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)