Kepler's Universe

The artifact shown in pictures is a model of the Universe as imagined by the German astronomer Johannes Kepler (1571 - 1630). He described this model in his book Mysterium Cosmographicum (The Cosmographic Mystery) which was published in 1596 and advocated in favour of the Copernican (heliocentric) model of the Universe.

Kepler claimed that planets circle around the Sun, and the orbit of a planet is the cross-section of a sphere. He also thought that those spheres could be inscribed into and circumscribed around regular polyhedra known as Platonic solids.

At the time when Kepler lived, six planets were known: Mercury, Venus, Earth, Mars, Jupiter, and Saturn. This fact allowed Kepler to construct a model of the Universe by using five Platonic solids as shown in a picture from his book. He ordered the solids in such a way that the spheres inscribed and circumscribed would correspond to the real (as they were known at that time, obviously) orbits of the planets.

Platonic solids

The faces of these solids, named after the ancient Greek philosopher Plato, are regular polygons, which means that their sides are equal, and the angles between the sides are equal. There are five Platonic solids: octahedron, icosahedron, dodecahedron, tetrahedron and cube.

Materials

1 mm thick veneer

3 mm thick plywood

thick paper

1 mm thick cardboard

thin fishing line (0.1 … 0.2 mm diameter)

white glue

paints: white, black, orange, brown, red, yellow, blue

Tools

Xacto knife

metal ruler

hard pencil

brush for glue

brush for paint

pin

sandpaper (a piece of sandpaper glued on a small flat wooden bar)

A regular octahedron is composed of eight equilateral triangles; the vertices of four triangles meet at each vertex of the octahedron.

The radius of the inscribed sphere is, approximately: R (in) = 0.408 a, where a is a side of the octahedron

The real orbit of Mercury is an ellipse with its semi-major axis equal to 57909050 km (see Reference). I considered it to be a circle with the radius of 57M km and scaled it down to 19 mm; thus, the resulting model would not be too big.

If R (in) is chosen to be 19 mm, then a = 46.5 mm (rounded to 47 mm).

Thus, the octahedron will be constructed of 8 equilateral triangles with their sides equal to 47 mm; the angle between two sides of such a triangle is 60 deg. The triangles were made of strips of veneer (3 mm wide, 1 mm thick) by means of the pattern shown in picture. Care should be taken when detaching the ready piece from the pattern, because the glued joint is not very strong. I also made a guide to cut the ends of strips at 60 deg for triangles and 72 deg for pentagons; in fact, it’s two rectangles of 3 mm thick plywood glued onto a board with 3 mm distance between them; one end of both rectangles is cut at 60 deg, the other - at 72 deg.

To make the pattern, I drew an equilateral triangle of required size on a sheet of thick paper and glued pieces of 1 mm thick cardboard to keep the strips of veneer in place when assembling the triangle.

Once the ready triangle is detached from the pattern, excessive parts of strips will be cut out to produce a regular triangular shape.

Four triangles will be joined by means of the pattern shown in picture, to create a half of the octahedron. A small hole will be made in the vertex of each figure (I used a pin to make those holes). Then, I inserted two pieces of thin fishing line (further called ‘threads’)into those holes and fixed each them with a knot. The octahedron will be suspended inside of the next figure, the icosahedron, by means of these threads.

After that, the two halves will be glued together to form the octahedron; an additional pattern should be made for this operation. In fact, this pattern is rather optional, but it would greatly facilitate the task of assembling the pieces. The ready octahedron will be painted dark-grey, the colour of Mercury (see Reference section). I mixed black paint with a bit of white to get the colour.

The radius of the circumscribed sphere is, approximately:
R (circ) = 0.707 a, where a = 47 mm Therefore, R (circ) = 32.9 mm (rounded to 33 mm). The sphere with this radius corresponds to the orbit of Venus and is inscribed into the next Platonic solid, which is the icosahedron.

A regular icosahedron is composed of 20 equilateral triangles. The radius of the sphere inscribed into an icosahedron is, approximately:

R (in) = 0.756 a, where a is the side of the icosahedron

R (in) is calculated in the previous section and equal to 33 mm; therefore, a = 43.7 mm (rounded to 44 mm). A face of this icosahedron will be an equilateral triangle with its side equal to 44 mm.

The proceeding to make these triangles is the same as one for the triangles of the octahedron; an appropriate pattern should be made for this operation. I constructed the icosahedron as an assembly of four groups each consisting of five triangles. A pattern like one shown in picture will be needed to join the five triangles of each group. One of the groups will be the top of the figure, the other group - the bottom.

After this, it’s necessary to glue ten lateral triangles to the bottom of the icosahedron; the technique is illustrated in pictures. I proceeded as follows:

temporarily fixed two triangles to the sides of the bottom which meet in one vertex; I used pieces of adhesive tape to fix the parts

put some glue on two sides of the triangle which would be installed between the two triangles of the previous step

inserted the third triangle and held it with my fingers until the glue was hard enough to keep the three triangles in place.

Fortunately, the white glue which I used sets in about two minutes enough to hold the parts; it’s not completely hardened, though, so, be careful when handling the assembly.

The remaining 7 triangles are installed by repeating the above mentioned steps.

After the glue is hardened, the pieces of adhesive tape will be removed, and some glue put on the joints of the sides where the tape was. As a result of the above mentioned operations you will have two parts of the icosahedron: one which looks like a pot, and the other which likes like its cover. Two small holes (I made them with a pin) will be made in the opposite vertices of the icosahedron. Then, you will pass the threads of the octahedron through these holes and glue the parts of the icosahedron together. After glue on the joint is hardened, you slightly tension the threads to place the octahedron more or less in the middle of the icosahedron; fix the free ends of the threads to the sides of the icosahedron by means of adhesive tape and put some glue on the holes to fix the threads in place.

Both parts of the icosahedron should be painted before their assembly with the octahedron; being that the icosahedron represents Venus, it will be painted white yellowish. White paint mixed with small amount of yellow would give this colour.

The radius of the circumscribed sphere is. approximately:
R (circ) = 0.951 a, where a = 44 mm. Thus, R (circ) = 41.8 mm (rounded to 42 mm). The sphere with this radius will represent the Earth’s orbit and be inscribed into the dodecahedron.

A regular dodecahedron is composed of 12 faces, each of which is a regular pentagon; the angle between its sides is 72 deg.

The radius of the sphere inscribed into a dodecahedron is, approximately: R (in) = 1.114 a, where a is the side of the dodecahedron R (in) is calculated in the previous section, it’s equal to 42 mm. Therefore, a = 37.7 mm (rounded to 38 mm).

Twelve pentagons with the side of 38 mm will be made by means of the pattern shown in picture.

The dodecahedron will be composed of two groups, each made of six pentagons. The technique to construct a half is as follows:

one pentagon is chosen to be the central one fix two pentagons to the sides meeting in a vertex of the central pentagon, use pieces of adhesive tape for this operation

raise these two pentagons to join their lateral sides and put some glue on the joint

after the glue on the lateral joint is set, remove the pieces of adhesive tape and put some glue on the joints

The 3rd, 4th and 5th pentagons of the group could be put in place without adhesive tape; however, it’s necessary to hold them manually until the glue sets (my experience tells that it’s about a minute and a half).

After the two halves of the figure are ready, you put them together (without gluing them, for the moment) and mark the opposite vertices. A small hole will be made in each vertex to pass the threads which fix the icosahedron inside the dodecahedron. After that, you pass the threads to each hole and glue the halves of the dodecahedron together; once the joint is strong enough, you tension the threads to place the icosahedron more or less n the middle of the dodecahedron.
Fix the free ends of the threads to the sides of the dodecahedron by means of adhesive tape and put some glue onto the holes to fix the threads in place.

The colour of the Earth would be light blue.

The radius of the circumscribed sphere is approximately calculated as: R (circ) = 1.401 a, where a = 38 mm Therefore, R (circ) = 53.2 mm (rounded to 53 mm). This sphere corresponds to the orbit of Mars and is inscribed into the tetrahedron.

A tetrahedron is composed of four equilateral triangles; in fact, it’s a triangular pyramid.

The radius of the sphere inscribed into the tetrahedron was calculated in the previous section; its value is 53 mm. The radius of the inscribed sphere which is tangent to the sides of the tetrahedron is approximately: R (in) = 0.204 a, where a the side of the tetrahedron. Thus, a = 259.6 mm (rounded to 260 mm).

The tetrahedron is made of strips of 3 mm thick plywood, their width being also 3 mm. A face of this figure is an equilateral triangle with its side 260 mm long; let it be the base of the tetrahedron. There are reinforcing triangular plates on the joints of the strips. The three remaining sides will be made of strips each 257 mm long, to keep the overall length of the sides equal to 260 mm. The ends of these lateral strips should be cut as shown in pictures VB (vertex bottom) and VT (vertex top). Once the sides are ready, you glued them to the base and between them at their common vertex. Reinforcing triangular plates made of veneer will be glued to each side of a vertex.

The colour of Mars is identified as reddish brown; I think there’s a lot of space for an artist’s fantasy here.

The dodecahedron will be suspended on a thread fixed to a vertex of the tetrahedron; the dodecahedron should touch the side of the tetrahedron which is opposite to this vertex. Now, let’s consider this side: it’s an equilateral triangle. You fix an end of the other thread to a vertex of this triangle, pass the thread above the bottom vertex of the dodecahedron, and fix the other end of the thread to the middle of the side of the triangle. Optionally, you could glue this vertex to the thread to fix the figure in place.

The radius of the sphere circumscribed around a tetrahedron is approximatively equal to 0.612 a, where ‘a’ is the side of the tetrahedron. The value of ‘a’ is known to be 160 mm, therefore, R (circ) = 159.2 mm (rounded to 160 mm, to ease the calculations).

This sphere (the orbit of Jupiter) is also inscribed into a cube.

A cube has six faces, each of them being a square. The radius of the sphere inscribed into a cube is calculated exactly as:

R (in) = 0.5 a, where a is the side of the cube. R (in) is equal to 160 mm, therefore, a = 320 mm.

The cube will be made of strips of 3 mm thick plywood, the strips being 4 mm wide. First, I assembled two 320 x 320 mm squares as shown in pictures. Each square is made of two strips 320 mm long, and two strips 312 mm long; thus, the exterior size is exactly 320 x 320 mm. There are reinforcing corners on the joints.

Then, four strips 314 mm long will be made to connect the squares; I glued two corners made of veneer on each end of a strip to facilitate its installation on the square. After the cube is assembled, and glue hardened, the excessive length of these corners will be cut and the joints will be sandpapered.

The tetrahedron will be suspended on two threads each of which forms a diagonal of a side of the cube. An end of a thread will be fixed to a corner of the side (I just attached the thread to the edge), the other end will be passed under the vertex of the tetrahedron and fixed to the opposite corner. The other thread will be installed the same way.
It’s important not to tension the threads too much, otherwise, the cube would be deformed; the tetrahedron should be somewhere in the middle of the cube. After the tetrahedron is in place, its vertex will be glued to the threads.

I simplified Jupiter’s colour and made it just orange.

The radius of the circumscribed sphere is approximately calculated as: R (circ) = 0.866 a. The value of ‘a’ being 320 mm, R (circ) would be 277 mm. The sphere with this radius corresponds to the orbit of Saturn; however, there’s no Platonic solid to represent this planet.

Reference

Kepler’s bio

https://en.wikipedia.org/wiki/Johannes_Kepler

Solar system

https://en.wikipedia.org/wiki/Solar_System

Platonic solids

https://en.wikipedia.org/wiki/Platonic_solid

Colours of planets

https://www.universetoday.com/33642/the-colors-of-the-planets/