Thread: [Solved] Hard space in justified text
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Old 12-14-2011, 09:54 PM
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Hi Steve,

For the 'z':
Find = _z_
Replace = _z^s
where the _ is a normal space character. unless you want to make a lot of work for yourself, though, there's nothing much that can be done to balance the size of this space with all the other spaces on the line. Even so, unless you're working with short lines, few people would notice the difference in spacing.

With expressions like "5 kg" or "12 PM", those usually don't have spaces (ie "5kg" and "12PM"). nevertheless, if you want to keep them with spaces you can do so with wildcard Find/Replace searches, where:
Find = ([0-9]) (kg)
Replace = \1^s^2
and:
Find = ([0-9]) ([AP]M)
Replace = \1^s^2

Something that might help minimise the problem is if you go to Word Options > Compatability Options > Layout Options and check the option to 'Do full justification the way Wordperfect 6.x for Windows does'. This allows Word to reduce the size of the normal space to fit an extra word on the line if it will fit without reducing the space too much.
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Paul Edstein
[Fmr MS MVP - Word]
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