Polygon area calculation using Latitude and Longitude generated from Cartesian space and a world file

Question

Given a series of GPS coordinate pairs, I need to calculate the area of a polygon (n-gon). This is relatively small (not larger than 50,000 sqft). The geocodes are created

Answer 1

Thank you Risky Pathak!

In the spirit of sharing, here's my adaptation in Delphi:

interface

uses 
  System.Math; 

TMapGeoPoint = record
  Latitude: Double;
  Longitude: Double;
end;


function AreaInAcres(AGeoPoints: TList<TMapGeoPoint>): Double;

implementation

function AreaInAcres(AGeoPoints: TList<TMapGeoPoint>): Double;
var
  Area: Double;
  i: Integer;
  P1, P2: TMapGeoPoint;
begin
 Area := 0;

 // We need at least 2 points
 if (AGeoPoints.Count > 2) then
 begin
   for I := 0 to AGeoPoints.Count - 1 do
   begin
     P1 := AGeoPoints[i];
     if i < AGeoPoints.Count - 1  then
       P2 := AGeoPoints[i + 1]
     else
       P2 := AGeoPoints[0];
     Area := Area + DegToRad(P2.Longitude - P1.Longitude) * (2 + 
        Sin(DegToRad(P1.Latitude)) + Sin(DegToRad(P2.Latitude)));
    end;

    Area := Area * 6378137 * 6378137 / 2;

  end;

  Area := Abs(Area); //Area (in sq meters)

  // 1 Square Meter = 0.000247105 Acres
  result := Area * 0.000247105;
end;

Answer 2

Adapted RiskyPathak's snippet to PHP

function CalculatePolygonArea($coordinates) {
    $area = 0;
    $coordinatesCount = sizeof($coordinates);
    if ($coordinatesCount > 2) {
      for ($i = 0; $i < $coordinatesCount - 1; $i++) {
        $p1 = $coordinates[$i];
        $p2 = $coordinates[$i + 1];
        $p1Longitude = $p1[0];
        $p2Longitude = $p2[0];
        $p1Latitude = $p1[1];
        $p2Latitude = $p2[1];
        $area += ConvertToRadian($p2Longitude - $p1Longitude) * (2 + sin(ConvertToRadian($p1Latitude)) + sin(ConvertToRadian($p2Latitude)));
      }
    $area = $area * 6378137 * 6378137 / 2;
    }
    return abs(round(($area));
}

function ConvertToRadian($input) {
    $output = $input * pi() / 180;
    return $output;
}

Answer 3

Based on the solution by Risky Pathak here is the solution for SQL (Redshift) to calculate areas for GeoJSON multipolygons (with the assumption that linestring 0 is the outermost polygon)

create or replace view geo_area_area as 
with points as (
    select ga.id as key_geo_area
    , ga.name, gag.linestring
    , gag.position
    , radians(gag.longitude) as x
    , radians(gag.latitude) as y
    from geo_area ga
    join geo_area_geometry gag on (gag.key_geo_area = ga.id)
)
, polygons as (
    select key_geo_area, name, linestring, position 
    , x
    , lag(x) over (partition by key_geo_area, linestring order by position) as prev_x
    , y
    , lag(y) over (partition by key_geo_area, linestring order by position) as prev_y
    from points
)
, area_linestrings as (
    select key_geo_area, name, linestring
    , abs( sum( (x - prev_x) * (2 + sin(y) + sin(prev_y)) ) ) * 6378137 * 6378137 / 2 / 10^6 as area_km_squared
    from polygons
    where position != 0
    group by 1, 2, 3
)
select key_geo_area, name
, sum(case when linestring = 0 then area_km_squared else -area_km_squared end) as area_km_squared
from area_linestrings
group by 1, 2
;

Answer 4

Adapted RiskyPathak's snippet to Ruby

def deg2rad(input)
  input * Math::PI / 180.0
end

def polygone_area(coordinates)
  return 0.0 unless coordinates.size > 2

  area = 0.0
  coor_p = coordinates.first
  coordinates[1..-1].each{ |coor|
    area += deg2rad(coor[1] - coor_p[1]) * (2 + Math.sin(deg2rad(coor_p[0])) + Math.sin(deg2rad(coor[0])))
    coor_p = coor
  }

  (area * 6378137 * 6378137 / 2.0).abs # 6378137 Earth's radius in meters
end

Answer 5

I am modifying a Google Map so that a user can calculate the area of a polygon by clicking the vertices. It wasn't giving correct areas until I made sure the Math.cos(latAnchor) was in radians first

So:

double xPos = (lon-lonAnchor)*( Math.toRadians( 6378137 ) )*Math.cos( latAnchor );

became:

double xPos = (lon-lonAnchor)*( 6378137*PI/180 ) )*Math.cos( latAnchor*PI/180 );

where lon, lonAnchor and latAnchor are in degrees. Works like a charm now.

Answer 6

1% error seems a bit high due to just your approximation. Are you comparing against actual measurements or some ideal calculation? Remember that there is error in the GPS as well that might be contributing.

If you want a more accurate method for doing this there's a good answer at this question. If you're going for a faster way you can use the WGS84 geoid instead of your reference sphere for converting to cartesian coordinates (ECEF). Here's the wiki link for that conversion.