Math question: How to compute the area of polygons

[Jim Hurley]

>
>  > Ben Rubinstein               |  Email: benr_mc at cogapp.com
>>  Cognitive Applications Ltd   |  Phone: +44 (0)1273-821600
>>  http://www.cogapp.com        |  Fax  : +44 (0)1273-728866
>>
>
>Here is a little routine I have written that will accurately calculate the
>area of irregular polygons so long as there are no crossed lines. You can
>have irregular shapes as funky as you like and this will work -- provided
>there are no lines which cross.
>
>The math behind this is fascinating.
>
>Area returned is in pixels. If you wish to use units as square cm, square
>inches, etc. it can be appropriately adjusted.
>
>function IrPolyArea i
>   put the points of graphic i into k
>   put 0,0 into n
>   repeat for each line i in k
>     add item 1 of i*item 2 of n-item 2 of i *item 1 of n to tsum
>     put i into n
>   end repeat
>   return abs(tsum/2)
>end IrPolyArea
>
>Raymond E. Griffith
>


Ray,

This  is really slick. I had to change the notation as follows before 
I could decipher your clever algorithm:

on function IrPolyArea i
   put the points of graphic i into tPoints
   put 0,0 into tPreviousLine
   repeat for each line tLine in tPoints
     add item 1 of tLine *item 2 of tPreviousLine - item 2 of tLine 
*item 1 of tPreviousLine to tsum
     put tLine into tPreviousLine
   end repeat
   return abs(tsum/2)
end IrPolyArea

Jim