Spherical Projection of Cube

Great Circles of the Spherical Hexahedron

great circles of the cube intersecting its vertices

The above image shows Great Circles projecting through the vertices of the Hexahedron.

The result of this is a network of spherical triangles. It is then from these spherical triangles that an actual model can be made up of the whole sphere. The projection of these triangles onto the surface of the sphere is known as spherical tessellation.

One might now ask themselves where does the geometry exist in this topic. In order to make this model one must be able to find the angle measure of all the spherical triangles needed for the spherical model.

To learn how to draw the shapes needed to make these spherical triangles it is felt that it is easiest to begin with the cube. Using the diagram shown below, of the cube circumscribed by a sphere, study carefully the shape of the spherical triangle (in red) shown on the surface of the sphere and also, the blue lines (1R, 2R, 0R) projecting from the common centre of the cube and the sphere to the vertices of the spherical triangle.

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Image A

Quadrirectangular Tetrahedron

The surface area of the triangle being projected onto the sphere is one eight of the total surface area of the top face which is broken into eight equal triangles each having a vertex at the centre of the face. The solid triangular polyhedron described in Image A by the blue lines within the cube is called a quadrirectangual tetrahedron because each of its four faces is a right-angled triangle. The whole cube is made up of forty-eight such solids, eight projecting from each face to the centre of the cube.

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Image B

Five of the forty-eight quadrirectangular tetrahedra contained in a cube are shown here.

Principles of Procedure

Referring to Image A it can be seen that Oo is a vertex of the cube. O1 is the mid point of an edge, O2 is an incentre of a face and O3 is the centre of the cube, that is at the same time the centre of the circumscribing sphere. Triangle Oo O1 O2 on the face of the cube can be projected by central or gnomic projection onto the surface of the sphere, generating the spherical triangle Q1 Q2 Q3. Notice that Qo and Oo are the same point. If the cube is now assigned an edge length of e = 2, then half of e equals 1 (one unit length). This is the length of Oo O1.

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Image A

It is from the spherical triangle Qo Q1 Q2 that a paper band can be designed for making a spherical model of the cube. Triangle Oo O1 O2 is omitted, and Image C shows how to draw the layout of the orthoscheme and how the angles a, b and c needed for the circular band are derived from it.

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Image C

a circular arc of radius oR = Oo O3 = Ö 3 is drawn first. The measure Ö 3 = 1.732 approx.. Note that oR is the radius of the circumsphere.

next draw a circle with Oo O3 as diameter

now open the compass to 1 unit of length and with Oo as centre mark the point O1 on the circle.

draw O3 O1 and produce it to Q1 on the circular arc. The arc Qo Q1 is now one side of the spherical triangle Qo Q1 Q2. Notice that the angle at O1 is a right angle since it is inscribed in a semi circle (theorem).

The same theorem from plane geometry is now used to locate the point O2 whose projection is Q2 as shown in previous drawing (Image C). Notice that Oo O2 = Ö 2 since it is half the diagonal of the square that is a face of the cube. This is derived from the theorem of Pythagoras.

So now open the compass to Ö 2 = 1.414 approximately and with Oo as centre mark the point Q2. Then O3 O2 produced gives Q2 which is the projection of the point O2 onto the surface of the sphere. The arc Qo Q2 thus becomes the second side of the spherical triangle Qo Q1 Q2

Line Oo O2 is named r because it is the radius of the small circle circumscribing a face of the cube

The third side of the spherical triangle Qo Q1 Q2 now follows as shown in Image D and Image E. The spherical model is shown in Image F below.