How to plot an Ellipse
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I want to plot an Ellipse. I have the verticles for the major axis: d1(0,0.8736) d2(85.8024,1.2157) (The coordinates are taken from another part of code so the ellipse must be on the first quadrant of the x-y axis) I also want to be able to change the eccentricity of the ellipse.
1 个评论
muhammad arfan
2019-6-18
dears!!!
i have asigned to write a matlab code for 8 point to fit it in ellipse by using least square method..
i am new in using matlab and try my best but my points are not fit on ellipse. i use annealing method so that i have satisfied my teacher by my work. please chk my work and help me.
thanks
arfan khan
clc;
clear all;
close all;
r1 = rand(1);
r2 = [1+rand(1)]; % r2>r1
x0 = 0;
y0 = 0;
N = 8;
n= 100;
x1 = 1;
x2 = 2;
y1 = 1;
y2 = 2;
for i = 1:n
x = x1 +(x2-x1).*rand(N,1);
y = y1 +(y2-y1).*rand(N,1);
f = ((((x./r1).^2) +(y./r2).^2)-1).^2;
[m,l] = min(f);
z =.001* exp(10*(1-i/n));
v = z/2;
% disp('v');
% disp(v)
x1 = x(l)*v;
x2 = x(l)*v;
% disp('x1')
% disp(x1)
% disp('x2')
% disp(x2)
% ay = v./y(l);
% by = v./y(l);
% disp(v);
%
% % hold on;
disp('f');
disp(f);
end
% plot(f,'or')
plot(x,y, '*b');
x=((x(i)-x0)*cos(z)) - ((y(i)-y0)*sin(z))
y=(x(i)-x0)*sin(z)-(y(i)-y0)*cos(z)
xa(i)=rand(1)
x(i)= a+(b-a)*rand(1);
y(i)= rand(1);
for
m(i) = ((((x).^2)/a^2) + (((y).^2)/b^2)-1).^2
end
hold on;
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更多回答(5 个)
Azzi Abdelmalek
2013-9-8
编辑:Azzi Abdelmalek
2015-6-12
a=5; % horizontal radius
b=10; % vertical radius
x0=0; % x0,y0 ellipse centre coordinates
y0=0;
t=-pi:0.01:pi;
x=x0+a*cos(t);
y=y0+b*sin(t);
plot(x,y)
3 个评论
Ivailo Ivanov
2016-1-14
It was very simple and comprehensible.
Cynthia Dickerson
2018-6-27
Thanks! This code worked for me perfectly. :)
Sandy M
2019-7-27
Hi, I do my ellipse graph
A=10;
B=7.5;
X=-10:.1:10;
Y=(7.5/10)*(1-x^2)^(1/2)
z=-(7.5/10)*(1-x^2)^(1/2)
Plot(x,y,x,z)
Its ok but i need it in cm units cause if i change properties of figure and paper to cm i get deference’s about 3 or 5 mm How can I justify the unit
Kommi
2022-11-24
1 个投票
For ellipse
>> clc
clear all
%length of major axis
a = input('please enter the length of major axis: ');
b = input('please enter the length of minor axis: ');
x1 = input('please input the x coordinate of the ellipse: ');
y1 = input('please enter the y coordinate of the ellipse: ');
t = -pi:0.01:pi;
x = x1+(a*cos(t));
y = y1+(b*sin(t));
plot(x,y);
Kate
2014-2-24
0 个投票
how would you plot a ellipse with only knowing some co-ordinates on the curve and not knowing the y radius and x radius?
4 个评论
Roger Stafford
2014-2-24
That involves more work, Kate. I'll show you a first, easy step and let you struggle through the rest.
It requires a minimum of five points to uniquely determine a conic section - that is, an ellipse, a parabola, or a hyperbola. One way to approach the problem is to find a set of values A, B, C, D, E, and F such that
A*x^2 + B*x*y + C*y^2 + D*x + E*y + F
is as close to zero as possible for the points (assuming the six constants have been normalized by having the sum of their squares be unity.) With five points it should be essentially zero for each of the points. For more points, it depends on how close they all are to a valid conic section. If they actually lie on some conic, then with the appropriate constants A, B, etc., the above would also be zero at each point.
These constants can be found using matlab's 'svd' singular value decomposition function. Let x be a column vector of all the points' x-coordinates and y a column vector of their corresponding y-coordinates. Do this:
[U,S,V] = svd([x.^2,x.*y,y.^2,x,y,ones(size(x))]);
Then the desired constants will be in the vector V(:,6), namely, A = V(1,6), B = V(2,6), C = V(3,6), D = V(4,6), E = V(5,6), and F = V(6,6). These are associated with the smallest singular value in S. If that value is zero, then there is a precise match with all points.
Probably the easiest way to proceed from here on in plotting the conic section, would be to solve for y as a function of x, (or possibly x as a function of y,) in the quadratic equation
A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0
The type of conic section is determined by the discriminant, B^2-4*A*C. If it is negative, the curve is an ellipse, if it is zero, the curve is a parabola, and if it is positive, it is a hyperbola. Recalling high school algebra, there will in general be two y solutions or none for each x, (and two x solutions or none for each y,) using the famous quadratic formula.
Making a successful plot would require a determination of the range of x (or of y) for which there are solutions and then generating a set of x values (or y values,) maybe with linspace, and making two separate plots of the two possible solutions for y (or for x).
Another approach to handling such a quadratic equation involves making the appropriate translation and rotations so as to place the equation in a standard form from which plots can readily be made. This, however, probably involves more effort than the above, though it is more informative.
Image Analyst
2014-2-25
Perhaps of some interest, if you need to find the ellipse points: http://www.ecse.rpi.edu/homepages/qji/Papers/ellipse_det_icpr02.pdf
Devi Satya Cheerla
2015-6-12
编辑:Walter Roberson
2022-8-26
in the equation of ellipse X2/a2 + Y2/b2 = 1. knowing the points on ellipse, can find a and b. then enter the code below to mathematically compute y and to plot x,y.
code:
x=(0:.01:a); # x value is from 0 to 'a' and discrete with 0.01 scale#
i=1:(a*100+1); # i is to calculate y at every discrete value. it should be for 1 i.e first x value to the last x value.. as it does not have a zero, add 1#
clear y # to clear any previous y value#
for i=1:(a*100+1)
y(i)=(b^2*(1-(x(i)^2)/a^2))^.5; #from the ellipse equation y=sqrt(b2(1-(x2/a2))#
end
plot(x,y)
hold on
plot(x,-y)
hold on
plot(-x,y)
hold on
plot(-x,-y)
Sandy M
2019-7-27
hi why u product the nmber with 100?
and, if i want the graph with cm units, what i do? cause i change garaph and paper properties but i still defreces about 4 mm when i prented it
Omar Maaroof
2019-5-13
0 个投票
you can use
Ellipse2d
1 个评论
Walter Roberson
2019-5-13
MATLAB does not offer Ellipse2d plotting directly. Instead, the Symbolic Toolbox's engine, MuPAD, offers plot::Ellipse2d https://www.mathworks.com/help/symbolic/mupad_ref/plot-ellipse2d.html which can only be used from within a MuPAD notebook . R2018b was intended to be the last release that included the MuPAD notebook, but it was carried on to R2019a as well.
Using this FEX download,
[x1,y1,x2,y2, e]=deal(1,2,20,8 ,0.85); %hypothetical input
a = 1/2*sqrt((x2-x1)^2+(y2-y1)^2);
b = a*sqrt(1-e^2);
center=[x1+x2,y1+y2]/2;
theta=atan2d(y2-y1,x2-x1); %rotation angle
obj=ellipticalFit.groundtruth([], center,[a,b], theta);
plot(obj); hold on;
plot([x1,x2],[y1,y2],'xk'); hold off;
axis padded

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