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One further iteration. Please let me know what you think of the
interface:
-\compoundSlur \with {
% offsets against the automatic control points
offsets =
#'((0 . -1.5) ; left starting point
(-2 . -1) ; second control point
(2 . -5) ; second-to-last control point
(0 . 0)) % right end point
inflection-ratio =
#'(0.6 . 0.65) % X/Y ratio of the inflection point
% calculated between the actual end points of the slur
% defaults to the center
% Slope of the line around the inflection point
% defaults to 1/1
inflection-slope = 5/3
% length of the "handles" around the inflection
% given as a ratio to the left or right baseline
% (distance between inflection point and slur's end point).
% If not given they default to the length of the respective handle
% on the left or right side
inflection-ratio-right = 0.5
inflection-ratio-left = 0.25
% Display control points
ann = ##t
}
Of course one could use a list of unnamed arguments or an alist
instead of the \with {} clause (I just happen to like them). I
have spiced up the "annotation" even more compared to the last
version (prints the original slur and its control points in light
gray in the background), and I have the impression this approach
to defining the inflection point is quite natural, but I'm open to
discussion.
One thing that is still missing from my personal wish list is a
flag to make the center symmetric again.
Urs
Am 19.09.2016 um 11:00 schrieb Urs
Liska:
Am 19.09.2016 um 00:32 schrieb David
Kastrup:
Urs Liska <address@hidden> writes:
Am 18.09.2016 um 20:54 schrieb David Kastrup:
Do you know how to split a bezier at a given ratio into equivalent
beziers? It's a comparatively simple operation and I think it's already
somewhere in the C++ code though without access from Scheme.
No, but I should be able to figure it out (if noone sends a pointer
before I manage to do so).
Well, METAFONT uses the notation
a[z1, z2]
for z1 + a*[z2-z1], mapping a range of 0..1 for a linearly between z1
and z2.
If we have points z1, z2, z3, z4 defining a Bezier, then the two split
beziers are defined with the points
z1, a[z1, z2], a[a[z1, z2], a[z2, z3]], a[a[a[z1, z2], a[z2, z3]],
a[a[z2, z3], a[z3, z4]]
and
a[a[a[z1, z2], a[z2, z3]], a[a[z2, z3], a[z3, z4]]],
a[a[z2, z3], a[z3, z4]], a[z3, z4], z4
Basically, calculation of a point a on an n-grade Bezier is done using a
recursive formula to depth n, and keeping the intermediate results will
give you the control points for the Bezier curves split at that point.
I think before diving into that I share what I currently have, so
we may discuss which approach should actually be continued.
The attached solution does the following:
- Apply offsets for the start/end points and to the second and
second-to-last control-points, based on the original points of
the non-compound slur
- Add an inflection point, which is specified as a point
between the (actual) end points of the slur, given X and Y
ratios (as a pair of numbers between 0 and 1
- Determine the length and slope of the line going through the
inflection point.
- Currently this is done through specifying one point
relative to the inflection point and mirroring it
symmetrically
- Instead I'd like to specify an angle and a length.
- It would be nice to have the angle relative to the slope
of the slur as a whole, but that may not be a good idea, as
we have actually two separate lines with different slopes
- Length should be given as a ratio, presumable relative to
the length of the line between the inflection point and the
respective end point.
- There should be one optional argument to enforce
symmetrical points here.
BTW I've spiced up the control points display a bit. I hope
it's self-explanatory.
I would like to integrate this with Janek's \shapeII functions
(https://github.com/openlilylib/snippets/tree/master/notation-snippets/shaping-bezier-curves)
as I think there'll be quite some code (and interface?) that can
be shared.
Opinions?
Urs
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compound-slur-relative.ly
compound-slur-relative.preview.pdf