曲柄滑块机构曲线

青青子衿

The curve of the slider-crank mechanism
\begin{cases}   
\begin{split}      
x&=r\sin\left(t\right)+\frac{L}{R}\sqrt{R^2-r^2\cos^2\left(t\right)}\\
y&=\left(1-\frac{L}{R}\right)r\cos\left(t\right)
\end{split}   
\end{cases}
\[x^4-2\left(L^2+r^2-\frac{\left(L^2+R^2\right)y^2}{\left(L-R\right)^2}\right)x^2+\left(L^2-r^2-\frac{\left(L^2-R^2\right)y^2}{\left(L-R\right)^2}\right)^2=0\]

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青青子衿

不知道能不能求出该封闭曲线的面积？
x

hbghlyj

Asymptote画图 $|OA|=1.6,|AC|=L=3.2,|AB|=R=6$
import graph;
unitsize(1cm);
real r = 1.6;
real L = 3.2;
real R = 6;

pair A = r*dir(60);
pair B = intersectionpoint(circle(A, R), (0,0)--(8,0));
pair C = intersectionpoint(circle(A, L), A--B);
draw((0,0)--A, orange);
draw(A--B, blue);
draw((0,0)--(9,0), Arrow, L=Label("$x$",EndPoint,E));
draw((0,0)--(0,3), Arrow, L=Label("$y$",EndPoint,N));
dot(A);dot(B);dot(C);
label("$A$", A, N);
label("$B$", B, S);
label("$C$", C, N);

pair f(real t) { return (r*sin(t) + (L/R)*sqrt(R^2 - r^2*cos(t)^2),(1 - L/R)*r*cos(t)); }
draw(graph(f, 0, 2*pi), dashed+red);

Mathematica数值计算面积为3.74618

1. L = 3.2;
2. R = 6;
3. Area[ParametricRegion[{r Sin[t] + (L/R) Sqrt[R^2 - r^2 Cos[t]^2], (1 -
4.        L/R) r Cos[t]}, {{t, 0, 2 Pi}, {r, 0, 1.6}}]]

复制代码

$L$从0增加到6的曲线族:
import graph;
unitsize(1cm);
real r = 1.6,L = 6,R=6;

draw((0,0)--(9,0), Arrow, L=Label("$x$",EndPoint,E));
draw((0,0)--(0,3), Arrow, L=Label("$y$",EndPoint,N));

for (;L>0;L-=0.2){draw(graph(new pair (real t) { return (r*sin(t) + (L/R)*sqrt(R^2 - r^2*cos(t)^2),(1 - L/R)*r*cos(t)); }, 0, 2*pi), red);}
$r$从0增加到4的曲线族:
import graph;
unitsize(1cm);
real r = 4,L = 3.2,R=6;

draw((0,0)--(9,0), Arrow, L=Label("$x$",EndPoint,E));
draw((0,0)--(0,3), Arrow, L=Label("$y$",EndPoint,N));

for (;r>0;r=r-.2){
  draw(graph(new pair (real t) { return (r*sin(t) + (L/R)*sqrt(R^2 - r^2*cos(t)^2),(1 - L/R)*r*cos(t)); }, 0, 2*pi), red);
}
$R$从1.6增加到3.2的曲线族:
import graph;
unitsize(1cm);
real r = 1.6,L = 3.2,R=1.6;

draw((0,0)--(9,0), Arrow, L=Label("$x$",EndPoint,E));
draw((0,0)--(0,3), Arrow, L=Label("$y$",EndPoint,N));

for (;R<3.2;R+=0.2){
  draw(graph(new pair (real t) { return (r*sin(t) + (L/R)*sqrt(R^2 - r^2*cos(t)^2),(1 - L/R)*r*cos(t)); }, 0, 2*pi), red);
}
$R$从3.2增加到6的曲线族:
import graph;
unitsize(1cm);
real r = 1.6,L = 3.2,R=3.2;

draw((0,0)--(9,0), Arrow, L=Label("$x$",EndPoint,E));
draw((0,0)--(0,3), Arrow, L=Label("$y$",EndPoint,N));

for (;R<6;R+=0.2){
  draw(graph(new pair (real t) { return (r*sin(t) + (L/R)*sqrt(R^2 - r^2*cos(t)^2),(1 - L/R)*r*cos(t)); }, 0, 2*pi), red);
}