Shade the region between a curve and the vertical axis

I have several velocity profiles in a figure, and I've been trying to play around with the area function to try and shade the region between the curve a vertical line. I'm looking to create a plot like this.

Here is the relavent parts of my code.

% V is the dependent variable, plotted on the x-axis
% Y is the independent variable, plotted on the y-axis
% My actual data is not attached but you can use tan(x) -1<x<1 or tan^-1(x) mirrored about y = x
Y = linspace(-1,1,50);
V = atan(Y);
% R is an array for the radial positions along the wing, the zero location has to move with each radial position
R = 1:4;    % for this case
% plot data
plot(V+(R(1)),Y);    % only using the first R value for this example
% draw vertical line for reference
plot(R(1)*ones(1,50),Y,'k')

Now if I try to shade in the region, let's say from a base of 0 this is the result.

area(V+R(1),Y,1,'facealpha',0.5)

As expected the area function fills in the region from Y = 1 to V. I tried switching X and Y as some have suggested, but I don't get the results I would like. The area function does what's it's supposed to and shades in the region up to V = atan(X)+1 from Y = 1.

I would like to shade along the X axis (instead of Y) up to the curve.

This last figure shows what I am actually trying to accomplish but with my actual data instead of with V = atan(Y) as I did above.

Answer 1

[Star Strider]

I am not certain what you want.

Try this:

Y = linspace(-1,1,50);
V = atan(Y);
% R is an array for the radial positions along the wing, the zero location has to move with each radial position
R = 1:4;    % for this case
% plot data
plot(V+(R(1)),Y);    % only using the first R value for this example
% draw vertical line for reference
hold on
plot(R(1)*ones(1,50),Y,'k')
Vidx = (V+R(1)) > 1;
Vones = ones(size(V));
patch([V(Vidx)+R(1) fliplr(V(Vidx)+R(1))], [Y(Vidx) Vones(Vidx)], [0.1 0.3 0.5], 'FaceAlpha',0.3)   % ‘patch’ Call
hold off

It uses the patch function to define the area I believe you want to shade. Increase the number of elements in ‘Y’ to increase the resolution of the patch object.

EDIT —

Added plot image —

[Nathaniel H Werner]

No I don't think this is what I am looking for. But I understand what you did and I think I may be able to work with it.

I can define the region I want to be shaded like this. I apologize if I was unclear in the original question. Recall the x-axis is the dependent variable and it "restarts" from zero at each black line, so I will be calling it the V-axis. The y-axis is the independent variable but actually represents the vertical height so I will still call it the y-axis.

(y-axis)

(x-axis, 0 restarts at each black line which is why I use V+R instead of V)

You can see why the code you provided does not work if I make the following changes.

Yidx = (Y>-0.4 & Y<0.6);
patch([V(Yidx)+R(1) fliplr(V(Yidx)+R(1))], [Y(Yidx) Vones(Yidx)], [0.6 0.5 0.2], 'FaceAlpha',0.3)   % ‘patch’ Call

If I limit the values in y-axis between -0.4 and 0.6 I get the second region. The y-axis should be at the black line at 1, because that is where V = 0. The blue area I've filled in by hand corresponds to what I am looking for in this case. In this case V(y) has a region on the left and right of 1.

[Star Strider]

The initial description was confusing.

Try this:

Y = linspace(-1,1,50);
V = atan(Y);
% R is an array for the radial positions along the wing, the zero location has to move with each radial position
R = 1:4;    % for this case
% plot data
plot(V+(R(1)),Y);    % only using the first R value for this example
% draw vertical line for reference
hold on
plot(R(1)*ones(1,50),Y,'k')
Vidx1 = ((V+R(1)) >= 1) & (Y <= 0.6);
Vidx2 = ((V+R(1)) < 1) & (Y >= -0.4);
Vones = ones(size(V));
patch([V(Vidx1)+R(1) fliplr(V(Vidx1)+R(1))], [Y(Vidx1) Vones(Vidx1)*0.6], [0.1 0.3 0.5], 'FaceAlpha',0.3)       % ‘patch’ Call
patch([V(Vidx2)+R(1) fliplr(V(Vidx2)+R(1))], [Y(Vidx2) Vones(Vidx2)*(-0.4)], [0.5 0.3 0.5], 'FaceAlpha',0.3)    % ‘patch’ Call
plot(xlim, [1 1]*0.6, '--r')                                                                                    % Dashed Horizontal Red Line
plot(xlim, [1 1]*(-0.4), '--r')                                                                                 % Dashed Horizontal Red Line
hold off

That appears to be closer to what you just now described:

The patch calls are straightforward. The key to defining them are ‘Vidx1’ and ‘Vidx2’, both being ‘logical index’ vectors. To understand how they work, see the documentation on Matrix Indexing.

I am not certain what the dashed red lines indicate. I added them here. To delete them, just eliminate the two plot calls that create them.

[Nathaniel H Werner]

I see, yes this would be what I am looking for if V has both negative and positive regions at the specific black line representing the y-axis. I had thought about using two patches but I wasn't sure if there was a better way.

If V does not cross the axis (V > 0 for all Y in the grey region from the original question) I can use one patch. I will try to eliminate the white region around the black line by using <= and refining the grid.

The red lines were to show where y = -0.4, and 0.6 were. They are not intended to be in final plot.

[Star Strider]

I was not able to do that with one patch call because the conditions defining the two regions are not the same. I doubt that is possible with one patch call.

[Nathaniel H Werner]

I just spent a few hours trying this out, and I wasn't able to get it to match what I wanted. I ended up just making a bunch of patches in the for loop after identifying the appropriate indices. It slows matlab down a bit but it works. Should I leave that as another answer to the question, or unaccept yours first since it really doesn't seem to work.

[Star Strider]

My Answer works with the problem you presented.

I have no idea what you want, since the problem you presented is apparently a proxy for the actual problem you want to solve.

I keep the Reputation Points regardless of your Unaccepting my Answer or my deleting it. I suspect that if others felt comfortable posting an Answer, they would have, since that’s the way Answers works. Posting the actual problem is likely the key to getting a more efficient solution to it (if one exists).

[Nathaniel H Werner]

That makes sense. I'll provide the solution I found as an answer later when I have time. Thank you for your help, and I apologize for the confusion. This is a surprisingly difficult problem.

[Star Strider]

As always, my pleasure.

While I would like to see the actual problem, itwould be best to post a detailed description of the problem — and your solution (and all other necessary code) — as a new Question if you want specific help with it.

It would be best to add a Comment with a link to the new Question here, for continuity.