plotting a simple Graph
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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
2 个评论
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
采纳的回答
Star Strider
2019-12-24
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:
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.
9 个评论
Shimon Katzman
2019-12-24
Shimon Katzman
2019-12-24
Star Strider
2019-12-24
My pleasure.
The contour function is another option.
Try this:
alpha=2.2;
Mx0=34.8262;
My0=15.7563;
Mx=linspace(0,50,25);
My=linspace(0,50,25);
[MX,MY] = meshgrid(Mx,My);
figure
contour(Interaction_Function(MX,MY), [0 0])
That should work. The contour function has been around as long as I can remember. It should allow the definition of the single contour at [0 0] in R2015a. (This imay be what the fimpllicit function does as well. It appears to produce the same plot.)
Shimon Katzman
2019-12-24
Star Strider
2019-12-24
As always, my pleasure!
I do not have ‘M_x’ and ‘M_y’, so I cannot run your latest code.
Shimon Katzman
2019-12-24
Star Strider
2019-12-24
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
Star Strider
2019-12-25
As always, my pleasure!
I very much appreciate your compliment!
更多回答（1 个）
Image Analyst
2019-12-24
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.
2 个评论
Shimon Katzman
2019-12-24
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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