Elliptical Arch Clearance for a Wide Load
An engineering team is checking whether a proposed pedestrian arch can carry a one-way flatbed truck. The opening is modeled as the upper half of an ellipse with a clear span of 16 m and a maximum height of 4 m at the centre; the road surface is the major axis. The truck is 2.40 m wide, and with its cargo it reaches 3.75 m above the road. The team first checks the truck centred in the lane, then checks the same truck after a 0.90 m lateral drift, and finally compares the ellipse with a parabolic arch of the same span and height, y = 4(1 − x²/64). The surveyor warns that the comparison must use the truck’s top outer corner, not its centreline, because that corner controls clearance. The team also wants to know how much lateral drift would leave less than 5 cm of clearance. All dimensions are in metres, and the arch is symmetric about the centreline.
1. Calculate the vertical clearance when the truck is centred and when it has drifted 0.90 m sideways.
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Take the centre of the road as x = 0 and write the ellipse as x²/8² + y²/4² = 1, so y = 4√(1 − x²/64). The controlling point is the truck’s top outer corner.
Centred, the outer corner is at x = 1.20 m: y = 4√(1 − 1.44/64) = 4√0.9775 = 3.955 m. Clearance = 3.955 − 3.75 = 0.205 ≈ 0.20 m.
With 0.90 m drift, the limiting corner is at x = 0.90 + 1.20 = 2.10 m: y = 4√(1 − 4.41/64) = 4√0.9311 = 3.860 m. Clearance = 3.860 − 3.75 = 0.110 ≈ 0.11 m.
The drift roughly halves the clearance, but the truck still fits.
2. Find the lateral drift at which the clearance falls to 5 cm.
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A 5 cm clearance means the arch height at the outer corner must be 3.75 + 0.05 = 3.80 m.
Set 3.80 = 4√(1 − x²/64). Then: (3.80/4)² = 0.9025 = 1 − x²/64 x²/64 = 0.0975 x² = 6.24 x = 2.498 m.
The allowed drift is d = x − 1.20 = 1.298 ≈ 1.30 m. At that drift the cargo corner just uses up the 5 cm margin. By symmetry, the same limit applies on the other side.
3. Compare the elliptical arch with the parabolic arch y = 4(1 − x²/64) at the centred position and at 0.90 m drift. Which shape is safer for the wide load?
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For the parabola, at the centred corner x = 1.20: y = 4(1 − 0.0225) = 3.910 m, clearance = 0.160 m. This is 4.5 cm less than the ellipse’s 0.205 m.
At 0.90 m drift, x = 2.10: y = 4(1 − 0.0689) = 3.724 m, clearance = 3.724 − 3.75 = −0.026 m. The cargo would strike the arch.
The ellipse is safer near the edge even though both shapes have the same span and crown height. The parabola falls away more quickly from the crown, so it gives less clearance exactly where a wide vehicle’s outer corner is controlling.