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Posts Tagged ‘John James’

All the while I wanted to know the exact number of the cogs / cusps / scallops / teeth  surrounding the Chartres labyrinth.
As well I noticed quite different values  in some new built replicas of the Chartres labyrinth.

If there is no unequivocal opinion about sense and purpose of the same, at least the exact number of these elements should be clear. This is important for everybody which wants to copy an original Chartres labyrinth and at the same time a sign how exactly they have looked at the original. Finally, we can admire the original more than 800 years in the Cathedral Notre Dame of Chartres in France.

Graphic of the labyrinth

Graphic of the labyrinth

The information in the literature about the exact number is not so unequivocal as one could mean. It is the speech of 112 or 113. However, it seems to be clear that round a closed labyrinth 114 elements would exist and that one element was left out to form the entrance. By no means two elements would be necessary for that, because one fits perfectly. Therefore we get 113 lunations. Exact counting on good graphics or photos also amounts to 113.

One have speculated a lot why these are just 113 or 114 elements. The centre of the labyrinth exists of 6 “petals” which are open to the middle. The 6, multiplied by 19 (number of perfection, the sun, the flower of the life) results in 114 similarly constructed “rays” for the border of the labyrinth. The sum of the digits 114 is 6, like the number of the elements in the middle.

Photo of the entrance

Photo of the entrance

Graphic of the entrance

Graphic of the entrance

Only, how should one count, actually? The lunations around the labyrinth are made of circular parts which like gear wheels are built of jags and bowl-shaped deepening which are cut off in the upper part. I propose to call a gear wheel with a quarter circle on both sides as one element. Single components in the labyrinth are constructed like this, as the joints in the photos show (e.g., element 58 middle  part at top).

The blue drawn unity should be counted as one element. If one applies this to the complete labyrinth, 114 elements arise in the closed labyrinth. If one cuts out one for the entrance, we have 113 remaining elements.
One could also count the “bowls” instead of the “peaks”. Then we would have 112 complete and two half parts (at the entrance). However, all together we would have again 113 items.

Photo of the middle part at top

Photo of the middle part at top

Graphic of the middle part at top

Graphic of the middle part at top

Something else has struck me when I constructed the periphery of the lunations. The circumference of 40.511 m (given by a mean value for the diameter of 12.895 m) can be divided in 114 equal parts, this would be 0.355 m for each. The position of every single element should thereby result from that. If one begins with the division in the middle of the omitting element for the entrance (114 in the graphic), then the tooth on top (56 in the graphic) would not lie on the middle axis as it is the case quite obviously in the original.
The exact place of this part would be about 7 cm farther on the right. If one liked to keep up the exact number of the teeth, one must diminish the distances in the left half (element 1 – 56) and increase it in the right half(element 56 – 113). I do not know how the master builders of the Middle Ages handled this. Maybe they had quite different points of view?

Finally once again the statement: The Chartres labyrinth has 113 visible lunations.

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There is different information about that in the literature. The length can be  hardly measured directly. A pedometer or even a tape measure are inexpedient for it.
In my opinion it can only be calculated with a mathematical model, so with an as accurately as possible reconstruction of the labyrinth. And there it could become difficult.
There is a book of John and Odette Ketley-Laporte with the title “CHARTRES – le labyrinthe déchiffré” in French in which quite precisely looking measures are stated. I have taken them into account as far as possible. But I have also corresponded with Jeff Saward, the best labyrinth expert worldwide who has to say a lot also to the Chartres labyrinth. He has written an own article about that on his website: The Chartres Cathedral Labyrinth – FAQ’s.
The best picture of the Chartres labyrinth which I know is this stone by stone exact drawing of Jeff Saward.

Drawing by kind permission of Jeff Saward (labyrinthos.net)

Drawing by kind permission of Jeff Saward (labyrinthos.net)

Just compare the joints in the drawing to those in the photos.

The entrance

Part in the middle

Before the centre

Upper part of the centre

Most upper part

Lunations

Although already (almost) everything is said, I dare, nevertheless to give some own considerations and to come along to answer the question: How long is then now the path in the Chartres labyrinth?
To this end I use a mostly exact design drawing of the labyrinth. I make this in the kind in which I have already provided several labyrinth drafts: with CAD (computer-aided design) in AutoCAD.The labyrinth has 11 circuits and a bigger middle. The six petals of the middle are not necessary for the calculation of the length of the path because the way ends in the very centre.
For the path width I attain 34.35 cm, for the limitations (walls) 8.25 cm. These are mean values, the joints are included. This results in a dimension between axes important for the design of 42.6 cm for one circuit, so 22 x 42.6 cm = 9.372 m for all the circuits. For the middle I take a dimension between axes of 3.035 m.Therefore I get an average diameter (measured in the axis of the walls) of 12.407 m for the labyrinth without the lunations. If I take the walls in addition, I come on 12.490 m.
I get an overall diameter of 12.895 m including the lunations.
Ariadne's Thread inside the Chartres Labyrinth

Ariadne’s Thread inside the Chartres Labyrinth

In this so drawn outline I construct the path axis (the red thread, Ariadne’s thread). I get a polyline with straight lines and curve pieces which collide free of sharp bends. Normally the path axis lies in the middle of the path. However the crossings of the straight lines in the middle part causes some difficulty. There is a sort of curve widening, the width is bigger than 34 cm. This concerns 6 places in the labyrinth. There the axis can be defined differently. From there one could also calculate different path lengths.

Ariadne's Thread

Ariadne’s Thread

According to my calculated alignment I get a path length of 263.05 m. The beginning of the line is the intersection of the path axis with the outside radius of the lunations.

John James stated 261.5 m as path length, Robert Ferré states 261.55 m, Jeff Saward has calculated 262.4 m (with the beginning further inside), Hermann Kern reported 294 m.

I would like to say the path length is 263 m. This is an integer. And the sum of the digits is 11, such as the 11 circuits of the labyrinth.

For my English speaking visitors I want to say it in feet: 862.86 ft.

The design drawing

The design drawing

Here you may check, print or copy the design drawing with the most important measurements as a PDF file.

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