How to draw a Taylor diagrams

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Is there any package or a way to draw Taylor diagrams (From this paper) easily?



Any help will be appreciated.



enter image description here










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  • Welcome to TeX.Stackexchange!
    – samcarter
    2 hours ago






  • 2




    I'm not sure if there is a dedicated package for only these diagrams, but I think you could just do them with polar axes that come with pgfplots, particularly see section 5.10.6 Partial Polar Axes of the manual as well as 5.11. SMITH CHARTS.
    – marmot
    2 hours ago















up vote
1
down vote

favorite












Is there any package or a way to draw Taylor diagrams (From this paper) easily?



Any help will be appreciated.



enter image description here










share|improve this question









New contributor




Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • Welcome to TeX.Stackexchange!
    – samcarter
    2 hours ago






  • 2




    I'm not sure if there is a dedicated package for only these diagrams, but I think you could just do them with polar axes that come with pgfplots, particularly see section 5.10.6 Partial Polar Axes of the manual as well as 5.11. SMITH CHARTS.
    – marmot
    2 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is there any package or a way to draw Taylor diagrams (From this paper) easily?



Any help will be appreciated.



enter image description here










share|improve this question









New contributor




Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Is there any package or a way to draw Taylor diagrams (From this paper) easily?



Any help will be appreciated.



enter image description here







tikz-pgf






share|improve this question









New contributor




Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




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edited 3 hours ago









Sebastiano

7,78741654




7,78741654






New contributor




Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









Mark

61




61




New contributor




Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Welcome to TeX.Stackexchange!
    – samcarter
    2 hours ago






  • 2




    I'm not sure if there is a dedicated package for only these diagrams, but I think you could just do them with polar axes that come with pgfplots, particularly see section 5.10.6 Partial Polar Axes of the manual as well as 5.11. SMITH CHARTS.
    – marmot
    2 hours ago

















  • Welcome to TeX.Stackexchange!
    – samcarter
    2 hours ago






  • 2




    I'm not sure if there is a dedicated package for only these diagrams, but I think you could just do them with polar axes that come with pgfplots, particularly see section 5.10.6 Partial Polar Axes of the manual as well as 5.11. SMITH CHARTS.
    – marmot
    2 hours ago
















Welcome to TeX.Stackexchange!
– samcarter
2 hours ago




Welcome to TeX.Stackexchange!
– samcarter
2 hours ago




2




2




I'm not sure if there is a dedicated package for only these diagrams, but I think you could just do them with polar axes that come with pgfplots, particularly see section 5.10.6 Partial Polar Axes of the manual as well as 5.11. SMITH CHARTS.
– marmot
2 hours ago





I'm not sure if there is a dedicated package for only these diagrams, but I think you could just do them with polar axes that come with pgfplots, particularly see section 5.10.6 Partial Polar Axes of the manual as well as 5.11. SMITH CHARTS.
– marmot
2 hours ago











1 Answer
1






active

oldest

votes

















up vote
5
down vote













OK. So, I won't do all the work for you. But, I will do enough so that you should be able to figure out how to add everything else that remains to be done.



enter image description here



What I show you here is as follows:



  • How to clip a portion of a picture

  • How to draw concentric circles about a given center

  • How to label along a curved path

  • How to place a label at a point along a path

  • How label at the end points of a line

  • How to get different styles of dashed and dotted lines

  • How to plot points

  • Different approaches to scaling the text

And, I believe that should allow you to complete the rest of the picture.



Here's the code to generate the diagram:



documentclass[tikz,border=6pt]standalone
usepackageamsmath
usepackagetikz
usetikzlibrarycalc
usetikzlibrarydecorations.text
pagestyleempty
begindocument

begintikzpicture[
x=(4cm,0),
y=(0,4cm),
radius=4cm,
]
%% notation (<angle>:<radius>) gives polar coordinates
%% (<x-coor>,<y-coord>) Euclidean coordinates

%% draw semi-circles clipped by the 1st quadrant arc
%% `scope` prevents the entirety of the remainder of
%% picture from being clipped.
beginscope
draw[clip] (0,0) -- (1,0) arc (0:90:1) -- cycle;
foreach myr in 1,2,...,5

%% draw a circle centered at (5/7,0)
%% arc starts at point (myr/6,0) and proceeds through an
%% angle of 180 degrees with a radius of myr/6 units
draw[blue,densely dotted] (5/7,0) ++ (myr/6,0) arc (0:180:myr/6);

endscope

%% place a label along one of the above clipped arcs.
path (5/7,0) ++ (2/6,0) arc (0:180:2/6) node[pos=0.75,
rotate=0.75*180-90,
anchor=south,
inner sep=4pt,
scale=0.5,
blue] (A) $0.25$;

%% label the correlation coefficient values
pgfkeys/pgf/number format/precision=1
foreach myp in 0,1,2,...,11

%% calculate the angle myangle from the integer myp
pgfmathsetmacromyangle90*(12-myp)/12
pgfmathparsemyp/10
pgfmathroundtozerofillpgfmathresult
pgfmathsetmacromylabelpgfmathresult
%% handle labels that don't follow the previous pattern
ifdimmylabel pt=1.0pt%%
defmylabel0.95
fi
ifdimmylabel pt=1.1pt%%
defmylabel0.99%%
fi
%% draw the dashed lines from orgin to arc with labels outside arc
draw[gray!50,
dash pattern=on 1.25pt off 0.5pt,
line width=0.1pt] (myangle:1) node[black,
anchor=180+myangle]
$scriptscriptstylemylabel$ -- (0,0);


%% labeling along a curve
%% a fancy approach that's necessary for labels conforming to
%% a curved path. Not necessary for labels along straight lines.
path[postaction=decoration=text along path,
text align=center,
text=Correlation Coefficient,
decorate]
(0,1.15) arc (90:0:1.15);

%% some data points
node[blue] at (35:4/7+1/2*1/7) $diamond$;
node[red,scale=0.5] at (32:5/7) $pmbtriangle$;

%% draw the quarter circle in the first quadrant
draw (0,0) -- (1,0);
draw (0,0) -- (1,0) arc (0:90:1) -- cycle;

endtikzpicture

enddocument





share|improve this answer






















  • +1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
    – marmot
    1 hour ago










  • @marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
    – A.Ellett
    1 hour ago










  • @marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
    – A.Ellett
    1 hour ago










  • In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
    – marmot
    1 hour ago










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote













OK. So, I won't do all the work for you. But, I will do enough so that you should be able to figure out how to add everything else that remains to be done.



enter image description here



What I show you here is as follows:



  • How to clip a portion of a picture

  • How to draw concentric circles about a given center

  • How to label along a curved path

  • How to place a label at a point along a path

  • How label at the end points of a line

  • How to get different styles of dashed and dotted lines

  • How to plot points

  • Different approaches to scaling the text

And, I believe that should allow you to complete the rest of the picture.



Here's the code to generate the diagram:



documentclass[tikz,border=6pt]standalone
usepackageamsmath
usepackagetikz
usetikzlibrarycalc
usetikzlibrarydecorations.text
pagestyleempty
begindocument

begintikzpicture[
x=(4cm,0),
y=(0,4cm),
radius=4cm,
]
%% notation (<angle>:<radius>) gives polar coordinates
%% (<x-coor>,<y-coord>) Euclidean coordinates

%% draw semi-circles clipped by the 1st quadrant arc
%% `scope` prevents the entirety of the remainder of
%% picture from being clipped.
beginscope
draw[clip] (0,0) -- (1,0) arc (0:90:1) -- cycle;
foreach myr in 1,2,...,5

%% draw a circle centered at (5/7,0)
%% arc starts at point (myr/6,0) and proceeds through an
%% angle of 180 degrees with a radius of myr/6 units
draw[blue,densely dotted] (5/7,0) ++ (myr/6,0) arc (0:180:myr/6);

endscope

%% place a label along one of the above clipped arcs.
path (5/7,0) ++ (2/6,0) arc (0:180:2/6) node[pos=0.75,
rotate=0.75*180-90,
anchor=south,
inner sep=4pt,
scale=0.5,
blue] (A) $0.25$;

%% label the correlation coefficient values
pgfkeys/pgf/number format/precision=1
foreach myp in 0,1,2,...,11

%% calculate the angle myangle from the integer myp
pgfmathsetmacromyangle90*(12-myp)/12
pgfmathparsemyp/10
pgfmathroundtozerofillpgfmathresult
pgfmathsetmacromylabelpgfmathresult
%% handle labels that don't follow the previous pattern
ifdimmylabel pt=1.0pt%%
defmylabel0.95
fi
ifdimmylabel pt=1.1pt%%
defmylabel0.99%%
fi
%% draw the dashed lines from orgin to arc with labels outside arc
draw[gray!50,
dash pattern=on 1.25pt off 0.5pt,
line width=0.1pt] (myangle:1) node[black,
anchor=180+myangle]
$scriptscriptstylemylabel$ -- (0,0);


%% labeling along a curve
%% a fancy approach that's necessary for labels conforming to
%% a curved path. Not necessary for labels along straight lines.
path[postaction=decoration=text along path,
text align=center,
text=Correlation Coefficient,
decorate]
(0,1.15) arc (90:0:1.15);

%% some data points
node[blue] at (35:4/7+1/2*1/7) $diamond$;
node[red,scale=0.5] at (32:5/7) $pmbtriangle$;

%% draw the quarter circle in the first quadrant
draw (0,0) -- (1,0);
draw (0,0) -- (1,0) arc (0:90:1) -- cycle;

endtikzpicture

enddocument





share|improve this answer






















  • +1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
    – marmot
    1 hour ago










  • @marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
    – A.Ellett
    1 hour ago










  • @marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
    – A.Ellett
    1 hour ago










  • In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
    – marmot
    1 hour ago














up vote
5
down vote













OK. So, I won't do all the work for you. But, I will do enough so that you should be able to figure out how to add everything else that remains to be done.



enter image description here



What I show you here is as follows:



  • How to clip a portion of a picture

  • How to draw concentric circles about a given center

  • How to label along a curved path

  • How to place a label at a point along a path

  • How label at the end points of a line

  • How to get different styles of dashed and dotted lines

  • How to plot points

  • Different approaches to scaling the text

And, I believe that should allow you to complete the rest of the picture.



Here's the code to generate the diagram:



documentclass[tikz,border=6pt]standalone
usepackageamsmath
usepackagetikz
usetikzlibrarycalc
usetikzlibrarydecorations.text
pagestyleempty
begindocument

begintikzpicture[
x=(4cm,0),
y=(0,4cm),
radius=4cm,
]
%% notation (<angle>:<radius>) gives polar coordinates
%% (<x-coor>,<y-coord>) Euclidean coordinates

%% draw semi-circles clipped by the 1st quadrant arc
%% `scope` prevents the entirety of the remainder of
%% picture from being clipped.
beginscope
draw[clip] (0,0) -- (1,0) arc (0:90:1) -- cycle;
foreach myr in 1,2,...,5

%% draw a circle centered at (5/7,0)
%% arc starts at point (myr/6,0) and proceeds through an
%% angle of 180 degrees with a radius of myr/6 units
draw[blue,densely dotted] (5/7,0) ++ (myr/6,0) arc (0:180:myr/6);

endscope

%% place a label along one of the above clipped arcs.
path (5/7,0) ++ (2/6,0) arc (0:180:2/6) node[pos=0.75,
rotate=0.75*180-90,
anchor=south,
inner sep=4pt,
scale=0.5,
blue] (A) $0.25$;

%% label the correlation coefficient values
pgfkeys/pgf/number format/precision=1
foreach myp in 0,1,2,...,11

%% calculate the angle myangle from the integer myp
pgfmathsetmacromyangle90*(12-myp)/12
pgfmathparsemyp/10
pgfmathroundtozerofillpgfmathresult
pgfmathsetmacromylabelpgfmathresult
%% handle labels that don't follow the previous pattern
ifdimmylabel pt=1.0pt%%
defmylabel0.95
fi
ifdimmylabel pt=1.1pt%%
defmylabel0.99%%
fi
%% draw the dashed lines from orgin to arc with labels outside arc
draw[gray!50,
dash pattern=on 1.25pt off 0.5pt,
line width=0.1pt] (myangle:1) node[black,
anchor=180+myangle]
$scriptscriptstylemylabel$ -- (0,0);


%% labeling along a curve
%% a fancy approach that's necessary for labels conforming to
%% a curved path. Not necessary for labels along straight lines.
path[postaction=decoration=text along path,
text align=center,
text=Correlation Coefficient,
decorate]
(0,1.15) arc (90:0:1.15);

%% some data points
node[blue] at (35:4/7+1/2*1/7) $diamond$;
node[red,scale=0.5] at (32:5/7) $pmbtriangle$;

%% draw the quarter circle in the first quadrant
draw (0,0) -- (1,0);
draw (0,0) -- (1,0) arc (0:90:1) -- cycle;

endtikzpicture

enddocument





share|improve this answer






















  • +1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
    – marmot
    1 hour ago










  • @marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
    – A.Ellett
    1 hour ago










  • @marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
    – A.Ellett
    1 hour ago










  • In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
    – marmot
    1 hour ago












up vote
5
down vote










up vote
5
down vote









OK. So, I won't do all the work for you. But, I will do enough so that you should be able to figure out how to add everything else that remains to be done.



enter image description here



What I show you here is as follows:



  • How to clip a portion of a picture

  • How to draw concentric circles about a given center

  • How to label along a curved path

  • How to place a label at a point along a path

  • How label at the end points of a line

  • How to get different styles of dashed and dotted lines

  • How to plot points

  • Different approaches to scaling the text

And, I believe that should allow you to complete the rest of the picture.



Here's the code to generate the diagram:



documentclass[tikz,border=6pt]standalone
usepackageamsmath
usepackagetikz
usetikzlibrarycalc
usetikzlibrarydecorations.text
pagestyleempty
begindocument

begintikzpicture[
x=(4cm,0),
y=(0,4cm),
radius=4cm,
]
%% notation (<angle>:<radius>) gives polar coordinates
%% (<x-coor>,<y-coord>) Euclidean coordinates

%% draw semi-circles clipped by the 1st quadrant arc
%% `scope` prevents the entirety of the remainder of
%% picture from being clipped.
beginscope
draw[clip] (0,0) -- (1,0) arc (0:90:1) -- cycle;
foreach myr in 1,2,...,5

%% draw a circle centered at (5/7,0)
%% arc starts at point (myr/6,0) and proceeds through an
%% angle of 180 degrees with a radius of myr/6 units
draw[blue,densely dotted] (5/7,0) ++ (myr/6,0) arc (0:180:myr/6);

endscope

%% place a label along one of the above clipped arcs.
path (5/7,0) ++ (2/6,0) arc (0:180:2/6) node[pos=0.75,
rotate=0.75*180-90,
anchor=south,
inner sep=4pt,
scale=0.5,
blue] (A) $0.25$;

%% label the correlation coefficient values
pgfkeys/pgf/number format/precision=1
foreach myp in 0,1,2,...,11

%% calculate the angle myangle from the integer myp
pgfmathsetmacromyangle90*(12-myp)/12
pgfmathparsemyp/10
pgfmathroundtozerofillpgfmathresult
pgfmathsetmacromylabelpgfmathresult
%% handle labels that don't follow the previous pattern
ifdimmylabel pt=1.0pt%%
defmylabel0.95
fi
ifdimmylabel pt=1.1pt%%
defmylabel0.99%%
fi
%% draw the dashed lines from orgin to arc with labels outside arc
draw[gray!50,
dash pattern=on 1.25pt off 0.5pt,
line width=0.1pt] (myangle:1) node[black,
anchor=180+myangle]
$scriptscriptstylemylabel$ -- (0,0);


%% labeling along a curve
%% a fancy approach that's necessary for labels conforming to
%% a curved path. Not necessary for labels along straight lines.
path[postaction=decoration=text along path,
text align=center,
text=Correlation Coefficient,
decorate]
(0,1.15) arc (90:0:1.15);

%% some data points
node[blue] at (35:4/7+1/2*1/7) $diamond$;
node[red,scale=0.5] at (32:5/7) $pmbtriangle$;

%% draw the quarter circle in the first quadrant
draw (0,0) -- (1,0);
draw (0,0) -- (1,0) arc (0:90:1) -- cycle;

endtikzpicture

enddocument





share|improve this answer














OK. So, I won't do all the work for you. But, I will do enough so that you should be able to figure out how to add everything else that remains to be done.



enter image description here



What I show you here is as follows:



  • How to clip a portion of a picture

  • How to draw concentric circles about a given center

  • How to label along a curved path

  • How to place a label at a point along a path

  • How label at the end points of a line

  • How to get different styles of dashed and dotted lines

  • How to plot points

  • Different approaches to scaling the text

And, I believe that should allow you to complete the rest of the picture.



Here's the code to generate the diagram:



documentclass[tikz,border=6pt]standalone
usepackageamsmath
usepackagetikz
usetikzlibrarycalc
usetikzlibrarydecorations.text
pagestyleempty
begindocument

begintikzpicture[
x=(4cm,0),
y=(0,4cm),
radius=4cm,
]
%% notation (<angle>:<radius>) gives polar coordinates
%% (<x-coor>,<y-coord>) Euclidean coordinates

%% draw semi-circles clipped by the 1st quadrant arc
%% `scope` prevents the entirety of the remainder of
%% picture from being clipped.
beginscope
draw[clip] (0,0) -- (1,0) arc (0:90:1) -- cycle;
foreach myr in 1,2,...,5

%% draw a circle centered at (5/7,0)
%% arc starts at point (myr/6,0) and proceeds through an
%% angle of 180 degrees with a radius of myr/6 units
draw[blue,densely dotted] (5/7,0) ++ (myr/6,0) arc (0:180:myr/6);

endscope

%% place a label along one of the above clipped arcs.
path (5/7,0) ++ (2/6,0) arc (0:180:2/6) node[pos=0.75,
rotate=0.75*180-90,
anchor=south,
inner sep=4pt,
scale=0.5,
blue] (A) $0.25$;

%% label the correlation coefficient values
pgfkeys/pgf/number format/precision=1
foreach myp in 0,1,2,...,11

%% calculate the angle myangle from the integer myp
pgfmathsetmacromyangle90*(12-myp)/12
pgfmathparsemyp/10
pgfmathroundtozerofillpgfmathresult
pgfmathsetmacromylabelpgfmathresult
%% handle labels that don't follow the previous pattern
ifdimmylabel pt=1.0pt%%
defmylabel0.95
fi
ifdimmylabel pt=1.1pt%%
defmylabel0.99%%
fi
%% draw the dashed lines from orgin to arc with labels outside arc
draw[gray!50,
dash pattern=on 1.25pt off 0.5pt,
line width=0.1pt] (myangle:1) node[black,
anchor=180+myangle]
$scriptscriptstylemylabel$ -- (0,0);


%% labeling along a curve
%% a fancy approach that's necessary for labels conforming to
%% a curved path. Not necessary for labels along straight lines.
path[postaction=decoration=text along path,
text align=center,
text=Correlation Coefficient,
decorate]
(0,1.15) arc (90:0:1.15);

%% some data points
node[blue] at (35:4/7+1/2*1/7) $diamond$;
node[red,scale=0.5] at (32:5/7) $pmbtriangle$;

%% draw the quarter circle in the first quadrant
draw (0,0) -- (1,0);
draw (0,0) -- (1,0) arc (0:90:1) -- cycle;

endtikzpicture

enddocument






share|improve this answer














share|improve this answer



share|improve this answer








edited 7 mins ago

























answered 1 hour ago









A.Ellett

35.6k1064164




35.6k1064164











  • +1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
    – marmot
    1 hour ago










  • @marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
    – A.Ellett
    1 hour ago










  • @marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
    – A.Ellett
    1 hour ago










  • In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
    – marmot
    1 hour ago
















  • +1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
    – marmot
    1 hour ago










  • @marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
    – A.Ellett
    1 hour ago










  • @marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
    – A.Ellett
    1 hour ago










  • In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
    – marmot
    1 hour ago















+1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
– marmot
1 hour ago




+1 (I guess that the angles between the segments should not coincide, but I also see why you would not want to go through the semi-clear paper linked in the question to look for the formula. ;-)
– marmot
1 hour ago












@marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
– A.Ellett
1 hour ago




@marmot Well, I don't want to do all of the OP's work. :-) But, I did want to give the OP and idea of how to approach each detail of the picture.
– A.Ellett
1 hour ago












@marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
– A.Ellett
1 hour ago




@marmot Actually, unless absolutely necessary, I probably would not use the same formula, but instead I'd use a logarithmic function to be place those labels.
– A.Ellett
1 hour ago












In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
– marmot
1 hour ago




In the paper, the author stresses that he is using the cosine law to motivate the diagram, but he fails to provide the formula for the labels. I guess that they are computed from the erf, but I'm not sure of that. Analytic approximations to erf can be found on this site. But I think that the least thing the OP should do is to specify this.
– marmot
1 hour ago










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