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Descartes' Law of Closure Deficit
Introduction Euler's formula and the properties of Platonic bodies can be derived systematically from Descartes' Law of closure deficit. Using Descartes' concept of angular deficit, here this law is derived from first principles ,
the concept of
Angular Deficit
Descartes' Law of
Closure Deficit

Descartes' law of closure deficit

Sometime between 1619 and 1621 Descartes wrote the Elementary Treatise on Polyhedra, which, however, he did not publish. After his sudden death in the year 1650 in Stockholm, the manuscript was inventoried and brought to Paris. Here it was - The Strange History of Descarte's Treatise - Gottfried Wilhem Freiherr von Leibniz who made in 1675/76 a copy of the manuscript. This work of Descartes was finally brought to public attention only in the middle of the 19th century, long after the corresponding work of Euler was published, Communications of the St. Petersburg Acadeny, 1752/53 .
  1. A prerequisite: the angular deficit (also called the angular defect).
    The angular deficit was introduced by Descartes as a measure for curvature in three dimensions. It is based on the relationship between the angles of a set of adjacent polygons in a plane.

    The angle at the corner of a regular polygon with n sides is given by

    (1-2/n) 180°.

    If k polygons are arranged to form a vertex, an angular deficit D is defined as

    D = 360° - k(1-2/n) 180°.

    In the example there are five regular triangles forming the vertex, The angular deficit is therefore D = 360° - 5/3 180° = 60°.

    If D>0, the planar arrangement of k polygons may be folded to form a vertex in three-dimensional space.

    In the example, the three-dimensional vertex consists of five regular triangles, which is the typical vertex of the icosahedron.

    If D=0, the vertex is planar and can not be folded.

    In the example, six regular triangles form a hexagon. The angular deficit is D=0 and the hexagon is necessarily planar.

  2. The fundamental law of René Descartes (1596-1650) may be formulated as:
    The sum of the angular deficits of all the vertices of a closed convex body is equal to 720° . In the symmetric case, i.e. for the regular polyhedra, the angular deficits of all the V vertices are equal and it holds
    V*D = 720°.

    Now the essential question has to be answered: Why is the sum of the angular deficits of a closed convex body exactly D=720°?

a sphere
Click on the selected Platonic sphere (also the tiny ones) and drag the mouse.
angular deficit
builds up the
curvature of
a half-sphere.

angular deficit
builds down the
curvature of the
other half-sphere

The total curvature Angle 720°

We take the unit sphere as the reference. The curvature builds up to the equator of the sphere and than builds down from the equator.
  1. The octahedron may be considered to be a 0-appromiximation to the sphere. We consider the special case of the octahedron, cut it into two halves along its equator. The lower halve of the octahedron is considered to be responsible for the build-up and the upper halve for the build-down of the total curvature. This implies that 1+½4=3 vertices are used for the build-up and again 3 vertices for the build-down. Since for the octahedron there are 4 equilateral triangles forming a vertex, each vertex of the octahedron carries an angular deficit of 120° and 3 vertices carry a deficit of 360°. Thus, for this special case the total deficit is indeed 720° as stated by Descartes' law.

  2. From the fact that the total of 360° of a planar vertex is needed for the build-up of the half-sphere'scurvature and again 360° is required to remove the curvature on the other side of the equator, it follows that the process of build-up and build-down cannnot depend on the distribution of curvature onto participating vertices. Thus, the total deficit is equal to 720° for any convex closed body. The total angular deficit is equal to the total curvature of the sphere.

  3. Expressed in radians, the total curvature 720° of the sphere is equal to 4pi. This provokes the question: Why is the total curvature of a sphere equal to its surface area 4 pi R2? This appears to be a direct consequence of the folding process: In folding we take a radius vector of dimension [cm] of the planar structure and rotate it upwards out of the plane. Depending on the angular deficit, the fraction (2*pi - D)R of the circumference will move upwards until the the angular deficit D is used up.This shows that folding is not a process of dimension [cm] but of dimension [cm2]. Strictly speaking, it is not the angular deficit, which is transformed into curvature. The process of build-up of curvature is only monitored by the angular deficit. If the deficit is used up, no further folding is possible. In conclusion, there is no mystery in the surprising occurrence of dimension [cm2]: the surface of a sphere is just equal to the steric arc of its total curvature!

  4. A last remark: Here it is assumed that angular deficits are measured with respect to the axis through the vertices pointing towards the center of the sphere. Taking a coordinate system outside of the sphere, the sign of the build-up curvature would be opposite from the sign of the build-down curvature and thus in such a system the total curvature would be zero.

Two explicit Examples for Closure Deficits

These examples are intended to show, that the total closure deficit does not depend on the number of vertices involved. We consider only the build-up process, which requires a closure deficit of 360°.
Example #1:
One half
of an
Octahedral Sphere
3 vertices

Example #1: One half of an Octahedral Sphere, 3 vertices.

We consider half of an octahedron enclosed by a half-sphere. There are the following vertices: The full vertex at the south pole and 4 half vertices at the equator of the octahedron, i.e. there are 1+½4=3 vertices. The angular deficit thus is D=360°/3=120°. As seen from the enclosing half-sphere, the vertex at the south pole builds up part of the curvature, the rest of the curvature is supplied by the 4 half vertices of the octahedron. At the equator, the surface normal vector of the sphere is perpendicular to the surface normal vector at the south pole of the sphere.The angular deficit of a vertex is 360°-4*60°=120°. The total curvature, being equal to the sum of the angular deficits of the 1+4/2=3 vertices is thus 360° as required.
Example #2:
One half
of an
Octahedral Sphere
9 Vertices

Example #2: One half Octahedral Sphere, 9 vertices.

Again we consider half of the octahedron, but now position between each of the vertices an additional one. The original vertices had the x,y,z-coordinates being either zero or ±1, Now, inserting new vertices exactly half-way inbetween, the x,y,z-coordinates of the new interstial vertices become either 0 or ±0.7071. Naturally, for all vertices the sum of the coordinate squares adds up to unity. As demonstrated by the figure to the left (rotate it for even better inspection), there are 1+4+½8=9 vertices belonging to the half octahedron. We require that each vertex gives the identical contribution to the total angular deficit. The angular deficit of one vertex is thus D=360°/9=40°.

For the vertex at the south pole, the four-fold symmetry of the octahedral vertex remains and the angular deficit is realized by 360°-4*80°=40°. By construction, this contribution to total curvature is much less than in example #1.

The new interstial vertices are formed each by six triangles. There results a quasi hexagonal structure. The six participating triangles are of two different kinds. Four participating triangles result from triangles which stem from two vertices of fourfold symmetry and thus contribute each (180°-80°)/2=50°. The other two angles of the hexagonal vertex are thus each equal to (360°-4*50°-40°)/2=60°.

In summary, the angular deficit of the hexagonal vertices is realized by the six angles as 360°-2*60°-4*50°=40° and the contribution of each these vertices is exactly equal to the vertex at the southpole with fourfold symmetry. The 8 vertices at the equator of the sphere alternate in the two types just considered. Thus, each of the nine vertices of the half-octaedron contributes 40° and the total angular deficit is 360°, as required.

The full
Octahedral Sphere

The full octahedral sphere corresponding to Example #2: 18 vertices.

By the classification of Buckminster Fuller, the Platonic bodies are 1-frequency structures. After the first tessellation, the 2-frequency structure results, after the second tessellation a 3-frequency structure and so on. For completeness we show here for the example #2 the corresponding full 2-frequency octahedral sphere. The 4-fold symmetric vertices and hexagonal vertices alternate. The corresponding symmetry may be viewed best by looking head on onto these two different types of polygon.
Surface Area of the Sphere

Surface Area of the Sphere

The surface area of a Platonic body may be considered to be the 0th order approximation of the surface area of the sphere. In the same way as for the octahedron, triangulation may be performed for any one of the Platonic bodies. Further triangulation steps leading to higher frequency bodies will result in a better and better approximation to the sphere.
A nice animation on triangulation has been given in a related page of Kirby Urner.
             # of      # of       surface    surface   -->   surface*
           vertices   faces        area     area /4 pi      area /4 pi 

Tetrahedron    4   4 triangles   4.618802   0.367526   -->  1.000000
Octahedron     6   8 squares     6.928203   0.587025   -->  1.000000
Hexahedron     8   6 triangles   8.000000   0.636620   -->  1.000000
Icosahedron   12  20 triangles   9.574541   0.761918   -->  1.000000
Dodecahedron  20  12 pentagons  10.514622   0.836727   -->  1.000000
  Sphere                        12.566371   1.000000 

                    *after a sufficient number of triangulation steps
The tetrahedron shows the highest, the dodecahedron the lowest curvature per vertex. The total angular deficit is 4 pi in both cases, the 0th order approximation of the surface of the sphere differs, however, already by a factor 2.3 .
For a high degree of triangulation, any one of the polyhedra becomes arbitrarily close to a sphere and its total surface area approaches that of a sphere. On the other side, the well known formula of spherical geometry on the relation between the sum of the three angles r,g, and b of a spherical triangle T and its area
degenerates in case of arbitrarily high tessellation to the formula for a planar triangle.
A computer programm for generating a triangle mesh approximating a sphere by recursive subdivision has been provided by Jon Leech.
Any Sphere carries
the fine structure
of a Platonic body.

A New Look at the Sphere

Apparently, by the point of view taken here, a sphere results from the angular deficits of 2N vertices, where N vertices are responsible for the buildup of curvature and N vertices are responsible for the build-down of curvature. Each vertex contributes a deficit of 360°/N to the curvature. With N-->oo, a sphere may be approximated arbitrarily close.
Even the finest grained sphere carries in its fine structure marks of the underlying Platonic body. Among the "infinite" number of triangles in a hexagonal arrangement, there are a very small number of triangles forming a vertex with the symmetry of the face of one of the Platonic bodies. The triangles in the hexagonal arrangement are not equilateral, since in that case the corresponding vertices would not contribute to the total curvature deficit.
Surface area
the sphere.

Why is the Surface of the Sphere
equal to 4pi ?

Independent on the kind of tessellation of the sphere, we have seen that the total angular deficit of the half-sphere is equal to 360° steredian and that of the full sphere equal to 720° steredian. Under the assumption, that we know 720° to be equal to 4pi, it is evident, that the surface area of the unit sphere, radius R=1, is equal to 4pi. In conclusion, the area of the sphere may be obtained without any kind of calculus at all, it is a direct consequence of the concept of Descartes' angular deficit. In fact, it is a tautology to say, "the area of the sphere is 4pi" and "Descartes' law of closure deficit is true". One of the statements is sufficient, each implies the other.
On the other hand, if we assume the value of pi to be not known, this value may be obtained by use of Descartes' concept of angular deficit by finer and finer tessellation of one of the Platonic bodies.
a sphere
Click on the selected Platonic sphere (also the tiny ones) and drag the mouse.
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other pages
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