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The Curvature of the Parthenon.
(an excercise in projective geometry)

(Photo SC, 1972)
 The Parthenon, completed over 2400 years ago, was the temple of Athena Parthenos (virgin), patron goddess of Athens.  As the photograph shows, its base  was constructed with a distinct upward curve.

 Accurate information is difficult to obtain, but it is believed that this curve was deliberately created so that from the correct vantage point, the base would appear to be straight.

 This excercise  is structured to help you gain an insight into the nature of visual perception, and, using points in a projective plane, formulate some principles and thus determine models which interpret the Parthenon phenomenon and develop your understanding further.

Definitions

Projective geometry puts the eye of the observer at the origin O, of IR³ and considers paths of rays of light approaching O as they intersect with (are embedded in) a plane, like a screen between the observer and the objects being observed:

points in the standard embedding plane

1) The "standard embedding plane" is the plane z = 1.
2) Lines through the origin are called projective points because they intersect at a point with the standard embedding plane. (Unless they are in the plane z = 0)
3) Planes through the origin are called projective lines because their intersection with the standard embedding plane is a straight line.  (Apart from the plane z = 0)

The Problem

1) Why could a rectangular building appear not to have straight edges to someone standing in front of it?

(2) An observer is standing at the point [0,0,0], and facing [0,0,1].  Determine a  formula for how the face of a building with vertices at the projective points [-1, 0.5, 1],    [1, 0.5, 1], [1,-0.5, 1], [-1, -0.5,1] might appear  to this observer.  Plot your results on a graph.

(3)      Determine formulae for curves in the projective plane to replace the lines which form the edges of the face of the building, so they would appear to be straight to your observer.  Plot your results on a graph.

(4) In mappings between projective planes, straight lines are always mapped to other straight lines.  People refer to the    "standard embedding plane".  Solving this problem has effectively  necessitated an alternative to this.  Explain!  Comment upon the realism of   your model.

Hints

(1) In a total eclipse of the sun, the moon just covers the face of the sun.  We can see the corona but we do not get a 'ring doughnut' effect.  Effectively, the moon appears to be the same size as the sun, even though it is much smaller.

 This is because the angle subtended by both to our eye is the same.  On the other hand, an object receding from us appears to get smaller.  This is because the angle subtended to our eye gets smaller.  The apparent size of something can be modelled as being proportional to the angle subtended to the pupil of our eye.  Hence, a spider close to us could eclipse a tree, and would appear much bigger than a spider further away.

(2)  Try mapping the [x,y,1] coordinates in the projective plane, to (qx,qy) angular displacement coordinates in the observer's visual frame of reference.  Firstly, what does [0,0,1] map to?

 Follow up problems

(1) If the observer is facing [-1,0,1], what would the (qx,qy) coordinates of our building be now?  How would the apparent shape of the building change?

(2) Suppose the vertices of the face of our building were now at  [0,0.5,1], [2, 0.5, 2],  [2,-0.5, 2], [0,-0.5,1].

 (i) Sketch the perspective of this in the standard embedding plane.
 (ii) Sketch the appearance to the observer in (qx,qy) coordinates.
   Compare the results.

(3) What is the approximate angle of your range of vision?  For what  proportion of your range do you think that a perspective in the projective plane is a good approximation to what you see?  Why?

Partial Solution

(1) As the centre of the building is closer to us than the edges, features subtend larger angles and hence appear bigger, as in the exaggerated diagram below:


(This is an approximation, one would appear to get sideways "bulge" too.)

(2)  Following the hint, and the diagram below.

 For the general projective point: [x,y,1]; OP² = 1 + x² +y²    => OP = Ö(1 + x² +y²)

sinq = PPx /OP = y/Ö(1 + x² +y²);  sinqx = PPy /OP = x/Ö(1 + x² +y²).

Hence: q = arcsin(x/Ö(1 + x² +y²)), and q =arcsin(y/Ö(1 + x² +y²)).

 The top and bottom of the building are the projective lines  y = ±½z, Hence:

[x, ±½, 1]  is mapped to (qx,qy) = (arcsin(x/Ö(1 + ¼ + x²)), ±arcsin(1/Ö(1 + ¼ + x²))

(see graph, below):

(3)   You need to find the inverse transformation of, say, (qx, const), mapping onto   [x, y, 1] for values of qx between ±p/4.

 To have a constant apparent  height, we need  q = h radians, where sin(h) = s.

 This implies that  s = y/Ö(1 + x² +y²),  => y² = s²(1 + x² +y²).  Hence:

 y²(1- s²) = s²(1 + x²); =>  y = ±(s/Ö(1- s²))Ö(1 + x²), = ±tan(h)Ö(1 + x²)

(4) Instead of embedding in a plane, the (qx , qy ) system means we are  embedding and projecting onto a unit sphere.  As our retinas are curved surfaces, this is not an unreasonable assumption.

Graph of solution to part 2
apparent curvature of a rectangular building

The values were calculated from the formulae above with the Lotus 1-2-3 spreadsheet and are tabulated below.  (as before, this diagram is only an approximation as vertical lines would  appear similarly curved.)

x     qx         qy         -qy
-1     -0.6667        0.3333     -0.3333
-0.9   -0.6270        0.3484     -0.3484
-0.8   -0.5819        0.3637     -0.3637
-0.7   -0.5307        0.3790     -0.3790
-0.6   -0.4729        0.3941     -0.3941
-0.5   -0.4082        0.4082     -0.4082
-0.4   -0.3369        0.4211     -0.4211
-0.3   -0.2592        0.4319     -0.4319
-0.2   -0.1761        0.4402     -0.4402
-0.1   -0.0891        0.4454     -0.4454
0       0             0.4472     -0.4472
0.1     0.0891        0.4454     -0.4454
0.2     0.1761        0.4402     -0.4402
0.3     0.2592        0.4319     -0.4319
0.4     0.3369        0.4211     -0.4211
0.5     0.4082        0.4082     -0.4082
0.6     0.4729        0.3941     -0.3941
0.7     0.5307        0.3790     -0.3790
0.8     0.5819        0.3637     -0.3637
0.9     0.6270        0.3484     -0.3484
1       0.6667        0.3333     -0.3333

qx and qy are both subjects of parametric equations with x as the variable.
The plotted points would be equidistant in the real world!

Graph of solution to part 3

The values in this partial solution were calculated from the formulae above with the Lotus 1-2-3 spreadsheet and are tabulated below.

       qx                      x            y            -y        qy(±)
-0.70000     -0.84229     0.47726     -0.47726      0.35000
-0.60000     -0.68414     0.44228     -0.44228      0.35000
-0.50000     -0.54630     0.41595     -0.41595      0.35000
-0.40000     -0.42279     0.39631     -0.39631      0.35000
-0.30000     -0.30934     0.38209     -0.38209      0.35000
-0.20000     -0.20271     0.37245     -0.37245      0.35000
-0.10000     -0.10033     0.36686     -0.36686      0.35000
0.00000       0.00000     0.36503     -0.36503      0.35000
0.10000       0.10033     0.36686     -0.36686      0.35000
0.20000       0.20271     0.37245     -0.37245      0.35000
0.30000       0.30934     0.38209     -0.38209      0.35000
0.40000       0.42279     0.39631     -0.39631      0.35000
0.50000       0.54630     0.41595     -0.41595      0.35000
0.60000       0.68414     0.44228     -0.44228      0.35000
0.70000       0.84229     0.47726     -0.47726      0.35000

The plotted points have equal angular displacement vertically (with the arbitrary value of 0.35 radians, and are plotted at equal angular intervals horizontally.  Similar curves would be necessary in the vertical lines (assuming our model is correct).

  Thinking point

 We have effectively assumed that our retina is a section of a sphere  This is unlikely to be true exactly, hence our model can be only an approximation.  Also, most of us have two eyes, and the brain interpolates, so what we perceive and what our eyes "see" will not be the same.

© C. A. Clarkson 22/10/1998 All rights reserved!
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