Calculating new longitude, latitude from old + n meters

I want to create 2 new longitude and 2 new latitudes based on a coordinate and a distanc开发者_开发知识库e in meters, I want to create a nice bounding box around a certain point. It is for a part of a city and max ±1500 meters. I therefore don't think the curvature of earth has to be taken into account.

So I have 50.0452345 (x) and 4.3242234 (y) and I want to know x + 500 meters, x - 500 meters, y - 500 meters, y + 500 meters

I found many algorithms but almost all seem to deal with the distance between points.

The number of kilometers per degree of longitude is approximately

(pi/180) * r_earth * cos(theta*pi/180)

where theta is the latitude in degrees and r_earth is approximately 6378 km.

The number of kilometers per degree of latitude is approximately the same at all locations, approx

(pi/180) * r_earth = 111 km / degree 

So you can do:

new_latitude  = latitude  + (dy / r_earth) * (180 / pi);
new_longitude = longitude + (dx / r_earth) * (180 / pi) / cos(latitude * pi/180);

As long as dx and dy are small compared to the radius of the earth and you don't get too close to the poles.

The accepted answer is perfectly right and works. I made some tweaks and turned into this:

double meters = 50;

// number of km per degree = ~111km (111.32 in google maps, but range varies
// between 110.567km at the equator and 111.699km at the poles)
//
// 111.32km = 111320.0m (".0" is used to make sure the result of division is
// double even if the "meters" variable can't be explicitly declared as double)
double coef = meters / 111320.0;

double new_lat = my_lat + coef;

// pi / 180 ~= 0.01745
double new_long = my_long + coef / Math.cos(my_lat * 0.01745);

Hope this helps too.

For latitude do:

var earth = 6378.137,  //radius of the earth in kilometer
    pi = Math.PI,
    m = (1 / ((2 * pi / 360) * earth)) / 1000;  //1 meter in degree

var new_latitude = latitude + (your_meters * m);

For longitude do:

var earth = 6378.137,  //radius of the earth in kilometer
    pi = Math.PI,
    cos = Math.cos,
    m = (1 / ((2 * pi / 360) * earth)) / 1000;  //1 meter in degree

var new_longitude = longitude + (your_meters * m) / cos(latitude * (pi / 180));

The variable your_meters can contain a positive or a negative value.

I had to spend about two hours to work out the solution by @nibot , I simply needed a method to create a boundary box given its center point and width/height (or radius) in kilometers:

I don't fully understand the solution mathematically/ geographically. I tweaked the solution (by trial and error) to get the four coordinates. Distances in km, given the current position and distance we shift to the new position in the four coordinates:

North:

private static Position ToNorthPosition(Position center, double northDistance)
{
    double r_earth = 6378;
    var pi = Math.PI;
    var new_latitude = center.Lat + (northDistance / r_earth) * (180 / pi);
    return new Position(new_latitude, center.Long);
}

East:

private static Position ToEastPosition(Position center, double eastDistance)
{
    double r_earth = 6378;
    var pi = Math.PI;
    var new_longitude = center.Long + (eastDistance / r_earth) * (180 / pi) / Math.Cos(center.Lat * pi / 180);
    return new Position(center.Lat, new_longitude);
}

South:

private static Position ToSouthPosition(Position center, double southDistance)
{
    double r_earth = 6378;
    var pi = Math.PI;
    var new_latitude = center.Lat - (southDistance / r_earth) * (180 / pi);
    return new Position(new_latitude, center.Long);
}

West:

private static Position ToWestPosition(Position center, double westDistance)
{
    double r_earth = 6378;
    var pi = Math.PI;
    var new_longitude = center.Long - (westDistance / r_earth) * (180 / pi) / Math.Cos(center.Lat * pi / 180);
    return new Position(center.Lat, new_longitude);
}

Have you checked out: How do I find the lat/long that is x km north of a given lat/long ?

These calculations are annoying at best, I've done many of them. The haversine formula will be your friend.

Some reference: http://www.movable-type.co.uk/scripts/latlong.html

Posting this method for sake of completeness.

Use this method "as it is" to:

• Move any (lat,long) point by given meters in either axis.

Python method to move any point by defined meters.

def translate_latlong(lat,long,lat_translation_meters,long_translation_meters):
    ''' method to move any lat,long point by provided meters in lat and long direction.
    params :
        lat,long: lattitude and longitude in degrees as decimal values, e.g. 37.43609517497065, -122.17226450150885
        lat_translation_meters: movement of point in meters in lattitude direction.
                                positive value: up move, negative value: down move
        long_translation_meters: movement of point in meters in longitude direction.
                                positive value: left move, negative value: right move
        '''
    earth_radius = 6378.137

    #Calculate top, which is lat_translation_meters above
    m_lat = (1 / ((2 * math.pi / 360) * earth_radius)) / 1000;  
    lat_new = lat + (lat_translation_meters * m_lat)

    #Calculate right, which is long_translation_meters right
    m_long = (1 / ((2 * math.pi / 360) * earth_radius)) / 1000;  # 1 meter in degree
    long_new = long + (long_translation_meters * m_long) / math.cos(lat * (math.pi / 180));
    
    return lat_new,long_new

Working Python code to offset coordinates by 10 metres.

def add_blur(lat, long):
meters = 10
blur_factor = meters * 0.000006279
new_lat = lat + blur_factor
new_long = long + blur_factor / math.cos(lat * 0.018)
return new_lat, new_long

if you don't have to be very exact then: each 10000 meters is about 0.1 for latitude and longitude. for example I want to load locations 3000 meters around point_A from my database:

double newMeter =  3000 * 0.1 / 10000;
double lat1 = point_A.latitude - newMeter;
double lat2 = point_A.latitude + newMeter;
double lon1 = point_A.longitude - newMeter;
double lon1 = point_A.longitude + newMeter;
Cursor c = mDb.rawQuery("select * from TABLE1  where lat >= " + lat1 + " and lat <= " + lat2 + " and lon >= " + lon1 + " and lon <= " + lon2 + " order by id", null);

public double MeterToDegree(double meters, double latitude)
{
    return meters / (111.32 * 1000 * Math.Cos(latitude * (Math.PI / 180)));
}

var meters = 50;
var coef = meters * 0.0000089;
var new_lat = map.getCenter().lat.apply() + coef;
var new_long = map.getCenter().lng.apply() + coef / Math.cos(new_lat * 0.018);
map.setCenter({lat:new_lat, lng:new_long});

See from Official Google Maps Documentation (link below) as they solve on easy/simple maps the problems with distance by countries :)

I recommended this solution to easy/simply solve issue with boundaries that you can know which area you're solving the problem with boundaries (not recommended globally)

Note:

Latitude lines run west-east and mark the position south-north of a point. Lines of latitude are called parallels and in total there are 180 degrees of latitude. The distance between each degree of latitude is about 69 miles (110 kilometers).

The distance between longitudes narrows the further away from the equator. The distance between longitudes at the equator is the same as latitude, roughly 69 miles (110 kilometers) . At 45 degrees north or south, the distance between is about 49 miles (79 kilometers). The distance between longitudes reaches zero at the poles as the lines of meridian converge at that point.

Original source 1 Original source 2

Official Google Maps Documentation: Code Example: Autocomplete Restricted to Multiple Countries

See the part of their code how they solve problem with distance center + 10 kilometers by +/- 0.1 degree

function initMap(): void {
  const map = new google.maps.Map(
    document.getElementById("map") as HTMLElement,
    {
      center: { lat: 50.064192, lng: -130.605469 },
      zoom: 3,
    }
  );
  const card = document.getElementById("pac-card") as HTMLElement;
  map.controls[google.maps.ControlPosition.TOP_RIGHT].push(card);
  const center = { lat: 50.064192, lng: -130.605469 };

  // Create a bounding box with sides ~10km away from the center point
  const defaultBounds = {
    north: center.lat + 0.1,
    south: center.lat - 0.1,
    east: center.lng + 0.1,
    west: center.lng - 0.1,
  };

  const input = document.getElementById("pac-input") as HTMLInputElement;
  const options = {
    bounds: defaultBounds,
    componentRestrictions: { country: "us" },
    fields: ["address_components", "geometry", "icon", "name"],
    origin: center,
    strictBounds: false,
    types: ["establishment"],
  };

This is what I did in VBA that seems to be working for me. Calculation is in feet not meters though

Public Function CalcLong(OrigLong As Double, OrigLat As Double, DirLong As String, DirLat As String, DistLong As Double, DistLat As Double)
    Dim FT As Double
    Dim NewLong, NewLat As Double
    FT = 1 / ((2 * WorksheetFunction.Pi / 360) * 20902230.971129)
    
    If DirLong = "W" Then
        NewLat = CalcLat(OrigLong, OrigLat, DirLong, DirLat, DistLong, DistLat)
        NewLong = OrigLong - ((FT * DistLong) / Cos(NewLat * (WorksheetFunction.Pi / 180)))
        CalcLong = NewLong
    Else
        NewLong = OrigLong + ((FT * DistLong) / Math.Cos(CalcLat(OrigLong, OrigLat, DirLong, DirLat, DistLong, DistLat) * (WorksheetFunction.Pi / 180)))
        CalcLong = NewLong
    End If
    
End Function


Public Function CalcLat(OrigLong As Double, OrigLat As Double, DirLong As String, DirLat As String, DistLong As Double, DistLat As Double) As Double
    Dim FT As Double
    Dim NewLat As Double
    
    FT = 1 / ((2 * WorksheetFunction.Pi / 360) * 20902230.971129)
    
    If DirLat = "S" Then
        NewLat = (OrigLat - (FT * DistLat))
        CalcLat = NewLat
    Else
        NewLat = (OrigLat + (FT * DistLat))
        CalcLat = NewLat
    End If
    
End Function

Original poster said: "So I have 50.0452345 (x) and 4.3242234 (y) and I want to know x + 500 meters..."

I will assume the units of the x and y values he gave there were in meters (and not degrees Longitude, Latitude). If so then he is stating measurements to 0.1 micrometer, so I will assume he needs similar accuracy for the translated output. I also will assume by "+500 meters" etc. he meant the direction to be due North-South and due East-West. He refers to a reference point: "2 new latitudes based on a coordinate"; but he did not give the Longitude and Latitude, so to explain the procedure concretely I will give the Latitudes and Longitudes for the corners of the 500 meter box he requested around the point [30 degrees Longitude,30 degrees Latitude].

The exact solution on the surface of the GRS80 Ellipsoid is given with the following set of functions (I wrote these for the free-open-source-mac-pc math program called "PARI" which allows any number of digits precision to be setup):

\\=======Arc lengths along Latitude and Longitude and the respective scales:
dms(u)=[truncate(u),truncate((u-truncate(u))*60),((u-truncate(u))*60-truncate((u-truncate(u))*60))*60];
SpinEarthRadiansPerSec=7.292115e-5;\
GMearth=3986005e8;\
J2earth=108263e-8;\
re=6378137;\
ecc=solve(ecc=.0001,.9999,eccp=ecc/sqrt(1-ecc^2);qecc=(1+3/eccp^2)*atan(eccp)-3/eccp;ecc^2-(3*J2earth+4/15*SpinEarthRadiansPerSec^2*re^3/GMearth*ecc^3/qecc));\
e2=ecc^2;\
b2=1-e2;\
b=sqrt(b2);\
fl=1-b;\
rfl=1/fl;\
U0=GMearth/ecc/re*atan(eccp)+1/3*SpinEarthRadiansPerSec^2*re^2;\
HeightAboveEllipsoid=0;\
reh=re+HeightAboveEllipsoid;\
longscale(lat)=reh*Pi/648000/sqrt(1+b2*(tan(lat))^2);
latscale(lat)=reh*b*Pi/648000/(1-e2*(sin(lat))^2)^(3/2);
longarc(lat,long1,long2)=longscale(lat)*648000/Pi*(long2-long1);
latarc(lat1,lat2)=(intnum(th=lat1,lat2,sqrt(1-e2*(sin(th))^2))+e2/2*sin(2*lat1)/sqrt(1-e2*(sin(lat1))^2)-e2/2*sin(2*lat2)/sqrt(1-e2*(sin(lat2))^2))*reh;
\\=======

I then plugged the reference point [30,30] into those functions at the PARI command prompt and had PARI solve for the point +/- 500 meters away from it, giving the two new Longitudes and two new Latitudes that the original poster asked for. Here is the input and output showing that:

? dms(solve(x=29,31,longarc(30*Pi/180,30*Pi/180,x*Pi/180)+500))
cpu time = 1 ms, real time = 1 ms.
%1172 = [29, 59, 41.3444979398934670450280297216509190843055]
? dms(solve(x=29,31,longarc(30*Pi/180,30*Pi/180,x*Pi/180)-500))
cpu time = 1 ms, real time = 1 ms.
%1173 = [30, 0, 18.6555020601065329549719702783490809156945]
? dms(solve(x=29,31,latarc(30*Pi/180,x*Pi/180)+500))
cpu time = 1,357 ms, real time = 1,358 ms.
%1174 = [29, 59, 43.7621925447500548285775757329518579545513]
? dms(solve(x=29,31,latarc(30*Pi/180,x*Pi/180)-500))
cpu time = 1,365 ms, real time = 1,368 ms.
%1175 = [30, 0, 16.2377963202802863245716034907838199823349]
?