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I want to place some elements on my page for which I need to calculate their size.

For my example lets say three of those elements should exactly occupy a linewidth when put side by side without space in between.

The easiest possibility might be to just give their width as 0.3linewidth -- but that might be a bit too small generating a gap somewhere. It is however possible to just use 0.3333333333333linewidth -- but that is much to write for a seemingly simple fraction [and it strikes my pedanticism as it's not exactly one third].

If I want to get an exact value of one third, I may use

newlengthonethirdlinewidth

onethirdlinewidth=linewidth

divideonethirdlinewidth by 3

that might be the best way if I use this length multiple times but might be a bit much to type for a one-shot use.

My question is: Is there any simple possibility to get a length of one third (or seven eighths) of a given length?

Let's try this plain TeX test:

newdimentest

newdimenfixed



fixed=345pt



defdotest#1{test=#1fixed thetest -- numbertest par}



dotest{0.3}

dotest{0.33}

dotest{0.333}

dotest{0.3333}

dotest{0.33333}

dotest{0.333333}

dotest{0.3333333}

dotest{0.33333333}

dotest{0.333333333}



test=fixed dividetest by 3 thetest -- numbertest



test=dimexprfixed/3relax thetest -- numbertest



fixed=101pt



bigskip



dotest{0.3}

dotest{0.33}

dotest{0.333}

dotest{0.3333}

dotest{0.33333}

dotest{0.333333}

dotest{0.3333333}

dotest{0.33333333}

dotest{0.333333333}



test=fixed dividetest by 3 thetest -- numbertest



test=dimexprfixed/3relax thetest -- numbertest



bye

As you see, adding more decimal digits doesn't improve the result: from five digits on the result is the same. The last column is the value in scaled points.

The method with divide truncates, whereas dimexpr rounds. There may be a difference of 1 scaled point with the two methods (in this case).

Is 0.33333fixed better or worse than dimexprfixed/3relax? If you add flexible glue to fill the gaps the difference cannot be seen with the naked eye, because in the first case it amounts to 115 scaled points, which is less than 0.00062 mm.

You can try with fixed=maxdimen and the difference between the 0.333333 and divide methods is 5462sp, less than 0.03 mm.

Six digits versus five can make a small difference, depending on the number. But using seven or more is irrelevant.

I described how TeX inputs dimensions and handles units in https://tex.stackexchange.com/a/231281 and Why pdf file cannot be reproduced? and possibly at other locations, including some comments which are not always read.

I am using Plain TeX but of course it works exactly the same in LaTeX.

newdimenfixed



fixed 1pt



newdimentestA



newdimentestB



testA 0.33333587646484374fixed



testB 0.33333587646484375fixed



ifdimtestA = testB

  The two dimensions are equal

else

  The two dimensions are not equal

fi



bye

Outputs:

One needs 17 fractional digits to be certain that the dimension stabilizes (of course you get only 1sp possible difference after 5 fractional digits, because 1/10^5 < 1/65536, here in this example where one multiplies 1pt). And some things are counterintuitive, for example 0.33333 is enough but 0.22222 is not although it looks closer to 0.222222 than 0.33333 was to 0.333333.

It goes without saying that Knuth has programmed it exactly to fetch 17 fractional digits and not one more, because the theorem is that it will never change after that.