The Wunderkreis has often been the subject of this blog. Today I would like to bring some basic remarks to it.

As is known, the Wunderkreis consists of labyrinthine windings and a double spiral in the center. Thus, there is no center to reach as usually in the labyrinth and, in addition, an extra exit, but it can also be formed together with the entrance in a branching.

This makes it more difficult to represent all this in a pattern. Also the usual path sequence (or circuit sequence) with the alternating odd and even numbers does not work properly anymore.

Therefore, I suggest to designate the spiral-shaped circuits with letters. This also gives the possibility to better describe the different types.

Here is the smallest Wunderkreis in my opinion:

A 3 circuit (normal) labyrinth with a double spiral. The path sequence, starting to the left, would be: 0-1-2-a1-a2-3-0. If I move to the right first, the result is: 0-3-a2-a1-2-1-0.

General note on “0”. This always means the area outside the labyrinth. Even if “0” does not appear on the drawings.

Now I can either increase the outer circuits or only the double spiral or both.

This is one more course for the double spiral. The path sequence to the left: 0-1-2-a1-b2-b1-a2-3-0. To the right: 0-3-a2-b1-b2-a1-2-1-0.

And now:

The double spiral as in the first example, the outer circuits increased by two. This creates a path sequence with (to the left): 0-3-2-1-4-a1-a2-5-0. Or to the right: 0-5-a2-a1-4-1-2-3-0.

Now follows:

In addition to the previous example, the double spiral is also enlarged. This results in: 0-3-2-1-4-a1-b2-b1-a2-5-0. And: 0-5-a2-b1-b2-a1-4-1-2-3-0.

In the circuit sequences I recognize the regularities as they occur also in the already known classical corresponding labyrinths. And if I omit the double spiral, I also end up with these labyrinths.

There is already something on this blog for a 3 or 7 circuit labyrinth. But not yet for a 5 circuit one.

As is well known, there are eight possibilities for a 5 circuit labyrinth (see Related Posts below). The best one for the purpose here seems to me to be the variant with the path sequence 0-5-2-3-4-1-6. Because in this case there are no crossing lines and it has only two turning points. That is, it consists of a single line. That is why it is best suited to be laid with a rope.

This is how the 5 circuit classical labyrinth in Knidos-style (with a larger center) might present itself:

Below are some notes on the more precise construction method. For this, I have assumed an axis dimension of 50 cm (corresponding to the path width) and chosen four times this for the center. This results in a total diameter of 14 x 0.50 m = 7.00 m. Here are the main elements first:

There are therefore a total of 3 midpoints around which the lines run in different radii. These must be determined first. Because they determine the appearance of the labyrinth. The entrance, the center and the orientation of the central axis.

Here are the associated dimensions for defining the three midpoints:

With this, starting from the center around M1 from M2 to M3 (or vice versa), the line can now be marked out, or the rope laid out.

The costruction drawing once again contains all the dimensions, as well as the radii of the various arch elements.

Now, if it’s about a certain labyrinth at a certain place, the dimensions can easily be changed. I can make the labyrinth larger or smaller. For this I have to calculate a scaling factor. How this is done will be explained in more detail. If the labyrinth shall have a diameter of about 9.00 m, I calculate the scaling factor with 9.00 : 7.00 = 1.2857142. By multiplying with this factor I can determine all other dimensions. For the axis dimension (= path width), I would then have 0.50 x 1.2857142 = 0.6428571. This would also be the minimum radius for the curved sections. This is not very clever. 0.65 would be better, wouldn’t it? So I calculate a new factor with 0.65 : 0.50 = 1.3. Then I would have 7.00 x 1.3 = 9.10 as diameter and 67.75 x 1.3 = 88.075 as lines, or rope length. All other dimensions in the construction drawing would then have to be recalculated with this factor.

But if I have, for example, only one rope of about 55 m length, I would have to reduce the whole. The factor would be 55.00 : 67.75 = 0.8118081. The path width would then be 0.50 x 0.8118081 = 0.405904. This is again not so happy. I prefer to use 0.8 as a factor and get 67.75 x 0.8 = 54.2 m. The diameter would then be 7.00 x 0.8 = 5.60. Again, all other dimensions have to be recalculated accordingly.

So I can perform calculations according to different points of view.

Andreas recently brought into play the sand drawing “Luan” on Malekula, which Hermann Kern rejected as a labyrinth. It represents an uninterrupted line, but without an entrance or an access to the center.

But it attracted me to try to make a “real” labyrinth out of it. To do this, there must be a beginning and an end. This is easily done by cutting the unbroken line at one point. And then you bend the end piece towards the center. I took the lowest point of the outer line.

This is how the drawing will look as a labyrinth figure in concentric form:

The figure may look quite different from the original at first glance, but the lines are identical. The labyrinth has five circuits and four axes with three double barriers and passes through four sectors. The entrance to the labyrinth is on the first circuit, as is the entrance to the center. The path moves in serpentines towards and away from the center. It is a sector labyrinth and is reminiscent of the Roman labyrinths in serpentine form.

From a design point of view, I don’t really like the entrance to the labyrinth on the first circuit. By the way, an entrance to the center from the last circuit is not very happy either. Both are often seen in newly designed labyrinths. How can this be changed? The easiest way to do this is to choose only two double barriers instead of three, thus obtaining three sectors.

This is how it looks then:

The entrance to the labyrinth is on the 5th circuit and the entrance to the center is again from the 1st circuit as in the four-axis labyrinth.

Now we want to work a bit more on the reshaping. What would it look like if I arranged only 3 circuits instead of the 5?

Now this is very reminiscent of labyrinths shown earlier in this blog (see related posts below), especially the 3 circuit Chartres labyrinth.

Very topically to this Denny Dyke offers a necklace with pendant with exactly this labyrinth on his website:

This shows once again how interesting the subject of labyrinths can be.

Where does a labyrinth belong? And what relatives does it have? How do I actually sort the related labyrinths in a group? What kind of relationships are there? Or: How do I find the related ones in a group?

If I want to know something more, I first take an arbitrary labyrinth and generate the further relatives of a group by counting backwards and completing the numbers of the circuit sequences. It doesn’t matter whether I “catch” the basic labyrinth by chance or any member of the group.

As an example, I’ll take the 11 circuit labyrinth chosen as the second suggestion in my last post. Here it can be seen in a centered version in Knidos style:

The level sequence is: 0-7-2-5-4-3-6-1-8-11-10-9-12. The entrance to the labyrinth is on the 7th circuit, the entrance to the center is from the 9th circuit. This is the reason to name it 7_9 labyrinth.

By counting backwards (and swapping 0 and 12), I create the transpose labyrinth to it: 0-9-10-11-8-1-6-3-4-5-2-7-12.

The entrance to the labyrinth is on the 9th circuit, and the entrance to the center is on the 7th circuit.

Now I complete this circuit sequence 9-10-11-8-1-6-3-4-5-2-7 to the number 12 of the center, and get the following level sequence: 0-3-2-1-4-11-6-9-8-7-10-5-12. This results in the corresponding complementary version.

Now a labyrinth is missing, because there are four different versions for the non-self-dual types. The easiest way to do this is to count backwards again (so I form the corresponding transpose version) and get from the circuit sequence 0-3-2-1-4-11-6-9-8-7-10-5-12 the circuit sequence: 0-5-10-7-8-9-6-11-4-1-2-3-12. Alternatively, however, I could have produced the complementary copy by completing the digits of the path sequence of the first example above to 12.

The entrance to the labyrinth is made on the 5th circuit, and the entrance to the center is made from the 3rd circuit.

Now I have produced many transpose and complementary copies. But which is the basic labyrinth and which the dual? And the “real” transpose and complementary ones?

Sorting is done on the basis of the circuit sequences. The basic labyrinth is the one that starts with the lowest digit: 0-3-2-1-4-11-6-9-8-7-10-5-12, in short: the 3_5 labyrinth, i.e. our third example above.

The next is the transpose, the 5_3 labyrinth, the fourth example above.

This is followed by the dual, the 7_9 maze, which is the first example above.

The fourth is the complementary labyrinth, the 9_7 labyrinth, the second example above.

The order is therefore: B, T, D, C. This is independent of how the labyrinth was formed, whether by counting backwards or by completing the circuit sequences.

To conclude a short excerpt from the work of Yadina Clark, who is in the process of working out basic principles about labyrinth typology:

Groups

Labyrinths related by Base-Dual-Transpose-Complement relationships

Any labyrinth in a group can be chosen as the base starting point to look at these relationships, but the standard arrangement of the group begins with the numerically lowest circuit sequence string in the base position.