Plotting implicit equation with fimplicit
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Hello
I have tried to plot this implicit equation. But when I tried it, the plot is showing empty.
Here is the code I used to plot. Could anyone help me with this.
Thanks in advance
clc
syms f(x,y)
n = 8;
a1 = 1.0086*y - 0.9216*(x - y);
b1 = 1.0107*(-x) - 1.0086*(y);
c1 = 0.9216*(x - y) - 1.0107*(-x);
h1 = 0.5877*(161.65);
I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6));
I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54));
th = acos(I3/(I2.^(3/2)));
v1 = ((2*th)+pi)/6;
an1 = (abs(2*cos(v1)))^n;
an2 = (abs(2*cos((2*th+3*pi)/6)))^n;
an3 = (abs(2*cos((2*th+5*pi)/6)))^n;
f(x,y) = ((3*I2).^(n/2)) * (an1 + an2 + an3) - (2*(189.32)^8);
fimplicit(f)
采纳的回答
Torsten
2019-2-11
0 个投票
function main
fimplicit (@(x,y)f(x,y))
end
function fun = f(x,y)
n = 8;
a1 = 1.0086*y - 0.9216*(x - y);
b1 = 1.0107*(-x) - 1.0086*(y);
c1 = 0.9216*(x - y) - 1.0107*(-x);
h1 = 0.5877*(161.65);
I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6));
I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54));
th = acos(I3./(I2.^(3/2)));
v1 = ((2*th)+pi)/6;
an1 = (abs(2*cos(v1))).^n;
an2 = (abs(2*cos((2*th+3*pi)/6))).^n;
an3 = (abs(2*cos((2*th+5*pi)/6))).^n;
fun = ((3*I2).^(n/2)).* (an1 + an2 + an3) - (2*(189.32)^8);
end
Resonable limits for plotting are required - no zeros are found in the default range [-5:5] for x and y.
6 个评论
Jay
2019-2-11
Stephan
2019-2-11
fsurf(@(x,y)f(x,y))
function fun = f(x,y)
n = 8;
a1 = 1.0086*y - 0.9216*(x - y);
b1 = 1.0107*(-x) - 1.0086*(y);
c1 = 0.9216*(x - y) - 1.0107*(-x);
h1 = 0.5877*(161.65);
I3 = ((a1.* b1.*c1)/(54))-((b1.*(h1.^2))/(6));
I2 = ((h1.^2)/(3))+((a1.^2 + b1.^2 + c1.^2)/(54));
th = acos(I3./(I2.^(3/2)));
v1 = ((2*th)+pi)/6;
an1 = (abs(2*cos(v1))).^n;
an2 = (abs(2*cos((2*th+3*pi)/6))).^n;
an3 = (abs(2*cos((2*th+5*pi)/6))).^n;
fun = ((3*I2).^(n/2)).* (an1 + an2 + an3) - (2*(189.32)^8);
end
Jay
2019-2-11
Torsten
2019-2-11
fimplicit (@(x,y)f(x,y),[-200 200 -150 150])
Torsten
2019-2-11
It "works" as long as the object is contained in the box defined by the specified limits for x and y.
更多回答（1 个）
John D'Errico
2019-2-11
编辑：John D'Errico
2019-2-11
Easy enough. Try this, for example.
vpasolve(f(1,y))
ans =
-80.224189505722446658042301607259
vpasolve(f(-20,y))
ans =
63.634253282860063957543062643774
Hmm. So [1,-80] is roughly a solution. That should be a good hint as to where to have fimplicit look.
fimplicit(f,[-150,150,-150,150])
axis equal
grid on
The problem was fimplicit looks by default in a rather narrow set of limits on x and y. It cannot know where it SHOULD be looking, and computer programs can sometimes be so clueless. Since fimplicit just found no solutions at all in the domain it was looking by default, you saw an empty figure. Sometimes you need to give even a computer a nudge in the right direction.
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