MATLAB Answers

## Plotting Multiple Curves on the Same Graph

Asked by Hollis Williams

### Hollis Williams (view profile)

on 16 Jul 2019
Latest activity Commented on by Star Strider

### Star Strider (view profile)

on 18 Jul 2019
Accepted Answer by Star Strider

### Star Strider (view profile)

I am plotting a graph of radial velocity against varying radius, but there is also a parameter in the velocity expression similar to the Reynolds number which can have various values, so I would like to set the value of the parameter to 0, 1 and 0.2 and to plot three curves on the same graph, what would be the easiest way to do this?

Adam

### Adam (view profile)

on 16 Jul 2019
doc hold
Set e.g.
hold( hAxes, 'on' )
for your axes with handle hAxes, then you can add as many plots as you like.
Alternatively if you have all the results together in a matrix (assuming they are the same length for each parameter) you can just plot the matrix all in one go using
doc plot
and making sure you have it oriented the right way.

Sign in to comment.

## 1 Answer ### Star Strider (view profile)

Answer by Star Strider

### Star Strider (view profile)

on 16 Jul 2019
Accepted Answer

It depends on your function and what you want to do:
t = linspace(0, 2*pi);
freq = [1 5 9];
ampl = [1 2 3];
s = bsxfun(@times, ampl(:), sin(freq(:)*t));
figure
plot(t, s)
grid

Star Strider

### Star Strider (view profile)

on 17 Jul 2019
... so are you sure that the y-axis is the function ur?
The original version of ‘ur’ that you provided had parentheses around ‘bp’ and in creating the function version of it I got confused by the extra parentheses:
ur = (1 + (bp)/R - (1 + bp)/R.^3);
since I was not sure where the parentheses belonged. However, that also resulted in the function being different.
Try this corrected version:
bp = @(Kn) -(60*pi + 345*pi*Kn + 450*(1+pi)*Kn.^2)./(40*pi + 270*pi*Kn + 18*(25*pi + 18)*Kn.^2 + 648*Kn.^3);
ur = @(R,Kn) (1 + bp(Kn)./R - (1 + bp(Kn))./R.^3);
Knv = [0, 0.2, 1];
Rv = 1:10;
[Rm,Knm] = meshgrid(Rv,Knv);
figure
plot(Knv, ur(Rm,Knm))
grid
xlabel('K_n')
lgdc = sprintfc('R = %2d', Rv);
legend(lgdc, 'Location','SE')
figure
plot(Rv, ur(Rm,Knm))
grid
xlabel('R')
lgdc = sprintfc('K_n = %3.1f', Knv);
legend(lgdc, 'Location','SE')
figure
surfc(Rm,Knm,ur(Rm,Knm))
grid on
xlabel('R')
ylabel('K_n')
zlabel('ur')
This appears to do what you want.
Hollis Williams

### Hollis Williams (view profile)

on 18 Jul 2019
Many thanks for your answer, I have tried something slightly different but getting a bit of an unexpected error message. So now I am trying:
bt = @(Kn) -sqrt(pi/2)*(30*Kn.^2 + 180*(1+(1./pi))*Kn.^3)./(20*pi + 135*pi*Kn + 9*(25*pi +18)*Kn.^2 +324*Kn.^3);
theta = @(R,Kn) (bt./(R.^2));
Knv = [0, 0.2, 1];
Rv = 1:10;
[Rm,Knm] = meshgrid(Rv,Knv);
figure
plot(Rv, theta(Rm,Knm))
grid
%xlabel('R')
lgdc = sprintfc('K_n = %3.1f', Knv);
legend(lgdc, 'Location','SE')
I seem to get an error message related saying that the ./ in the expression for theta is an undefined operator for input arguments of type 'function_handle', although I have used it elsewhere with no problems.
Star Strider

### Star Strider (view profile)

on 18 Jul 2019
As always, my pleasure!
Remember that in my code, ‘bt’ is a function, so it needs to be evaluated in any calculation using it, and that requires that it be supplied with a numeric argument so that it can return a numeric result:
theta = @(R,Kn) (bt(Kn)./(R.^2));
The complete code is now:
bt = @(Kn) -sqrt(pi/2)*(30*Kn.^2 + 180*(1+(1./pi))*Kn.^3)./(20*pi + 135*pi*Kn + 9*(25*pi +18)*Kn.^2 +324*Kn.^3);
theta = @(R,Kn) (bt(Kn)./(R.^2));
Knv = [0, 0.2, 1];
Rv = 1:10;
[Rm,Knm] = meshgrid(Rv,Knv);
figure
plot(Rv, theta(Rm,Knm))
grid
%xlabel('R')
lgdc = sprintfc('K_n = %3.1f', Knv);
legend(lgdc, 'Location','SE')
I tested that to be sure it works. It does.

Sign in to comment.