PERIMETER OF A 3D POLYGON WHICH DOES NOT LIE IN THE X-Y PLANE

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Francesca Danielli
编辑： Torsten 2022-7-5
Hi everyone,
I have the spatial coordinates (x,y,z) of a cloud of points.
To date, I am able to calculate the area of the polygon in space obtained interpolating these points. Any suggestions abou how to calculate the perimeter?
Thank you in advance
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Francesca Danielli
Francesca Danielli 2022-7-5
yes
I used this function
h=fnplt(cscvn(p(:,[1:end 1])),'g',2)
where p is a 3x10 array (number of row 3 for spatial coordinates x,y,z and coloumn 10 for the available points)
It returns h, a 3xN array (with N number of additionally points for a better interpolation)
Bjorn Gustavsson
Bjorn Gustavsson 2022-7-5
@Torsten: In 3-D it isn't obvious that a cloud of points are flat. There would still be a "natural" area from the convex hull, but I wasn't sure there would be a unique perimeter. But now I see that the points seem to lie in a plane - or near enough.

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Torsten
Torsten 2022-7-5
Ran in:
The code should apply to your case:
x = 3*cos(0:pi/20:2*pi);
y = 3*sin(0:pi/20:2*pi);
fn = cscvn([x;y]);
% Compute enclosed area
fnprime = fnder(fn);
Kofs = @(s) [1 -1]*(fnval(fn,s) .* flipud(fnval(fnprime,s)));
A = 1/2*integral(Kofs,fn.breaks(1),fn.breaks(end))
A = 28.2726
pi*3^2
ans = 28.2743
% Compute curve length
Lfun = @(s) sqrt(sum(fnval(fnprime,s).^2,1));
L = integral(Lfun,fn.breaks(1),fn.breaks(end))
L = 18.8491
2*pi*3
ans = 18.8496
  2 个评论
Francesca Danielli
Francesca Danielli 2022-7-5
I tried and it works for the perimeter! I compared the obtained results with an expected one from raw analytical calculations!
I am currently having problems with the calcualtion of the area you suggested, but I am trying to face it! In anycase, I have another method for the area calculation!
Thanks so much for your help
Torsten
Torsten 2022-7-5
编辑：Torsten 2022-7-5
I just noticed that the area calculation only works for closed curves in the x-y plane.
For closed curves in 3d lying in a plane, things will become more complicated.
But you said that you have already determined the enclosed area :-)

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