Makers’ marks

Attachment 31025

I have been intrigued by the purpose of makers’ marks for over 30 years. Carpenters’ marks were discussed a lot in the early days of the Carpenters’ Fellowship. People would find curious incisions in medieval timber frames, say, on the lintel above a hearth. This would be described “apotropaic” ie protection against witchcraft, but to my mind they might equally have been the marks of the men who built it, as a kind of celebration, like topping-out. (One example above, reproduced in the Fellowship’s “Mortise & Tenon”
I read somewhere that in some trades, masons especially, their mark had to be proven. Now what does this mean? Probably that the apprentice or journeyman has reached an acceptable standard to be received into the guild. But I have also read that their mark had to be geometrically proven, with compasses and straightedge, in the way that a theorem must be proven, so that the candidate’s mark also demonstrates competence in geometry, well, a mason would have to, wouldn’t he? This idea I seem to have derived from a novel by Alan Garner, “The Stone Book Quartet”.
These traditions have largely died out, but if anybody knows anything about them, please let us know.
I have a mark, but nobody authorised it. It was a persistent doodle, and it stuck. But it is a regular geometric pattern, not a star, but similar!

The geometrically proven mark served much like a trade mark does today. Another tradesmen might make a mark which looked almost identical but once examined it would be found out not to be the mark of the master whose mark had been proven. This copy tradesmen may not have known of the other's mark and his may be a proven mark as well thereby having his name in the journal along with his mark. Otherwise if he was less than an upstanding tradesmen and he was trying to capitalize on another master's work, much the same way counterfeiters tried and more often failed to recreate an exact reproduction of a secrete identifying mark on a painting or bank note only the original master or the records keepers knew the true geometric dimensions of the marks. he would eventually be exposed as a fraud.

Thanks, Frank, this is a very plausible explanation.
If anybody copies my work, I should be flattered. It’s not likely to occur. And I can’t remember the geometry, either!

So, a maker's mark could not include a regular septagon or nonagon (7 or 9 sides) since it's mathematically provable that neither of those figures could be constructed with only compass and straightedge.

A brief diversion from my mathematical bag of tricks...

Everyone knows that a regular hexagon can be easily constructed in a circle using only a compass and a straightedge. With the compass set to the radius of the circle, simply walk it around the circle striking off points on the circumference. There will be exactly six points. Connect these points with straight lines and the result will be a regular hexagon.

This raises the question of what other regular polygons can be constructed using only compass and straightedge? Obviously, three and four sides are easy and we know from above that six is possible. What about five or seven?

Karl Friederich Gauss, the German mathematical prodigy and genius, solved the generalized problem. He proved that a regular n-gon can be constructed with compass and straightedge if n is the product of a power of 2 and any number of *distinct* Fermat primes (including none).

A Fermat prime is a Fermat number of the form

Fk = 2^(2^k)+1

that is also a prime. The known Fermat primes are:

k Fk

0 3
1 5
2 17
3 257
4 65537

Not all Fermat numbers are prime. The list above includes all the currently known Fermat primes. For example, for n = 5, we have the Fermat number 4,294,967,297 which has the factors 641 and 6,700,417 and so is not prime.

Also, the restriction to *distinct* primes is important. A 9-sided polygon cannot be constructed. 9 has prime factors 3*3 and, while 3 is a Fermat prime, ALL the factors of 9 must be DISTINCT Fermat primes for it to be constructible.

So, based on the above, a regular septagon cannot be constructed, but a pentagon can.

While I don't think this qualifies a maker's mark. . .it does identify a carpenter. My house was completed in 1902. Behind an original bathroom mirror is "E Jenkins 1902" scratched into the wood panel backing of the frame stiles. I'm certain this mark was made with point of a nail or scratch awl. Not being guild or union member, I don't think I've come across anything distinct as a provable mark.
Maker marks are common for silver & goldsmiths, some glass makers, tailors and others in custom 'bespoke' works, typically ornate stamps. Rare for production items unless further embellished, such as firearm engravings. They tended to use their name, ie Gustave Young, Felix Funken, or greatest of all Czesław Słania.

Attachment 31034
Here’s one method, not fully accurate, of constructing a pentagon. There is another method, using only 2 instruments, which I can’t find at present. If interested, I’ll draw it for you.
This method uses a square piece of paper.
Attachment 31035

Here’s a way to construct a heptagon, using 7 equal length rods. The page shows a less accurate method of construction, using a cord.Attachment 31036

Here’s an enneagon:
Attachment 31037

These diagrams are from a book by Miranda Lundy, entitled “sacred geometry”. Your knowledge of maths, far exceeds mine, Marv

This is a nice visual of one of the construction methods using only compass and straightedge...

http://www.mathopenref.com/constinpentagon.html

There are a few instances of the use of pentagons in metalworking fields. Pentagonal valve actuators on fire hydrants are commonly used to prevent tampering and many police shields are five-pointed.

The only septagonal item I've encountered is seven-pointed police shields. Some jurisdictions seem sensitive about offending Jews by using an hexagonal form.

I had a set of hubcaps for a ford that had 7 faux lug nut protrusions they drove people crazy trying to figure out when Ford started using a 7 lug pattern on their trucks

Quote:

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Originally Posted by Frank S

I had a set of hubcaps for a ford that had 7 faux lug nut protrusions they drove people crazy trying to figure out when Ford started using a 7 lug pattern on their trucks

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I think I owned the same set it was for my ford econoline work van. I bought them off ebay for like $40 for the set. I opened the box and I immediately noticed 7 lugs. They must of been a factory mistake

Here's a simple way to construct an approximate heptagon...

https://www.youtube.com/watch?v=cErccoHui9g

The true length of one side of a heptagon inscribed in a circle of radius R is given by:

C = 2 * R * SIN (180/7)

which for the unit circle (R = 1) evaluates to 0.86777

In the video he constructs the line labeled QM and uses that for the pentagon side. The length of QM can be found by noting that QO = R and MO = R/2, so:

QM = SQRT ( R^2 - R^2/4) = R * SQRT (3/4) = R * 0.5 * SQRT (3)

which for R = 1 evaluates to 0.86603

Close, but not perfect. Gauss' conclusions are not threatened.

This is a simpler method than the one in my book. Not sure how to write this: but I do not understand SQRT, or the asterisks. But certainly 0.866025 is almost exactly the square root of 3, divided by 2, which, following Pythagoras is my understanding of QM. There is another similar method on YouTube.

Quote:

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Originally Posted by Philip Davies

This is a simpler method than the one in my book. Not sure how to write this: but I do not understand SQRT, or the asterisks. But certainly 0.866025 is almost exactly the square root of 3, divided by 2, which, following Pythagoras is my understanding of QM. There is another similar method on YouTube.

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SQRT = square root
The asterisks denote multiplication; the circumflex (^) denotes exponentiation. All standard computer programming symbology.

Yes, the QMO triangle is a 30/60/90 triangle so QM = QO * cos (30) = R * 0.5 * SQRT (3). I simply chose to derive the length using the Pythagorean theorem rather than trigonometry.

Nevertheless, as I pointed out in my previous post, R * 0.5 * SQRT (3) is NOT the correct length of one side of a regular heptagon; the correct formula is the one I gave.