2D Section of 3D coordinates

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fauer781
fauer781 2016 年 12 月 14 日
コメント済み: fauer781 2016 年 12 月 15 日
Hello, I have a 3D plot and the x y z coordinates of corresponding vertices as vectors n:1, m:1 and p:1. There is no function f(x,y,z). From theses vertices I want to get a 2D profile of an arbitrary sectionplane. It's possible that the points are not part of the plane. Is there a function for this problem?

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Matt J
Matt J 2016 年 12 月 14 日
編集済み: Matt J 2016 年 12 月 14 日
I think you can do it all just by using coordinate transformations. Transform your x,y,z coordinates so that the origin and the x,y axes lie in the plane of interest. Then throw away all (x,y,z) data whose distance from the plane (as measured by the new z coordinates) are too large in magnitude. Then, make a 2D plot of the (x,y) in the data that remain.
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fauer781
fauer781 2016 年 12 月 15 日
Thank you, I will try that. It's a good idea.

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Kelly Kearney
Kelly Kearney 2016 年 12 月 14 日
The slice function may be of use here.
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fauer781
fauer781 2016 年 12 月 14 日
Hello, thank you for answering. I checked the slice function, but I don't have the m-by-n-by-p volume array. Only the x,y,z positions are available. How can I get the volume array with this information? There is no function depending on x,y,z.

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