2018-06-23
How to Draw Circles in Perspective like a Geometer
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There are a lot of tutorials and guides in the net on the subject
of drawing ellipses, but they — invariably to my knowledge — follow
the same formula: project a square in perspective, then sketch an
ellipse inside it with guides.
The ghost of Euclid cries in agony every time someone draws an
ellipse like that. This guide will tell you how to do it precisely
using the classic tools: a ruler, a compass, and additionally a
piece of string and two pins, to render the ellipse itself.
We'll start with a basic setup to project objects in perspective
from a top and a side view:
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Under the canvas is the top view, and on the right the side view,
each with viewpoints (VP). The bottom and right segments of the
canvas are used as the 2D screen to project onto. The vanishing
point (marked with an X) and the horizon can be found by drawing
lines parallel to the projection lines from the viewpoints.
Let's start by projecting a rectangle we've drawn around the
circle. This step is a red herring when using our exact method, but
it'll work as an introduction to projecting in perspective and also
as a reality check when we finish constructing the ellipse.
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This is pretty simple. Draw rays from the viewpoints through the
points you want to project. Where the rays cross the projection
line (bottom and right), draw perpendicular lines across the
canvas. The points can be found in the intersections of the
relevant lines.
Let's put the square away and take the first steps of our method.
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First we'll find out the leftmost and rightmost points of the
ellipse on the canvas by projecting the tangent points Tleft and
Tright. Draw rays from the top viewpoint on the projection line
across the tangent point. (You can find the tangent points by using
the fact that the angle Ctop - Tleft - VPtop is a straight angle.
You know Ctop - Tleft and Ctop - VPtop, so you can find out Tleft -
VPtop using Thales's_theorem.)
Measure the lengths of the tangent points from the projection line
(Tleft - Tleftx and Tright - Trightx) and use them to find out the
distances of the tangent points (T''right, T''left) on the side
view. Project, perpendiculars, crossings, ready. We've found the
tangent points and one conjugate_diameter. We need another of
those, so on to the next step.
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Here we go about finding T'far. T'left and T'right are the tangent
points, so the vertical lines across them are the tangents. The
smaller conjugate diameter has to be parallel to these, so it must
be somewhere along the line that crosses Cparal. Now we "project
backwards": draw that vertical line, and where it crosses the
projection line, draw to VPtop. Now we found Tfar and Tnear. We
only need to find one of their distance from the projection plane
(Tfar - Tfarx) to get both points, since they are the same distance
from Cparal.
We can do a sanity check by doing the same a bit harder way. You
can skip this step. Honestly it's here because I didn't see the
easier method before doing the graphics...
As you might have noticed, all points that go through a ray from
VPtop, are projected in the same vertical line on the canvas. So,
points that lie on some other ray will form a line that is parallel
to the first one. Imagine drawing a bunch of rays from a single
point and taking a photograph from the exact point where they
converge. The rays will seem like a bunch of parallel vertical
lines on the photo.
Turns out that any rays projected from some other point (like point
K here) that is equally far from the projection line as your
viewpoint, any rays drawn from that point will seem parallel on the
canvas. We use this fact to find out T'far.
Draw a ray from Tright through Tleft to a line that goes through
VPtop and is parallel to the projection line. Where it crosses, is
our point K. We want to find tangent line T'far that is parallel
with T'left - T'right, so we draw a ray from K to Tfar (that we
found using the right angle triangle as before).
Now we have the conjugate diameters, and must find the major and
minor axes of the ellipse using a procedure described in
http://www3.ul.ie/~rynnet/swconics/EP's.htm. (Another choice would
be Rytz's_construction).
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The blue segments are the diameters from the previous step.
From one of the ends of the shorter diameter (T'far) draw a segment
(T'far - A) that is as long as half of the longer diameters,
perpendicular to the longer diameter.
Then a segment from A to the center (Cparal). Find the middle of
this segment, call it B.
Draw a circle with the center in B, with radius A-B == B-Cparal.
Then T'far to C on the circle. Mark the point where this line first
crosses the circle as D. The illustration is a bit messy, so here's
a detail:
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Now from C to Cparal. This is the direction of the minor axis.
Get the length of T'far to D. This is the length of half of the
minor axis. T'far to A is the length of half of the major axis,
and they are of course perpendicular to each other. Now we should
have both the axes.
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Now just take half the length of the major axis (Cparal to
Major'right or Major'left), and draw a circle from one of the small
axis points (Minorfar or Minornear). Where it crosses the major
axis (Focusleft, Focusright) are the foci. Take your pins, stick
them into these points. Measure a loop of string such that it spans
Focusleft, Focusright, and Minorfar. Stick your pen inside and draw
an ellipse. Enjoy the feeling of accomplishment, shed the shame of
drawing potatoes and calling them circles in perspective. May the
ghost of Euclid smiles at you.
When we superimpose the square we projected in the first step, it
indeed looks sane:
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Now, as simple as this might seem in hindsight, this bugged me for
years. There are lots of assumptions you might make that turn out
wrong. I learned this the hard way. Here are all the things I tried
superimposed before I found the Truth:
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Everything on this page was drawn with GeoGebra. You can download
the source from Google_Drive to play with. Try moving the vanishing
point, viewpoints and the circle!
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