Irregular polygons - Equal area under horizontal line


by jamnitzer
Tags: equal, horizontal, irregular, line, polygons
jamnitzer
jamnitzer is offline
#1
Mar11-12, 09:49 AM
P: 1
I've been trying to find an algorithm for finding a line passing through any 2-dimensional polygon that will divide the shape horizontally or vertically into two sections that each have the exact same area, or rather exactly half the area of the whole. I've prepared some simple examples below.

My question is if someone has heard of this or done something similar, and/or if anyone knows where to find more information.

Using the shoelace method of measuring area and/or after calculating the center of mass, I want to find a way to find either a horizontal line that will divide the top and bottom of a shape or a vertical line that will divide left and right such that the area of the shape on either side will be equal.

So the half-way line in terms of area, but only from one direction. I'm hoping there's already a name for this somewhere and I just haven't been able to find it. If this is more of a calculus question let me know. Really any information will be appreciated.


Some simple examples:

These are two shapes defined with their coordinates (x,y), repeating the first at the end.
{(4,0),(6,0),(6,2),(8,4),(4,4),(4,6),(2,4),(0,4),(4,0)} Area = 20
{(1,0),(1,1),(7,1),(4,0),(8,0),(8,1),(6,4),(4,2),(2,4),(6,6),(8,8),(0,8 ),(0,4),(1,0)} Area = 40.5

For the first shape, the vertical line evenly dividing left and right the area is x= 4.
The horizontal line dividing top and bottom is y= 4/sqrt(2).

In the second shape, the vertical line is x= -(sqrt(109)-20)/3
The horizontal line is y= (sqrt(2)sqrt(5.5)+6)/2
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