Distance between a point(x,y,z) and a surface(X,Y,Z)

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Aurea94
Aurea94 2021 年 3 月 22 日
編集済み: Ali Grysah 2021 年 4 月 2 日
I am trying to calculate the minimum distance between a point(x,y,z) and a surface defined by points (X;Y,Z). I now the normal to the point (x,y,z) from which I want to calculate distance, however I am not able to figure out the way of interpolating my surface so that the minimum distance is obtained, no matter the number of points in my surface.
Right now my option is to calcualte the distance from my point (x,y,z) to all the points in surface (X,Y,Z) and keep just the smallest one. However, I think there must be a better way of doing so.
Thank you very much for your help.
  3 件のコメント
Aurea94
Aurea94 2021 年 3 月 22 日
It should be D0 precise. For the moment with the simplification I am doing I know I am calculating D1.
Ali Grysah
Ali Grysah 2021 年 4 月 2 日
編集済み: Ali Grysah 2021 年 4 月 2 日
hi i need matlab 2008 windos7 /.bt32
can you help me,,and thunks

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採用された回答

darova
darova 2021 年 3 月 23 日
編集済み: darova 2021 年 3 月 23 日
THe best idea i have: refine close region using interp2 and just find closest distance (blue)
  1 件のコメント
Aurea94
Aurea94 2021 年 3 月 23 日
Thank you very much! I think it is the best option I can get!

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その他の回答 (1 件)

David Hill
David Hill 2021 年 3 月 22 日
Seems like a math question rather than a Matlab question. The minnum distance between a plane and a point is just the absolute value of the dot product between the point and the unit normal vector of the plane. The normal vector of the plane can be easily found with the 3 points given and is just the cross product between two vectors lying in the plane. For example:
p=[1 -2 0;0 -1 2;3 1 4];%three points given on the plane
d=diff(p);%two vectors lying in the plane
N=cross(d(1,:),d(2,:));
n=N/norm(N);%unit normal vector of plane
You do the rest
  3 件のコメント
Star Strider
Star Strider 2021 年 3 月 22 日
Perhaps the surfnorm functon could be helpful?
Aurea94
Aurea94 2021 年 3 月 23 日
Thats a good option I didn't knw t obtain surface normals! Thanks

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