|
The new thread on reverse engineering, and the large
number who have asked me to explain how I calculate dimensions
from perspective pictures (such as the Kratzer dial
at Oxford), prompt me to post the answer, rather than replying
privately.
Given only a perspective painting or photograph of a 3D
object, you can easily find the true dimensions of the object. All you need is
three lines, each at right angles to the others in the 3D world. For instance,
for a horizontal dial use the 12 o'clock and 6am-6pm lines and a vertical.
For a vertical, such as a stained glass dial, the 12 o'clock line is
normally vertical, the top of the dial horizontal and you need to find another
horizontal line normal to the window, such as a wall. Measure the angles, in the
picture, between these lines, apply one of my formulae and you get the scale
factors. You can now easily reverse engineer the dial.
I found the equations from first principles. They are
very simple, so probably well-known. In fact, I can offer two sets of
equations. The first give the relative scale factors by which perspective has
reduced dimensions in each direction. The second are slightly more complex but
give the true scale factors.
First, the maths for the simplest case: Given a point O at
which three equal-length orthogonal lines OA, OB and OC meet,
their apparent lengths in a perspective picture are in the
proportions:
OA/sqrt(sin(2*BOC)) = OB/sqrt(sin(2*COA)) =
OC/sqrt(sin(2*AOB))
or
OA^2 / sin(2*BOC) = OB^2 / sin(2*COA) = OC^2 / sin(2*AOB)
where OA^2 means OA squared, sqrt means square
root, and BOC is the angle in the picture between OB and OC and therefore
opposite OA. If this is greater than 90 degrees, first subtract 90 from it or
subtract it from 180. If greater than 180, first subtract 180.
Although these equations give the relative proportions in
the three directions, you might sometimes want the true dimensions, especially
if you have two sets of orthogonal axes. By 'true' dimensions, I mean the
dimensions the picture of the object would have if seen face-on instead of in
perspective. For this, you can use:
OA^2 = 1 - cot(AOB)*cot(AOC)
OB^2 = 1 - cot(AOB)*cot(BOC)
OC^2 = 1 - cot(AOC)*cot(BOC)
where cot is cotangent (1/tangent). If all angles are
greater than 90 degrees, you may subtract them from 180 for convenience (but you
must not subtract 90 from them for these equations to work).
Note that the three orthogonal lines do not need to meet
at a point. If they don't, just measure the angles of the triangle they enclose.
If you label this triangle ABC, dimensions in the BC direction, for instance,
have been compressed by sqrt(1 - cot(ACB)*cot(ABC))
Next, how you use them:
- Measure the angles in the picture between three
orthogonal lines such as edges. Ideally, these should meet at a point. If not,
choose lines that are near one another in the picture (so the triangle is
small).
- Subtract from 180, or subtract 180 as needed to reduce
the angles to the range 0 to 90.
- Double each angle and calculate the square root of the
sine of the doubled angle.
You now have three numbers that are the relative scale
factors that represent how much dimensions have been compressed in the picture
due to perspective.
-optionally, calculate the cot of each angle and apply the
second set of equations.
The true proportions of the original 3D object can be
calculated by dividing any measured length by the compression factor. But be
aware that the factors vary across the picture and from top to bottom, unless it
was taken with a telephoto lens. So, for most accuracy, repeat the calculations
at all the points where you can identify three orthogonal lines.
Here's a concrete example, using a print of the photo on
the MHS Web site of the Kratzer dial in Oxford. I don't think I can
attach pictures on this mailing list, and I can't copy the photo without
permission, but can send an explanatory diagram to anyone who asks off
list.
Through the top left corner of the vertical south-facing dial, I drew a line which would be horizontal and
north-south on the octagonal west face. I measured the angles between
the vertical, East-West and South-North lines meeting there. The angles
opposite these lines are 155, 103 and 102 degrees
respectively.
Subtract 90 from these to give 65, 13 and 12
degrees. (Or, subtract each from 180, to give 25, 77 and 78, if you
prefer.)
For each, calculate sqrt(sin(2 * angle)), to give, approximately:
0.875, 0.66, 0.64
I measured the three lines as 44mm, 45mm and
77mm respectively.
Dividing these by the scale factors gives
44/0.875 = 50 units, 45/0.66 = 68
units,77/0.64 = 120 units
So the height of the south-facing vertical dial
is 50/68 or 74% of its width.
And the width of the octagonal West-facing dial is 2.4
times the length of its vertical edges.
Applying the second set of equations to the same angles
gives:
sqrt(1-cot(103)*cot(102)) = 0.975
sqrt(1-cot(102)*cot(155)) = 0.738
sqrt(1-cot(155)*cot(103)) = 0.711
so the true dimensions in the scale of the picture, if the respective lines
were viewed face-on, would be:
44/0.975 = 45mm
45/0.738 = 61mm
77/0.711 = 108mm
These are all about 10% smaller than the first method. They are the true
figures, but the first method gives correct relative results and is easier to
use so may be preferred for its simplicity.
As the camera was quite close to the subject, the
perspective changes over the picture, so for greater accuracy you could do the
same calculation at several other corners and average the results.
When I did this for the end of the polyhedral dial in The
Ambassadors, I found that it isn't square. Its proportions are, from memory,
about 0.85 : 1 which seems quite a big error, but perhaps results from Holbein's
inaccuracy. To the naked eye it looks square, I think.
Hope this is helpful to you
Regards
Chris Lusby Taylor
51.4N, 1.3W
|
- Calculating dimensions from perspective pictures; rever... Chris Lusby Taylor
