The shape that carries itself, and the arch that is its reflection

A chain hanging between two points has no choice about its shape. It cannot resist bending, so the only configuration available is the one in which every part of it is in pure tension along its own length — and that shape is determined entirely by the loads.

Turn the whole thing upside down. Every tension becomes a compression of the same magnitude, and the inverted shape carries the identical load in pure compression, with no bending anywhere. That is an arch, and the relationship is exact rather than analogical.

Why bending disappears #

A cable has no bending stiffness. It cannot carry a moment because it would have to be stiff to do so, and it is not.

That sounds like a limitation and is the whole mechanism. Since the cable can only carry force along itself, the shape it settles into must be one in which the loads are balanced by axial force alone. If any bending were required at some point, the cable would move until none was — and where it stops moving is the funicular shape for that load.

So the cable is a form-finder. It does not resist the load in a shape chosen by a designer; it finds the shape in which the load can be resisted without bending, and adopts it.

The shape depends on the load, which is the part most often glossed. Under a load uniform along the horizontal — a suspension bridge deck — the shape is a parabola. Under a load uniform along the cable itself — a chain carrying only its own weight — it is a catenary, the hyperbolic cosine. Under a set of point loads it is a polygon with a vertex at each load.

The two curves are genuinely different functions and they are very hard to tell apart at ordinary sags. Galileo asserted the hanging chain was a parabola; it took Huygens, Leibniz and Johann Bernoulli, in 1691, to establish that it is not.

Inverting it #

Hooke published the principle in 1675 as an anagram, having found it by drawing rather than by summing forces, which decoded reads: as hangs the flexible line, so but inverted will stand the rigid arch.

The inversion works because equilibrium does not care about sign. A set of forces in equilibrium remains in equilibrium if every force is reversed, so a shape that balances a load in tension balances the same load in compression when turned over. Nothing about the geometry changes.

That makes the hanging model a design instrument, and it was used as one. Poleni analysed the cracked dome of St Peter’s in 1748 by hanging a chain loaded to represent the dome’s weight and checking that the inverted curve lay within the masonry. Gaudí spent ten years on a hanging model of strings and small bags of shot for the Colònia Güell chapel, photographing it and turning the photographs upside down to obtain the elevations.

What the supports have to do #

A cable pulls inward and downward on whatever holds it. An arch pushes outward and downward. Neither can be supported by a vertical reaction alone, and that horizontal component is the whole difficulty of both.

The horizontal force is the same everywhere along the cable — it has to be, since nothing horizontal is applied between the ends — and it is inversely proportional to the sag. A shallow cable carrying the same load has a large horizontal pull; a deep one has a small one.

$$H = \frac{wL^2}{8d}$$

for a uniform load $w$ over span $L$ with sag $d$. Halving the sag doubles the thrust.

That is the same cable drawn twice, and the two drawings are the whole of the trade. The deeper curve above sags 170 units across a 520-unit span; the one below sags 85 across the same span, which is exactly half.

The two curves carry the same load over the same span and differ by a factor of two in what they demand of the ground, which is why sag is chosen by the abutments rather than by the cable. A designer who flattens a cable to gain headroom has moved the problem into the foundations and not removed it.

For an arch that thrust has to be resisted at the springing, and the history of arch construction is largely the history of what resists it: massive abutments cut into rock, flying buttresses transferring it down a pier, or a tie across the base turning the whole thing into a self-contained unit. A tied arch needs no horizontal reaction at all, at the cost of a member in tension across the span — which is a truss, approached from the other direction.

The thrust line, and a lower bound #

The funicular idea gives masonry its assessment method, and the method is unusual in that it proves a structure is safe without knowing what it is actually doing.

Masonry has no useful tensile strength. It stands because compression finds a path through it. The thrust line is the funicular polygon for the actual loads: the line along which the compressive resultant travels from the crown down to the foundation.

If a thrust line can be drawn that stays inside the masonry everywhere, then a set of internal forces exists that is in equilibrium with the loads and requires no tension. That is enough to conclude the arch can stand. It does not say the arch is carrying load along that particular line — it almost certainly is not — only that at least one safe possibility exists.

That is the safe theorem of plasticity, and it is why Heyman’s analysis of Gothic cathedrals works with a straightedge. It also explains the middle-third rule: if the thrust line stays within the middle third of the section, no tension is required anywhere across it.

The corollary is more interesting. A masonry arch that has cracked has not failed — it has moved to a geometry in which a thrust line fits. Cracks in old arches are usually evidence of successful adjustment rather than of impending collapse, which is the opposite of the intuition.

Two curves, from one free body #

The claim that a uniform horizontal load gives a parabola and a uniform load along the cable gives a catenary is usually asserted. Both fall out of the same free body, and doing it once shows precisely where the two part company.

Cut the cable at its lowest point and again at some station along it, and keep the piece between. Three things act on that piece: the horizontal force $H$ at the low point, the tension along the cable at the far cut, and the weight of the piece itself. There is nothing horizontal applied anywhere in between, so the horizontal component of the tension is $H$ at every station — the horizontal force in a cable is constant along it, which is the single most useful fact about cables and comes from one force sum.

The vertical component $V$ at the cut is simply the weight of the piece hanging below. And since the cable can carry no moment, its tangent must be along the resultant of $H$ and $V$:

$$\frac{dy}{dx} = \frac{V}{H}.$$

Everything now depends on how the weight of the piece accumulates.

Load uniform along the horizontal. A suspension bridge’s cable carries a deck, and the deck’s weight is spread evenly in plan, so $V = wx$. Then $y' = wx/H$, and integrating gives $y = wx^2/2H$ — a parabola, exactly.

Load uniform along the cable. A chain carries only itself, so its weight accumulates with arc length rather than with horizontal distance: $dV = w\,ds = w\sqrt{1 + y'^2}\,dx$. That square root is the entire difference between the two problems, and it turns a one-line integration into a differential equation whose solution is $y = (H/w)\cosh(wx/H)$ — the catenary.

The square root is small when the cable is shallow, which is why the two curves are so nearly identical at ordinary sags and why Galileo’s assertion survived as long as it did. It is also a good illustration of a general habit: the difference between two structural models is often one term that is negligible in the common case and decisive in the extreme one. A chain hanging almost straight down is nothing like a parabola.

Inverted, but not equivalent #

Hooke’s principle is exact for equilibrium. It is not exact for anything else, and the asymmetry it conceals is the most important thing about arches.

A hanging cable is stable. Disturb it and it returns, because any departure from the funicular shape produces forces that push it back — the shape is a minimum of potential energy, and the cable finds it without being told. This is why a hanging model is a form-finding instrument: it does the optimisation itself, and it cannot be persuaded into a wrong answer.

An arch is the same geometry with every force reversed, and reversing the forces reverses the stability. Compression members buckle; tension members do not. An arch nudged out of its funicular shape develops forces that push it further out, and whether it recovers depends on its bending stiffness and its geometry — precisely the properties the funicular argument said were unnecessary.

So the inversion transfers the equilibrium and leaves the stability behind. A cable needs no stiffness at all and works. An arch derived from it needs enough stiffness to remain stable, which is why arch design always contains a buckling check that has no counterpart in cable design, and why a very shallow arch — whose thrust is enormous and whose stiffness against snap-through is small — is a dangerous form rather than merely an inefficient one.

The general principle is worth carrying beyond arches, because it recurs. Equilibrium is symmetric under reversing every force; stability is not. Any argument that establishes a structural result by inverting or mirroring something has established a statement about equilibrium only, and the stability question has to be asked again from the beginning.

What the shape costs #

A structure that carries its load without bending sounds strictly better than one that does not. Three things are being paid for it.

Foundations that can take thrust. A beam needs the ground to push up. An arch needs it to push sideways as well, and sideways is the direction soil is worst at and most expensive to improve. A great many arch bridges are the shape they are because of what the rock at the abutments would take, and a tied arch — which converts the thrust into a tension member across the span — exists purely to remove the requirement, at the cost of a tie that must not fail, ever, under any circumstance.

Stiffness for the loads it was not shaped for. The funicular is funicular for one load case, so every other load case produces bending in a member chosen for having none. Suspension bridges answer this with a stiffening truss, arches with depth in the arch rib, and cable roofs with prestress — and in each case the added material is a substantial fraction of the whole. The elegance of the funicular argument recovers less of the material than it appears to, because most of a real structure is dealing with the loads the shape does not suit.

Geometry that has to be built accurately. A beam built fifty millimetres off line is a beam. An arch built fifty millimetres off its funicular curve has a thrust line displaced by that amount, and the resulting moment is the thrust times the error — which, for a shallow arch with a large thrust, is not small. Funicular structures are demanding to set out and demanding to keep, and settlement of a support changes the geometry after the fact.

None of which is an argument against the form. It is an argument that the saving is not free, and that it is paid at the supports, in the stiffening and on site, rather than in the member where the saving appeared.

What a funicular structure is not #

The shape carries one load case with no bending. Change the load and the funicular shape changes, but the built structure cannot.

A suspension bridge shaped for its own dead weight is not funicular for a train crossing one half of it. The asymmetric load wants a different shape, and since the cable cannot take one, the deck has to resist the difference in bending. That is why suspension bridges have stiffening trusses, and why the Tacoma Narrows deck — which had a shallow plate girder instead — behaved as it did.

The same applies to arches. A masonry arch under asymmetric load has its thrust line move, and if it moves outside the masonry the arch hinges. Four hinges make a mechanism, and that is how arches actually collapse: not by crushing, but by turning into a four-bar linkage.

What the shape does to the supports #

A funicular shape carries its load without bending and pays for it at the ends.

The thrust is $wL^2/8d$, which is the same expression as a truss chord force with the sag in place of the depth. Halving the sag doubles the pull on the abutments, and for a masonry arch that is the difference between a buttress and a collapse.

Equilibrium at the supports is where the difference between a beam and an arch actually shows. A beam delivers a vertical reaction; an arch delivers a thrust, and the history of arch construction is the history of what resists it. The reaction of a funicular structure is not vertical and cannot be made vertical: it lies along the curve’s own tangent at the springing, leaning inward for a cable and outward for an arch, and its horizontal component is the $H$ that is constant along the whole length. Whatever holds the end has to take that component, and the flatter the curve arrives, the larger the share of the reaction it is.

How much the two curves actually differ #

The catenary and the parabola are set out above as two answers to two load distributions, which leaves an obvious question unanswered: on a real cable, which carries its weight along its arc, how wrong is the parabola anybody actually uses?

Solve both at a stated sag ratio and compare the horizontal tension. Writing $u = L/2a$, a catenary has $f/L = (\cosh u - 1)/2u$ and an arc length $S = 2a\sinh u$; spread that same weight uniformly along the span and take $H = wL^2/8f$:

At the tenth that suspension bridges use, the parabola is high by one part in eighty — smaller than the uncertainty in the deck weight, and on the safe side. At a quarter it is seven per cent, which is a number worth carrying rather than ignoring.

The same comparison is worth seeing at the proportion a bridge is actually built to rather than at the one that draws well. The figure near the top of this page sets a catenary against a parabola at a sag of 150 across a 520 span, which is deeper than one in four; here the sag is 52 across the same span, which is one in ten.

Two curves that cannot be told apart by eye at the proportion everything is built to, and that part decisively when the proportion is extreme, is the ordinary condition of structural models rather than a curiosity about cables. The useful question is never whether two models differ but whether they differ where the structure sits.

So the distinction is real and it is not usually a design distinction. What matters at ordinary sags is not which curve was assumed but whether the load is where it was assumed to be, and that is the sensitivity the rest of this page is about.

Where the model stops #

No bending stiffness. True for a cable and false for every arch ever built. A real arch has a section with a second moment of area, which is exactly what lets it tolerate a thrust line that has wandered off the funicular curve — at the cost of the bending the funicular shape was chosen to avoid.

No self-weight, or self-weight only. The two clean solutions are the parabola and the catenary, and each assumes a specific load distribution. A real arch carries both its own weight and a deck, and its funicular is neither curve exactly.

Small deflections. A cable changes shape under load substantially, so the usual first-order assumption does not apply to it at all. Cable analysis is genuinely nonlinear, and the linear methods used elsewhere on this site do not apply to it.

Rigid supports. An arch’s thrust spreads its abutments, which flattens it, which increases the thrust — a load making itself worse, in masonry. Several medieval arches have this recorded in their geometry.

The figures carry a specific distortion worth naming: the sag is drawn at perhaps a fifth of the span, and a real suspension bridge cable sags about a tenth while a shallow arch is far flatter still. Since the thrust goes inversely with the sag, every figure on this page understates the horizontal force by a factor of two or more. The shape is honest; the consequence at the supports is drawn smaller than it is.

The ladder from here #

Later rungs: the catenary derived, and why the hyperbolic cosine appears. The parabola for a uniform horizontal load. The funicular polygon and the pole diagram. Thrust lines in masonry and the middle-third rule. Heyman’s safe theorem. The three-pinned arch, determinate by construction. Tied arches. Suspension bridges and the stiffening truss. Cable nets and prestress. And form-finding by computer, which is the hanging model rewritten as an optimisation and used to design roofs nobody could have drawn.

Hooke’s anagram was published in 1675 and the solution only after his death. He had the principle, and he did not have the mathematics to find the curve.