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.