Am 13.12.2008 um 16:11 schrieb Yue Wang:
there is *no* conflict with the line above.
It is common scene that a paragraph should not be set solid (like 10pt/10pt leading).
of course not!
["conflict" means the given interlinespace gets bigger in this line]
Thus, two questions:
1) Why does it has to be like this?
Cannot reproduce your problem. see the attachment
the problem occurs in footnotes only, see attachment
2) As it is not possible to choose a bigger interlinespace, is it possible to adjust the high/depth of the superset math element "^0" ?
Yes, see "It's in the details" document. Maybe you also need grid typesetting.
the problem is grid independent.
\usetypescript[postscript][\defaultencoding] \setupbodyfont[postscript,10pt] \setupinterlinespace[line=12pt] \def\setnotebodyfont {\let\setnotebodyfont\relax \restoreglobalbodyfont \switchtobodyfont[rm,9pt]\setupinterlinespace[line=9.8pt,height=. 79,depth=.21]\setupalign[block,hanging,hz]\parskip2pt}% FN-Text \starttext right\footnote{\input zapf V\high{0} \input zapf\par} wrong\footnote{\input zapf $V^0$ \input zapf\par} \stoptext
Am 13.12.2008 um 19:19 schrieb Yue Wang:
Hi,
the problem is grid independent.
It's obvious: V\high{0} \neq $V^0$. you can make two boxes, and it is clear that their heights are different. I can send you pictures to demonstrate their difference privately.
I see the pictures and they demonstrate how weird the problem is: In the case of "V\high{0}" the "0" is higher than in "$V^0$", and yet the interline space gets NOT disturbed!!! Or put it the other way round: You use $V^0$, which doesn't move the "0" as high as "\high{0}" would do, and nevertheless the distance to the line above gets increased!!! 1) This doesn't make sense, does it? 2) Why gets the line space increased even though the "0" is not that high?! 3) How can the behaviour of "$V^0$" be adjusted in such way as "\high {0}" does (i.e. set the "0" but dont move away from the line above) Thanks, Steffen
I see the pictures and they demonstrate how weird the problem is: In the case of "V\high{0}" the "0" is higher than in "$V^0$", and yet the interline space gets NOT disturbed!!!
Or put it the other way round: You use $V^0$, which doesn't move the "0" as high as "\high{0}" would do, and nevertheless the distance to the line above gets increased!!!
TeX did this quite right, I think. The bounding box of the inline formula looks ok. Concerning your \high stuffs: you can read the code and it is quite clear that the box dimension has nothing to do with those within \high{}, no matter where the 0 is placed, the bounding box of "V\high{0}" is still the same: \def\dodohighlow {\ifx\fontsize\empty \ifmmode \ifnum\fam<0 \tx \else \holamathfont \fi \else \tx \fi \else \tx \fi} \def\dohighlow#1#2#3#4#5% todo, named fontdimens {\dontleavehmode \bgroup \scratchdimen\ifdim\fontexheight\textfont2=1ex #2\textfont2\else #3ex\fi \advance\scratchdimen #4ex \kern.1ex \setbox\scratchbox\hbox{#1\scratchdimen\hbox{\dodohighlow#5}}% \ht\scratchbox\strutheight \dp\scratchbox\strutdepth \box\scratchbox \egroup} \unexpanded\def\high{\dohighlow\raise\mathsupnormal{.86}{0}} \unexpanded\def\low {\dohighlow\lower\mathsubnormal{.48}{0}}
1) This doesn't make sense, does it? 2) Why gets the line space increased even though the "0" is not that high?! 3) How can the behaviour of "$V^0$" be adjusted in such way as "\high{0}" does
if it is quite right, why adjust it?
(i.e. set the "0" but dont move away from the line above)
participants (2)
-
Steffen Wolfrum -
Yue Wang