Re: How to get X/Y-extent of a bezier-curve?

[BB]

Just for fun tried to undertie all the text with one unertie

\markup \line {

\undertie {

\underline "underlined"

\undertie "undertied"

\override #'(offset . 5)

\override #'(thickness . 1)

\undertie "undertied"

\override #'(offset . 1)

\override #'(thickness . 5)

\undertie "undertied"

"Eng" \undertie "ele" "en"

}

}

It is only possible to override the vertical coordinate? It is  not possible to override start and end coordinate to produce inclined ties?

Regards

On 07.10.2015 12:14, Thomas Morley wrote:

2015-10-06 14:49 GMT+02:00 David Kastrup <address@hidden>:

Thomas Morley <address@hidden> writes:

Hi all,

I'm going to write a generic bow-stencil.
Below you'll find a boiled down example.

The main problem: how to determine the correct extents.
Looks like I need to calculate the actual X/Y-extents of the resulting
bezier-curve.
Though, obviously my maths-skills are not sufficient.

Oh, that's a nuisance.

Any hints?

I'd just call make-path-stencil and use the bounding box results from
that.  No need to reinvent the wheel.

Yep.
Using make-path-stencil is much more straight-forward.
Thanks.

If someone interested, I'll attach an image with an excerpt of my test-suite.

I plan to replace make-parenthesis-stencil and to implement
https://sourceforge.net/p/testlilyissues/issues/3088/

While this particular wheel could likely profit from a do-over with more
of a view towards efficiency and numerical robustness, there is no point
in code duplication.

I think we could replace the body of current
make-bezier-sandwich-stencil with make-path-stencil.
But obviously you think about some deeper modification.
Could you give some details?

Cheers,
  Harm

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