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Golden Section as a method for finding regular and bold complements

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Posted

I understand (the maths at least) from Lucas de Groot's site how an optical "halfway" weight can be established by b² = ac (where a = regular stem, b = medium stem and c = bold stem thickness), although I misquoted it as a² + b² = c² which is Pythagoras' theorem because I wasn't really concentrating!

I also understand that individual letters (such as e) impose crazy complications on the process. I struggle getting bolds and heavies to look right. The heavier it gets the harder it becomes. I also understand that there is a problem with this whole process at the thin and, particularly, the fat ends of a typeface family.

Nevertheless, the real question was "How much thicker a bold weight should be compared to its regular?"

What I noticed from my 'research' (which I know was not thorough) was that as I added more regular/bold examples to the list, the closer the average got to Φ.

James, I completely agree with you and the Duke that “If it looks good, it is good”. But what I found was that when I used Φ, it looked better than the proportions that I was using. I need a lot of help and if maths helps me I'm happy to use it. I want to know how to make it look good/better, I have no problem with your assertion, it's getting there that I don't know how to do!

Incidentally, I realise there was a slight flaw in my explanantion: It turns out that the ratios (regular:bold) varied between 0.52 (not enough difference to be effective) and 0.67 (too much difference), but the average was 0.6216. Of course, the smaller the ratio (regular:bold), the larger the difference between the weights (so the two comments in parentheses above are the wrong way round).

Bill, using your example figures from Minion, if the regular width was 1, then I assume you mean that the semibold is 1.39 and the bold was 1.65. It follows that the ratio I am looking at for regular to bold = 1/1.62 = 0.617 (which is very near Φ). In these terms, the ratio of the semibold to the bold = 1.39/1.62 = 0.858 * (nowhere near Φ and far too large to provide a sufficient contrast to be used as a regular/bold complement together in text).

*The relation of Minion semibold and bold is not 39/62 but 1.39/1.62

Posted

>not 39/62 but 1.39/1.62

Yeah, I was comparing the excess over the regular, which is probably meaningless--hence the smiley.

But my question concerned the utility of the Adobe Semi-Bold vs the Bold. To me the Semi-Bold looks better within running text--a side head, a term to be defined, beginning of a list, etc.--and that doesn't have the golden ratio, but significantly less contrast.

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