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

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

[Bjorn Gustavsson]

What perimeter do you want? If the points form a "northern half-sphere" we can select the equatorial boundary as a perimeter, or one curve over the pole and across the bottom. If you have "a natural" normal-vector for your point-cloud it seems to me that the sensible way to proceed would be to transform your coordinates to a system with one coordinate parallel to that normal-direction and two coordinates perpendicular. Then your perimeter might be the one in the 2 perp coordinates.

[Torsten]

@Bjorn Gustavsson

Isn't the perimeter just the sum of the side lengths of the polygon ?

[Francesca Danielli]

Maybe the image could help.

I have the spatial coordinates (black dots) and I was able to calculate the area contained in the interpolant (green line). But I am facing problems with the perimeter

[Torsten]

So you want the perimeter of the green line, not of the polygon ?

If yes: How is the green line given ? By a parametric equation ?

[Francesca Danielli]

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]

@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.

Answer 1

[Torsten]

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

[Francesca Danielli]

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]

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 :-)