Perspective, projective geometry and the golden ratio myth
Tonight's idea
Put your eye at the origin, with \(x\) to the right, \(y\) up and \(z\) straight ahead, and hang a sheet of glass at distance \(d\) in front of you. A point \((X, Y, Z)\) in the world is seen where the line from your eye to it pierces the glass. By similar triangles that is
Take a railway track running straight away from you, rails at \((\pm 1, -h, Z)\) where \(h\) is the height of your eye above the ground. Their images are \((\pm d/Z, -dh/Z)\). As \(Z\) grows both coordinates shrink toward 0: the rails, which never meet in the world, meet in the picture at \((0, 0)\), the vanishing point.
This works for any family of parallel lines that are not parallel to the glass. A line through \(P\) in direction \((a, b, c)\), with \(c \ne 0\), is the set of points \(P + t(a, b, c)\). For large \(t\) its image heads to \(\left(\dfrac{da}{c}, \dfrac{db}{c}\right)\), which does not depend on \(P\). Horizontal directions have \(b = 0\), so their vanishing points all lie on \(y = 0\): the horizon, at eye level.
And the golden ratio? Claims that the Parthenon was built around \(\varphi \approx 1.618\) do not survive measurement. Where \(\varphi\) really lives is the regular pentagon (the stretch).
Watch one: the floor tiles
Eye height \(h = 1.6\) m, glass at \(d = 1\) m. A line across the floor at depth \(Z = 4\) m has \(Y = -1.6\), so its image is at \(y = -1.6/4 = -0.4\). At \(Z = 8\): \(y = -0.2\). At \(Z = 16\): \(y = -0.1\). Doubling the distance halves the drop below the horizon: equally spaced floor lines bunch up as they recede.
A 2 m fence post at \(Z = 4\) runs from \(Y = -1.6\) to \(0.4\), so its image runs from \(-0.4\) to \(0.1\): half a metre tall on the glass. At \(Z = 8\) it is a quarter of a metre. Size is inversely proportional to distance, and the top is above the horizon because the post is taller than you.
Your turn
- With \(d = 1\), find the image of \((3, 2, 6)\), of \((3, 2, 12)\) and of \((3, 2, 24)\). They lie on a line parallel to the \(z\)-axis. Where are their images heading?
- Find the vanishing point (\(d = 1\)) of the lines in direction \((1, 0, 2)\); of the lines in direction \((1, 1, 1)\); of the lines in direction \((1, 0, 0)\). Which of these lie on the horizon, and what happens to the third family?
- Railway sleepers lie across the track (as in the figure) at depths \(Z = 3, 6, 9, 12\) m. Your eye is 1.5 m above the rails and \(d = 1\). Find the four image heights \(y\) and the three gaps between consecutive images. What happens to equal spacing in the picture?
- A website draws a rectangle around a face in a painting, measures it at 10.0 cm by 6.2 cm, and announces the golden ratio. (a) What ratio did they get? (b) Their measurements are good to \(\pm 2\) mm. What are the smallest and largest ratios consistent with that? Does the data prefer \(\varphi = 1.618\) over \(1.6\) or \(5/3\)?
Stretch: where φ really lives
A regular pentagon has side 1 and diagonal \(d\). Triangle \(ACD\) has angles 36°, 72°, 72°, and the bisector of the angle at \(C\) meets \(AD\) at \(Q\). Find a triangle in the figure similar to \(ACD\), explain why, and deduce \(DQ = 1/d\). Then explain why \(AQ = 1\), deduce \(d = 1 + 1/d\) and solve it.
Answers
- \((0.5, 0.333)\), \((0.25, 0.167)\), \((0.125, 0.083)\). Each doubling of depth halves both coordinates; the images head to \((0, 0)\), the vanishing point of all lines parallel to the \(z\)-axis.
- \((1, 0, 2)\): \((0.5, 0)\), on the horizon. \((1, 1, 1)\): \((1, 1)\), above the horizon since the lines rise as they recede. \((1, 0, 0)\) has \(c = 0\): no vanishing point. Those lines are parallel to the glass and their images stay parallel and horizontal.
- \(y = -1.5/Z\): \(-0.5, -0.25, -0.167, -0.125\). Gaps: \(0.25, 0.083, 0.042\). Each gap is \(1.5\left(\frac1Z - \frac1{Z+3}\right)\), shrinking roughly like \(1/Z^2\). Equal spacing becomes spacing that bunches up toward the horizon.
- (a) \(10.0/6.2 \approx 1.613\). (b) Smallest \(9.8/6.4 \approx 1.53\), largest \(10.2/6.0 = 1.70\). That range contains 1.6, \(\varphi\), 5/3 and a lot else; the measurement cannot tell them apart. And the person drawing the rectangle chose where its edges went, which painting to use, and which of hundreds of possible rectangles to measure. With that much freedom you can find a golden rectangle in almost any picture, which is why finding one proves nothing.
Stretch. Triangle \(CDQ\) has angles 36° (at \(C\), half of 72°), 72° (at \(D\)) and so 72° at \(Q\): similar to \(ACD\). Side \(CD = 1\) corresponds to \(AC = d\), so \(\dfrac{DQ}{CD} = \dfrac{CD}{AD}\), giving \(DQ = 1/d\). Triangle \(ACQ\) has 36° at \(A\) and 36° at \(C\), so \(AQ = CQ\); and \(CQD\) has two 72° angles, so \(CQ = CD = 1\). Then \(d = AD = AQ + QD = 1 + 1/d\), so \(d^2 - d - 1 = 0\) and \(d = \dfrac{1 + \sqrt5}{2} = \varphi\). The diagonal of a regular pentagon is \(\varphi\) times its side. That is real; the Parthenon is not.