Question

126

Calculating Polygon Area

rated 0 times [ 133] [ 7] / answers: 1 / hits: 17335 / 11 Years ago, mon, april 29, 2013, 12:00:00

So I've gotten this code in javascript to calculate irregular polygon area from the net.

function polygonArea(X, Y, numPoints)  

{    

area = 0;  // Accumulates area in the loop   

j = numPoints-1;  // The last vertex is the 'previous' one to the first



  for (i=0; i<numPoints; i++)

  { area = area +  (X[j]+X[i]) * (Y[j]-Y[i]); 

      j = i;  //j is previous vertex to i

  }   

  return area/2; 

}



var xPts = [3, 3, 2, 2, 3, 3, 6, 6, 9, 9, 4, 4 ];

var yPts = [2, 4, 4, 5, 5, 6, 6, 5, 5, 3, 3, 2];



var a = polygonArea(xPts, yPts, 4); 

alert(Area  =  + a);

The results seems to be correct. if the vertex traced by clock-wise direction, it will shows positive results however it will become negative if i traced vertex in anti clockwise direction. Why is that so?

How does this algorithm work? i really want to know what is the mathematical explanation behind it, because i am still having a hard time to understand the explanation on the net.

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brookelynshaem

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2 Years ago

answered 11 Years ago raymondd · Accepted Answer

Imagine drawing horizontal lines from each vertex to the Y axis; for each edge, this will describe a trapezoid:

Y-axis

^

|

|--------o (X[j], Y[j])

|         

|          

|           

|------------o (X[i], Y[i])

|

+----------------------------> X-axis

The formula (X[j]+X[i]) * (Y[j]-Y[i]) in the inner loop computes twice the area of this trapezoid if Y[i] <= Y[j], or negative twice the area if Y[i] >= Y[j].

For a closed polygon, this naturally subtracts the area to the left of the upgoing edges from the area to the left of the downgoing edges. If the polygon is clockwise, this neatly cuts out the exact (doubled) area of the polygon; if counterclockwise, you get the negative (doubled) area.

To compute the area of a given polygon,

Y-axis

^

|

|        o------o

|        |       

|        |        

|        o         

|                  o                  

|                 /

|                /

|               /

|              /

|              o

|

+-------------------------> X-axis

take the downgoing area:

Y-axis

^

|

|--------o------o

|                

|                 

|        o         

|                   o                  

|                  /

|                 /

|                /

|               /

|--------------o

|

+-------------------------> X-axis

minus the upgoing area:

Y-axis

^

|

|--------o      o

|        |

|        |

|        o

|                  o                  

|          

|           

|            

|             

|--------------o

|

+-------------------------> X-axis

Although the above example uses a convex polygon, this area computation is correct for arbitrary polygons, regardless of how many up- and down- paths they may have.