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#### Spherical Models

**By Magnus J. Wenninger**

**Dover Publications, Inc.**

**Copyright © 1999 Magnus J. Wenninger**

All rights reserved.

ISBN: 978-0-486-14365-1

All rights reserved.

ISBN: 978-0-486-14365-1

CHAPTER 1

**I. The regular spherical models**

The basic symmetry groups of the five regular polyhedrons actually reduce themselves to three - the tetrahedral, the octahedral, and the icosahedral. You may in fact already be aware that the tetrahedron, the octahedron, and the icosahedron all have faces that are equilateral triangles. Projecting their edges onto the surface of their respective circumscribing spheres generates in each case a set of 4, 8, and 20 equilateral spherical triangles. Introducing the respective planes of symmetry for each yields the complete set of 24,48, and 120 right spherical triangles. These are illustrated in Photos 1, 2-3, and 4-5.

**The spherical hexahedron or cube**

To learn how to make these spherical models you will find it easiest if you begin with the cube (see **Fig.1**). Study carefully the shape shown in heavy lines on the interior of the cube, which is shown with its circumscribing sphere. This shape is a special type of tetrahedron called an orthoscheme or a quadrirectangular tetrahedron because each of its four faces is a right-angled triangle. The cube has forty-eight such tetrahedral cells in mirror-image pairs. In **Fig. 1** you see that *O0* is a vertex of cube, *O1* is the midpoint of an edge, *O2* is the in center of a face, and *O3* is the center of the cube that is at the same time the center of the circumscribing sphere. Triangle *O*0*O*1*O*2 on the face of the cube can be projected by central or gnomonic projection onto the surface of the sphere, generating the spherical triangle *O*0*O*1*O*2 Notice that Q0 and *O*0 are the same point. If the cube is now assigned an edge length of *e*=2, then half of e equals 1 (one unit of length). This is the length of *O0O1*.

It is from the spherical triangle *Q0Q1Q2* that a paper band can be designed for making a spherical model of the cube. Triangle *Q0Q1Q2* is omitted, and **Fig.2a** shows how to draw the layout of the orthoscheme and how the angles χ, ?, ψ needed for the circular band, are derived from it.

The procedure is as follows: A circular arc of radius 0*R* = *O0O0* = [??]3 is drawn first. The measure [??]3 = 1.732 approximately. Note that 0*R* is the radius of the circumsphere. Next draw a circle with *O*0*O*3 as diameter. Now open the compass to 1 unit of length and, with *O*0 as center, mark the point *O*1, on this circle. Draw *O*3*O*1 and produce it to Q1 on the circular arc. The arc *Q*0*Q*1 is now one side of the spherical triangle *Q*0*Q*1*Q*2. Notice that the angle at *O*1 is a right angle, since it is inscribed in a semicircle.

The same theorem from plane geometry is now used to locate the point *O*2, s whose projection is *Q*2, as shown on the left in **Fig.2a**. Notice that *O*0*O*2 = [??]2, since it is half the diagonal of the square that is a face of the cube. This value is derived from the well-known theorem of Pythagoras. In fact [??]3 is itself derived from the same theorem, given that the edge length of the cube is 2 units.

So now open the compass to [??]2 = 1.414 approximately and, with O0 as center, mark the point *O*2. Then *O*3*O*2 produced gives *Q*2, which is the projection of the point *O*2 onto the surface of the sphere. The arc *Q*0*Q*2 thus becomes the second side of the spherical triangle *Q0Q1Q2*.

Line *O*0*O*2 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 *Q*0*Q*1*Q*2 now follows as shown on the right in **Fig. 2a**. A semicircle must be drawn with *01O2* as diameter, so that another right angle may be inscribed in it. *01O2* is 1 unit of length, as may be seen from Fig. 1. Open the compass to this length and with *O*1, as center, mark the point *O*2 and project it to obtain *Q*2. Arc Q*Q*1*Q*2 is the third side of the spherical triangle *Q*0*Q*1*Q*2.

Line *O*1*O*2 is named *a* because it is an apothem, namely a line drawn from the incenter of a face to a midpoint of an edge. Since regular polyhedrons all have regular polygons for faces, the apothem is always a perpendicular bisector of an edge.

The angle measures associated with the spherical triangle *Q*0*Q*1*Q*2 can now easily be determined by ordinary trigonometry from the respective right-angled triangles. Thus sin ? = 1/[??]3, sin χ = [??]2/[??]3, and sin ψ = 1/[??]2. Hence the arc length *Q*0*Q*0Q1 = 35.264°, the arc length *Q*0*Q*2 = 54.736°, and the arc length *Q*1*Q*2 = 45°.

Consideration of linear measures may now be dropped and the circular band of paper may be made using any convenient radius. This band is to be used as a template to multiply parts. A recommended linear or chord length of about 1 in. or 2.54 cm between radial lines is suggested and a width of about three- eighths of an inch or 1 cm for the band. Actual model making shows this gives best results with paper. The layout of the circular band for the spherical cube is shown in **Fig. 2b**.

To multiply the parts the following procedure may be used. Place the template over three, four, or five sheets of card or tag that have been stapled together. Outline the template by pencil and prick through the card at the radial lines and through the points marked by heavy dots in **Fig. 2b**. Now cut with scissors along the curved pencil outline as carefully as you can without allowing the paper to move. No scoring is needed because the paper folds easily along the pricked holes in the band. The parts are folded up for right-handed parts, and down for left-handed parts, or vice versa, whichever you decide on. A tab is left at one end of each band. This is used to glue the bands so as to form small spherical triangles. For the spherical cube, eight of these small spherical triangles are joined together, band to band, so that the tab joints disappear between the bands.

What emerges is a spherical dome whose edges form a spherical square. Six such domes are needed to complete the model of a spherical cube. Beyond all doubt before you complete the model you will see it coming out as a spherical octahedron as well, the dual with respect to the cube. The same feature appears in all the other regular and semiregular spherical models. In this respect the dual is most simply described as the polyhedron generated by great circle arcs joining the incenters of the faces of a given spherical polyhedron. In other words, here the polyhedron and its dual are both projected onto the surface of the same circumscribing sphere.

**General instructions for making models**

A word of caution is in place here for the dedicated model maker. The templates must be very accurately drawn to achieve even moderately good results. If in each case you make your own drawings, you will learn first of all how careful you must be about every detail. Then if things do not turn out well, you have only yourself to blame. It is a good idea to check out linear and angular measures against each other, so that calculation will always verify the correctness of geometrical drawing and vice versa. Even so you may find that slight adjustments may be needed as radial lines are folded. Make only a few parts at first, then start assembling them to see how they fit before you continue with the rest. Only experience can teach you what adjustments are best made to achieve desired results.

Do not be dismayed if at first the parts seem to fit rather poorly. Slight differences tend to disappear or compensate each other at vertex points in the completed models. You may find that these vertex points are never perfectly closed. You can observe this on the models shown in the photographs. But as the work moves toward completion the various parts begin to exert their structural tensions relative to each other. Somehow the end result always seems satisfactory, where misgivings or doubts may have previously tempted you to abandon the work. Remember that no model can be perfect. The scale at which the work is done as well as the material used has its limitations. A model is a model, and its purpose is to serve insight and understanding. From this point of view the per-fect model can exist only in your mind. Or if you want to espouse the mystical approach of Johann Kepler, you would say it exists only in the mind of God the Creator.

**The spherical octahedron**

Having completed the spherical cube you can satisfy yourself that it is the spherical octahedron as well, not only by looking at the model, but by going through the process of drawing the layout (**Fig. 3b**). The circular arc for the octahedron must have a radius of [??]2. This is easy to see if you notice that the octahedron has a set of four edges forming a square that lies on a plane of symmetry through the center of the polyhedron. Obviously the distance from one of these vertices to the center of the polyhedron must be half the diagonal of this square. If the edge length *e* = 2, the diagonal is 2[??]2, thus making the radius [??]2. You see now that the only way in which the circular band for the spherical cube differs from that of the spherical octahedron, once you drop linear measurements and compare only the arc lengths, is in an interchange in the position of the central angles ? and ψ, whereas χ remains the same. Plates 1 and 2 show heavy lines to give emphasis to the arcs belonging to the spherical cube and the spherical octahedron, respectively. If you make only one model your mind must supply the emphasis, or to put this another way, your eye sees first one shape, then the other, depending on what you want to see.

**The spherical tetrahedron**

The spherical tetrahedron is illustrated in **Plate 3**. The tetrahedron, as you may know, is the simplest of all uniform polyhedrons, in fact of polyhedrons of any kind, having only four faces, six edges, and four vertices. It is unique in another way. It is the only uniform polyhedron that does not have vertices in diametrically opposite pairs. This means that its planes of symmetry each cut through only one vertex. The spherical model quickly shows that its dual is simply another spherical tetrahedron. Figure 4a gives the layout needed for deriving the circular band in this case; the band is shown in **Fig. 4b**.

For the tetrahedron 0*R* = 1.225 approximately. This is the radius of the circumscribing sphere. Here and in all subsequent spherical models the radius of the circumscribing sphere will be given. In almost every instance it is a rather tricky affair to derive this radius in terms of an edge length. Euclid does a very elaborate job on this in the Thirteenth Book of his classic work *The elements*. If you are interested in the mathematical derivations, you may consult the more modem works listed in the References at the end of this book.

For the spherical tetrahedron it is best to keep the scale small, so that the paper bands will have less tendecy to become bowed or bent. The very simplicity of this model perhaps keeps if from being very attractive, or at least it seems to be less tetrahedron attractive than the other spherical models. Beauty usually demands some complexity, because it arises from the harmonious relationship of parts. But it is a good model to have on hand for inspection; further consideration is given to it later on in the book.

**The spherical icosahedron and dodecahedron**

The spherical icosahedron is the third of the regular spherical models. It turns out to be the most attractive, perhaps because the so-called golden ratio t appears in its linear measurements. The same beauty manifests itself in all the polyhedrons that belong to the icosahedral symmetry group. The dual of the icosahedron is the dodecahedron. Here too, as in the case of the cube and the octahedron, one model serves for both. Look again at Photo 4-5 and observe this.

You can make this model by using either of the layouts shown in **Fig.5a** or **5b**. If you assemble triangle faces, you must arrange them as the faces of an icosahedron. If you assemble pentagon faces, it is the dodecahedral arrangement that is needed. In practice the latter seems preferable. For triangle faces, *r* and *a* have already been calculated. You can find r and a for the pentagon by using ordinary trigonometry applied to **Fig. 5c**. In comparing the two layouts you see that ? and ITLψITL are interchanged, whereas χ remains the same - this is the same situation as was found previously for the cube and the octahedron. The drawings of Plates 4 and 5 show heavy lines to emphasize the spherical icosahedron and the spherical dodecahedron. The bands are shown in **Figs. 5d** and **5e**.

The beauty of these models will be greatly enhanced by an appropriate use of color. If the paper used is colored tag, all the left-handed parts may be white, and all the right-handed ones colored. Color effects can be used in all the spherical models of this book. It will be left for you to work out whatever you may choose. No further references will be made to color in what follows.

**The polyhedral kaleidoscope**

If you have now made the three regular spherical models, you are in a position to use them for attaining a better understanding of the polyhedral kaledioscope. The basic idea here involves reflection of a point in a line. A point *P*' is said to be the reflection or image of a point *P* in a line *l* when the line segment *PP*' is perpendicular to *l* and the distances of *P* from *l* and of *P*' from *l* are equal; that is, line *l* is the perpendicular bisector of line segment *PP*' (see **Fig.6**).

A plane polygon can be thought of as being generated by repeated reflections of a given point *P* in a set of lines all radiating from a fixed point *O*, the lines having suitable angular distances between them. Reflection may be replaced by the idea of rotation. Reflection and rotation are types of mathematical transformations. To simplify matters the discussion here will limit itself to mathematical reflection. **Figure 6** illustrates the idea of reflection applied to an equilateral triangle, a square, and a pentagon.

If two mirrors are used and set upright on a desk so that the angle between them is 120°, and if a small object is used to mark a point and this object is placed midway between the mirrors, the object with its images marks the vertices of an equilateral triangle. If the angle between the mirrors is changed to 90°, the result is a square. If the angle is 72°, the result is a pentagon. This arrangement of mirrors is called a kaleidoscope.

The idea of a point and its images can now be applied very effectively to a spherical surface, where the point is any given point on the spherical surface and where the lines are geodesics or great circle arcs.

Three mirrors are needed for a polyhedral kaleidoscope. They must be cut as circle sectors whose central angles are precisely the values of χ, ?, ψ already calculated for making the spherical models in paper. You can save yourself a lot of trouble by forgetting about the use of mirrors, except possibly to try them and see for yourself that the results are not completely satisfactory. The paper models you now have on hand should give you the same results in a far more satisfactory way.

*(Continues...)*

Excerpted fromSpherical ModelsbyMagnus J. Wenninger. Copyright © 1999 Magnus J. Wenninger. Excerpted by permission of Dover Publications, Inc..

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