ABSTRACT (original article): This article examines the shape of a surface obtained by a hanging flexible, inelastic material with prescribed area and boundary curve. The shape of this surface, after being turned upside down, is a model for cupolas (or domes) under the simple hypothesis of compression. Investigating the rotational examples, we provide and illustrate a novel design for a roof that has the extraordinary property that its shape, although natural, is modeled by a surface of revolution whose axis of rotation is horizontal. CITATION (original article): Rafael López (2023) What Is the Shape of a Cupola?, The American Mathematical Monthly, 130:3, 222-238, https://doi.org/10.1080/00029890.2022.2154557.

Introduction

The aim of this project is to discuss three mathematical approaches of the shape of a dome, beginning with the ideal model and comparing with two known surfaces, the catenary rotation surface and the paraboloid. The surface that models a rotational dome is called a rotational tectum. In Figure 1 we see two domes which can be viewed as surfaces of revolution with respect to a vertical axis. Under ideal conditions, the rotational tectum is the surface with the lowest center of gravity among all surfaces with the same area and the same boundary curve. In this work, we compare numerically how far are the energies between the ideal model and the other two surfaces [4].​​
​
​​Figure 1. Left: the Dome of the Rock, Jerusalem. Right: the Bundestag, Berlin ([1]).

The Rotational Tectum and the Two Non-Ideal Models

It is well known that the shape of a hanging flexible chain suspended from its endpoints is the catenary. The two-dimensional analogue is a bounded piece S of homogeneous flexible material of given area
A
0
hanging from its boundary curve
Γ
and suspended by its weight. In a position of static equilibrium, the tension forces exerted on the surface S only have tangential components [2,3]. Therefore, if we turn S upside down, these forces become only internal compressive forces. In consequence these surfaces are models of domes and roofs and with a consequent reduction of collapse. Under ideal conditions in the construction of domes, the energy is the weight of the surface and it is expressed by the height of the center of gravity of the dome. The Lagrange conditions are the area
A
0
of the dome together its boundary curve
Γ
. If S is given as the graph z=u(x,y), then this energy isJ[S]=
∫
S
σu
1+|Du
2
|
+λ
∫
S
1+|Du
2
|
,where the first summand is the height of the center of gravity, σ is the area density, |Du|
2
​
=
2
(
∂
x
u)
+
2
(
∂
y
u)
and the second summand is the Lagrange constraint given by the area of S. Assuming σ=1 and after a vertical translation, the Euler-Lagrange equation is ​div
Du
1+|Du
2
|
=
1
u
1+|Du
2
|
​​We now consider a dome or cupola of rotational shape. Assume that the z-axis is the rotational axis and that the dome is parametrized by X(r,θ)=(r cosθ,r sinθ,u(r)). Consider a bounded piece S of a dome bounded by a horizontal circle
Γ
R
of radius R >0 and height z=u(R). This surface, a rotational tectum, is determined by the generating curve r ↦ (r,0,u(r)) contained in the xz-plane and the Euler-Lagrange writes now as ​(1)
′′
u
1+
′2
u
+
′
u
r
=
1
u
​where r∈ [0,R]. This equation cannot solved explicitly by quadratures and we need to use numerical methods to get solutions. On the other hand, there another two surfaces of revolution which, not solving (1), may be candidates to the model of a rotational dome. The first surface is the catenary rotation surface. This surface is obtained by rotating a vertical catenary
z(x)=
Cosh(cx)
c
+m
with respect to its rotation axis. The reason to study this surface is because the catenary is the model of an arch, so rotating about its symmetry axis, we obtain a surface of revolution that could be model of a dome. The second candidate is the paraboloid, the rotational surface obtained by rotating a vertical parabola z(x)=c
2
x
+m
with respect to its symmetric axis. The motivation has its origin in the well-known dispute between parabolas and catenaries in the construction of arches. Furthermore, the paraboloid has a clear advantage from the architectural viewpoint because this surface is defined with polynomials, in contrast to the catenary that uses exponential functions.

Implementation with Mathematica

Consider u=u(r) a solution of (1) with initial conditions ​(2) u(0)=u0, u’(0)=0,​ defined in the interval [0,R]. Let T denote the rotational tectum obtained by rotating z=u(r) about the z-axis. The boundary curve of T is the horizontal circle ​ (3)
Γ
R
= {(x,y,u(R)):
2
x
+
2
y
=
2
R
}, where u(R)>0 is the height of
Γ
R
. The area
A
0
of T is​(4)
A
0
=
2π
R
∫
0
x
1+
′
u
2
(x)
x
​ and the height
h
T
of its center of gravity is ​(5)
h
T
=
2π
A
0
R
∫
0
xu(x)
1+
′
u
2
(x)
x
​ Let now the catenary rotation surface with the same area and boundary curve. The generating catenary is​(6)
z
c,m
[x]=
Cosh[cx]
c
+m
, x∈ [0,R], where c and m and real numbers and c>0. Let
C
c,m
denote the corresponding rotational catenary obtained when we rotate
z
c,m
with respect to the z-axis. We will choose the values c and m in (6) so the area of
C
c,m
is
A
0
and its boundary is
Γ
R
. The last condition implies ​(7)
m+
Cosh[cR]
c
=u[R]
. The area of
C
c,m
is obtained by quadratures, ​(8)
A
0
=
2π
R
∫
0
x
1+
′
z
c,m
2
(x)
x=
2π(cRsinh[cR]-cosh[cR]+1)
2
c
​ Finally, the height
h
C
of the center of gravity of
C
c,m
is ​(9)
h
c
=
2π
R
∫
0
x
z
c,m
(x)
1+
′
z
c,m
(x)^2
x
A
0
=
π(2
2
c
2
R
+2cRsinh[2cR]-cosh[2cR]+1)
4
3
c
A
0
+ m It is important to point out that the derivation of (8) and (9) are obtained directly from Mathematica with the function Integrate. To derive the parameters c and m from (7) and (8), it is enough with the command FindRoot. In the case of paraboloids, the methodology is analogous. Consider the parabola ​
p
c,m
(x) = c x^2 + m, x∈ [0,R], where c and m are real numbers, c>0, and situated in the coordinate xz-plane. Let
P
c,m
be the paraboloid obtained by rotating
p
c,m
about the z-axis. Let
h
P
be the height of the center of gravity. Since the boundary is
Γ
R
, we have the condition​(10) c R^2 + m = u(R). The area of
P
c,m
is​(11)
A
0
=
2π
R
∫
0
x
1+
′
p
c,m
2
(x)
x=
π
3/2
(4
2
c
2
R
+1)
-1
6
2
c
​ and the height
h
P
of its center of gravity is​(12)
h
P
=
2π
R
∫
0
x
p
c,m
(x)
1+
′
p
c,m
(x)^2
x
A
0
=
π(6
2
c
2
R
-1)
3/2
(4
2
c
2
R
+1)
+1
60
3
c
A
0
+ m As it is expectable, all expressions for the paraboloid are polynomial in the variables c and R which, from the computational and architectural viewpoint, is much manageable. Again, the function Integrate allows to solve both integrals. We show the values of
h
P
in Table 1. ​We use Mathematica to compute the area (4) and the center of gravity (5) of the rotational tectum. The steps to follow are:
1
.
Fix
u
0
>0 in (2), the lowest point of T, and by numerical methods, we solve (1)-(2).
2
.
Fix R>0 and let [0,R] be the domain of u which determines the rotational tectum T. With this value for R, we prescribe
Γ
R
, the boundary curve of T, which it is a data in the problem.
3
.
Compute the area
A
0
using (4).
Next, compute the height of the center of gravity of the rotational tectum using (5):
In Mathematica, we use the function FindRoot to find c and m:
The Mathematica codes for the paraboloids are similar.
​
In Figures 2 and 3 we show some pictures of rotational tectums, catenary rotational surfaces and paraboloids with the same boundary and area.
​
Plotting the rotational tectum and the paraboloid: Figure 3

Results

2
.
Both deviation percentages attain a maximum at R=12 and next decreases.

Conclusions

As far the author knows, catenary rotation surfaces and paraboloids have not been tested in relation with the centers of gravity nor in comparison with the ideal model of the rotational tectum. It is known that the paraboloid is as an example of a quadric employed in architecture but without to clearly state what is its utility other than they are more tractable in the computational treatment of the design in construction (which is not a minor matter). From our conclusions, we give the following consequence in plain language and easy to understand:
​
​Catenary rotation surfaces and paraboloids are surfaces that ‘adjust extremely well' to the ideal model of a rotational dome, being the paraboloid better than the catenary rotation surface.

References

[1] The photo of the Dome of the Rock is obtained from Wikipedia (commons.wikimedia.org/wiki/File : Jerusalem - 2013 (2) - Temple Mount - Dome of the Rock (SE exposure).jpg). The second image is obtained from https : // www.bundestag.de/en/visittheBundestag/dome/registration - 245686, copyright by Deutscher Bundestag/Neuhauser.
[2] Böhme, R., Hildebrandt, S., Taush, E.: The two-dimensional analogue of the catenary. Pacific J. Math. 88, 247- 278 (1980)
[3] López, R.: What is the Shape of a Cupola? Amer.Math.Monthly, to appear.
[4] López, R.: Comparison analysis between catenary rotation surfaces, paraboloids and surfaces subjected to compression forces, preprint (2022)

CITE THIS NOTEBOOK

What Is the Shape of a Cupola?​
by Rafael López​
Wolfram Community, STAFF PICKS, March 14, 2022
​https://community.wolfram.com/groups/-/m/t/2490216