How to convert latitude or longitude to meters?

Question

Hi,

If I have a latitude or longitude reading in standard NMEA format is there an easy way / forumla to convert that reading to meters, which I can then implement in Java (J9)?

Edit: Ok seems what I want to do is not possible /easily/, however what I really want to do is:

Say I have a lat and long of a way point and a lat and long of a user is there an easy way to compare them to decide when to tell the user they are within a /reasonably/ close distance of the way point? I realise reasonable is subject but is this easily do-able or still overly maths-y?

Thanks,

Adam

Answer 1

You may find this document helpful: http://www.johndcook.com/lat_long_details.html

Answer 2

One nautical mile (1852 meters) is defined as one arcminute of longitude at the equator. However, you need to define a map projection (see also UTM) in which you are working for the conversion to really make sense.

Answer 3

Latitudes and longitudes specify points, not distances, so your question is somewhat nonsensical. If you're asking about the shortest distance between two (lat, lon) points, see this Wikipedia article on great-circle distances.

Answer 4

There are many tools that will make this easy. See monjardin's answer for more details about what's involved.

However, doing this isn't necessarily difficult. It sounds like you're using Java, so I would recommend looking into something like GDAL. It provides java wrappers for their routines, and they have all the tools required to convert from Lat/Lon (geographic coordinates) to UTM (projected coordinate system) or some other reasonable map projection.

UTM is nice, because it's meters, so easy to work with. However, you will need to get the appropriate UTM zone for it to do a good job. There are some simple codes available via googling to find an appropriate zone for a lat/long pair.

Answer 5

    'below is from
'http://www.zipcodeworld.com/samples/distance.vbnet.html
Public Function distance(ByVal lat1 As Double, ByVal lon1 As Double, _
                         ByVal lat2 As Double, ByVal lon2 As Double, _
                         Optional ByVal unit As Char = "M"c) As Double
    Dim theta As Double = lon1 - lon2
    Dim dist As Double = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + _
                            Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * _
                            Math.Cos(deg2rad(theta))
    dist = Math.Acos(dist)
    dist = rad2deg(dist)
    dist = dist * 60 * 1.1515
    If unit = "K" Then
        dist = dist * 1.609344
    ElseIf unit = "N" Then
        dist = dist * 0.8684
    End If
    Return dist
End Function
Public Function Haversine(ByVal lat1 As Double, ByVal lon1 As Double, _
                         ByVal lat2 As Double, ByVal lon2 As Double, _
                         Optional ByVal unit As Char = "M"c) As Double
    Dim R As Double = 6371 'earth radius in km
    Dim dLat As Double
    Dim dLon As Double
    Dim a As Double
    Dim c As Double
    Dim d As Double
    dLat = deg2rad(lat2 - lat1)
    dLon = deg2rad((lon2 - lon1))
    a = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) + Math.Cos(deg2rad(lat1)) * _
            Math.Cos(deg2rad(lat2)) * Math.Sin(dLon / 2) * Math.Sin(dLon / 2)
    c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a))
    d = R * c
    Select Case unit.ToString.ToUpper
        Case "M"c
            d = d * 0.62137119
        Case "N"c
            d = d * 0.5399568
    End Select
    Return d
End Function
Private Function deg2rad(ByVal deg As Double) As Double
    Return (deg * Math.PI / 180.0)
End Function
Private Function rad2deg(ByVal rad As Double) As Double
    Return rad / Math.PI * 180.0
End Function

Answer 6

There are quite a few ways to calculate this. All of them use aproximations of spherical trigonometry where the radius is the one of the earth.

try http://www.movable-type.co.uk/scripts/latlong.html for a bit of methods and code in different languages.

Answer 7

The earth is an annoyingly irregular surface, so there is no simple formula to do this exactly. You have to live with an approximate model of the earth, and project your coordinates onto it. The model I typically see used for this is WGS 84. This is what GPS devices usually use to solve the exact same problem.

NOAA has some software you can download to help with this on their website.

Answer 8

You need to convert the coordinates to radians to do the spherical geometry. Once converted, then you can calculate a distance between the two points. The distance then can be converted to any measure you want.

Answer 9

If its sufficiently close you can get away with treating them as coordinates on a flat plane. This works on say, street or city level if perfect accuracy isnt required and all you need is a rough guess on the distance involved to compare with an arbitrary limit.