0

I'm looking at page 433 of the TeXbook with the list of \fontdimen parameters. So the interword space equals \fontdimen2 plus \fontdimen3 minus \fontdimen4 where the latter two parameters are the interword stretch and interword shrink, respectively. But the extra space added at the end of a sentence equals just \fontdimen7 The added inter-sentence spacing does not shrink or stretch is proportion to the stretch or shrink of the interword space. This is bad typography IMO (I am a novice I admit). I feel that the intersentence spacing should stretch or stretch in proportion to its increased size. For example, if the intersentence spacing of a font equals 1.33 times its interword spacing, then the total intersentence stretch/shrink should also equal 1.33 times the interword stretch/shrink.

0

1 Answer 1

1

"Why" questions are impossible to answer unless you ask the code designer but what you suggest (multiplying all three components equally) would make end of sentence space have larger shrink than normal spaces, which could have some weird effects.

If \spaceskip and \xspaceskip are not set then

if f is the current space factor, the space used is

f =1000: \fontdimen2 plus \fontdimen3 minus \fontdimen4

f <2000: \fontdimen2 plus (f/1000)\fontdimen3 minus (1000/f)\fontdimen4

f ≥2000: (\fontdimen2+\fontdimen7) plus (f/1000)\fontdimen3 minus (1000/f)\fontdimen4

which seems to work well enough in practice

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .