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 January 24th, 2015, 10:45 PM #1 Newbie   Joined: Jan 2015 From: Washington, United States Posts: 11 Thanks: 1 Perimeter of a Convex Hull I came upon this problem while working with one that I had been toying with for awhile: Is there an elegant/simple way to write an equation that sets the perimeter of the convex hull of 'n' points, where one of the points has coordinates (x,y) and the rest are constants, equal to a constant called 's'? There may be a couple of ways to do it as relating to computer algorithms for convex hulls, but I don't want to over complicate things Another way of putting it is that I'm looking for a way to do this: Per(Con(A))=s Where Per(m) returns the perimeter of a polygon 'm' Con(k) computes the convex hull of the set of points 'k' and 'A' is the set of 'n' points that I'm using.
 January 29th, 2015, 06:13 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The only way I can see is to calculate the individual side lengths and add. If one vertex is at (x, y) and we label the other vertices [math](a_1, b_1)[math] up to $\displaystyle (a_{n-1}, b_{n-1})$ counterclockwise from (x, y), so that the points $\displaystyle (a_1, b_1)$ and $\displaystyle (a_{n-1}, b_{n-1})$ are joined to (x, y), then the lengths of those two edges is $\displaystyle \sqrt{(x- a_1)^2+ (y- b_1)^2}$ and $\displaystyle \sqrt{(x- a_{n-1})^2+ (y- b_{n-1})^2}$. The distance between $\displaystyle (a_i, b_i)$ and $\displaystyle (a_{i+1}, b_{i+1})$ is, of course, $\displaystyle \sqrt{(a_i- a_{i+1})^2+ (b_i- b_{i+1})^2}$, for i= 1 to n- 2. You could do that in a computer routine by $\displaystyle S= sqrt((x- a_1)^2+ (y- b_1)^2)$ $\displaystyle S= S+ sqrt((x- a_{n-1})^2+ (y- b_{n-1})^2)$ for i= 1 to n- 2 do $\displaystyle S= S+ sqrt(sqrt{(a_i- a_{i+1})^2+ (b_i- b_{i+1})^2)$

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