How to Calculate the Classical 7 Circuit Labyrinth

The question of a formula or table for calculating the construction elements in the labyrinth has arisen several times. For me, I solved the problem by designing and constructing the various labyrinths using a drawing program (AutoCAD). This creates drawings that contain all elements geometrically and mathematically exactly.
However, I do not print these on a specific scale, but adjust the size of the drawing so that it always fits on a sheet in A4 format.
Only the dimensions are decisive for the implementation of the labyrinth in the location. If possible, I also try not to use “crooked” measurements, but simple units, usually the meter.
The dimensions are therefore suitable to be scaled and so labyrinths can be constructed in different sizes.
The drawings thus represent a kind of prototype. Since the axes of the lines are always given, the widths of the boundary lines and the path can be varied.

The construction elements of the lines in the labyrinth mostly consist of arcs and straight lines. In the program used (AutoCAD), these individual elements can be combined into so-called polylines and their total length is then calculated.

The length specifications for the boundary lines and the path (the Ariadne thread) in the drawing are thus created. The boundary lines consist of 2 straight lines and 22 arcs (24 elements in total). The Ariadne thread consists of 1 straight line and 25 arcs (26 elements in total). The entire labyrinth consists of 50 individual elements.

It would be possible to calculate all of this in a table with the corresponding formulas, but it would be more cumbersome and extensive.

The scaling factor makes it easier to calculate variants in different sizes. The labyrinth calculator is something like a summary and general instructions for use. Here especially for the well-known Classical 7 circuit labyrinth.
However, this method has also been described for other types of labyrinths in this blog.

The Labyrinth Calculator

The Labyrinth Calculator

Here you can view, print or download the drawing as a PDF file

Here are some comments on copyright:
All drawings and photos in this blog are either mine or Andreas Frei, unless otherwise stated, and are subject to license CC BY-NC-SA 4.0
This means: You may use or change the drawings and photos without having to ask us if you name our names as authors, if you do not use the drawings and photos for commercial purposes and if you publish or distribute them under the same license. A link to this blog would be nice and we would be happy, but it is not a requirement.

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The Classical 7 Circuit Labyrinth with Crossed Axis

Andreas recently brought here a posting to the Jericho labyrinth St. Gallen which probably shows the historically first crossing of the main axis in a classical labyrinth.

I have written about that already. But now I would like to do it once again. Because for me it seems to be an element to design the labyrinth which I have not seen anywhere.

I became aware of that when I have no longer drawn the labyrinth from a seed pattern, but from the path sequence. And, besides, have noted that there also are different possibilities to connect the lines.

Using the example of the classical 7 circuit labyrinth I will explain this once again. How many possibilities to cross the axis are there, and how does it look like?

First the original labyrinth, however round and with a bigger middle.

The classical 7 circuit labyrinth

The classical 7 circuit Labyrinth

The last path sequence into the middle lies on the vertical main axis. The entrance lies on the left side of the main axis, leads to the third circuit, and turns to the left at first. The entry into the middle takes place from the fifth circuit from the right side, and faces the entrance.


How often can I now traverse the axis?
At two positions: From the first to the fourth circuit, and from the fourth to the seventh circuit. This can happen at each position alone or at both positions together. The result are three variations.

Here the first version:

Crossing the axis from the 1st to the 4th circuit

Crossing the axis from the 1st to the 4th circuit

By crossing the main axis from the first to the fourth circuit I do not change the direction of movement as in the original labyrinth. I am still turning to the left in the fourth circuit.
However, thereby I also reach the middle from the left side, so to speak I have laid this entry on the other side of the main axis.
The main entrance slides a little further to the left, and the two lower turning points also move to the left.


The second version:

Crossing the axis from the fourth to the seventh circuit

Crossing the axis from the fourth to the seventh circuit

Here the change from the first to the fourth circuit remains like in the original, however, from the fourth to the seventh circuit I maintain the “spin”.
The entry into the middle is executed from the left side as it is in the original. However, the main entrance slides to the right side. Both lower turning points are shifted to the right.


The third version:

Crossing the axis from the first to the fourth, and from the fourth to the seventh circuit

Crossing the axis from the first to the fourth, and from the fourth to the seventh circuit

The vertical main axis is crossed twice as in the previous versions, now together.
Through that the  lines are displaced considerably. Everything moves to the left. The main entrance lies again on the left side, the entry into the middle is made from the right side.


Here the path as Ariadne’s thread:

Ariadne's thread in familiar manner

Ariadne’s thread in familiar manner

Ariadne's thread with the axis crossed twice

Ariadne’s thread with the axis crossed twice

 

 

 

 

 

 

 

 

 

 

Maybe one can dismiss that as unnecessary? It would be nice, nevertheless to try it out in practice. Above all how it feels to experience another change of direction than in the original.

Maybe the opportunity arises in a big sandtable exercise? On a sandy beach for example? Where one can simply scratch the lines into the sand, and allows the flood to erase them out leniently.

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The Cretan Labyrinth: The Long and Winding Way to the Middle

Many are surprised how long the way in a labyrinth can be, especially if they walking labyrinth for the first time. And many who want to build a labyrinth, e.g., with stones or with candles, are astonished how much material they need.

Under the heading “Principles of Form” in his book Labyrinths Hermann Kern writes about the >tortuous path principle<:

– if the path fills the entire interior space by wending its way in the most circuitous fashion possible

If I stand ahead of a labyrinth, the middle, the goal is within  my reach. However, only when walking in I get to know how winding and complicated the way is in reality. But yet, this way, the red thread or Ariadne’s Thread is the continuous line in the labyrinth, without crossroads or junctions.

Ariadne's Thread

From A to Z: The long and the short path

In the drawing I call “A” the beginning of the path and “Z” the goal, the center or middle. In many labyrinths I could reach directly the middle with a few steps across all limitations. But this is not really what is intended with a labyrinth.

Now I compare for a 7 circuit labyrinth with a diameter of about 15 m the short way (direct connection between A and Z) with the long way along the Ariadne’s Thread. The length of the short way amounts to 6.33 m, the long way has a length of 154.62 m. Or differently expressed: The long way is 24.4 times longer than the short way (154.62: 6.33 = 24.4).
One could also see in this a formula for the labyrinth. To calculate how powerful is the  layout for example. Or how wended is the way? Or from what minimal surface area I can extract which maximal length?
Maybe one could call this value in honour of Hermann Kern “detour factor” 24.4?

If I handle this thread at the beginning and at the end and pull it apart, I will get a straight line which reaches from “A” to” Z” and is as long as the way inside the labyrinth, i.e. 154.62 m.
I can arrange this to a circle. The perimeter corresponds to the straight line of 154.62 m. The resulting diameter would be 49.22 m.
I can also make a square with the same extent from it. Then this would have four side lengths of 38.65 m.

The following drawing, yet not true to scale, illustrates the different figures and the true ratios among each other:

Ariadne's Thread unrolled

Ariadne’s Thread unrolled

Easter Labyrinth

Beginning to live again? I can do that each day. But at this time we are particularly reminded of it. Symbol for new life is amongst others the egg – and the labyrinth. Can that be combined?
I could however paint a labyrinth on a egg. As I am not so talented manually (or perhaps only too comfortably?) I made that digitally.

Labyrinthegg

Labyrinthegg

Happy Easter!