera, a 50mm lens has a FOV of 46°. But on a crop-sensor camera the FOV is a hair under 29°. This loss of over a third of the FOV means that on crop-sensor cameras the EF 50mm f/1.8 lens has distinct limitations as to how much you can fit in the frame.
Fortunately, on my EOS 5D the 50mm behaves as expected. Because the 5D is a full frame camera, its sensor is the same size as a 35mm film frame. So I get 46° out of my EF 50mm, just as nature intended. 
The "nifty fifty" on crop-sensor cameras is often described as a "portrait lens". With the crop factor, the 50mm lens behaves like an 80mm lens, and 80mm is ideal for portraits. But if you want to shoot anything larger than a head-and-shoulders portrait with the EF 50 1.8 on your Rebel XT or 30D, you'd better have a lot of room behind you, because you are going to need to back up... a lot.
But how much?
The Trigonometry
Well that's where the trig comes in (you can skip this section if you don't want to see how I figured it out). In order for me to say how much, I needed to be able to reliably compute the distance necessary to view an object of a given width. But how? I started by drawing a diagram like this one:
V is my viewing angle. Okay it's not 29° (or 28.98333° which is the actual FOV of the nifty fifty on a crop sensor), but close enough. The legs of the triangle extending out from V represent the edges of my FOV as the distance to the subject (marked by the dashed line, d) grows. The base of the triangle (marked as w) is the width of the field of view at the distance d. Basically this is a representation of the wedge or cone of that falls within a particular FOV, in this case 30°.
I can pick any distance I want for d, but what I really need is a way to say what d should be to accomodate a subject of a certain width. In other words, to fit a subject 10 feet wide in my viewfinder, how far back do I need to stand with the nifty fifty on my EOS 30D camera? I supposed that given a formula for that, I could solve the formula for the width so that one could also compute the maximum width viewable given a distance.
The triangle depicted above is an isoceles triangle, as both the legs are the same length, and consequently the angles where the legs meet the base is also to the same. I spent some time looking online for computations for isoceles triangles, but what I was looking for didn't appear (namely, given the length of the base, and the angle of the peak, what is the height or altitude of an isoceles triangle?)
I studied trig over 20 years ago so I remember very little of it, but I did remember there were a lot of simple equivalences for right triangles (that is, triangles where one of the angles is 90°). And I realized while looking at my diagram that the line I had drawn to represent the distance, bisected V and split the triangle into 2 right triangles, each of which looked like this:
Bisecting V gives me a 15° angle (V'), and a base width exactly half of what it was before (w'). So if I could take a given distance d and come up with a formula for w', then I should be able to solve that formula for either d or w', keeping in mind that V' is V/2 and w' is w/2.
Doing a quick check online I found the two rudimentary trignometric equivalences for right triangles: for either of the angles other than the 90° one, the sin of that angle is equal to the length of the opposite side divided by the length of the hypoteneuse, and the cos of that angle is equal to the length of the adjacent side divided by the length of the hypoteneuse. Here are those equivalences for the right triangle above:
Sin V' and cos V' I can get with a pocket calculator, and I'm going to pick a value for either d or w' and solve for the other. I can solve the equation on the left for w' [w' = (sin V') * h] and I can solve the equation on the right for d [d = (cos V') * h], but both of these solutions require me to know what the hypoteneuse of this triangle is.
But in order to get w' from d or d from w' I need to do more work, mostly because I am not going to know what the hypoteneuse is. I'm only going to be starting with either V' and w' or V' and d. So what I need to do is solve one of the equations for h, and then plug that into the other equation. That should give me a formula I that I can use to solve for either d in terms of w' and V' or w' in terms of d and V'. So I picked the equation on the right. Solving that for h gives h = d / (cos V').
So I should be able to substitute d / (cos V') in the equation on the left, like so:
Now I'm good. I know what V' is, I can get sin V' or cos V' from my calculator, and I am going to pick either d or w'. So now I can solve for either one, like so:
Done, right? Well, yes, if I want to know what the appropriate distance is for half the width of my subject using a lens with half the field of view. Now I want to substitute in the equivalences that w' = w/2 and V' = V/2. In the equation on the right that will put w/2 on the left of the equal sign, so I will multiply both sides by 2 to solve the equation for w. That gives me:
Okay they probably aren't the cleanest formulas in the world, but they work and let you get the height of an isoceles triangle from its base width and peak angle, or vice versa. Using these formulas I could handily compute the needed distance for a given width in a given field of view, and this allowed me to present something more concrete than "gee whiz, that EF 50mm 1.8 is awfully confining on a crop sensor camera."
Back to the Argument
So how confining is that nifty fifty?
5 feet wide = 9.7 feet away
10 feet wide = 19.3 feet away
15 feet wide = 29 feet away
20 feet wide = 38.7 feet away
25 feet wide = 48.4 feet away
Pretty confining! If you are trying to capture 3 people sitting on a couch which is 8 feet long all in one shot, you need to stand 15 feet 6 inches away. Better have a big living room, or one where there isn't a TV 10 feet from the couch. Or maybe if you moved the couch outside... that would be cool for an album cover, but for Aunt Bea, Uncle Joe, and Granny, it is probably less so.
Working with the EF 50mm f/1.8 is a good exercise though for learning how to push a lens to do what you need, and it's plain old good exercise, because you're going to be backing up a lot. You can get that 8 foot couch in shot if you shoot from an angle, but then you will need to stop your aperture down to widen up the depth of field so that everyone will be in focus... which means you can't shoot low light anymore so you might need lamps or a flash. Or you could give up on that shot and shoot the people individually.
Or, you could simply not get the EF 50mm 1.8 in the first place, if you are not planning to shoot primarily portraits. If you want to consider the traditional FOV that the great 35mm film artists shot with, you need a lens that gives a FOV on a crop-sensor camera similar to a 50mm lens on a 35mm film camera (i.e. 46°). The closest bet would be a 28mm lens, like the EF 28mm f/1.8 or EF 28mm f/2.8. These have a crop-sensor FOV of 47.25°. With one of these lenses the width to distance figures look like this:
5 feet wide = 5.7 feet away
10 feet wide = 11.4 feet away
15 feet wide = 17.1 feet away
20 feet wide = 22.8 feet away
25 feet wide = 28.6 feet away
Much more reasonable. And quite interesting how the distance to subject is almost the same as the width of the subject. No surpise that the 50mm lens became the standard on the cameras of old.
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BoltBait
Nov-16 05:47 PM (EST)
Circuit City wanted $899 for the camera. I was informed by the staff that the camera would come with a "temporary battery" and a small charger, but that neither of those components were standard and that I would need to by the standard ones for serious camera use. This would be another $140 or so. Then of course I needed a memory card and it had to be a high-speed (133x) card, which was a mere $50 more. Add a camera bag to keep my equipment safe ($50), and Circuit City's 4-year warranty ($100) and you were talking about $1250... then add another $63 for sales tax. Ouch. I could never afford all this, so I took the information and went home to talk to my wife about it.
