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How do I get a profile from a point cloud?

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Joshua Knicely
Joshua Knicely 2021 年 11 月 10 日
コメント済み: Lyhour Chhay 2022 年 1 月 23 日
I have a point cloud of data. It's a 3D scan of a cratering experiment. I want to get a bunch of profiles from the center of the crater to the edge of the data at multiple angles. This is so I can later extract some metrics about the crater.
My problem is that I have no idea how to get the profile. My first thought is to somehow transform the data from a bunch of points into a gridded data set, but even if I did that, I don't know how to get the profile extracted from the grid at arbitrary angles.
I'm not sure it matters, but I also need to later determine the volume of the displaced material from the crater. My thought with that is simply to subtract the original 3D surface (as a grid or mesh or whatever) from the cratered surface. I only bring this up as I imagine it could be important whether I transform the point cloud to a gridded data set or a mesh or something else I don't know about.
So, how would I go about getting a profile from a point cloud?

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Matt J
Matt J 2021 年 11 月 11 日
編集済み: Matt J 2021 年 11 月 12 日
Can the z-coordinates of the point cloud data be considered a function of (x,y)? If so, then,
F=scatteredInterpolant(x,y,z);
Xp=linspace(x1,x2,100); %profile x-coordinates
Yp=linspace(y1,y2,100); %profile y-coordinates
Zp=F(Xp,Yp); %profile z-coordinates
The volume underneath whichever surface can be computed with integral2,
underVolume=integral2(@(x,y) F(x,y) ,xmin,xmax,ymin,ymax))
  2 件のコメント
Joshua Knicely
Joshua Knicely 2021 年 11 月 11 日
That works fantastically! Thanks, Matt.
Lyhour Chhay
Lyhour Chhay 2022 年 1 月 23 日
Dear Matt J,
I see you again here. I hope that you remember me when we discuss about the rotation of point cloud. However, I am interesting in the question above. I use your suggestion coding in this forum. I can get the profile. However, I want to know that is it possible to calculate the distance from peak point to peak point ? I will show you in the figure. Thank you very much sir.

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