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# Because I am a well-rounded geek....

While driving to Topeka today (and therefore unable to just look this up), I wondered how to figure out the depth of an object when using a one-point perspective with a horizon line. After all, while pretty much every How To Draw book ever made has a page on perspective, artist books tend to be kinda light on math, and you're supposed to just eyeball it (or use two and three point perspective to help with depth). It was, of course, a bit difficult to do while driving, but I finally had a breakthrough when I realized that the assumed distance from viewer to object was a required variable. From there, it was pretty simple, all told.

Okay, so let's say you have an object of real dimensions w width, h height and d depth (I'm going to avoid subscripts and more complicated variables here). Furthermore, it is assumed to be x away from the viewer. w, h, d and x are all in whatever "real" units you like, be they meters or feet or rods. So if you want to draw a person 2m tall, 50cm wide and 30cm deep that needs to look like they're standing 10m away, then: x = 10m, h = 2m, w = 50cm, d = 30cm.

The object you draw on the paper will be have width W, height H, and really be X away from the viewer. So the person you draw might be only 25cm away from the viewer looking at the page. So the first step, even before addressing my original question, would be to decide how tall they need to be so that it appears like you're looking through a window at a person 10 m away. By some basic use of similar triangles, x/X = h/H = w/W. So if you want the image to look 40 times farther away than it really is (10m apparent, 25cm to page), then the image has to be 1/40th as big as the real thing it represents. That would give you a drawing of a person 5cm tall...a 5cm tall object 25cm away looks as big as a 2m tall object 10m away.

Similarly, if you know how big you want your image, you can figure out the apparent distance by working the other way. If you want to draw your figure at 12cm tall, then h/H = 16.7, and the apparent distance x will be 16.7 times X, or about 4.2 meters away if you assume a 25cm distance from drawing to viewer's eye.

Good so far? Okay, moving on.

In simple one-point perspective, things below eye level drop lower on the page as they get closer, while things above eye level move higher on the page as they get closer. Let F be the vertical distance on the page between the horizon line and a point on the front of your object. (This is an approximation for objects near the center of the page, by the way. You really want the distance along a line between the horizon and the viewer, but that's a lot harder to calculate, and for anything in the middle half of the page this is good enough.) Now let B be how far from the horizon line a point on the back of the object, directly behind the point used for F, needs to be.

If d=x, then the back of the object is twice as far away from the front, should look half as big, and should be halfway between the front of the object and the vanishing point. One point perspective means that the size of an object is proportional to how far it is from the vanishing point, again going back to similar triangles. A little reasoning along those lines leads to this fairly simple equation:

B = F / 2d/x

So, let's say you have a car that's 4m long, is supposed to be 10m from the viewer, and the center of the front bumper is 6cm below your horizon line. F = 6cm, d = 4m, x = 10m. So B = 6cm / 20.4 = 4.5 cm. The center of the rear bumper should be 4.5 cm below your horizon line, in order to give the car the proper depth.
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