# Surface Area

• 12-06-2005
Dazzer
Surface Area
Heres an interesting one that I'm having a lot of difficulty with (i'm sure you maths boffins out there will scoff at my strugling with such a simple problem :wink: ) .

Given a set of lat\lon co-ordinates for a closed polygon shape how can I work out the surface area of that shape in sq miles or sq km (or feet or meters or something sensible!).

I've searched the internet but the solutions I find are all looking for the ultimate in accuracy and far to complex for me to understand! I'm really just looking at this from a 'Flat Earth' scenario it doesn't need to be extremly accurate (unless its just as easy to make it more accurate of course).
• 12-06-2005
Wilfried
Hi,

No so simple but not really difficult also :) I try to explain. Take a pencil and draw a irregular polygon.

First step is, you take first point and draw a line to a second point skipping 1 point. What you now have is a triangle. Easy to calculated surface area of a triangle.

go on with a line to next point (skipping also just 1 point), do over and over again until you have the last resulting triangle.
• 12-06-2005
Dazzer
I think I understand what your saying and I may be missing something but for the map shown here http://img384.imageshack.us/img384/1...acearea3kt.jpg

would the triangle shown with the red line not be outside the surface area of the shape?
• 12-06-2005
DrExcitement

Here's how to do it with (x,y) vertices: http://astronomy.swin.edu.au/~pbourk...etry/polyarea/. This works for concave polygons like the one you showed. To use with (lat,long) vertices, I guess you could convert each vertex to an (x,y) in meters (or whatever), based on a local coordinate system, and assuming the world is locally flat. Clearly there are some approximation errors involved here.
• 12-06-2005
Wilfried
Hi,

Quote:

I think I understand what your saying and I may be missing something but for the map shown here http://img384.imageshack.us/img384/1...acearea3kt.jpg
would the triangle shown with the red line not be outside the surface area of the shape?
Yes that's what I meant in 'no so simple'. There are some complications here and there, but that's what we programmers do like eh dont we ?
• 12-08-2005
Dazzer
Thanks for you help Wilfred, DrExcitement, it appears that this isn't as simple as I hoped (I didn't really think it would be to be honest!)

Looking at DrExcitements link it appears that this could be something I could use in my application if I had a way of converting Lat, Lon to some local Grid system, does anyone know of a way to do this?
• 12-08-2005
Wilfried
Hi,

Just use lat/long pairs. eg multiply them with 10000 or so and make them integers. Using 1/10000 will give you accuratie around 11 meter. 1 minute is exact 1 sea mile.

You will have (small) calculation errors because 1 minute in longitude is 1 seamile on the equator and less of course if go more distance away for it. You can correct it by multiply it by Cos(Latitude).
• 12-08-2005
DrExcitement
A slight refinement of the reply above is to first compute the mean (lat,long) of all your vertices. This will be the origin of your local coordinate system, called (lat0,long0). Now for each (lat,long) vertext, the corresponding (x,y) is y = pi*earth_radius_meters*(lat - lat0)/180 and x = pi*earth_radius_meters*cos(lat0)*(long - long0)/180. This assumes (lat,long) is in degrees. earth_radius_meters is approximately 1000.0 * 6371.01. This will give you coordinates in meters, so your area will be in square meters. If you want another square area unit, use the corresponding linear unit for the radius of the earth.
• 12-08-2005
Dazzer
Excellent, thank you both for your help, I should have the basis for a working algorithm now. I'll post a reply containing the class for any one else whos interested in this when i'm finished.
• 01-05-2006
Dazzer
Not really a class for doing this but heres how I implemented this in my code (VB .NET), should give someone some ideas towards writing a class of there own. In this I have my coordinates held in the table 'Shapes' each shape has a unique ID.

Code:

```    Imports System.Math     Private Function CalculateArea&#40;ByVal ID As String&#41; As String         Dim con As New System.Data.SqlClient.SqlConnection&#40;ConnString&#41;         con.Open&#40;&#41;         Dim cmd As New System.Data.SqlClient.SqlCommand         Dim da As New System.Data.SqlClient.SqlDataAdapter         Dim ds As New DataSet         Dim dr As DataRow         Dim MyItem As ListViewItem         Dim xCoord&#40;&#41; As Double         Dim yCoord&#40;&#41; As Double         Dim n As Integer         Dim x, y, dist As Double         cmd = con.CreateCommand         cmd.CommandText = "SELECT * FROM Shapes WHERE ID = '" & ID & "' ORDER BY point"         cmd.CommandType = CommandType.Text         da.SelectCommand = cmd         da.Fill&#40;ds, "Coords"&#41;         ReDim xCoord&#40;ds.Tables&#40;"Coords"&#41;.Rows.Count&#41;         ReDim yCoord&#40;ds.Tables&#40;"Coords"&#41;.Rows.Count&#41;         For n = 0 To UBound&#40;xCoord&#41;             If n = UBound&#40;xCoord&#41; Then                 xCoord&#40;n&#41; = xCoord&#40;0&#41;             Else                 xCoord&#40;n&#41; = ds.Tables&#40;"Coords"&#41;.Rows&#40;n&#41;.Item&#40;"Lat"&#41;             End If             If n = UBound&#40;yCoord&#41; Then                 yCoord&#40;n&#41; = yCoord&#40;0&#41;             Else                 yCoord&#40;n&#41; = ds.Tables&#40;"Coords"&#41;.Rows&#40;n&#41;.Item&#40;"Lon"&#41;             End If         Next         For n = 0 To UBound&#40;xCoord&#41;             x += xCoord&#40;n&#41;         Next         x = x / &#40;UBound&#40;xCoord&#41; + 1&#41;         dist = DistanceFrom&#40;x, x, 0, 1&#41;         For n = 0 To UBound&#40;xCoord&#41;             xCoord&#40;n&#41; = xCoord&#40;n&#41; * 69.11         Next         For n = 0 To UBound&#40;yCoord&#41;             yCoord&#40;n&#41; = yCoord&#40;n&#41; * dist         Next         CalculateArea = Format&#40;AreaByCoordinates&#40;xCoord, yCoord&#41;, "#00.000 sqMiles"&#41;     End Function     Function AreaByCoordinates&#40;ByVal Xcoord&#40;&#41; As Double, ByVal Ycoord&#40;&#41; As Double&#41; As Double         Dim I As Long         Dim Xold As Double         Dim Yold As Double         Dim Yorig As Double         Dim ArrayUpBound As Long         Dim x, y As Double         ArrayUpBound = UBound&#40;Xcoord&#41;         Xold = Xcoord&#40;ArrayUpBound&#41;         Yorig = Ycoord&#40;ArrayUpBound&#41;         Yold = 0.0#         For I = LBound&#40;Xcoord&#41; To ArrayUpBound             x = Xcoord&#40;I&#41;             y = Ycoord&#40;I&#41; - Yorig             AreaByCoordinates = AreaByCoordinates + &#40;Xold - x&#41; * &#40;Yold + y&#41;             Xold = x             Yold = y         Next         AreaByCoordinates = Abs&#40;AreaByCoordinates&#41; / 2     End Function     Function DistanceFrom&#40;ByVal lat1 As Double, ByVal lat2 As Double, ByVal lon1 As Double, ByVal lon2 As Double&#41; As Double         Dim theta, dist As Double         theta = lon1 - lon2         dist = Sin&#40;deg2rad&#40;lat1&#41;&#41; * Sin&#40;deg2rad&#40;lat2&#41;&#41; + Cos&#40;deg2rad&#40;lat1&#41;&#41; * Cos&#40;deg2rad&#40;lat2&#41;&#41; * Cos&#40;deg2rad&#40;theta&#41;&#41;         dist = Acos&#40;dist&#41;         dist = rad2deg&#40;dist&#41;         DistanceFrom = dist * 60 * 1.1515     End Function     Function deg2rad&#40;ByVal deg As Double&#41; As Double         deg2rad = CDbl&#40;deg * PI / 180&#41;     End Function     Function rad2deg&#40;ByVal rad As Double&#41; As Double         rad2deg = CDbl&#40;rad * 180 / PI&#41;     End Function```
I got the figure 69.11 from using this site

http://www.zodiacal.com/tools/lat_table.htm

I don't really have time to explain all of the details write now and this isn't entirely accurate due to the complications involved in making it so, but hopefully this should help anyone looking to do anything similar and perhaps they can expand on it and improve it and post there results back here. (Also if i've made any obvious mistakes could someone let me know :wink: !)