On Sat, 2021-09-04 at 22:21 +0200, Jean Abou Samra wrote:
> Le 04/09/2021 à 20:39, Richard Shann a écrit :
> > (I'm trying to write code that will draw a box round A4 or letter
> > paper
> > with just a change of some scaling factors - I can work around it
> > but
> > it's distressing not to
Le 04/09/2021 à 20:39, Richard Shann a écrit :
(I'm trying to write code that will draw a box round A4 or letter paper
with just a change of some scaling factors - I can work around it but
it's distressing not to understand what is going on).
Is this the kind of thing you're looking for?
On Sat, 2021-09-04 at 20:44 +0200, David Kastrup wrote:
> Richard Shann writes:
>
> > Can someone educate me as to why this works:
> >
> > \markup \path #0.4 #'((moveto 0 0) (lineto -30 50) (lineto -20 80)
> > (closepath))
> >
> > but this does not:
> >
> > \markup \path #0.4 #'((moveto 0 0)
On 2021-09-04 11:39 am, Richard Shann wrote:
Can someone educate me as to why this works:
\markup \path #0.4 #'((moveto 0 0) (lineto -30 50) (lineto -20 80)
(closepath))
but this does not:
\markup \path #0.4 #'((moveto 0 0) (lineto -30 (* 2 25)) (lineto -20
80) (closepath))
The difference
Richard Shann writes:
> Can someone educate me as to why this works:
>
> \markup \path #0.4 #'((moveto 0 0) (lineto -30 50) (lineto -20 80)
> (closepath))
>
> but this does not:
>
> \markup \path #0.4 #'((moveto 0 0) (lineto -30 (* 2 25)) (lineto -20 80)
> (closepath))
>
> The difference being
Can someone educate me as to why this works:
\markup \path #0.4 #'((moveto 0 0) (lineto -30 50) (lineto -20 80) (closepath))
but this does not:
\markup \path #0.4 #'((moveto 0 0) (lineto -30 (* 2 25)) (lineto -20 80)
(closepath))
The difference being I replace 50 with (* 2 25)
(I'm trying to
