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

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Francesca Danielli on 5 Jul 2022
Edited: Torsten on 5 Jul 2022
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?
Bjorn Gustavsson on 5 Jul 2022
@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.

Torsten on 5 Jul 2022
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 CommentsShowHide 1 older comment
Torsten on 5 Jul 2022
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 :-)