How to convert latitude or longitude to meters?

Question

If I have a latitude or longitude reading in standard NMEA format is there an easy way / formula 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?

Answer 1

Here is a javascript function:

function measure(lat1, lon1, lat2, lon2){  // generally used geo measurement function
    var R = 6378.137; // Radius of earth in KM
    var dLat = lat2 * Math.PI / 180 - lat1 * Math.PI / 180;
    var dLon = lon2 * Math.PI / 180 - lon1 * Math.PI / 180;
    var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
    Math.cos(lat1 * Math.PI / 180) * Math.cos(lat2 * Math.PI / 180) *
    Math.sin(dLon/2) * Math.sin(dLon/2);
    var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
    var d = R * c;
    return d * 1000; // meters
}

Explanation: https://en.wikipedia.org/wiki/Haversine_formula

The haversine formula determines the great-circle distance between two points on a sphere given their longitudes and latitudes.

Answer 2

Given you're looking for a simple formula, this is probably the simplest way to do it, assuming that the Earth is a sphere with a circumference of 40075 km.

Length in meters of 1° of latitude = always 111.32 km

Length in meters of 1° of longitude = 40075 km * cos( latitude ) / 360

Answer 3

For approximating short distances between two coordinates I used formulas from http://en.wikipedia.org/wiki/Lat-lon:

m_per_deg_lat = 111132.954 - 559.822 * cos( 2 * latMid ) + 1.175 * cos( 4 * latMid);
m_per_deg_lon = 111132.954 * cos ( latMid );

.

In the code below I've left the raw numbers to show their relation to the formula from wikipedia.

double latMid, m_per_deg_lat, m_per_deg_lon, deltaLat, deltaLon,dist_m;

latMid = (Lat1+Lat2 )/2.0;  // or just use Lat1 for slightly less accurate estimate


m_per_deg_lat = 111132.954 - 559.822 * cos( 2.0 * latMid ) + 1.175 * cos( 4.0 * latMid);
m_per_deg_lon = (3.14159265359/180 ) * 6367449 * cos ( latMid );

deltaLat = fabs(Lat1 - Lat2);
deltaLon = fabs(Lon1 - Lon2);

dist_m = sqrt (  pow( deltaLat * m_per_deg_lat,2) + pow( deltaLon * m_per_deg_lon , 2) );

The wikipedia entry states that the distance calcs are within 0.6m for 100km longitudinally and 1cm for 100km latitudinally but I have not verified this as anywhere near that accuracy is fine for my use.

Answer 4

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 5

Here is the R version of b-h-'s function, just in case:

measure <- function(lon1,lat1,lon2,lat2) {
    R <- 6378.137                                # radius of earth in Km
    dLat <- (lat2-lat1)*pi/180
    dLon <- (lon2-lon1)*pi/180
    a <- sin((dLat/2))^2 + cos(lat1*pi/180)*cos(lat2*pi/180)*(sin(dLon/2))^2
    c <- 2 * atan2(sqrt(a), sqrt(1-a))
    d <- R * c
    return (d * 1000)                            # distance in meters
}

Answer 6

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 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

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 9

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 10

    '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 11

To convert latitude and longitude in x and y representation you need to decide what type of map projection to use. As for me, Elliptical Mercator seems very well. Here you can find an implementation (in Java too).

Answer 12

Here is a MySQL function:

SET @radius_of_earth = 6378.137; -- In kilometers

DROP FUNCTION IF EXISTS Measure;
DELIMITER //
CREATE FUNCTION Measure (lat1 REAL, lon1 REAL, lat2 REAL, lon2 REAL) RETURNS REAL
BEGIN
-- Multiply by 1000 to convert millimeters to meters
RETURN 2 * @radius_of_earth * 1000 * ASIN(SQRT(
    POW(SIN((lat2 - lat1) / 2 * PI() / 180), 2) +
    COS(lat1 * PI() / 180) *
    COS(lat2 * PI() / 180) *
    POW(SIN((lon2 - lon1) / 2 * PI() / 180), 2)
));
END; //
DELIMITER ;

Answer 13

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.

Answer 14

Here is a version in Swift:

func toDegreeAt(point: CLLocationCoordinate2D) -> CLLocationDegrees {
    let latitude = point.latitude  
    let earthRadiusInMetersAtSeaLevel = 6378137.0
    let earthRadiusInMetersAtPole = 6356752.314
    
    let r1 = earthRadiusInMetersAtSeaLevel
    let r2 = earthRadiusInMetersAtPole
    let beta = latitude

    let earthRadiuseAtGivenLatitude = (
      ( pow(pow(r1, 2) * cos(beta), 2) + pow(pow(r2, 2) * sin(beta), 2) ) /
      ( pow(r1 * cos(beta), 2) + pow(r2 * sin(beta), 2) )
    )
    .squareRoot()
      
    let metersInOneDegree = (2 * Double.pi * earthRadiuseAtGivenLatitude * 1.0) / 360.0
    let value: CLLocationDegrees = self / metersInOneDegree
    return value
  }

Answer 15

Based on average distance for degress in the Earth.

1° = 111km;

Converting this for radians and dividing for meters, take's a magic number for the RAD, in meters: 0.000008998719243599958;

then:

const RAD = 0.000008998719243599958;
Math.sqrt(Math.pow(lat1 - lat2, 2) + Math.pow(long1 - long2, 2)) / RAD;

Answer 16

If you want a simple solution then use the Haversine formula as outlined by the other comments. If you have an accuracy sensitive application keep in mind the Haversine formula does not guarantee an accuracy better then 0.5% as it is assuming the earth is a sphere. To consider that Earth is a oblate spheroid consider using Vincenty's formulae. Additionally, I'm not sure what radius we should use with the Haversine formula: {Equator: 6,378.137 km, Polar: 6,356.752 km, Volumetric: 6,371.0088 km}.

Answer 17

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.