Finding the area, algebraic curve and jaggedness of an arbitrary shape

Question

I start with a photograph of a shape (physically made by the flow of a liquid into another), of which I can extract the border, manually or using Mathematica's feature detect feature :

Using the method described here, I extract some points of the shape, in a standard list form ({{x,y},{x,y}, ...}). Here is the example plot of such data :

How to :

1. Obtain the shape's area (area of the enclosed zone)
2. How to get an algebraic fit of the shape (and/or part of it)?
3. How to compare one shape against another, on their respective "jaggedness".

Question 3 is more of a mathematical question, but I'm just searching for an approximate comparison tool, more in the spirit of the following example than something absolute:

I expected CornerFilter to work, but it seems to give no result whatsoever. As for 2, I can fit small part of the curve using Fit[], but the general shape has multiple point with the same x, which forbid this.

Answer 1

You can use Mathematica's image processing functions for questions 1 and 3. Here's how:

1: Area

img = Binarize@Import["http://i.stack.imgur.com/I2gkK.jpg"] ~Erosion~ 1;
(m = MorphologicalComponents[img]) // Colorize

To get the area of the pink part in sq. pixels, use ComponentMeasurements:

2 /. ComponentMeasurements[{m, img}, "Area"]
(* 25168.1 *)

3: Jaggedness

Given two shapes with a similar area, the more jagged one will have a larger perimeter. So a simple way to approximate "jaggedness" would be to use the same function, ComponentMeasurements and compare the perimeters (provided the areas are similar)

2 /. ComponentMeasurements[{m, img}, "PerimeterLength"]
(* 1352 *)

If the areas are not the same, you could find the "EquivalentDiskRadius" (which gives you the radius of the circle with the same area as the shape) for each shape and see by what percentage the corresponding perimeter lengths are off from that of the respective circles.

Answer 2

For computing the area of a non-intersecting polygon, I propose:

polyarea = 
 Compile[{{v, _Real, 2}}, 
   Block[{x, y},
    {x, y} = Transpose[v]; 
    Abs[x.RotateLeft[y] - RotateLeft[x].y]/2
   ]
 ]

Example:

Needs["ComputationalGeometry`"]
poly = #[[ConvexHull @ #]] & @ RandomReal[100, {15, 2}];

Graphics[Polygon @ poly]

polyarea @ poly

5429.68