Parabolic arch bridge: clearance, focus and model checking
A road passes under a masonry arch that springs from ground level on both sides. The clear span is $12.0\ \text{m}$ and the crown is $5.0\ \text{m}$ above the road at the centre of the span. The arch profile was designed as a parabola, with the origin taken on the road surface directly beneath the crown. A haulage company wants to send a rigid truck through, 3.0 m wide and 4.2 m tall including its load, and needs to know how accurately it must be steered. Separately, a surveyor checking the arch for earthquake assessment measures the height at a horizontal distance of 3.0 m from the crown as 3.75 m, and argues that a circular arc through the same three points (both springing points and the crown) would fit the data just as well. The council engineer has to decide whether the parabolic model has any real predictive value before signing off the truck route.
1. Derive the equation of the parabolic arch and write it in the form $x^2 = -4p(y-k)$. State the coordinates of the focus and the equation of the directrix, and explain what the focus means for this structure.
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Setting up. Take the crown at $(0, 5)$ and the springing points at $(\pm 6, 0)$. A downward-opening parabola with vertex $(0,5)$ has the form $y = 5 - ax^2$.
Fixing $a$. At $x = 6$, $y = 0$: $0 = 5 - 36a$, so $a = \tfrac{5}{36}$ and
$$y = 5 - \tfrac{5}{36}x^2 .$$
Standard form. Rearranging, $\tfrac{5}{36}x^2 = 5 - y$, so
$$x^2 = -\tfrac{36}{5}(y - 5) = -7.2(y - 5).$$
Comparing with $x^2 = -4p(y - k)$ gives $k = 5$ and $4p = 7.2$, so $p = 1.8$.
Focus and directrix. The parabola opens downward, so the focus lies $p$ below the vertex: focus $(0,\ 3.2)$, directrix $y = 5 + 1.8 = 6.8$.
Meaning. Every point on the arch is equidistant from the focus and the directrix. For a parabolic arch this is a structural statement about how the load path is distributed, and it is also the property that makes such a curve useful for reflectors and acoustic shells.
2. The truck is 3.0 m wide and 4.2 m tall. Determine how far its centreline can be from the crown line and still clear the arch. Show the reasoning, not just a number.
Read the worked answer
Clearance condition. The truck's roof corners are its highest points, at horizontal offsets $|x| = |c| + 1.5$ from the centreline if the centreline sits $c$ metres from the crown. The truck clears where the arch height at that $x$ is at least 4.2 m:
$$5 - \tfrac{5}{36}x^2 \ge 4.2 \ \Rightarrow\ \tfrac{5}{36}x^2 \le 0.8 \ \Rightarrow\ x^2 \le 5.76 \ \Rightarrow\ |x| \le 2.4\ \text{m}.$$
Translate to steering tolerance. The outermost roof corner must stay within 2.4 m of the centre, so the centreline must satisfy
$$|c| \le 2.4 - 1.5 = 0.9\ \text{m}.$$
Interpretation. A $0.9\ \text{m}$ tolerance is tight for a 3 m vehicle but workable if the lane is marked down the middle. If the lane centre is instead placed, say, 1.5 m off the crown line, the truck cannot pass at 4.2 m at all - the constraint is a usable-lane question, not merely a vehicle question. A sensible operational answer is to require the lane to be marked within $\pm 0.9\ \text{m}$ of the crown line, or to impose a height limit that leaves a clearance margin (for 0.3 m clearance, $|x| \le 1.90$ and the tolerance drops to $0.40\ \text{m}$).
3. Check the surveyor's 3.75 m measurement against your parabola. Then find the circular arc through the two springing points and the crown, and evaluate its height 3.0 m from the crown. Decide which model the survey supports and justify the decision numerically.
Read the worked answer
Parabola check. At $x = 3$: $y = 5 - \tfrac{5}{36}(9) = 5 - 1.25 = 3.75\ \text{m}$, exactly the surveyed value.
Circular model. For a circle with chord $c = 12$ and rise $h = 5$, the radius is $R = \dfrac{h}{2} + \dfrac{c^2}{8h} = 2.5 + \dfrac{144}{40} = 2.5 + 3.6 = 6.1\ \text{m}$. The centre sits $6.1$ m below the crown, at $(0, -1.1)$, so the circle's upper arc is $y = \sqrt{6.1^2 - x^2} - 1.1$.
At $x = 3$: $y = \sqrt{37.21 - 9} - 1.1 = \sqrt{28.21} - 1.1 = 5.311 - 1.1 = 4.21\ \text{m}$.
Decision. The two models differ by $4.21 - 3.75 = 0.46\ \text{m}$ at the survey station - about half a metre, far outside plausible survey error on a masonry arch. The measurement matches the parabola to the centimetre and misses the circle by nearly 0.5 m, so the parabolic model is supported and the circular alternative is rejected. With three points alone a circle and a parabola cannot be distinguished, so the extra measurement at $x = 3$ is exactly what makes the model choice evidence-based rather than a matter of taste.