You could speed it up a bit more and simplify a bit like so:
put line 1 of pPts into oldL
repeat for each line L in line 2 to -1 of pPts
add (item 1 of L - item 1 of oldL) * ( item 2 of oldL + item 2 of
L) / 2 to tArea
put L into oldL
end repeat
On Sat, Nov 7, 2015 at 7:46 PM, Roger Guay <i...@mac.com> wrote:
> Beautiful, Alex. I was working on this idea myself, but your scripts is so
> much more elegant!
>
> Thanks,
>
> Roger
>
>
>
> > On Nov 7, 2015, at 4:43 PM, Alex Tweedly <a...@tweedly.org> wrote:
> >
> > Actually, you should shorten the maths - makes it a bit more efficient
> >
> >
> > function getArea pPts
> > if line 1 of pPts <> line -1 of pPts then return 0
> > put 0 into tArea
> > put empty into oldL
> > repeat for each line L in pPts
> > if oldL is not empty then
> > -- put item 1 of L - item 1 of oldL into dx
> > -- put item 2 of oldL + (item 2 of L - item 2 of oldL) /
> 2 into avgY
> > -- put dx * avgY into thisArea
> > -- add thisArea to tArea
> > add (item 1 of L - item 1 of oldL) * ( item 2 of oldL + item 2
> of L) / 2 to tArea
> > end if
> > put L into oldL
> > end repeat
> > return abs(tArea)
> >
> > end getArea
> >
> > -- Alex.
> >
> > On 08/11/2015 00:09, Alex Tweedly wrote:
> >> Roger,
> >>
> >> try this (I've done some minimal testing, so do please check it ...)
> >> it assumes the polygon is closed
> >>
> >> function getArea pPts
> >> if line 1 of pPts <> line -1 of pPts then return 0
> >> put 0 into tArea
> >> put empty into oldL
> >> repeat for each line L in pPts
> >> if oldL is not empty then
> >> put item 1 of L - item 1 of oldL into dx
> >> put item 2 of oldL + (item 2 of L - item 2 of oldL) / 2 into
> avgY
> >> put dx * avgY into thisArea
> >> add thisArea to tArea
> >> end if
> >> put L into oldL
> >> end repeat
> >> return abs(tArea)
> >> end getArea
> >>
> >> NB you could shorten it, but I left each math step separate to make it
> easier to debug / check
> >>
> >> -- Alex.
> >>
> >> On 07/11/2015 22:29, Roger Guay wrote:
> >>> Thanks very much, Scott. I do have a long points list that I would
> have to manipulate even before apply this technique. I’m not looking
> forward to that, but I may have to go this route.
> >>>
> >>> Cheers,
> >>>
> >>> Roger
> >>>
> >>>
> >>>> On Nov 7, 2015, at 12:05 PM, Scott Rossi <sc...@tactilemedia.com>
> wrote:
> >>>>
> >>>> It looks like this could translate to LC if you have the point
> locations
> >>>> of your polygon:
> >>>>
> >>>> https://www.mathsisfun.com/geometry/area-irregular-polygons.html
> >>>>
> >>>>
> >>>> Regards,
> >>>>
> >>>> Scott Rossi
> >>>> Creative Director
> >>>> Tactile Media, UX/UI Design
> >>>>
> >>>>
> >>>>
> >>>>
> >>>> On 11/7/15, 2:01 PM, "use-livecode on behalf of Roger Guay"
> >>>> <use-livecode-boun...@lists.runrev.com on behalf of i...@mac.com>
> wrote:
> >>>>
> >>>>> Can anyone think of an easy way of getting the area of an Irregular
> >>>>> polygon?
> >>>>>
> >>>>> Thanks,
> >>>>>
> >>>>> Roger
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