Mixing
Exact Fraction of a length
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?
calculations lengths
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
|
show 3 more comments
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?
calculations lengths
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
1
dimexprlinewidth/3anddimexpr7linewidth/8work. However TeX uses scaled integers to represent dimensions, so technically0.3333333333333linewidthis more precise than TeX's representation oflinewidth/3. When you input a long decimal chain like that TeX will truncate that to a value it can represent.
– Phelype Oleinik
Jan 29 at 12:47
Actually, I didn't think about dimexpr. Mind putting that in an answer?
– David Woitkowski
Jan 29 at 12:51
@PhelypeOleinik "a long decimal chain like that TeX will truncate that to a value it can represent" is misleading, it could be that no truncation can be exactly represented as an integer multiple of pt/65536.
– user4686
Jan 29 at 12:58
1
the most precise way to multiply by a fraction isdimexprnumexpr A*<dimen>/B sprelax(where<dimen>is likelinewidthbut not10ptthen usedimexpr10ptrelaxin place of<dimen>)
– user4686
Jan 29 at 13:00
@jfbu Agreed, unfortunate choice of words. On your second comment, why is that? Do you mind writing the answer explaining your comment, please?
– Phelype Oleinik
Jan 29 at 13:03
|
show 3 more comments
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?
calculations lengths
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
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?
calculations lengths
calculations lengths
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
asked Jan 29 at 12:45
David WoitkowskiDavid Woitkowski
898513
898513
1
dimexprlinewidth/3anddimexpr7linewidth/8work. However TeX uses scaled integers to represent dimensions, so technically0.3333333333333linewidthis more precise than TeX's representation oflinewidth/3. When you input a long decimal chain like that TeX will truncate that to a value it can represent.
– Phelype Oleinik
Jan 29 at 12:47
Actually, I didn't think about dimexpr. Mind putting that in an answer?
– David Woitkowski
Jan 29 at 12:51
@PhelypeOleinik "a long decimal chain like that TeX will truncate that to a value it can represent" is misleading, it could be that no truncation can be exactly represented as an integer multiple of pt/65536.
– user4686
Jan 29 at 12:58
1
the most precise way to multiply by a fraction isdimexprnumexpr A*<dimen>/B sprelax(where<dimen>is likelinewidthbut not10ptthen usedimexpr10ptrelaxin place of<dimen>)
– user4686
Jan 29 at 13:00
@jfbu Agreed, unfortunate choice of words. On your second comment, why is that? Do you mind writing the answer explaining your comment, please?
– Phelype Oleinik
Jan 29 at 13:03
|
show 3 more comments
1
dimexprlinewidth/3anddimexpr7linewidth/8work. However TeX uses scaled integers to represent dimensions, so technically0.3333333333333linewidthis more precise than TeX's representation oflinewidth/3. When you input a long decimal chain like that TeX will truncate that to a value it can represent.
– Phelype Oleinik
Jan 29 at 12:47
Actually, I didn't think about dimexpr. Mind putting that in an answer?
– David Woitkowski
Jan 29 at 12:51
@PhelypeOleinik "a long decimal chain like that TeX will truncate that to a value it can represent" is misleading, it could be that no truncation can be exactly represented as an integer multiple of pt/65536.
– user4686
Jan 29 at 12:58
1
the most precise way to multiply by a fraction isdimexprnumexpr A*<dimen>/B sprelax(where<dimen>is likelinewidthbut not10ptthen usedimexpr10ptrelaxin place of<dimen>)
– user4686
Jan 29 at 13:00
@jfbu Agreed, unfortunate choice of words. On your second comment, why is that? Do you mind writing the answer explaining your comment, please?
– Phelype Oleinik
Jan 29 at 13:03
1
1
dimexprlinewidth/3 and dimexpr7linewidth/8 work. However TeX uses scaled integers to represent dimensions, so technically 0.3333333333333linewidth is more precise than TeX's representation of linewidth/3. When you input a long decimal chain like that TeX will truncate that to a value it can represent.– Phelype Oleinik
Jan 29 at 12:47
dimexprlinewidth/3 and dimexpr7linewidth/8 work. However TeX uses scaled integers to represent dimensions, so technically 0.3333333333333linewidth is more precise than TeX's representation of linewidth/3. When you input a long decimal chain like that TeX will truncate that to a value it can represent.– Phelype Oleinik
Jan 29 at 12:47
Actually, I didn't think about dimexpr. Mind putting that in an answer?
– David Woitkowski
Jan 29 at 12:51
Actually, I didn't think about dimexpr. Mind putting that in an answer?
– David Woitkowski
Jan 29 at 12:51
@PhelypeOleinik "a long decimal chain like that TeX will truncate that to a value it can represent" is misleading, it could be that no truncation can be exactly represented as an integer multiple of pt/65536.
– user4686
Jan 29 at 12:58
@PhelypeOleinik "a long decimal chain like that TeX will truncate that to a value it can represent" is misleading, it could be that no truncation can be exactly represented as an integer multiple of pt/65536.
– user4686
Jan 29 at 12:58
1
1
the most precise way to multiply by a fraction is
dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)– user4686
Jan 29 at 13:00
the most precise way to multiply by a fraction is
dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)– user4686
Jan 29 at 13:00
@jfbu Agreed, unfortunate choice of words. On your second comment, why is that? Do you mind writing the answer explaining your comment, please?
– Phelype Oleinik
Jan 29 at 13:03
@jfbu Agreed, unfortunate choice of words. On your second comment, why is that? Do you mind writing the answer explaining your comment, please?
– Phelype Oleinik
Jan 29 at 13:03
|
show 3 more comments
1 Answer
1
active
oldest
votes
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.
As another random example consider this
numberdimexpr 0.824440000ptrelax
numberdimexpr 0.824440003ptrelax
bye
which produces
showing that 0.824440003 gives distinct result from 0.82444.
We can confirm this also in a rôle as <factor> :
documentclass[a4paper]{article}
usepackage{geometry}
newlength{mylength}
begin{document}
setlength{mylength}{0.82444linewidth}
verb|0.82444linewidth| gives themylength.
setlength{mylength}{0.824440003linewidth}
verb|0.824440003linewidth| gives themylength.
These two things differ!
I hope this will dispel some misunderstandings\
about ``five fractional digits suffice''. Wrong.
end{document}
Notice that above linewidth is 418.25368pt so 0.000000003linewidth is in truth 0.00000125476104pt well below the TeX "error".
THE DIFFERENCE IS AMPLIFIED BY A FACTOR GREATER THAN 5000 !
As per the actual question, here is my comment
the most precise way to multiply by a fraction is dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
add a comment |
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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.
As another random example consider this
numberdimexpr 0.824440000ptrelax
numberdimexpr 0.824440003ptrelax
bye
which produces
showing that 0.824440003 gives distinct result from 0.82444.
We can confirm this also in a rôle as <factor> :
documentclass[a4paper]{article}
usepackage{geometry}
newlength{mylength}
begin{document}
setlength{mylength}{0.82444linewidth}
verb|0.82444linewidth| gives themylength.
setlength{mylength}{0.824440003linewidth}
verb|0.824440003linewidth| gives themylength.
These two things differ!
I hope this will dispel some misunderstandings\
about ``five fractional digits suffice''. Wrong.
end{document}
Notice that above linewidth is 418.25368pt so 0.000000003linewidth is in truth 0.00000125476104pt well below the TeX "error".
THE DIFFERENCE IS AMPLIFIED BY A FACTOR GREATER THAN 5000 !
As per the actual question, here is my comment
the most precise way to multiply by a fraction is dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
add a comment |
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.
As another random example consider this
numberdimexpr 0.824440000ptrelax
numberdimexpr 0.824440003ptrelax
bye
which produces
showing that 0.824440003 gives distinct result from 0.82444.
We can confirm this also in a rôle as <factor> :
documentclass[a4paper]{article}
usepackage{geometry}
newlength{mylength}
begin{document}
setlength{mylength}{0.82444linewidth}
verb|0.82444linewidth| gives themylength.
setlength{mylength}{0.824440003linewidth}
verb|0.824440003linewidth| gives themylength.
These two things differ!
I hope this will dispel some misunderstandings\
about ``five fractional digits suffice''. Wrong.
end{document}
Notice that above linewidth is 418.25368pt so 0.000000003linewidth is in truth 0.00000125476104pt well below the TeX "error".
THE DIFFERENCE IS AMPLIFIED BY A FACTOR GREATER THAN 5000 !
As per the actual question, here is my comment
the most precise way to multiply by a fraction is dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
add a comment |
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.
As another random example consider this
numberdimexpr 0.824440000ptrelax
numberdimexpr 0.824440003ptrelax
bye
which produces
showing that 0.824440003 gives distinct result from 0.82444.
We can confirm this also in a rôle as <factor> :
documentclass[a4paper]{article}
usepackage{geometry}
newlength{mylength}
begin{document}
setlength{mylength}{0.82444linewidth}
verb|0.82444linewidth| gives themylength.
setlength{mylength}{0.824440003linewidth}
verb|0.824440003linewidth| gives themylength.
These two things differ!
I hope this will dispel some misunderstandings\
about ``five fractional digits suffice''. Wrong.
end{document}
Notice that above linewidth is 418.25368pt so 0.000000003linewidth is in truth 0.00000125476104pt well below the TeX "error".
THE DIFFERENCE IS AMPLIFIED BY A FACTOR GREATER THAN 5000 !
As per the actual question, here is my comment
the most precise way to multiply by a fraction is dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)
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.
As another random example consider this
numberdimexpr 0.824440000ptrelax
numberdimexpr 0.824440003ptrelax
bye
which produces
showing that 0.824440003 gives distinct result from 0.82444.
We can confirm this also in a rôle as <factor> :
documentclass[a4paper]{article}
usepackage{geometry}
newlength{mylength}
begin{document}
setlength{mylength}{0.82444linewidth}
verb|0.82444linewidth| gives themylength.
setlength{mylength}{0.824440003linewidth}
verb|0.824440003linewidth| gives themylength.
These two things differ!
I hope this will dispel some misunderstandings\
about ``five fractional digits suffice''. Wrong.
end{document}
Notice that above linewidth is 418.25368pt so 0.000000003linewidth is in truth 0.00000125476104pt well below the TeX "error".
THE DIFFERENCE IS AMPLIFIED BY A FACTOR GREATER THAN 5000 !
As per the actual question, here is my comment
the most precise way to multiply by a fraction is dimexprnumexpr A*<dimen>/B sprelax (where <dimen> is like linewidth but not 10pt then use dimexpr10ptrelax in place of <dimen>)
edited Jan 29 at 18:06
answered Jan 29 at 16:21
user4686
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
add a comment |
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
@jfbu And it would be great if you would consider to stay on TeX.SE. Things happen, tone can get rough, and it can be healed by understanding on both sides and cleaning up comments or texts that led to it.
– Stefan Kottwitz♦
Jan 30 at 18:40
add a comment |
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1
dimexprlinewidth/3anddimexpr7linewidth/8work. However TeX uses scaled integers to represent dimensions, so technically0.3333333333333linewidthis more precise than TeX's representation oflinewidth/3. When you input a long decimal chain like that TeX will truncate that to a value it can represent.– Phelype Oleinik
Jan 29 at 12:47
Actually, I didn't think about dimexpr. Mind putting that in an answer?
– David Woitkowski
Jan 29 at 12:51
@PhelypeOleinik "a long decimal chain like that TeX will truncate that to a value it can represent" is misleading, it could be that no truncation can be exactly represented as an integer multiple of pt/65536.
– user4686
Jan 29 at 12:58
1
the most precise way to multiply by a fraction is
dimexprnumexpr A*<dimen>/B sprelax(where<dimen>is likelinewidthbut not10ptthen usedimexpr10ptrelaxin place of<dimen>)– user4686
Jan 29 at 13:00
@jfbu Agreed, unfortunate choice of words. On your second comment, why is that? Do you mind writing the answer explaining your comment, please?
– Phelype Oleinik
Jan 29 at 13:03