Straight and inclined pipes

mauro sacchetto

In a serif like Garamond (but the problem is general), the thick straight pipes (as in /I/) have a certain width, but then it is necessary to build the inclined pipes (the right for the /M/, those for the /W/ etc.).

Now, if the straight pipes are simply inclined (as in certain automatisms allowed by the editor for fonts), they are decidedly thinner, and the thinner the more inclined.

Is there a sort of "algorithm" to calculate exactly the thickness of such inclined pipes, in relation to straight ones, according to the degree of inclination?
Thamk you

Andreas Stötzner

The appropriate width of any upright or inclined stem is measured best by the judgement of your eyes. Usually the leaning stems (A, M, N, V, W, X, Y Z) tend to be a tiny bit thicker than the upright ones, in order to appear equivalent.

mauro sacchetto

Yes, I see that they have to be thicker. Only, maybe it would be possible to apply some geometry about triangles... PS Stems, not pipes, you're right.....

RichardW

Well, there's a simple geometric rule that for the same physical thickness, the thickness along the x-axis has to be proportional to the secant of the offset from the vertical.  There are all sorts of reasons for that not to be good enough.

K Pease

Yes, there is a trigonometric relation. The proportion by which it is thinned is the secant of the skew angle, and the proportion by which it must be enlarged to make it the same again is the cosine.

But these are small changes, and in the end you're probably working with whole UPM, so it doesn't much bear thinking about, when in practice you'll just add a unit or two and measure what you get.

mauro sacchetto

Thank you for your scientific clarification.

As always in this area, I seem to understand that formulas alone do not solve everything and must then be adapted according to the eye.

I studied some trigonometry many years ago and frankly I don't remember anything.

I am not working on various weights of the same family, so I do not modify the UPM.

Simply, where I have an /I/, I have to produce for example the /V/, and proportionally adapt the left stem.

Is there a simple formula I can apply?

Adam Jagosz

> I am not working on various weights of the same family, so I do not modify the UPM.

Why would you?

You're working with FontForge, right? Hit the R key for the ruler tool, then measure the stem's thickness (its actual thickness, perpendicular to the stem). Start with the same as a vertical stem, then go a bit thinner (or thicker) until it's right.

Btw I wouldn't call an element's horizontal dimension thickness. When a fat man lies down, does he become suddenly tall?

Ray Larabie

When dealing with manually adjusting extra-light, ultra-light or hairline obliques I sometimes slant a vertical line at 30 degrees. I manually adjust and measure until the thickness is correct and make a note of the horizontal offset. I can use the offset to quickly figure out how much I need to adjust 15 degrees, 10 degrees, 20 degrees etc. I'm no good with mathematics but the correlation between angles and offsets seems to be linear. Using this technique, I can guess how many arrow key taps I'll need to match the thickness of the vertical. While it's true that you should rely on your eyes, there are cases where you might want more precision or speed. I find that to be especially true with very light obliques.

RichardW

@Mauro Sacchetto  Obviously we've been too elegant in our statements.  If the x-coordinates of a simple stem differ by a, then for a 'stem' slanting at angle θ and of the same thickness (shortest distance from one side to the other), its x-coordinates at the same y-coordinate will differ by a/cos θ.  (The secant of an angle is the reciprocal of its cosine.)

mauro sacchetto

Scrivi il tuo commThanks for your answers, as always of the utmost competence.
Now I try to work both by eye and "trigonometrically", to see the differences between the two proceduresento