Re: fitting a picture to the available space

["Thomas A. Schmitz" <thomas.schmitz@uni-bonn.de>]

On Sep 11, 2008, at 7:59 PM, Thomas A. Schmitz wrote:

>
> On Sep 11, 2008, at 6:20 PM, Thomas A. Schmitz wrote:
>
>>> \starttext
>>>
>>> \start
>>>
>>> \setbox\scratchbox\vbox{\externalfigure[mill]}
>>>
>>> \dimen0=\wd\scratchbox
>>> \dimen2=\ht\scratchbox
>>>
>>> \framed[frame=on,strut=no,width=8cm,height=2cm]
>>> {\dimen1=\hsize \divide\dimen1 by \dimen0
>>> \dimen3=\vsize \divide\dimen3 by \dimen2
>>> \ifdim\dimen1>\dimen3
>>>   \externalfigure[mill][height=\vsize]
>>> \else
>>>   \externalfigure[mill][width=\hsize]
>>> \fi}
>>>
>>> \framed[frame=on,strut=no,width=2cm,height=8cm]
>>> {\dimen1=\hsize \divide\dimen1 by \dimen0
>>> \dimen3=\vsize \divide\dimen3 by \dimen2
>>> \ifdim\dimen1>\dimen3
>>>   \externalfigure[mill][height=\vsize]
>>> \else
>>>   \externalfigure[mill][width=\hsize]
>>> \fi}
>>>
>>> \stop
>>>
>>> \stoptext
>>>
>>> Wolfgang
>>
>> Wolfgang,
>>
>> as always, you're a source of wisdom and knowledge... Just I
>> understand this correctly and can adapt it to my macro: \dimen1=
>> \hsize: here \hsize refers to the size of the \framed inside which
>> we're operating, right?
>
> And another question: I get an error "! Illegal unit of measure (pt
> inserted)." Is it really possible to divide a dimension by another
> dimension? Not according to what I read here: http://www.tug.org/utilities/plain/cseq.html#divide-rp
>  "must be a nonzero integer." See, I'm far from being a native
> speaker of TeX...
>
I now have something along these lines (ugly code ahead!):

\setbox\scratchbox\vbox{\externalfigure[mill]%
        \dimen0=\wd\scratchbox%
        \def\@WD{\withoutpt{\the\dimen0}}%
        \def\@@WD{\integerrounding{\@WD}}%
        \dimen4=\textwidth \divide\dimen4 by \@@WD%
        \dimen2=\ht\scratchbox%
        \def\@HT{\withoutpt{\the\dimen2}}%
        \def\@@HT{\integerrounding{\@HT}}%
        \dimen6=\textheight \divide\dimen6 by \@@HT%
        \ifdim\dimen4>\dimen6%
	  \def\Myheight{\textheight}\def\Mywidth{}%
        \else%
	  \def\Mywidth{\textwidth}\def\Myheight{}%
        \fi

and then use height=\Myheight,width=\Mywidth. Bu there's still a  
problem... Anyway, does it make sense in general?

Thomas
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