plotting a simple Graph

Question

Shimon Katzman 2019-12-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/497944-plotting-a-simple-graph

评论： Star Strider 2019-12-25

采纳的回答： Star Strider

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Hi everyone,

Trying to plot a graph unsucssesfully :((

alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Mx=linspace(0,50,0.0001);
Interaction_Curve=(Mx./Mx0).^alpha+(My./My0).^alpha-1
plot(Interaction_Curve,Mx)

Thank You Very much

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madhan ravi 2019-12-24

You didn’t define My and you haven’t used the linspace() properly for Mx.

Shimon Katzman 2019-12-24

编辑：Shimon Katzman 2019-12-24

So what is the correct way?

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Answer 1

Star Strider 2019-12-24

1
链接

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https://ww2.mathworks.cn/matlabcentral/answers/497944-plotting-a-simple-graph#answer_407637

在 MATLAB Online 中打开

One problem is that ‘My’ is missing. Beyond that, the linspace arguments resulted in an empty vector for ‘Mx’.

It might be easier to plot this as an implicit function:

alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
figure
fimplicit(Interaction_Function, [0 50  0 30])
ylim([0 30])

producing:

1plotting a simple Graph - 2019 12 24.png

That seems to produce the sort of plot you want. Make appropriate changes to the second agrument in the fimplicit call to get the result you want.

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Star Strider 2019-12-24

在 MATLAB Online 中打开

Not immediately obvious, so it took some time.

Try this:

alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Mx=linspace(0,50,25);
My=linspace(0,50,25);
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
MxdivMy = @(Mx,My) Mx-3*My;
[MX,MY] = meshgrid(Mx,My);
figure
hc1 = contour(Mx, My, Interaction_Function(MX,MY), [0 0])
hc2 = contour(Mx, My, MxdivMy(MX,MY), [0 0])
xy1 = hc1(:,2:end);                                             % ‘Interaction Function’
xy2 = fliplr(hc2(:,2:end));                                     % ‘MxdivMy’
[xy2u,ix] = unique(xy2(1,:), 'stable');                         % Unique ‘MxdivMy’
xy2i = interp1(xy2u(1,:), xy2(2,ix), xy1(1,:));                 % Interpolate To Same x-Coordinates
xy2i = [xy1(1,:); xy2i];
xint = interp1(xy1(2,:)-xy2i(2,:), xy1(1,:), 0);                % Interpolate To Find X-Intercept
yint = interp1(xy1(1,:), xy1(2,:), xint);                       % Interpolate To Find Y-Intercept
alpha=2.2;
% Mx0=M_x;
% My0=M_y;
Mx = linspace(0,50, 1000);
My = linspace(0,50, 1000); 
Interaction_Function = @(Mx,My) (Mx./Mx0).^alpha+(My./My0).^alpha-1;
figure
hold on
ez1 = ezplot(Interaction_Function,[0 40  0 20]);
MxdivMy = @(Mx,My) Mx-3*My;
ez2 = ezplot(MxdivMy, [0 40  0 20]);
plot(xint, yint, 'pr', 'MarkerSize',10, 'MarkerFaceColor','g')  % Plot Intercept
grid on;
hold off

That calculates and plots the intercept.

Shimon Katzman 2019-12-25

Wow Star. Thank you so much!!

You Are The Best!

Star Strider 2019-12-25

As always, my pleasure!

I very much appreciate your compliment!

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Answer 2

Image Analyst 2019-12-24

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/497944-plotting-a-simple-graph#answer_407634

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Try this:

alpha = 2.2;
Mx0 = 34.8262;
My0 = 15.7563;
Mx = linspace(0,50, 1000);
My = linspace(0,50, 1000);  % Not sure what My should be!!!
Interaction_Curve = (Mx./Mx0).^alpha+(My./My0).^alpha-1
plot(Mx, Interaction_Curve, 'b-', 'LineWidth', 2)
grid on;

Be sure to define My because I just guessed incorrectly.

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Shimon Katzman 2019-12-24

Hi, it doesnt plot the right graph :(

Image Analyst 2019-12-24

I know. Because I don't have the value of the My variable. That's why I asked you to define it. What is it? But doesn't matter since it looks like Star figured it out.

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