Coloring areas with TikZ
5
Consider the following MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
foreach x in 1,1.2,...,3
pgfmathsetmacroa10*x
fill[blue!a,shift=(-1,-2)] (x,-.3) rectangle (x+1,2.3);
fill[white] (0,0) .. controls (1,1) and (1.5,-1) .. (3,0) -- (3,.4) -- (0,.4) -- cycle;
fill[white] (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2) -- (3,-2.3) -- (0,-2.3) -- cycle;
draw (0,0) .. controls (1,1) and (1.5,-1) .. (3,0);
draw (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2);
beginscope[yshift=4cm]
foreach x in 0,.1,...,.8
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ([yshift=-1cm,xshift=.5cm]$(0,0)!x!(2,.5)$) -- ([yshift=.11cm,xshift=.5cm]$(0,0)!x!(2,.5)$);
draw[shorten >=.1cm,thick] (0,0) -- (2,.5);
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
pgfmathsetmacrobx+.1
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ($(0,-1)!x!(2,-.5)$) -- ($(0,-1)!b!(2,-.5)$) -- ($(0,0)!b!(2,.5)$) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endscope
endtikzpicture
enddocument
As you can see in the above part (see the first image) I can usetikzlibrarycalc to achieve a convient coloring of the areas. But in the above part (see second image) I had to manually overlay areas which may not get "painted". I tried here the calc library, too, but it calulates the points just as the are at a straight line, not as a curve with some band angle(s). My question is: How can I use the calc library to have a more ellegant code for the second part?
tikz-pgf color
asked Aug 24 at 17:18
current_user
3,1451436
add a comment |
5
Consider the following MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
foreach x in 1,1.2,...,3
pgfmathsetmacroa10*x
fill[blue!a,shift=(-1,-2)] (x,-.3) rectangle (x+1,2.3);
fill[white] (0,0) .. controls (1,1) and (1.5,-1) .. (3,0) -- (3,.4) -- (0,.4) -- cycle;
fill[white] (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2) -- (3,-2.3) -- (0,-2.3) -- cycle;
draw (0,0) .. controls (1,1) and (1.5,-1) .. (3,0);
draw (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2);
beginscope[yshift=4cm]
foreach x in 0,.1,...,.8
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ([yshift=-1cm,xshift=.5cm]$(0,0)!x!(2,.5)$) -- ([yshift=.11cm,xshift=.5cm]$(0,0)!x!(2,.5)$);
draw[shorten >=.1cm,thick] (0,0) -- (2,.5);
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
pgfmathsetmacrobx+.1
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ($(0,-1)!x!(2,-.5)$) -- ($(0,-1)!b!(2,-.5)$) -- ($(0,0)!b!(2,.5)$) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endscope
endtikzpicture
enddocument
As you can see in the above part (see the first image) I can usetikzlibrarycalc to achieve a convient coloring of the areas. But in the above part (see second image) I had to manually overlay areas which may not get "painted". I tried here the calc library, too, but it calulates the points just as the are at a straight line, not as a curve with some band angle(s). My question is: How can I use the calc library to have a more ellegant code for the second part?
tikz-pgf color
asked Aug 24 at 17:18
current_user
3,1451436
1
+1 but why could you not just clip against the bounding curves and fill just rectangles that are large enough?
– marmot
Aug 24 at 17:24
@marmot: I've got such a version, too, I just use the "rectangle option" because I want to do something with it and thanks! ;)
– current_user
Aug 24 at 17:26
1
In the lower part I would just dofill[blue!a] ($(0,0)!x!(2,.5)$) -- ++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;. I really think it is so much simpler with relative coordinates.
– marmot
Aug 24 at 17:50
add a comment |
5
5
5
Consider the following MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
foreach x in 1,1.2,...,3
pgfmathsetmacroa10*x
fill[blue!a,shift=(-1,-2)] (x,-.3) rectangle (x+1,2.3);
fill[white] (0,0) .. controls (1,1) and (1.5,-1) .. (3,0) -- (3,.4) -- (0,.4) -- cycle;
fill[white] (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2) -- (3,-2.3) -- (0,-2.3) -- cycle;
draw (0,0) .. controls (1,1) and (1.5,-1) .. (3,0);
draw (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2);
beginscope[yshift=4cm]
foreach x in 0,.1,...,.8
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ([yshift=-1cm,xshift=.5cm]$(0,0)!x!(2,.5)$) -- ([yshift=.11cm,xshift=.5cm]$(0,0)!x!(2,.5)$);
draw[shorten >=.1cm,thick] (0,0) -- (2,.5);
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
pgfmathsetmacrobx+.1
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ($(0,-1)!x!(2,-.5)$) -- ($(0,-1)!b!(2,-.5)$) -- ($(0,0)!b!(2,.5)$) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endscope
endtikzpicture
enddocument
As you can see in the above part (see the first image) I can usetikzlibrarycalc to achieve a convient coloring of the areas. But in the above part (see second image) I had to manually overlay areas which may not get "painted". I tried here the calc library, too, but it calulates the points just as the are at a straight line, not as a curve with some band angle(s). My question is: How can I use the calc library to have a more ellegant code for the second part?
tikz-pgf color
asked Aug 24 at 17:18
current_user
3,1451436
Consider the following MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
foreach x in 1,1.2,...,3
pgfmathsetmacroa10*x
fill[blue!a,shift=(-1,-2)] (x,-.3) rectangle (x+1,2.3);
fill[white] (0,0) .. controls (1,1) and (1.5,-1) .. (3,0) -- (3,.4) -- (0,.4) -- cycle;
fill[white] (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2) -- (3,-2.3) -- (0,-2.3) -- cycle;
draw (0,0) .. controls (1,1) and (1.5,-1) .. (3,0);
draw (0,-2) .. controls (1,-1) and (1.5,-3) .. (3,-2);
beginscope[yshift=4cm]
foreach x in 0,.1,...,.8
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ([yshift=-1cm,xshift=.5cm]$(0,0)!x!(2,.5)$) -- ([yshift=.11cm,xshift=.5cm]$(0,0)!x!(2,.5)$);
draw[shorten >=.1cm,thick] (0,0) -- (2,.5);
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
pgfmathsetmacrobx+.1
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ([yshift=-1cm]$(0,0)!x!(2,.5)$) -- ($(0,-1)!x!(2,-.5)$) -- ($(0,-1)!b!(2,-.5)$) -- ($(0,0)!b!(2,.5)$) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endscope
endtikzpicture
enddocument
As you can see in the above part (see the first image) I can usetikzlibrarycalc to achieve a convient coloring of the areas. But in the above part (see second image) I had to manually overlay areas which may not get "painted". I tried here the calc library, too, but it calulates the points just as the are at a straight line, not as a curve with some band angle(s). My question is: How can I use the calc library to have a more ellegant code for the second part?
tikz-pgf color
tikz-pgf color
asked Aug 24 at 17:18
current_user
3,1451436
asked Aug 24 at 17:18
current_user
3,1451436
asked Aug 24 at 17:18
current_user
3,1451436
asked Aug 24 at 17:18
current_user
3,1451436
asked Aug 24 at 17:18
current_user
3,1451436
3,1451436
1
+1 but why could you not just clip against the bounding curves and fill just rectangles that are large enough?
– marmot
Aug 24 at 17:24
@marmot: I've got such a version, too, I just use the "rectangle option" because I want to do something with it and thanks! ;)
– current_user
Aug 24 at 17:26
1
In the lower part I would just dofill[blue!a] ($(0,0)!x!(2,.5)$) -- ++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;. I really think it is so much simpler with relative coordinates.
– marmot
Aug 24 at 17:50
add a comment |
1
+1 but why could you not just clip against the bounding curves and fill just rectangles that are large enough?
– marmot
Aug 24 at 17:24
@marmot: I've got such a version, too, I just use the "rectangle option" because I want to do something with it and thanks! ;)
– current_user
Aug 24 at 17:26
1
In the lower part I would just dofill[blue!a] ($(0,0)!x!(2,.5)$) -- ++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;. I really think it is so much simpler with relative coordinates.
– marmot
Aug 24 at 17:50
1
1
+1 but why could you not just clip against the bounding curves and fill just rectangles that are large enough?
– marmot
Aug 24 at 17:24
+1 but why could you not just clip against the bounding curves and fill just rectangles that are large enough?
– marmot
Aug 24 at 17:24
@marmot: I've got such a version, too, I just use the "rectangle option" because I want to do something with it and thanks! ;)
– current_user
Aug 24 at 17:26
@marmot: I've got such a version, too, I just use the "rectangle option" because I want to do something with it and thanks! ;)
– current_user
Aug 24 at 17:26
1
1
In the lower part I would just do
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;. I really think it is so much simpler with relative coordinates.– marmot
Aug 24 at 17:50
In the lower part I would just do
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;. I really think it is so much simpler with relative coordinates.– marmot
Aug 24 at 17:50
add a comment |
2 Answers
2
active
oldest
votes
6
Here another option, using nodes in a path, defining certain points in a path and drawing a line from them that changes the color.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
%Iterative tricks
foreach x in 1,2,...,30
path
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0) node[pos=x/30](ax);
draw[line width = 4pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0)
(1,-2)
.. controls +(1,1) and +(-1,-1) .. ++(3,0);
%Another path
foreach x in 1,2,...,50
path
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endtikzpicture
enddocument
UPDATE:
Added another line shifted in preactión modifier and shorten the lines terminations to hide the rectangles; but is not good solution if you work with some backgrounds; best option is to work with clip, with shades, or even rectangles.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc,patterns
begindocument
begintikzpicture[
ShiftLine/.style=
preaction=
transform canvas=
shift=(#1),
,
draw=white,
line width=3pt,
]
fill [blue!30](0.5,-8) rectangle (12.5,0.5);
pattern [pattern=checkerboard,pattern color=blue!10](0.5,-8) rectangle (12.5,0.5);
%Avoid use clip
foreach x in 1,2,...,49
path
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x,shorten <=-2pt, shorten >=-2pt] (ax.center) -- ++ (0,-2);
draw[line width = 1pt,ShiftLine=0,2pt,shorten <=2pt, shorten >=2pt]
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 1pt,ShiftLine=0,-2pt,shorten <=2pt, shorten >=2pt]
(1,-2)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
%using Clip and shades
beginscope
clip
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++ (0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
shade[right color=blue!50, left color=white]
(1,-3) rectangle ++ (4.5,-4);
draw[line width = 2pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 2pt]
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Using Rectangles in a foreach shifting.
beginscope[shift=(4,2)]
clip
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++(0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
foreach k in 0,1,...,50
fill[blue!k]
(3+k*0.1,-3) rectangle ++(0.5,-4);
draw[line width=2pt]
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width=2pt]
(3,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Some labels
draw[font=tiny,fill opacity=0.2,text opacity=1,align=center]
(3.5,-1.8) node[anchor=center,fill=yellow]verb+Shift preaction white line +\ verb+foreach lines from nodes in a path+
(3.5,-5.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+and shades+
(9.5,-3.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+rectangles in foreach shifting+;
endtikzpicture
enddocument
answered Aug 24 at 18:52
J Leon V.
7,824530
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
1
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
add a comment |
4
I don't know if that's what you're asking but one can simplify the shaded contours quite a bit by working with relative coordinates. No manual shift is needed then.
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) --
++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endtikzpicture
enddocument
answered Aug 24 at 17:54
marmot
86.4k499184
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
1
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
Here another option, using nodes in a path, defining certain points in a path and drawing a line from them that changes the color.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
%Iterative tricks
foreach x in 1,2,...,30
path
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0) node[pos=x/30](ax);
draw[line width = 4pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0)
(1,-2)
.. controls +(1,1) and +(-1,-1) .. ++(3,0);
%Another path
foreach x in 1,2,...,50
path
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endtikzpicture
enddocument
UPDATE:
Added another line shifted in preactión modifier and shorten the lines terminations to hide the rectangles; but is not good solution if you work with some backgrounds; best option is to work with clip, with shades, or even rectangles.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc,patterns
begindocument
begintikzpicture[
ShiftLine/.style=
preaction=
transform canvas=
shift=(#1),
,
draw=white,
line width=3pt,
]
fill [blue!30](0.5,-8) rectangle (12.5,0.5);
pattern [pattern=checkerboard,pattern color=blue!10](0.5,-8) rectangle (12.5,0.5);
%Avoid use clip
foreach x in 1,2,...,49
path
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x,shorten <=-2pt, shorten >=-2pt] (ax.center) -- ++ (0,-2);
draw[line width = 1pt,ShiftLine=0,2pt,shorten <=2pt, shorten >=2pt]
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 1pt,ShiftLine=0,-2pt,shorten <=2pt, shorten >=2pt]
(1,-2)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
%using Clip and shades
beginscope
clip
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++ (0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
shade[right color=blue!50, left color=white]
(1,-3) rectangle ++ (4.5,-4);
draw[line width = 2pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 2pt]
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Using Rectangles in a foreach shifting.
beginscope[shift=(4,2)]
clip
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++(0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
foreach k in 0,1,...,50
fill[blue!k]
(3+k*0.1,-3) rectangle ++(0.5,-4);
draw[line width=2pt]
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width=2pt]
(3,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Some labels
draw[font=tiny,fill opacity=0.2,text opacity=1,align=center]
(3.5,-1.8) node[anchor=center,fill=yellow]verb+Shift preaction white line +\ verb+foreach lines from nodes in a path+
(3.5,-5.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+and shades+
(9.5,-3.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+rectangles in foreach shifting+;
endtikzpicture
enddocument
answered Aug 24 at 18:52
J Leon V.
7,824530
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
1
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
add a comment |
6
Here another option, using nodes in a path, defining certain points in a path and drawing a line from them that changes the color.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
%Iterative tricks
foreach x in 1,2,...,30
path
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0) node[pos=x/30](ax);
draw[line width = 4pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0)
(1,-2)
.. controls +(1,1) and +(-1,-1) .. ++(3,0);
%Another path
foreach x in 1,2,...,50
path
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endtikzpicture
enddocument
UPDATE:
Added another line shifted in preactión modifier and shorten the lines terminations to hide the rectangles; but is not good solution if you work with some backgrounds; best option is to work with clip, with shades, or even rectangles.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc,patterns
begindocument
begintikzpicture[
ShiftLine/.style=
preaction=
transform canvas=
shift=(#1),
,
draw=white,
line width=3pt,
]
fill [blue!30](0.5,-8) rectangle (12.5,0.5);
pattern [pattern=checkerboard,pattern color=blue!10](0.5,-8) rectangle (12.5,0.5);
%Avoid use clip
foreach x in 1,2,...,49
path
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x,shorten <=-2pt, shorten >=-2pt] (ax.center) -- ++ (0,-2);
draw[line width = 1pt,ShiftLine=0,2pt,shorten <=2pt, shorten >=2pt]
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 1pt,ShiftLine=0,-2pt,shorten <=2pt, shorten >=2pt]
(1,-2)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
%using Clip and shades
beginscope
clip
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++ (0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
shade[right color=blue!50, left color=white]
(1,-3) rectangle ++ (4.5,-4);
draw[line width = 2pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 2pt]
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Using Rectangles in a foreach shifting.
beginscope[shift=(4,2)]
clip
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++(0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
foreach k in 0,1,...,50
fill[blue!k]
(3+k*0.1,-3) rectangle ++(0.5,-4);
draw[line width=2pt]
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width=2pt]
(3,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Some labels
draw[font=tiny,fill opacity=0.2,text opacity=1,align=center]
(3.5,-1.8) node[anchor=center,fill=yellow]verb+Shift preaction white line +\ verb+foreach lines from nodes in a path+
(3.5,-5.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+and shades+
(9.5,-3.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+rectangles in foreach shifting+;
endtikzpicture
enddocument
answered Aug 24 at 18:52
J Leon V.
7,824530
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
1
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
add a comment |
6
6
6
Here another option, using nodes in a path, defining certain points in a path and drawing a line from them that changes the color.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
%Iterative tricks
foreach x in 1,2,...,30
path
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0) node[pos=x/30](ax);
draw[line width = 4pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0)
(1,-2)
.. controls +(1,1) and +(-1,-1) .. ++(3,0);
%Another path
foreach x in 1,2,...,50
path
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endtikzpicture
enddocument
UPDATE:
Added another line shifted in preactión modifier and shorten the lines terminations to hide the rectangles; but is not good solution if you work with some backgrounds; best option is to work with clip, with shades, or even rectangles.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc,patterns
begindocument
begintikzpicture[
ShiftLine/.style=
preaction=
transform canvas=
shift=(#1),
,
draw=white,
line width=3pt,
]
fill [blue!30](0.5,-8) rectangle (12.5,0.5);
pattern [pattern=checkerboard,pattern color=blue!10](0.5,-8) rectangle (12.5,0.5);
%Avoid use clip
foreach x in 1,2,...,49
path
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x,shorten <=-2pt, shorten >=-2pt] (ax.center) -- ++ (0,-2);
draw[line width = 1pt,ShiftLine=0,2pt,shorten <=2pt, shorten >=2pt]
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 1pt,ShiftLine=0,-2pt,shorten <=2pt, shorten >=2pt]
(1,-2)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
%using Clip and shades
beginscope
clip
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++ (0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
shade[right color=blue!50, left color=white]
(1,-3) rectangle ++ (4.5,-4);
draw[line width = 2pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 2pt]
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Using Rectangles in a foreach shifting.
beginscope[shift=(4,2)]
clip
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++(0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
foreach k in 0,1,...,50
fill[blue!k]
(3+k*0.1,-3) rectangle ++(0.5,-4);
draw[line width=2pt]
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width=2pt]
(3,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Some labels
draw[font=tiny,fill opacity=0.2,text opacity=1,align=center]
(3.5,-1.8) node[anchor=center,fill=yellow]verb+Shift preaction white line +\ verb+foreach lines from nodes in a path+
(3.5,-5.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+and shades+
(9.5,-3.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+rectangles in foreach shifting+;
endtikzpicture
enddocument
answered Aug 24 at 18:52
J Leon V.
7,824530
Here another option, using nodes in a path, defining certain points in a path and drawing a line from them that changes the color.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
%Iterative tricks
foreach x in 1,2,...,30
path
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0) node[pos=x/30](ax);
draw[line width = 4pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,0)
.. controls +(1,1) and +(-1,-1) .. ++(3,0)
(1,-2)
.. controls +(1,1) and +(-1,-1) .. ++(3,0);
%Another path
foreach x in 1,2,...,50
path
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x] (ax.center) -- ++ (0,-2);
draw[line width = 1pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endtikzpicture
enddocument
UPDATE:
Added another line shifted in preactión modifier and shorten the lines terminations to hide the rectangles; but is not good solution if you work with some backgrounds; best option is to work with clip, with shades, or even rectangles.
RESULT:
MWE:
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc,patterns
begindocument
begintikzpicture[
ShiftLine/.style=
preaction=
transform canvas=
shift=(#1),
,
draw=white,
line width=3pt,
]
fill [blue!30](0.5,-8) rectangle (12.5,0.5);
pattern [pattern=checkerboard,pattern color=blue!10](0.5,-8) rectangle (12.5,0.5);
%Avoid use clip
foreach x in 1,2,...,49
path
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0) node[pos=x/50](ax);
draw[line width = 3pt,blue!x,shorten <=-2pt, shorten >=-2pt] (ax.center) -- ++ (0,-2);
draw[line width = 1pt,ShiftLine=0,2pt,shorten <=2pt, shorten >=2pt]
(1,0)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 1pt,ShiftLine=0,-2pt,shorten <=2pt, shorten >=2pt]
(1,-2)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
%using Clip and shades
beginscope
clip
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++ (0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
shade[right color=blue!50, left color=white]
(1,-3) rectangle ++ (4.5,-4);
draw[line width = 2pt]
(1,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width = 2pt]
(1,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Using Rectangles in a foreach shifting.
beginscope[shift=(4,2)]
clip
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0)
-- ++(0,-2)
.. controls +(-1.5,-3) and +(1,0.5) .. ++(-4.5,0)
-- cycle;
foreach k in 0,1,...,50
fill[blue!k]
(3+k*0.1,-3) rectangle ++(0.5,-4);
draw[line width=2pt]
(3,-3.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
draw[line width=2pt]
(3,-5.5)
.. controls +(1,0.5) and +(-1.5,-3) .. ++(4.5,0);
endscope
%Some labels
draw[font=tiny,fill opacity=0.2,text opacity=1,align=center]
(3.5,-1.8) node[anchor=center,fill=yellow]verb+Shift preaction white line +\ verb+foreach lines from nodes in a path+
(3.5,-5.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+and shades+
(9.5,-3.3) node[anchor=center,fill=yellow]verb+clip shape +\ verb+rectangles in foreach shifting+;
endtikzpicture
enddocument
answered Aug 24 at 18:52
J Leon V.
7,824530
edited Aug 24 at 21:30
answered Aug 24 at 18:52
J Leon V.
7,824530
answered Aug 24 at 18:52
J Leon V.
7,824530
answered Aug 24 at 18:52
J Leon V.
7,824530
7,824530
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
1
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
add a comment |
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
1
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
+1 Nearly perfect! Just one little detail: Could you edit your answer, such that the rectangles are not visible outside the Bézier-curve (just clipping against it). Then you will geht the check-mark!
– current_user
Aug 24 at 20:04
1
1
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
wait a bit, I think I can add other tricks xD
– J Leon V.
Aug 24 at 20:08
add a comment |
4
I don't know if that's what you're asking but one can simplify the shaded contours quite a bit by working with relative coordinates. No manual shift is needed then.
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) --
++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endtikzpicture
enddocument
answered Aug 24 at 17:54
marmot
86.4k499184
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
1
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
add a comment |
4
I don't know if that's what you're asking but one can simplify the shaded contours quite a bit by working with relative coordinates. No manual shift is needed then.
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) --
++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endtikzpicture
enddocument
answered Aug 24 at 17:54
marmot
86.4k499184
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
1
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
add a comment |
4
4
4
I don't know if that's what you're asking but one can simplify the shaded contours quite a bit by working with relative coordinates. No manual shift is needed then.
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) --
++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endtikzpicture
enddocument
answered Aug 24 at 17:54
marmot
86.4k499184
I don't know if that's what you're asking but one can simplify the shaded contours quite a bit by working with relative coordinates. No manual shift is needed then.
documentclass[border=5pt,tikz]standalone
usetikzlibrarycalc
begindocument
begintikzpicture
beginscope[shift=(0,-2)]
foreach x in 0,.1,...,.9
pgfmathsetmacroax*40
fill[blue!a] ($(0,0)!x!(2,.5)$) --
++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;
draw[shorten >=.2cm,thick] (0,0) -- (2,.5);
endscope
endtikzpicture
enddocument
answered Aug 24 at 17:54
marmot
86.4k499184
answered Aug 24 at 17:54
marmot
86.4k499184
answered Aug 24 at 17:54
marmot
86.4k499184
answered Aug 24 at 17:54
marmot
86.4k499184
86.4k499184
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
1
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
add a comment |
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
1
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
Nearly, but that is really interessting. Yes, I should use relative coordinates more often … +1
– current_user
Aug 24 at 20:05
1
1
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
@current_user I simply do not understand the question, I think... ;-)
– marmot
Aug 24 at 21:59
add a comment |
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1
+1 but why could you not just clip against the bounding curves and fill just rectangles that are large enough?
– marmot
Aug 24 at 17:24
@marmot: I've got such a version, too, I just use the "rectangle option" because I want to do something with it and thanks! ;)
– current_user
Aug 24 at 17:26
1
In the lower part I would just do
fill[blue!a] ($(0,0)!x!(2,.5)$) -- ++ (0,-1) -- ++ ($(0,0)!0.1!(2,.5)$) -- ++(0,1) -- cycle;. I really think it is so much simpler with relative coordinates.– marmot
Aug 24 at 17:50