

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 , 
From the concept of Angular Deficit to 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 .

Rotate a sphere 
Click on the selected Platonic sphere (also the tiny ones) and drag the mouse. 
An angular deficit D=360° builds up the curvature of a halfsphere. Another angular deficit D=360° builds down the curvature of the other halfsphere 
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.

Examples 
These examples are intended to show, that the total closure deficit
does not depend on the number of vertices involved. We
consider only the buildup process, which requires a closure
deficit of 360°.

Example #1: One half of an Octahedral Sphere 3 vertices 
We consider half of an octahedron enclosed by a halfsphere.
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 halfsphere, 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 
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,zcoordinates being either
zero or ±1, Now, inserting new vertices exactly halfway inbetween,
the x,y,zcoordinates 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°.

The full 2frequency Octahedral Sphere 
By the classification of Buckminster Fuller, the Platonic bodies are
1frequency structures. After the first tessellation, the 2frequency
structure results, after the second tessellation a 3frequency
structure and so on.
For completeness we show here for the example #2 the corresponding
full 2frequency octahedral sphere. The 4fold 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 
The surface area of a Platonic body may be considered
to be the 0^{th} 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.

Any Sphere carries the fine structure of a Platonic body. 
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 builddown
of curvature. Each vertex contributes a deficit of 360°/N
to the curvature. With N>oo, a sphere may be approximated
arbitrarily close.

Surface area of the sphere. 
Independent on the kind of tessellation of the sphere, we have seen
that the total angular deficit of the halfsphere 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.
equal to 4pi ? 
Rotate a sphere 
Click on the selected Platonic sphere (also the tiny ones) and drag the mouse. 
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