Skip to content
Coordinate Geometry · Mathematics

Parabolic Arch Bridge Simulation: Parabola vs Catenary

Why are some bridges parabolic — and why not elliptical or circular? In 3D: a river bridged by a parabola, semi-ellipse, hyperbola, circle or catenary with traffic on the deck and a boat testing the headroom, the NCERT arch problems, the push of a parabolic arch on its banks (H = wL²/8h), the same load bending every other shape (the rib turns red), a lorry whose thrust line kinks as it drives across, a chain held in a hand and flipped into an arch, own weight against deck load, and a Roman semicircle that stands only while its thrust line fits inside the stone.

Interactive virtual lab · no download · works on phones and laptops

Chapter
Conic Sections
For
Class 9–12 · JEE Main · JEE Advanced
Time
~30 min

How to use it

  1. 1Pick a module from the list (8, in three groups); set your level at the top right.
  2. 2Every scene runs by itself — drag the orange knobs (the crown, the boat, the lorry, the hand holding the chain, the ring's thickness) to drive it.
  3. 3Use Predict before you look, then Derive, Questions, Practise and Compare; the bending graphs sit under the stage.

What to notice

  • Arch problems: put the origin at the vertex () or the centre of the base ().
  • Under a load uniform along the deck a parabolic arch is in pure compression; its thrust on each bank is .
  • For any shape, : the further an arch strays from its thrust line, the more it bends — the semi-ellipse most.
  • A hanging chain is a catenary, ; inverted, it is the ideal arch for its own weight. Point loads give a polygon.
  • Stone can only push: a masonry arch stands if a thrust line fits inside it, which a semicircle manages if it is about 11 % of its radius thick.

For teachers

Teach it in 32 minutes

A lesson plan, questions to put to the class, a quick quiz you can project or print, and the formulas behind Bridge Arch 3D Lab.

  1. 1Hook

    ~2 min
    • Open the simulation full screen on the projector. Before touching anything, ask the class what they expect to see, and write two or three predictions on the board.

  2. 2Demonstrate

    ~18 min
    • Pick a module from the list (8, in three groups); set your level at the top right.
    • Every scene runs by itself — drag the orange knobs (the crown, the boat, the lorry, the hand holding the chain, the ring's thickness) to drive it.
    • Use Predict before you look, then Derive, Questions, Practise and Compare; the bending graphs sit under the stage.

  3. 3Pause and ask

    ~5 min
    • Have a student say this in their own words: Arch problems: put the origin at the vertex (x² = -4py) or the centre of the base (x²/((L/2)²) + y²/h² = 1).
    • Have a student say this in their own words: Under a load uniform along the deck a parabolic arch is in pure compression; its thrust on each bank is H = wL²/8h.
    • Have a student say this in their own words: For any shape, M = H (yT - y): the further an arch strays from its thrust line, the more it bends — the semi-ellipse most.
    • Have a student say this in their own words: A hanging chain is a catenary, y = acosh()/xa; inverted, it is the ideal arch for its own weight. Point loads give a polygon.
    • Have a student say this in their own words: Stone can only push: a masonry arch stands if a thrust line fits inside it, which a semicircle manages if it is about 11 % of its radius thick.

  4. 4Check

    ~7 min
    • Run the 7-question quiz below: project it and take answers by show of fingers (1–4), or print it as a worksheet.

About the Bridge Arch 3D Lab simulation

The Bridge Arch Lab asks why some bridges are parabolic, and why not elliptical or circular. Over a 3D river with traffic crossing the deck, the same span and rise can be bridged by a parabola, a semi-ellipse, a hyperbola, a circular arc or a catenary; a sailing boat tests the headroom near the banks, and the NCERT arch problems (a parabolic arch 10 m high and 5 m wide, a semi-elliptical arch 8 m wide and 2 m high) are worked on a board with draggable probes.

Then the engineering. A parabolic arch under a load spread evenly along its deck is in pure compression, and arrows show it pushing on each bank with H = wL²/8h sideways and wL/2 down — drag the crown down and the push soars. Give the same load to the other shapes and the rib is coloured by its bending moment, M = H × (distance from the thrust line): grey for the parabola, red for the semi-ellipse. A lorry driving across kinks the thrust line into a polygon. A chain held in a hand hangs in a catenary, y = a cosh(x/a), and can be flipped into the ideal arch for its own weight; sliding the share of load between the arch's own weight and the deck moves the ideal shape between the catenary and the parabola. Last, a semicircular Roman arch of stone blocks stands only while a thrust line fits inside the stone — about 11 % of its radius thick — so Roman bridges work because they are thick, not because the semicircle is ideal.

Also covers: parabolic arch, parabolic arch bridge, why are bridges parabolic, arch bridge parabola, semi elliptical arch, arch in the form of a parabola, arch in the form of a semi ellipse, ncert arch problem, catenary, catenary vs parabola, hanging chain, cosh, funicular, thrust line, line of thrust, three hinged arch, horizontal thrust wl^2/8h, bending moment arch, suspension bridge cable parabola, gateway arch catenary, roman arch, semicircular arch, masonry arch, applications of parabola, real life application of conic sections.

Frequently asked questions

Why are many arch bridges parabolic?

For an idealised arch carrying a load spread evenly along the horizontal (a heavy, uniform deck), the parabola is the funicular shape: the line of thrust lies exactly along the arch, so it is squeezed everywhere and does not bend. Stone and concrete are strong in compression and weak in bending, so following the thrust line is efficient. Other shapes with the same span and rise carry the same thrust but also bend by M = H × (their distance from the thrust line).

Is a parabolic arch always the best shape for a bridge?

No. The ideal shape depends on the load. A load uniform per horizontal metre gives a parabola; an arch carrying mainly its own weight (uniform per metre along the arch) is ideal as a catenary; point loads give a polygon. Moving traffic keeps changing the load, so real arches are built stiff enough to carry some bending, and the span, materials and site matter too.

What is the difference between a catenary and a parabola?

A catenary, y = a cosh(x/a), is the shape of a chain hanging under its own weight. A parabola is the shape of a cable whose load is uniform along the horizontal, such as a suspension bridge cable carrying a deck. For a shallow sag the two are almost identical (cosh u ≈ 1 + u²/2); they differ clearly when the sag is deep. Turned upside down, each becomes the ideal arch for the same load.

Why do semicircular Roman arches stand if the semicircle is not ideal?

Stone joints can only push, so a masonry arch stands if some line of thrust fits inside its thickness (Heyman's safe theorem). A semicircle is not the funicular shape, so its thrust line wanders across the ring; a ring thick enough — in this lab's model about 11 % of its radius for its own weight — leaves room for it. Roman arches were much thicker than that.

Try next

All simulations