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

[Thomas Morley]

2015-10-06 15:15 GMT+02:00 BB <address@hidden>:
> I do not really understand what you desire to get. May be my response is not
> helpful but silly.

The constructed bezier-curves should get correct bounding-boxes.
Which can be watched by doing
\markup \box \stencil #whatever-stencil

>
> #(define pts-list
>   '((12 . 8)
>     (5 . 8)
>     (2 . 2)
>     (15 . 2)))
>
> In your List you have the start point defined with ay (x.y)= (2,2) and the
> end point (15,2) - that is the last pair of coordinates.
>
> Looks like I need to calculate the actual X/Y-extents of the resulting
> bezier-curve.
>
> The x extent of your curve therefore is 15-2 = 13.

Ofc true for `pts-list', but not for the rotated derivates

> The y extent is harder to calculate as it depends on the parameters of the
> bezier definition. A good description is here:
> http://www.math.ucla.edu/~baker/149.1.02w/handouts/bb_bezier.pdf

Thanks for it.

>
> The line override does nothing?
>
>          \override #'(box-padding . 0)

Play around with different values.

>
> If you set (0,0) instead of (2.2) the starting point is at (0.0) and this is
> the rotation center. (You rotate in your example around this point.) Then
> you have to change the last corrdinate pair to (13,0) for to get the seme
> base length and to stay in the box.
>
> The first coordinate pair of the definition in
>
> #(define pts-list
>
> controls the steepness of the start and end of the curve and the form,
> please see the link above for details.
>
> May be I missed the point?


I followed David's hint with up to now convincing results.
`make-path-stencil' does correct bounding-boxes out of the box.

Cheers,
  Harm