tikz-pgf
I am looking for an easy way to draw an arc that goes e.g. from 45 deg to 90 deg and with 0,0 as center. The arc should start then at (0.707,0.707) and end at (0,1). The common syntax
\draw (0,0) arc [radius=1, start angle=45, end angle= 90];
always starts the arc at (in this case) (0,0) which makes it hard to draw several arcs with the same center but starting from different angles.
Or is there something I am missing?
Here is an answer, using the technique described here, with more detail,
\pgfmathsetmacro{\ex}{0}
\pgfmathsetmacro{\ey}{1}
\draw (\ex,\ey) -- ++(-15:1)
(\ex,\ey) -- ++(15:1);
\draw (\ex,\ey) ++(45:.8) arc (45:-45:.8);
The secret to this is in the last line,
\draw (\ex,\ey) ++(45:.8)
Jumps the draw cursor to "the 45 degree position on a circle of radius 0.8", without drawing anything (achieved by using only ++ and not any -- in the command)
Then, from there, we draw an arc
arc (45:-45:.8);
from 45 degrees to -45 degrees, of what would be a circle of radius .8.
Pretty roundabout way to do this, but still it works fine.
My finished eye:
\begin{tikzpicture}
%eye
\pgfmathsetmacro{\eyeSize}{1}
\pgfmathsetmacro{\ex}{0}
\pgfmathsetmacro{\ey}{1}
\pgfmathsetmacro{\eRot}{-10}
\pgfmathsetmacro{\eAp}{-55}
\draw[rotate around={\eRot:(\ex,\ey)}] (\ex,\ey) -- ++(-.5*\eAp:\eyeSize)
(\ex,\ey) -- ++(.5*\eAp:\eyeSize);
\draw (\ex,\ey) ++(\eRot+\eAp:.75*\eyeSize) arc (\eRot+\eAp:\eRot-\eAp:.75*\eyeSize);
% IRIS
\draw[fill=gray] (\ex,\ey) ++(\eRot+\eAp/3:.75*\eyeSize) % start point
arc (\eRot+180-\eAp:\eRot+180+\eAp:.28*\eyeSize);
%PUPIL, a filled arc
\draw[fill=black] (\ex,\ey) ++(\eRot+\eAp/3:.75*\eyeSize) % start point
arc (\eRot+\eAp/3:\eRot-\eAp/3:.75*\eyeSize);
\end{tikzpicture}
The trigonometric functions in PGF use degrees, so you'll have to use 360 instead of 2*pi, and you'll have to deduct the full circle from the angle, not the other way round.
This becomes clear if you make a minimal example and look at the angles. The delta angle of the arc is the end angle minus the start angle. In your case, you have a delta angle that's larger than 180°, so you get the large arc. By subtracting a full circle from the end angle, the absolute value of the difference angle becomes smaller than 180°:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\coordinate [label=right:O] (O) at (0,0);
\coordinate [label=below:A] (A) at (-1,-1.5);
\coordinate [label=left:B] (B) at (-2,1);
\draw (A) -- (O) -- (B);
\draw [-latex]
let
\p0 = (O),
\p1 = (A),
\p2 = (B),
\n1 = {atan2(\x1 - \x0, \y1 - \y0)},
\n2 = {atan2(\x2 - \x0, \y2 - \y0)},
\n3 = {1cm}
in
(\p0) + (\n1:\n3) arc [radius=\n3, start angle=\n1, end angle=\n2]
node [anchor=west, align=right] at (1.5,0) {\verb|\n1|: \pgfmathparse{\n1}\pgfmathprintnumber{\pgfmathresult}\\ \verb|\n2|: \pgfmathparse{\n2}\pgfmathprintnumber{\pgfmathresult}\\
delta: \pgfmathparse{\n2-\n1}\pgfmathprintnumber{\pgfmathresult}};
\end{tikzpicture}
\begin{tikzpicture}
\coordinate [label=right:O] (O) at (0,0);
\coordinate [label=below:A] (A) at (-1,-1.5);
\coordinate [label=left:B] (B) at (-2,1);
\draw (A) -- (O) -- (B);
\draw [-latex]
let
\p0 = (O),
\p1 = (A),
\p2 = (B),
\n1 = {atan2(\x1 - \x0, \y1 - \y0)},
\n2 = {atan2(\x2 - \x0, \y2 - \y0) - 360},
\n3 = {1cm}
in
(\p0) + (\n1:\n3) arc [radius=\n3, start angle=\n1, end angle=\n2]
node [anchor=west, align=right] at (1.5,0) {\verb|\n1|: \pgfmathparse{\n1}\pgfmathprintnumber{\pgfmathresult}\\ \verb|\n2|: \pgfmathparse{\n2}\pgfmathprintnumber{\pgfmathresult}\\
delta: \pgfmathparse{\n2-\n1}\pgfmathprintnumber{\pgfmathresult}};
\end{tikzpicture}
\end{document}
\documentclass[convert = false]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc, intersections, backgrounds, arrows}
\begin{document}
\tikzset{circle with radius/.style = {shape = circle, inner sep = 0pt,
outer sep = 0pt, minimum size = {2 * (#1)}}}
\begin{tikzpicture}[line join = round, line cap = round,
every label/.append style = {font = \scriptsize},
dot/.style = {inner sep = +0pt, shape = circle,
draw = black, label = {#1}},
small dot/.style = {minimum size = .05cm, dot = {#1}},
big dot/.style = {minimum size = .1cm, dot = {#1}},
]
\begin{scope}
%\clip ($(B) + (1.5, -2)$) rectangle ($(B) + (-3, 2)$);
\coordinate (O) at (0, 0);
\pgfmathsetmacro{\as}{3}
\pgfmathsetmacro{\bs}{2.25}
\pgfmathsetmacro{\c}{sqrt(\as^2 - \bs^2)}
\pgfmathsetmacro{\al}{3.75}
\pgfmathsetmacro{\bl}{2.9}
\pgfmathsetmacro{\cl}{sqrt(\al^2 - \bl^2)}
\pgfmathsetmacro{\xs}{abs(\c - \cl)}
\coordinate (O) at (0, 0);
\node[fill = black, big dot = {below left: \(F\)}] (F) at (\c, 0) {};
\path[name path global = line1] (\c, 0) -- ++(60:{\as} and \bs);
\path[name path global = line2] (\c, 0) -- ++(150:6cm);
\draw[name path global = ell1, blue] (O) ellipse
(\as cm and \bs cm);
\draw[name path global = ell2, red] (-\xs, 0)
ellipse (\al cm and \bl cm);
\path[name intersections = {of = line1 and ell1, by = P1}];
\node[fill = black, big dot = {right: \(A\)}] (A) at (P1) {};
\path[name intersections = {of = line2 and ell2, by = P2}];
\node[fill = black, big dot = {above: \(B\)}] (B) at (P2) {};
\begin{scope}[declare function = {doubleA = 5.8cm;},
]
\clip ($(A.center) + (1, 0)$) rectangle ($(B.center) + (0, 1)$);
\begin{pgfinterruptboundingbox}
\path let
\p1 = ($(A) - (F)$),
\p2 = ($(B) - (F)$),
\n1 = {veclen(\x1, \y1)},
\n2 = {veclen(\x2, \y2)}
in
(A) node[name path global = aCircle, circle with radius = doubleA-\n1]
{}
(B) node[name path global = bCircle, circle with radius = doubleA-\n2]
{}
(F) node[name path global = fCircle,
circle with radius = .5 * doubleA] {};
\tikzset{name intersections = {of = aCircle and bCircle, name = F'} }
\foreach \solA in {2} {
\path ($(F)!.5!(F'-\solA)$) coordinate (C'-\solA)
($(C'-\solA)!doubleA/2!(F)$) coordinate (xDir-\solA)
(F'-\solA) node[name path global/.expanded = f'Circle-\solA,
circle with radius = .5 * doubleA] {};
} %!?
\foreach \solA in {2} { %!?
\path[name intersections = {of = fCircle and f'Circle-\solA,
by = {yDir-\solA}}]
($(xDir-\solA)-(C'-\solA)$) coordinate (xDir'-\solA)
($(yDir-\solA)-(C'-\solA)$) coordinate (yDir'-\solA)
;
}
\end{pgfinterruptboundingbox}
\foreach \solA in {2}
\draw[x = (xDir'-\solA), y = (yDir'-\solA), name path global = traj]
(C'-\solA) circle [radius = 1];
\end{scope}
\path[name path = circ] (B) circle [radius = 1bp];
\draw[name intersections = {of = circ and traj}, -latex] (B) --
($(intersection-1)!1.25cm!(intersection-2)$) coordinate (P3);
\draw[name intersections = {of = circ and ell2}, -latex, red] (B) --
($(intersection-1)!2cm!(intersection-2)$) coordinate (P4);
\draw[-latex, name path = line3] (P3) -- (P4);
\draw[name path = line4] (B) -- ($(B)!2.1cm!-90:(F)$) coordinate (P5);
\path[name path = circ2] (P3) circle [radius = 1bp];
\path[name intersections = {of = circ2 and line3}, name path = line5]
(P3) -- ($(intersection-1)!2cm!(P3)$);
\path[name intersections = {of = line5 and line4, by = P6}];
\path[name intersections = {of = circ and traj}] (B) --
($(intersection-1)!1.5cm!(intersection-2)$) coordinate (P7);
\path[name intersections = {of = circ and ell2}] (B) --
($(intersection-1)!2.5cm!(intersection-2)$) coordinate (P8);
\draw[-latex] let
\p0 = (P6),
\p1 = (P5),
\p2 = (P4),
\n1 = {atan2(\x1 - \x0, \y1 - \y0)},
\n2 = {atan2(\x2 - \x0, \y2 - \y0)-360},
\n3 = {.75cm}
in (P6) +(\n1:\n3) arc[radius = \n3, start angle = \n1, end angle = \n2];
\end{scope}
\end{tikzpicture}
\end{document}
Best Answer
Perhaps this is what you want? Notice the black, red and blue arcs with center at
(2,2):For example to get an arc from
50to200with center at(3,2)and radius3, you could use