Is colormap have relation with mm?
Show older comments
Can somebody explained to me what is the colormap value stand for?

Fig.A 3D plot converted from 2D image
Answers (1)
KSSV
on 23 Jul 2020
0 votes
The plot is obtained using surf. To plot you need (X,Y) location matrices and a matrix of values Z. The colorbar shows the distribution of your values. For example, in your plot the red color shows the value of >220. So where ever you have red color in the plot you have values > 220. Blue color corresponds to values < 20, so wherever you have blue color in the plot, your values are < 20.
14 Comments
agung pratama
on 23 Jul 2020
Edited: agung pratama
on 23 Jul 2020
KSSV
on 24 Jul 2020
The colormap have relation with the distribution of Z matrix.....thats all. How this mm came into picture?
agung pratama
on 24 Jul 2020
Edited: agung pratama
on 24 Jul 2020
KSSV
on 24 Jul 2020
Yes...it has relation with objection height too.
agung pratama
on 24 Jul 2020
Image Analyst
on 24 Jul 2020
Why don't you ask the paper authors for their MATLAB code?
agung pratama
on 18 Aug 2020
Walter Roberson
on 18 Aug 2020
By default, the colorbar reflects the range of Z values that you have, and has no independent meaning. Your data is roughly 0 to 230 or 240, and the colorbar is the same (rounded up to the nearest 20).
Your colorbar does not itself mean mm: it just means that your data runs to whatever value.
If your data having value 200 means 200 mm then the colorbar color for 200 gives you information about how 200 mm appears in your plot.
If your data having value 200 means 200 microvolts of piezoelectric effect voltage generated due to strain, then the color associated with 200 in the colorbar relates to piezoelectric microvolts.
surf and colorbar do not know what real-world units your data refer to. 200 mm, 200 microvolts, 200 milliradians of twist, 200 lux of photon emissions triggered by strain... surf and colorbar do not know.
Any interpretation of 200 is beyond the scope of MATLAB and depends on the calculations and the data meaning.
Walter Roberson
on 18 Aug 2020
The z value looks like it might be
intensity = rgb2gray(TheImage) ;
z = intensity;
surf(z)
If I am correct then the plot is only Brightness encoded as uint8, and the 240ish upper limit has no direct relationship to physical units and just reflects RGB encoding. If the physical brightness had been half as much they probably would have opened the lens another f-stop and ended up with the same image.
agung pratama
on 18 Aug 2020
Walter Roberson
on 18 Aug 2020
You cannot.
agung pratama
on 18 Aug 2020
Edited: agung pratama
on 18 Aug 2020
Walter Roberson
on 18 Aug 2020
"But I see some paper using interferometer, codded light pattern, to measure the deflection of a metalic plate."
You do not have an interferometer image. Your image is some light projected on the underside of an object, and you have a clearer circular membrane or possibly lens in the material, and that more transparent area has a circular black rim.
The highest point in the image is somewhere related to the disk, but at the moment I cannot tell if the clearer area is slightly convex or concave. I suspect that the highest point might be the black rim itself.
You are looking through the center area towards whitish light. The orange reflects the result of the white light transmitted through the material (which is perhaps brown colored).
This is not a side view of an object. This is not light shining on a raised area. Look at how the black circle interrupts the bars: that is because the light is from the back and the black is blocking it.
The color information gives little information about height here. With constant brightness of light you could probably use the color of the orange lines to deduce something about the thickness of the material, since thicker material would block more of the light. But that would be about thickness, not about height.
agung pratama
on 19 Aug 2020
Categories
Find more on Color and Styling in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!