Best way to illustrate Keplers 2nd law with tikz
For the problem of filling the ellipse sectors, you can draw "bigger" triangles and then clip them to the elipse shape.
For this approach it is better to have the points A1, A2, B1, B7, C1 and C7 in polar coordinates. In fact, only the angle is important, since the radius will be made long enough to guarantee that the point is out of the ellipse. A radius of 5 is enough in this example.
The following code implements this idea:
% We define the orbit as a macro because we will use it twice, first for clipping and then
% to actually draw the ellipse. This way we avoid inconsistencies.
\def\orbit{(1.5,0) ellipse(2.5cm and 2cm)}
\begin{tikzpicture}
\fill (0,0) coordinate (O) circle (2pt) node[below =7pt] {sun};%
\coordinate (A1) at (50.992527:5);
\coordinate (A2) at (41.913511:5);
\coordinate (B1) at (136.450216:5);
\coordinate (B7) at (-150.524111:5);
\coordinate (C1) at (-23.749494:5);
\coordinate (C2) at (-18.581735:5);
\coordinate (P) at (3.42,1.28) ;%%
\fill (P) circle (1pt) node[above right] {planet};%
\begin{scope} % The blue shaded regions
\clip \orbit;
\filldraw[fill=blue,opacity=0.5] (O) -- (A1) -- (A2) -- cycle;
\filldraw[fill=blue,opacity=0.5] (O) -- (B1) -- (B7) -- cycle;%
\filldraw[fill=blue,opacity=0.5] (O) -- (C1) -- (C2) -- cycle;%
\end{scope}
% The ellipse
\draw \orbit;
\draw (1.5,0) coordinate (M) --node[above]{\footnotesize $a$} (4,0);
\fill (M) circle (1pt);
end{tikzpicture}
This is the result:
Update. Automate the finding of the sectors approximating them by tiangles or circular sectors
The following code implements some ideas, but the implementation is very hackish. These are the ideas:
- Given a pair of initial and final angles (given indeed by two points located in the outside of the ellipse) the macro
\ComputeAreacalculates the area of the triangle formed by the sun and the two points in the orbit at those angles. - Given any other point in the orbit, the macro
\ComputePointNextTofinds the next point in the orbit (in counterclockwise direction) which covers the same area computed before. In this case the sector is assumed to be a circular sector with center at the sun, instead of a triangle, to simplify the calculus.
In order to solve 1, I used the formula found here which gives the area of a triangle from the coordinates of its three vertex. In order to implement this in TikZ I had to first find the three points, which involves solving some intersections. The formula is implemented in a let...in path, and saved via \xdef in a macro called \area for later use.
In order to solve 2, I used the formula for the area of a circular sector of angle theta, which is area=(theta*r^2), given theta in radians. Finding for theta we have then: theta = 2*area/r^2. I implemented this formula again in a let...in path and from this value of theta I built a coordinate called (result) which lies in the appropiate angle at the outisde of the ellipse.
The complete code follows. In this case I kept the original figure with blue regions exactly as the one given by the OP, and added my computations.
The area of the "big" sector is computed, and the result is shown below the figure, for debugging purposes (the unit of length is pt, so the resulting area is in pt^2).
For each of the other blue sectors, I used the first point (A1) and (C1) as the "given point" and computed as described the other "next points" (A2) and (C2). The figure shows in two red lines over the blue sectors the directions in which the found points are.
As you can see, the approximation is good enough unless the figure has to be used to take precise measures on it.
Code:
\def\orbit{(1.5,0) ellipse(2.5cm and 2cm)}
\def\ComputeArea#1#2{
\path[name path=orbit] \orbit;
\path[name path=line1] (O) -- (#1);
\path[name path=line2] (O) -- (#2);
\path[name intersections={of=orbit and line1,by=aux1}];
\path[name intersections={of=orbit and line2,by=aux2}];
\path let \p1=(O),
\p2=(aux1),
\p3=(aux2),
\n1 = {abs(\x1*(\y2-\y3)+\x2*(\y3-\y1)+\x3*(\y1-\y2))/2.0}
in node[above] {\pgfmathparse{\n1}\xdef\area{\pgfmathresult}};
}
\def\ComputePointNextTo#1{
\path[name path=line1] (O) -- (#1);
\path[name intersections={of=orbit and line1,by=aux1}];
\path let \p1=($(aux1)-(O)$),
\n1 = {veclen(\p1)}, % r
\n2 = {atan2(\x1,\y1)}, % initial angle
\n3 = {deg(2*\area/\n1/\n1)} % angle to cover
in coordinate (result) at (\n2+\n3:5);
}
\usetikzlibrary{intersections,calc}
\begin{tikzpicture}
% Original figure (using the clipping technique)
\fill (0,0) coordinate (O) circle (2pt) node[below =7pt] {sun};%
\coordinate (A1) at (41.913511:5);
\coordinate (A2) at (50.992527:5);
\coordinate (B1) at (136.450216:5);
\coordinate (B7) at (-150.524111:5);
\coordinate (C1) at (-23.749494:5);
\coordinate (C2) at (-18.581735:5);
\coordinate (P) at (3.42,1.28) ;%%
\fill (P) circle (1pt) node[above right] {planet};%
\begin{scope}
\clip \orbit;
\filldraw[fill=blue,opacity=0.5] (O) -- (A1) -- (A2) -- cycle;
\filldraw[fill=blue,opacity=0.5] (O) -- (B1) -- (B7) -- cycle;%
\filldraw[fill=blue,opacity=0.5] (O) -- (C1) -- (C2) -- cycle;%
\end{scope}
\draw \orbit;
\draw (1.5,0) coordinate (M)
--node[above]{\footnotesize $a$} (4,0);
\fill (M) circle (1pt);
% Added. Trying to automatically find (A2) and (C2)
% from (A1) and (C1), such that the area is equal to the
% sector from (B1) to (B7)
\ComputeArea{B1}{B7}
\node[right] at (0,-2.3) {Area: \area}; % Show it, for debugging
\ComputePointNextTo{A1}
\draw[red] (O) -- (result);
\ComputePointNextTo{C1}
\draw[red] (O) -- (result);
\end{tikzpicture}
Result:
The basic struct PlanetaryMotion which handles the elliptic sector area calculations
is defined in asydef environment, and two examples of the illustration
are shown in two asy pictures.
kepler.tex:
\documentclass{article}
\usepackage{lmodern}
\usepackage{subcaption}
\usepackage[inline]{asymptote}
\usepackage[left=2cm,right=2cm]{geometry}
\begin{asydef}
import graph;
import gsl; // for newton() solver
size(200);
struct PlanetaryMotion{
real a,b,e;
real planetTime,sunR,planetR;
pair F0,F1;
guide orbit;
transform tr=scale(-1,-1); // to put the Sun in the left focus
pair ellipse(real t){
return (a*cos(t),b*sin(t));
}
real Area(real t){ // area 0..t
return a*b/2*(t-e*sin(t));
}
real calcArea(real t0,real t1){
return Area(t1)-Area(t0);
}
real AreaPrime(real t){
return 1/2*a*b*(1-e*cos(t));
}
real findTime(real areaToFit, real tstart){ // find time tend to fit areaToFit
real tend=newton(new real(real t){return calcArea(tstart,t)-areaToFit;}
,new real(real t){return AreaPrime(t);},tstart,tstart+2pi);
return tend;
}
void drawBG(){
draw(tr*orbit,darkblue);
filldraw(tr*shift(F0)*scale(sunR)*unitcircle,yellow,orange);
filldraw(tr*shift(ellipse(planetTime))*scale(planetR)*unitcircle,blue,lightblue);
//dot(tr*F1,UnFill);
label("$F_0$",tr*F0,3N);
//label("$F_1$",tr*F1,3N);
label("Sun",tr*F0,3S);
label("planet",tr*ellipse(planetTime),SW);
draw(((0,0)--(a,0)));
label("$a$",(a/2,0),N);
dot((0,0),UnFill);
}
void drawSector(real t0, real t1,pen p=blue+opacity(0.3)){
fill(tr*(F0--graph(ellipse,t0,t1)--cycle),p);
}
void operator init(real a, real b
,real planetTime
,real sunR=0.05a, real planetR=0.3sunR
){
this.a=a;
this.b=b;
this.planetTime=planetTime;
this.sunR=sunR;
this.planetR=planetR;
this.e=sqrt(a^2-b^2)/a;
this.F0=(a*e,0);
this.F1=(-a*e,0);
this.orbit=graph(ellipse,0,2pi);
}
}
\end{asydef}
\begin{document}
\begin{figure}
\captionsetup[subfigure]{justification=centering}
\centering
\begin{subfigure}{0.49\textwidth}
\begin{asy}
PlanetaryMotion pm=PlanetaryMotion(1,0.618,1.2pi);
pm.drawBG();
real t0,t1,t2,t3,t4,t5;
t0=-0.1pi;
t1= 0.1pi;
pm.drawSector(t0,t1);
real area0=pm.calcArea(t0,t1);
t2=0.7pi;
t3=pm.findTime(area0,t2);
pm.drawSector(t2,t3);
t4=1.5pi;
t5=pm.findTime(area0,t4);
pm.drawSector(t4,t5);
\end{asy}
\caption{}
\label{fig:1a}
\end{subfigure}
%
\begin{subfigure}{0.49\textwidth}
\begin{asy}
PlanetaryMotion pm=PlanetaryMotion(1,0.8,1.35pi,sunR=0.09);
pm.drawBG();
real t0,t1,t2,t3,t4,t5;
t0=-0.05pi;
t1= 0.17pi;
pm.drawSector(t0,t1);
real area0=pm.calcArea(t0,t1);
t2=0.4pi;
t3=pm.findTime(area0,t2);
pm.drawSector(t2,t3);
t4=1.7pi;
t5=pm.findTime(area0,t4);
pm.drawSector(t4,t5);
\end{asy}
\caption{}
\label{fig:1b}
\end{subfigure}
\caption{Illustration of Keplers 2nd law with \texttt{Asymptote}.}
\end{figure}
\end{document}
To process it with latexmk, create file latexmkrc:
sub asy {return system("asy '$_[0]'");}
add_cus_dep("asy","eps",0,"asy");
add_cus_dep("asy","pdf",0,"asy");
add_cus_dep("asy","tex",0,"asy");
and run latexmk -pdf kepler.tex.
The Keplerequation (link to German wikipedia, which is unusually more informative than the English one on this topic) has no algebraic/closed solution. There are good approximations, but if one has to be approximative from the beginning, one can also simulate the physics instead of doing the math:
\documentclass{standalone}
\usepackage{etoolbox}
\usepackage{tikz}
\gdef\myposx{10.0}
\gdef\myposy{0.0}
\gdef\vx{0.0}
\gdef\vy{4.6}
\gdef\forcefactor{150}
\gdef\deltat{0.01}
\gdef\smallmass{1}
\gdef\startone{100}
\gdef\endone{200}
\gdef\starttwo{1800}
\gdef\endtwo{1900}
\gdef\pathone{}
\gdef\pathtwo{}
\begin{document}
\begin{tikzpicture}[scale=0.2]
\filldraw(0,0)circle(0.1);
\foreach \n in {1,...,3625}
{
\pgfextra{%
\global\let\oldx\myposx
\global\let\oldy\myposy
\pgfmathsetmacro{\currentsquareddistance}{\myposx*\myposx+\myposy*\myposy}
\pgfmathsetmacro{\currentforce}{\forcefactor/\currentsquareddistance}
\pgfmathsetmacro{\currentangle}{atan2(\myposx,\myposy)}
\pgfmathsetmacro{\currentforcex}{-1*\currentforce*cos(\currentangle)}
\pgfmathsetmacro{\currentforcey}{-1*\currentforce*sin(\currentangle)}
\pgfmathsetmacro{\currentvx}{\vx+\deltat*\currentforcex/\smallmass}
\pgfmathsetmacro{\currentvy}{\vy+\deltat*\currentforcey/\smallmass}
\pgfmathsetmacro{\currentposx}{\myposx+\deltat*\currentvx}
\pgfmathsetmacro{\currentposy}{\myposy+\deltat*\currentvy}
\global\let\myposx\currentposx
\global\let\myposy\currentposy
\global\let\vx\currentvx
\global\let\vy\currentvy
\global\let\forcex\currentforcex
\global\let\forcey\currentforcey
\global\let\myangle\currentangle
\ifnumequal{\n}{\startone}{%
\global\let\startonex\oldx
\global\let\startoney\oldy
\xappto{\pathone}{(\oldx,\oldy)}
}{}
\ifnumequal{\n}{\starttwo}{%
\global\let\starttwox\oldx
\global\let\starttwoy\oldy
\xappto{\pathtwo}{(\oldx,\oldy)}
}{}
\ifnumequal{\n}{\endone}{%
\global\let\endonex\myposx
\global\let\endoney\myposy
\xappto{\pathone}{,(\myposx,\myposy)}
}{}
\ifnumequal{\n}{\endtwo}{%
\global\let\endtwox\myposx
\global\let\endtwoy\myposy
\xappto{\pathtwo}{,(\myposx,\myposy)}
}{}
}
% \draw[very thin,->](\oldx,\ol dy)--++(\forcex,\forcey);
\ifnumgreater{(\n-\startone)*(\endone-\n)}{-1}
{
\pgfextra{%
\xappto{\pathone}{,(\myposx,\myposy)}
}
}
{}
\ifnumgreater{(\n-\starttwo)*(\endtwo-\n)}{-1}
{
\pgfextra{%
\xappto{\pathtwo}{,(\myposx,\myposy)}
}
}
{}
\draw(\oldx,\oldy)--(\myposx,\myposy);
}
\begin{scope}[red]
\filldraw (0,0)%
\foreach \point in \pathone
{%
--\point
}--(0,0);
\filldraw (0,0)%
\foreach \point in \pathtwo
{%
--\point
}--(0,0);
\end{scope}
\end{tikzpicture}
\end{document}
The values at the beginning are in totally arbitrary units. Realism would do much better, because our planets have very inexcentric orbits, so there wouldn't be much to see.
What's still to do:
- compute the full period OR replace the
\foreach-loop by awhile-construct that terminates, when the ellipse is full - compute the start and end points out of the time phase
- replace all the global definitions by appropriate ones (I'm not good in TeX scoping)