#1
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Character Spacing Conversion to Inches
Hi everyone,
How many inches is one character space in Times New Roman, 12 pt (no other special settings, just plain roman type)? I'm being forced to make a hanging indent of "two spaces" in a table of contents, so that's why I'm asking. I'm guessing around 0.18"? I tried 0.13" and was told it was incorrect but not by how much or in which direction. I've asked my document reviewer for feedback but in the meantime I'm curious if there is an exact answer. Thank you! |
#2
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as a follow up, my best guess at this point is that one character space in Times New Roman 12 pt is equivalent to 1/12" (0.83"). I realize that point is not the same as pitch, but with the given font settings I hit the space bar 12 times and it was very very close to 1" on the ruler, so it may not be an absolute perfect value but it's what I went with for now. Word doesn't allow 3 decimal places in hanging indents from what I can see, so I had to round a bit.
If someone has a better solution I'd be curious for future reference but for now I had to make a decision. |
#3
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There is no standard character width for proportional fonts such as Times New Roman. Whilst a space character will have a nominal width, the use of paragraph justification can cause that to grow or shrink. If you input 100 consecutive 12pt Times New Roman spaces, their combined width will be 4.2in (302.4pt) when no justification is used, which implies a nominal character width of 0.042in (3.024pt).
__________________
Cheers, Paul Edstein [Fmr MS MVP - Word] |
#4
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Hi Paul, thanks for your response. Yeah, sorry, no justification was being used.
Nice idea to measure by 100. It's interesting how different that comes out from using 12 spaces (which gives me an average of 0.83 inches per space). For my particular project, I was working on a hanging indent of "two spaces" to comply with some style guidelines. I think the reviewers actually want me to insert manual line breaks in runover lines of text in the table of contents and manually hit the space bar twice, but I'm trying to develop a template for myself for future work so that's why I was looking for a measurement. Your measurement seems more accurate, but I have a feeling that they might be eyeballing the hanging indent based on the letters on the line above it (which is not going to give a consistent comparison of course, but I get the impression that that's what they're doing). I'm curious to see how the next document review comes back but this was really helpful discussion. Thanks again for your response. |
#5
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Quote:
__________________
Cheers, Paul Edstein [Fmr MS MVP - Word] |
#6
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Cross-posted at: http://answers.microsoft.com/en-us/o...9-06d71ad6f058
For cross-posting etiquette, please read: http://www.excelguru.ca/content.php?184
__________________
Cheers, Paul Edstein [Fmr MS MVP - Word] |
#7
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Quote:
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#8
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So why not do as I suggested and use a 'proper' layout so that you don't need the manual line breaks at all. A whole lot simple IMHO.
__________________
Cheers, Paul Edstein [Fmr MS MVP - Word] |
#9
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Hi Paul, Sorry for not being clear. I was agreeing with you. Using manual formatting doesn't make sense.
I'm formatting a document for a client, and there is a formal document-review process that is unfortunately based on required guidelines that seem about 20 years out of date. They seem to want me to enter two "spaces" manually, and doing that would result in your measurement of 0.042" X 2 = 0.084". But I had something close to that in my first submission to them, and they seem to want me to widen the measurement, which suggests to me that they are actually defining a "space" as the average width of a letter. The problem of course is that a spacebar keystroke, at least with the font settings defined above, is not the same as the width of an average letter. Also I apologize about the cross-post. I meant to put the link up and forgot to do so. There is more discussion there. Ultimately I did not go with the response on either forum, not because I think they're wrong, but I think I've come closer to what my document reviewers actually want, which conflicts with the language and examples they provide in their guide. Thank you again for your help. The feedback was informative and will be useful in the future for me and I'm sure others who have similar questions. Thanks again and have a great weekend! Last edited by cheech1981; 08-04-2012 at 01:38 AM. |
#10
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Quote:
__________________
Cheers, Paul Edstein [Fmr MS MVP - Word] |
#11
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right, the letter variation is yet another reason that this guideline is simply a poor choice. (incidentally, I just measured about twenty letters on a sample page and most of them are closer to 1/12th of an inch, especially lowercase letters other than i or l. Many capital letters are wider of course).
I think the formatting guide I'm dealing with was created a long time ago and it just was never really thought through since then. I'm developing a template that I'm going to recommend to them that discards the two-space setting and other problematic issues. for now though, unfortunately, I have to play trial and error with them. |
#12
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Attached is a table showing the widths of the std chracter set for 12pt TNR (I used a macro to get precise measurements - see below). As you can see (and as you can show your clients) specifying something as "two characters' width" is meaningless - 'W', for instance, is 3.375 times wider than 'i', when both are at the same point size. Start changing point sizes (as you might do when changing between headings & body text) and all you achieve is an even less meaningful designation, if such were possible.
Code:
Sub GetChrWidths() Application.ScreenUpdating = False Dim i As Long, l As Double, r As Double, x As Long, n As Long, StrTmp As String, StrChr As String n = 144 With ActiveDocument With .PageSetup .PageWidth = CentimetersToPoints(55.88) .LeftMargin = CentimetersToPoints(0) .RightMargin = CentimetersToPoints(0) .TopMargin = CentimetersToPoints(0) .BottomMargin = CentimetersToPoints(0) End With With .Range .ParagraphFormat.Alignment = wdAlignParagraphLeft With .Font .Size = 10 .Name = "Times New Roman" End With For x = 32 To 255 With .Paragraphs.Last.Range .InsertAfter vbCr l = .Characters.Last.Information(wdHorizontalPositionRelativeToTextBoundary) StrTmp = "" For i = 1 To n StrTmp = StrTmp & Chr(x) Next .InsertAfter StrTmp r = .Characters.Last.Information(wdHorizontalPositionRelativeToTextBoundary) StrChr = StrChr & Chr(x) & vbTab & Format((r - l) / n, "0.000") & vbTab .Text = vbNullString DoEvents End With Next .InsertAfter Left(StrChr, Len(StrChr) - 1) End With With .PageSetup .PaperSize = wdPaperA4 .Orientation = wdOrientLandscape .LeftMargin = CentimetersToPoints(2.5) .RightMargin = CentimetersToPoints(2.5) .TopMargin = CentimetersToPoints(2.5) .BottomMargin = CentimetersToPoints(2.5) End With .Range.Paragraphs.Last.Range.ConvertToTable Separator:=vbTab, Numrows:=16, NumColumns:=28 With .Tables(.Tables.Count) .LeftPadding = 0 .RightPadding = 0 .TopPadding = 0 .BottomPadding = 0 .AllowAutoFit = True With .Range.ParagraphFormat .SpaceAfter = 3 .SpaceBefore = 3 .Alignment = wdAlignParagraphCenter End With End With End With Application.ScreenUpdating = True End Sub
__________________
Cheers, Paul Edstein [Fmr MS MVP - Word] |
#13
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Excellent! That was very nice of you to take the time to do that Paul, and I think it will come in handy.
Also for whatever reason I wasn't seeing it before, but I see now that 6pt is essentially the same as 1/12th of an inch (i.e., 1/72 X 6 = 1/12). The conversions are starting to make more sense to me now. Thanks again and have a good one! |
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character spacing |
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