The shape of a bridge
Tonight's idea
On a suspension bridge the road deck hangs from the main cables by vertical rods called hangers. The weight of the deck is spread evenly along the span, and a cable loaded evenly like that hangs in a parabola.
Put the origin at the cable's lowest point, in the middle of the span. Measure \(x\) sideways from there and \(y\) upwards, both in metres. The parabola is then as simple as it can be.
\(y = ax^2\)
It is lowest at \(x = 0\). And \(x\) and \(-x\) give the same height, so the two halves are mirror images.
One measurement fixes \(a\). Say the tower tops are \(d\) metres out from the middle and \(h\) metres above the lowest point. Then \((d, h)\) is on the curve, so \(h = ad^2\) and \(a = \dfrac{h}{d^2}\).
Now the height of the cable is known at every \(x\). Add the gap between the lowest point and the deck, and you have the length of every hanger before any steel is cut.
A parabolic arch is the same curve turned upside down.
Watch one: a 200 m span
The towers are 100 m from the middle. Their tops are 25 m above the cable's lowest point. So \(a = \dfrac{25}{100^2} = 0.0025\) and \(y = 0.0025x^2\). The lowest point is 2 m above the deck, so a hanger is \(y + 2\) metres long. The left half is the same, by symmetry.
| \(x\) (m) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| \(y = 0.0025x^2\) | 0 | 1 | 4 | 9 | 16 | 25 |
| Hanger (m) | 2 | 3 | 6 | 11 | 18 | tower |
Your turn
- A cable follows \(y = 0.01x^2\) between towers at \(x = -50\) and \(x = 50\). (a) How high above its lowest point is the cable at \(x = 10\), \(x = 30\) and \(x = -30\)? (b) How high are the tower tops above the lowest point?
- A footbridge spans 80 m. Its cable rises 8 m from its lowest point to the tower tops. (a) Find \(a\). (b) The lowest point is 1 m above the deck. Find the hanger lengths at \(x = 0\), 10, 20 and 30.
- The main span of the Golden Gate Bridge is about 1,280 m. Its cables dip roughly 144 m from the tower tops to the middle. (a) Find \(a\). (b) How high above its lowest point is the cable 160 m, 320 m and 480 m from the middle? (c) Half way out to a tower, what fraction of the full rise has the cable climbed?
- A parabolic arch over a road is 8 m wide at the ground and 6 m high in the middle. Measured down from the top, the arch drops \(ax^2\) at a distance \(x\) from the centre line. (a) Find \(a\). (b) A truck 3 m wide and 4.5 m tall drives down the centre. Does it fit? (c) Now it must keep to one lane, with its near side 0.5 m from the centre line. Does it fit?
Stretch: the same shape every time
A cable rises \(h\) at towers a distance \(d\) from the middle. Show that a fraction \(t\) of the way out to a tower, the height is \(t^2 h\), whatever the bridge. At what fraction of the way out has the cable climbed half of \(h\)?
Answers
- (a) \(0.01 \times 100 = 1\) m, \(0.01 \times 900 = 9\) m, and 9 m again, by symmetry. (b) \(0.01 \times 2500 = 25\) m.
- (a) The towers are 40 m from the middle, so \(a = 8/40^2 = 8/1600 = 0.005\). (b) \(y = 0, 0.5, 2, 4.5\), so the hangers are 1 m, 1.5 m, 3 m and 5.5 m.
- (a) \(d = 640\), so \(a = 144/640^2 = 144/409600 \approx 0.00035\). (b) 9 m, 36 m and 81 m. (c) At 320 m the cable is 36 m up, and \(36/144 = \tfrac14\). Half way out, only a quarter of the climb is done. The cable is nearly flat in the middle and steep at the towers.
- (a) The ground is 4 m from the centre line and 6 m below the top, so \(6 = 16a\) and \(a = 0.375\). (b) The truck's sides are at \(x = 1.5\). There the arch is \(6 - 0.375 \times 1.5^2 = 5.16\) m high. That clears 4.5 m, so it fits. (c) Its far side is now at \(x = 3.5\), where the arch is only \(6 - 0.375 \times 3.5^2 = 1.41\) m high. It does not fit.
Stretch. A fraction \(t\) of the way out means \(x = td\). Then \(y = a(td)^2 = t^2 \times ad^2 = t^2 h\), because \(ad^2 = h\). For half the height, \(t^2 = \tfrac12\), so \(t = \dfrac{1}{\sqrt2} \approx 0.71\). The cable does half its climbing in the last 29% of the way.