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This labyrinth, while looking very similar to last year’s, has a changed line layout.
As we have seen (in part 1), the most different variants of the Wunderkreis can be created. Depending on which part is emphasized more or less, they then look like.
When creating a new labyrinth, of course, it also depends on the size of the available space and the purpose the labyrinth is to serve.
The path sequence, if we go first to the left: 0-3-2-1-4-a1-b2-c1-c2-b1-a2-5-0. To the right we have: 0-5-a2-b1-c2-c1-b2-a1-4-1-2-3-0.
With the digits we have the sequence with odd and even numbers, as we know it from a classical labyrinth.
With the letters, which designate the elements of the double spiral, we can also see a certain systematic: The letters come alternately one after the other. If two identical letters follow each other, we have reached the center of the spiral and the basic change of direction. The additions “1” designate the lower part and the addition “2” the upper part of a transition.
If we take a closer look at the circuit sequences, we can see that the second one (to the right) is opposite to the first one.
So we can say that here two different but related labyrinths of a group are united in one. Depending on which path we choose first.
How many circuits does this Wunderkreis actually have?
That is a little difficult to count. To do this, we divide the figure into three parts, the lower left quarter, the upper half, and the lower right quarter. Let’s start at the bottom left: There are the 3 “labyrinthine” circuits and 3 of the double spiral. At the top we have 4 “labyrinthine” circuits and the 3 of the double spiral. Bottom right: 5 “labyrinthine” circuits and the 3 of the double spiral. So, depending on the angle of view, we have 6, 7 or 8 circuits.
The type designation is the maximum number of “labyrinthine” turns plus the letter sequence for the turns of the double spiral. Adding both gives the number of total circuits. In this example “5 a-c” so 8 in total.
In the file name for the drawings I have tried to express this as well, additionally provided with the indication of the entrance and the exit of the labyrinth.
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.
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.
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.