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.

As I go from discussion forum to discussion form, I repeatedly get asked that question in some form or another. "What's 'Abacquer' mean?" or "What is a Plastered Dragon?" When that happens I get to tell a story that I've told many times.  And each time I tell it I embellish it a little more than the last time.  But writing it over and over again each time it happens is a bit silly.  So I've decided to put the latest incarnation here so that I can just refer people to this article in the future.  It will save me on typing.   Good friends of mine already know this story and have already heard some of these jokes before.  But if you are really curious, read on...

In a sort of story I wrote once (a D&D campaign really) there was a dragon who had a penchant for getting drunk.

As the story goes, many hundreds of years ago the residents of the little community of Wayside and many people from the surrounding towns and environs are gathered at the Wayside Inn for the annual Oktobrefest--a huge party which sees partygoers from all over their country--all manner of tradesman, adventurer, and race is sure to be there. And many get swilling drunk of course, particularly the gnomes (as is their custom) who prefer to drink more than anyone else (as is their custom) from a trough (as is their custom) and then beat each other senseless until they pass out (as is their custom).

So anyway as I said this party was going on one brisk Oktobre morn when all of a sudden this young bull drake appears in the sky, circles the town, and promptly lands right smack dab in the center of the town green. Nobody knew what to make of it. It was neither a metallic nor a chromatic dragon, so nobody was certain what its disposition would be, but it was also quite young as dragons go and would therefore be "easily" put down by the town guard if need be.

It turned out that need did not be, because the drake trotted to the trough of ale, stuck his head in, drank the whole thing in about 30 seconds flat, let out a loud burp and promptly passed out, to much admiration from the gnomes.  The party continued about the sleeping beast unabated. Apparently the dragon was simply another partygoer, Wayside's first of that particular variety. And he quickly became the life of the party once he woke up and started dancing to the fine music provided by the local musicians. And apart from the inevitable property damage, his performance was well received.

It was the beginning of a long friendship, and as the years went by, the half-seas-over reptile would return each Oktobre to drink himself silly and make merry during Oktobrefest. He explained to the townspeople (in a rare sober moment) that his name was Abacquer (that's pronounced AB-bah-kur) and that his unusual coloration was due to his curious pedigree. His father was a benevolent copper dragon of some fame known as "Morrich the Claw". His mother was a chromatic dragon (specifically a white dragon) that had been raised as a foundling by copper dragons, who went by the moniker "Tiarrel the Rime". By his own admission, as a White/Copper hybrid, that made him a "Whopper Dragon", a fact that brought him much amusement and was sure to produce a beery guffaw from him whenever he brought it up. And given that his white dragon descent included a frost breath weapon, he could immediately chill your drink for you, so most people put up with hearing the story over and over, for his company was a good one, even if a bit sozzled.

Abacquer took a liking to Perronian Pink Champagne in particular, and the Wayside Inn began to stock up on it for his annual appearance. Once word of the inebriated dragon began to spread, the Oktobrefest became even more of an attraction that helped put Wayside on the map like never before. People would come from all over New Irth just to catch a glimpse of the "plastered dragon".

Upon reaching adulthood, Abacquer, like all dragons, chose a title for himself. He chose "Abacquer the Belch", much to the consternation of his parents, in recognition of his uncontested claim to the longest running belch in recorded history (12 minutes 48 seconds).

Later in life he made his aerie on Sherenpate Pyke and became something of a protector of the surrounding communities and vineyards. Especially the vineyards. Go figure. In his massive ice cavern he welcomed visitors and even had one visit of note from the musicians of a nearby farming village. A painting produced by a local artist hangs in the town's music hall commemmorating this event. In the painting the band is seated and playing in Abacquer's aerie while nervously glancing up at the (clearly intoxicated) dragon as he dances upside down on the ceiling. Of course white dragons can walk on any frozen surface seeming without regard for gravity, but as a half-white, Abacquer's icewalking ability was, according to him, "a little spotty". According to records of the event, the Belch informed the musicians that they needn't worry, his icewalking only gave out when he was drunk. Somehow that wasn't very reassuring, but the event went off without a hitch although all further concerts in his honor were held in town.

Eventually the fame of his story became so great that the Wayside Inn was renamed Inne of Ye Plastered Dragon (or simply, the Plastered Dragon Inn), and the rest is history. His life was long and colorful and he did many great deeds in spite of being a complete sot. He was knighted by the gnomes of Perro (there's a switch, as Abacquer once noted, being a knight and a dragon both meant that if he ever took a damsel hostage he would have to slay himself and then marry her, and not being the marrying sort, that was probably best.) And it is said that he is up for canonization in the gnomish pantheon, which is kind of a big deal. All previously canonized gnomish saints were bartenders and Abacquer has no talent for mixing drinks anywhere but in his belly. But the gnomes of Perro say (as is their custom) that he is "noseworthy", and I guess when it comes to binge drinking, nobody would be a better judge than they. And besides, being the biggest single consumer of Perronian Pink Champagne in all of New Irth (200 barrels a month) there's a certain economic interest among the gnomes of Perro in staying on his good side. But I digress... as is my custom.

I've always loved the Abacquer character, and so over the years, my online name has always been Plastered Dragon, or PDragon, or some variant thereof. And that's the story behind it--at some great length!