6

I am trying to reproduce this figure but I have the impression that alpha does not correspond to the length of the path subpath (0,1.3) of p and that the value -pi/2 is not correct.

  \documentclass[border=5mm]{standalone}
  \usepackage{luatex85}
  \usepackage{luamplib}
  \begin{document}
  \mplibtextextlabel{enable}
  \begin{mplibcode}

  beginfig(1);

  numeric u, pi;
  u = 10mm;
  pi = 3.141592654;

  path C, T, xx, yy;

  C = fullcircle scaled 5u;

  xx = (point 4 of C -- point 0 of C) scaled 1.1;
  yy = (point 6 of C -- point 2 of C) scaled 1.1;

  T = ((xpart point 0 of C,200) -- (xpart point 0 of C,-200));

  numeric l;
  l = length subpath(0,1.3) of C;

  draw xx;
  draw yy;
  draw C withpen pencircle scaled 3/4;
  draw origin -- (point 1.2 of C) withcolor blue;
  draw  subpath(0,1.2) of C withcolor blue;
  draw T withcolor red;

 label.urt(btex $\pi$ etex,(xpart point 0 of C,u*(pi)));
 label.urt(btex $1$ etex,(xpart point 0 of C,u*1));
 label.urt(btex $2\pi$ etex,(xpart point 0 of C,u*(2pi)));
 label.urt(btex $\frac{\pi}{2}$ etex,(xpart point 0 of C,u*(pi/2)));
 label.urt(btex $-\frac{\pi}{2}$ etex,(xpart point 0 of C,u*(-pi/2)));
 label.urt("$M$", point 1.3 of C);
 label.llft("$O$", origin);
 label.lrt("$I$", point 0 of C);
 label.ulft("$J$", point 2 of C);
 label.urt(btex $\frac{\pi}{2}$ etex, point 2 of C);

 fill fullcircle scaled dotlabeldiam
 shifted (xpart point 1.2 of C, ypart point 1.2 of C)
 withcolor blue;

 endfig;
 \end{mplibcode}
 \end{document}

enter image description here

3

Just for comparison (or more honestly just for my own amusement), here is another version, that you might find instructive.

enter image description here

Source

Compile with lualatex.

\documentclass[border=5mm]{standalone}
\usepackage{luatex85}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
beginfig(1);
    numeric u, pi, alpha;
    u = 42;
    pi = 3.141592653589793;
    alpha = 1.2;

    % you might not want this grid
    numeric minx, miny, maxx, maxy;
    minx = -7/2; maxx = 1; miny = -2; maxy = 13/2;
    for x=minx step 1/2 until maxx:
        draw ((x, miny) -- (x, maxy)) scaled u withcolor 3/4 white;
    endfor
    for y=miny step 1/2 until maxy:
        draw ((minx, y) -- (maxx, y)) scaled u withcolor 3/4 white;
    endfor

    path C, ii, jj;
    C = fullcircle scaled 2u shifted (-u,0);
    ii = (point 4 of C -- point 0 of C) shifted - center C scaled 1.15 shifted center C;
    jj = (point 6 of C -- point 2 of C) shifted - center C scaled 1.15 shifted center C;

    path negative_d, positive_d;
    positive_d = origin -- (0, maxy) scaled u;
    negative_d = origin -- (0, miny) scaled u;

    draw negative_d withcolor 2/3 red;
    draw positive_d withcolor 1/2 green;
    label.lft("$\cal D$", 3/2 pi * u * up) withcolor 1/2 green;

    vardef mark_y_axis(expr value, name, shade) = 
        save p; pair p; p = value * u * up;
        draw (left--right) scaled 2 shifted p withcolor shade;
        label.rt(name, p shifted 2 right);
    enddef;

    mark_y_axis(1,      "$1$",      1/2 green);
    mark_y_axis(alpha,  "$\alpha$", 1/2 green);
    mark_y_axis(1/2 pi, "$\pi/2$",  1/2 green);
    mark_y_axis(pi,     "$\pi$",    1/2 green);
    mark_y_axis(2pi,    "$2\pi$",   1/2 green);
    mark_y_axis(-1/2 pi, "$-\pi/2$",  2/3 red);

    z.M = point alpha * 4 / pi of C;

    path trajet; 
    trajet = subpath (0, alpha * 4 / pi) of C .. {left} (minx, 4/3)*u;

    numeric a; pair p;
    a = arctime pi*u of trajet;
    p = point a of trajet;
    draw (down--up) scaled 2 rotated angle direction a of trajet shifted p withcolor 1/2 green;
    label.ulft("$\pi$", p);

    draw ii withpen pencircle scaled 1/4;
    draw jj withpen pencircle scaled 1/4;
    draw C;
    draw trajet withcolor 1/2 green;

    draw center C -- z.M withcolor 1/4 blue;
    label("$M$", z.M + (-2,8)); filldraw fullcircle scaled dotlabeldiam shifted z.M;

    vardef show_mapping(expr a, b) = 
        interim ahangle := 30;
        drawarrow a {left} .. b {dir 240}
            cutbefore fullcircle scaled 8 shifted a
            cutafter  fullcircle scaled 8 shifted b
            withcolor 3/4 blue;
    enddef;

    show_mapping(pi * u * up, p);
    show_mapping(alpha * u * up, z.M);

    label.llft("$O$", center C); 
    label.lrt("$I$", point 0 of C);
    label.llft("$J$", point 2 of C);

endfig;
\end{mplibcode}
\end{document}

Some notes

  • You can vary u to adjust the size of the whole thing (except the font sizes)

  • But it's probably not a good idea to vary the value of the variable pi. There is no constant built into MP, so I just defined it here, using the value that happens to be built into my editor. If you don't trust my constant you could try writing pi = 1/4 arclength (quartercircle scaled 16); which also works...

  • You can just remove the grid if you don't want it

  • I have put the point alpha and the corresponding point M in the mathematically correct places. To get from the scalar value of alpha to the correct point round the circle I used point alpha * 4 / pi of C. This is because the length of a semicircle is 4 units of MP time and pi units of length.

  • You can use arbitrary suffixes with the z notation, so z.M works just like z0. You need the . to separate the z from the suffix if the suffix is alphabetic.

  • arctime x of p finds the "time" t along path p such that length(subpath(0, t) of p) = x, which is exactly what we want to find the point that is pi units along our curved path.

  • the show_mapping definition shows a way to shorten a path using cutbefore and cutafter (and shows off nice thin arrow heads).

| improve this answer | |
  • Thank you very much, what a mastery of this great tool ! When you write u=42, what is the unit of measure ? – Fabrice Oct 9 '19 at 22:09
  • 42 points. Postscript points are the default unit. 1 in = 2.54 cm = 72 points. In TeX terms they are bp rather than pt. So 42pt is 7/12 in or about 1.48cm. 42 is of course always the answer. :-) – Thruston Oct 9 '19 at 22:21
7

For the labels on your y-axis, you use

label.urt(btex $\pi$ etex,(xpart point 0 of C,u*(pi)));
label.urt(btex $1$ etex,(xpart point 0 of C,u*1));
label.urt(btex $2\pi$ etex,(xpart point 0 of C,u*(2pi)));
label.urt(btex $\frac{\pi}{2}$ etex,(xpart point 0 of C,u*(pi/2)));
label.urt(btex $-\frac{\pi}{2}$ etex,(xpart point 0 of C,u*(-pi/2)));
label.urt("$M$", point 1.3 of C);`

First, this has a lot of repetitions which makes it hard to adjust, so let's move this into a macro:

vardef labeled (expr t, y) =
  save p; pair p;
  p = (xpart point 0 of C, y*u);
  label.urt(t, p);
enddef;

labeled(btex $\pi$ etex,pi);
labeled(btex $1$ etex,1);
labeled(btex $2\pi$ etex,2pi);
labeled(btex $\frac{\pi}{2}$ etex,pi/2);
labeled(btex $-\frac{\pi}{2}$ etex,-pi/2);

Now we can add little red lines to indicate the exact positions: Add

draw p shifted (-0.5mm, 0) -- p shifted (0.5mm, 0)  withcolor red;

to labeled. This gives enter image description here

Now we see why -pi/2 seemed to be in a odd position and not symmetric to pi/2: label.urt places the label in an upper right position and therefore moves all labels a bit up. We can use label.rt in labeled instead to avoid this: enter image description here

Now the axis still looks odd: The drawing seems to illustrate the correspondence between the length of the circular arc and the angle in radians, so it only works if the radius of the circle is 1 unit. So we change C = fullcircle scaled 5u; to C = fullcircle scaled 2u; (The diameter should be 2u) This make the entire diagram quite small, so we could also increase u. Our new figure is enter image description here

Now we want to add alpha. First we calculate alpha by asking for the length of the arc (arclength) of the path you are interested in (I use 1.2 instead of 1.3 because that is the path you actually used in your code):

alpha*u = arclength subpath(0,1.2) of C;

Then we can add

labeled(btex $\alpha$ etex, alpha);

enter image description here

It turns out that this arc has a length slightly below 1, so let's use subpath(0, 1.4) instead to make it look a bit more like your original: enter image description here

Now wwe can make everything a bit bigger and add some alignment to the labels to get:

\documentclass[border=5mm]{standalone}
\usepackage{luatex85}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}

beginfig(1);

numeric u, pi;
u = 20mm;
pi = 3.141592654;

path C, T, xx, yy;

C = fullcircle scaled 2u;

xx = (point 4 of C -- point 0 of C) scaled 1.1;
yy = (point 6 of C -- point 2 of C) scaled 1.1;

T = ((xpart point 0 of C,6.5u) -- (xpart point 0 of C,-2u));

alpha*u = arclength subpath(0,1.4) of C;

draw xx;
draw yy;
draw C withpen pencircle scaled 3/4;
draw origin -- (point 1.4 of C) withcolor blue;
draw subpath(0,1.4) of C withcolor blue;
draw T withcolor red;

vardef labeled (expr t, y) =
  save p; pair p;
  p = (xpart point 0 of C, y*u);
  label.rt(t, p);
  draw p shifted (-0.5mm, 0) -- p shifted (0.5mm, 0)  withcolor red;
enddef;

labeled(btex $\alpha$ etex,alpha);
labeled(btex \hbox to 1.5em{\hfill$\pi$} etex,pi);
labeled(btex \hbox to 1.5em{\hfill$1$} etex,1);
labeled(btex \hbox to 1.5em{\hfill$2\pi$} etex,2pi);
labeled(btex \hbox to 1.5em{\hfill$\frac{\pi}{2}$} etex,pi/2);
labeled(btex \hbox to 1.5em{\hfill$-\frac{\pi}{2}$} etex,-pi/2);
label.urt("$M$", point 1.4 of C);
label.llft("$O$", origin);
label.lrt("$I$", point 0 of C);
label.ulft("$J$", point 2 of C);
label.urt(btex $\frac{\pi}{2}$ etex, point 2 of C);

fill fullcircle scaled dotlabeldiam
shifted point 1.4 of C
withcolor blue;

endfig;
\end{mplibcode}
\end{document}

enter image description here

| improve this answer | |
  • Thank you very much, I start with Metapost and I still have a lot to learn. If I dared, because I'm short of time, how to add the trajectories of the points ? – Fabrice Oct 7 '19 at 19:51
  • In the macro p = (xpart point 0 of C, u*y); – Fabrice Oct 7 '19 at 20:34
  • I'm not sure which trajectories you mean. The green line continuing to the left? I'm not even quite sure what it represents, so I also can't really tell you how to draw it. – Marcel Krüger Oct 7 '19 at 20:39
  • "and add some alignment to the labels " How ? – Fabrice Oct 7 '19 at 20:49
  • @Fabrice Oh sorry, I forgot to add the final code. Fixed that now, but basically I aligned them by just putting them into a right aligned \hbox in TeX. – Marcel Krüger Oct 7 '19 at 20:54

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