Basic question about coloring of a 3d plot

Hi,

I have the following question:

I have a 3D plot which shows the dependencies of 3 variables. One of these variables is the wavelength of light. Is it possible to have one specified color for each wavelength along the wavelength axis of my plot?

For example: if the wavelength equals 530 nm, then I would want the plot to be green for all points where wavelength = 530nm.

I am looking forward to, and appreciate your answers!

Kind regards, Marc

Answer 1

[Kelly Kearney]

Some more details about your plot would help... are you plotting points? Lines? Surfaces? A scatter plot could help with points; a patch with facecolor and/or edgecolor set to interp could help with lines or surfaces.

[Kelly Kearney]

It would be helpful if you could comment on a specific answer rather than adding a new answer yourself. But in response to your clarification...

The fourth input to mesh specifies the color to plot the surface. For example, the following colors the mesh according to the x-coordinate, which I'm assuming represents wavelength:

z = peaks;
[x,y] = meshgrid(linspace(400,750,49), 1:49);
mesh(x,y,z,x);
colorbar;

Getting your colormap to reflect the visible spectrum is a little trickier using Matlab's native colormaps. Using one of my own functions (available here) along with this color palette table will accomplish it pretty easily though:

cptcmap('visspec', 'mapping', 'direct');

[Marc]

Hi Kelly,

I am sorry for having posted an answer and not commented on yours. I am new to this forum and did not know about the difference between the two. In the future, I will use the commenting and answer options accordingly.

I tried what you posted, setting the wavelength axis as the fourth input to mesh, but I am only getting a "red" plot.

I will read through the links you provided and try with that information - thank you very much!

If I don't get it to work, I'll let you know and ask for more specific help.

Marc

Answer 2

[Marc]

Hi Kelly,

I am using mesh(x,y,z), if that is what you mean.

Marc