A chain hanging freely takes the one shape in which it is in pure tension; turn that shape upside down and it is the shape of an arch in pure compression for the same loads.
free-hanging chain (catenary) ↔ ideal arch under its own weight
The problem of the oblong bay
Two semicircular barrels of different spans rise to different heights. To groin-vault an oblong bay the narrower barrel must be stilted or flattened, and the groins then wander out of plane: the difficulty that pushed builders towards ribs and pointed arches.
Romanesque builders avoided the problem by keeping their vault bays square: the “bound system” of Speyer or Mainz, in which one square nave bay corresponds to two square aisle bays, and the piers alternate strong and weak. Oblong bays, which give a nave a quicker, more even rhythm, became easy only once the builder could make each arch reach whatever height he chose.
In the model the narrower barrel has been given an elliptical section so that its crown matches the wider one; compare the rib vault, in which pointed transverse and wall arches do the same job cleanly.
Related rules
Draw a pointed arch from two centres on the springing line. With the centres at the springings, the arch is equilateral; with centres beyond them, a lancet; with centres between them, a drop arch. Late Gothic added arches of four centres and the reversed curve of the ogee.
r = s (equilateral) · r > s (lancet) · s⁄2 < r < s (drop)
Classify arches by the geometry of the intrados: the number and position of the centres (one for round and segmental, two for pointed, three or four for basket and Tudor arches), or a single continuous curve (ellipse, parabola, catenary), and distinguish true arches, with voussoirs, from corbelled “arches” of oversailing courses.
A round arch must rise half its span, so arches of different spans on one springing line have crowns at different heights. A pointed arch can be given any rise at least half its span, so arches of different widths can all be made to reach the same height.
round: f = s⁄2 (fixed) · pointed: r = (s²⁄4 + f²) ⁄ s for any f ≥ s⁄2
To find how thick the pier of an arch must be, divide the intrados into three equal parts; draw a line from the springing through the first division, and produce it beyond the springing by its own length. A vertical through the end of that line marks the outer face of the pier.
pier ≈ ¼ span (semicircle) · ≈ 0.23 span (equilateral) · less for a lancet
An arch cannot support itself until it is complete. Build it on a timber centring shaped to the intrados, strike the centring only when the keystone is set and the mortar has taken, and reuse it from bay to bay, which is one reason to keep the spans of an arcade equal.
Cite this page Chris Paton, “The problem of the oblong bay”, The Textbook of Architecture, https://textbookofarchitecture.com/rule/groin-oblong. Text, drawings and models licensed under CC BY-NC 4.0.