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?

Comestible

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

Do you mean to UTM? en.wikipedia.org/wiki/… – Myopic

What do you mean by converting a lat/long to meters? meters from where? Are you looking for a way to compute the distance along the surface of the earth from one coordinate to another? – Drabble

These seems like a copy of this question https://mcmap.net/q/137203/-java-convert-lat-lon-to-utm-closed – Myopic

Ok I think I maybe asked the wrong question.. I've updated it. – Comestible

I stumbled upon this question wanting to do SQL queries on latitude and longitude and found this great article with some Java code at the bottom. It might interest you as well. – Interlunation

possible duplicate of Simple calculations for working with lat/lon + km distance? – Urethra

possible duplicate of How do I calculate distance between two latitude-longitude points? – Batista

Most of the answers here are using simple spherical trigonometry, so the results are rather crude compared to the WGS84 ellipsoid distances used in the GPS system. Some of the answers refer to Vincenty's formula for ellipsoids, but that algorithm was designed for use on 1960s' era desk calculators and it has stability & accuracy issues; we have better hardware and software now. Please see GeographicLib for a high quality library with implementations in various languages. (I'm not sure what GDAL uses for its WGS84 work, I think it still uses Vincenty). – Blatant

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.

Maloney

For those looking for a library to convert between wgs and utm: github.com/urbanetic/utm-converter – Aphanite

Would be really grateful if somebody could add in some explanatory comments on the above code. Thanks in advance! – Oxfordshire

Found this which this comment seem to be an adoption of. The link also says its based on this article on distance calculation. So any unanswered questions should be found in the original link. :) – Blackheart

How do I add elevation into this calculation? – Lute

@dangalg, assuming lower distances where the floor is plane, you have also altitudes alt1 and alt2, and dm is the distance in meters (the result of measure function above). You can use the hypothenuse function of JS Math.hypot(x, y), where x is dm and y is max(alt1, alt2) - min(alt1, alt2). – Imitation

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 km of 1° of latitude = always 111.32 km

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

Sherlock

How does the longitude equation work? with a latitude of 90 degrees you'd expect it to show near 111km; but instead it shows 0; similarly, values close to it are also near 0. – Parch

Latitude is 0° at the equator and 90° at the pole (and not the opposite). For equator the formula gives 40075 km * cos(0°) / 360 = 111 km. For pole the formula gives 40075 * cos(90°) / 360 = 0 km. – Sherlock

I think this approach is simple especially as the question didn't ask for the exact distance between two points, but rather if they are "reasonably close enough".With these formulas we easily check if the user is within a square centered on the waypoint. It's much simpler to check for a square than for a circle. – Sherlock

this is far faster than using haversine if u're making an approx, works great thanks – Strychnic

Nope. If on the left side of the equation you have Km, you will have Km also on the right side, not meters – Chough

It's now corrected, the result is in km. – Sherlock

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.

Anchoress

Note that in 2017 the Wikipedia page has another (seems refined) formula. – Burglary

Yes, the formula in Wikipedia is slightly different, but it seem that the other Wikipedia formula is based on the similar results from this great SO answer, where someone actually went through the calculations. – Bridgeport

Keep in mind that in this equation "latMid" is in radians while "m_per_deg_lat" is for degrees. So if you want to compute this for a latitude of 30N (say), in the equation latMid = pi*30/180. – Underplay

I think you have a typo for this: m_per_deg_lon because inputs might need to be lon and not lat. – Ripp

@Ripp No, he is right, the transformation factor for longitude depends on the latitude, since the distance between meridians gets smaller until they meet at the poles, so the m_per_long_degree gets smaller too – Contrabass

Answer 4

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
}

Speedwell

Answer 5

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.

Sylph

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.

Hammons

Answer 7

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.

Longoria

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.

Twobit

No, the nautical mile is defined by international standard (v en.wikipedia.org/wiki/Nautical_mile) to be 1852m. Its relationship to the measurement of an arc on the surface of a spheroid such as the Earth is now both historical and approximate. – Marcum

Answer 9

    '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

Intermolecular

I see the link is full of broken. – Onehorse

Answer 10

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).

Carnify

Answer 11

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 ;

Sifuentes

Answer 12

Original poster asked "If I have a latitude or longitude reading in standard NMEA format is there an easy way / formula to convert that reading to meters"

I haven't used Java in a while so I did the solution here in "PARI". Just plug your point's latitude and longitudes into the equations below to get the exact arc lengths and scales in meters per (second of Longitude) and meters per (second of Latitude). I wrote these equations for the free-open-source-mac-pc math program "PARI". You can just paste the following into it and the I will show how to apply them to two made up points:

\\=======Arc lengths along Latitude and Longitude and the respective scales:
\p300
default(format,"g.42")
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;
\\=======

To apply that to your type of problem I will make up that one of your data points was at

[Latitude, Longitude]=[+30, 30]

and the other at

[Latitude, Longitude]=[+30:00:16.237796,30:00:18.655502].

To convert those points to meters in two coordinates:

I can setup a system of coordinates in meters with the first point being at the origin: [0,0] meters. Then I can define the coordinate x-axis as due East-West, and the y-axis as due North-South.

Then the second point's coordinates are:

? [longarc(30*Pi/180,30*Pi/180,((18.655502/60+0)/60+30)*Pi/180),latarc(30*Pi/180,((16.237796/60+0)/60+30)*Pi/180)]
%9 = [499.999998389040060103621525561027349597207, 499.999990137812119668486524932382720606325]

Warning on precision: Note however: Since the surface of the Earth is curved, 2-dimensional coordinates obtained on it can't follow the same rules as cartesian coordinates such as the Pythagorean Theorem perfectly. Also lines pointing due North-South converge in the Northern Hemisphere. At the North Pole it becomes obvious that North-South lines won't serve well for lines parallel to the y-axis on a map.

At 30 degrees Latitude with 500 meter lengths, the x-coordinate changes by 1.0228 inches if the scale is set from [0,+500] instead of [0,0]:

? [longarc(((18.655502/60+0)/60+30)*Pi/180,30*Pi/180,((18.655502/60+0)/60+30)*Pi/180),latarc(30*Pi/180,((16.237796/60+0)/60+30)*Pi/180)]
%10 = [499.974018595036400823218815901067566617826, 499.999990137812119668486524932382720606325]
? (%10[1]-%9[1])*1000/25.4
%12 = -1.02282653557713702372872677007019603860352
? 

The error there of 500meters/1inch is only about 1/20000, good enough for most diagrams, but one might want to reduce the 1 inch error. For a completely general way to convert lat,long to orthogonal x,y coordinates for any point on the globe, I would chose to abandon aligning coordinate lines with East-West and North-South, except still keeping the center y-axis pointing due North. For example you could rotate the globe around the poles (around the 3-D Z-axis)
so the center point in your map is at longitude zero. Then tilt the globe (around the 3-D y-axis) to bring your center point to lat,long = [0,0]. On the globe points at lat,long = [0,0] are farthest from the poles and have a lat,long grid around them that is most orthogonal so you can use these new "North-South", "East-West" lines as coordinate x,y lines without incurring the stretching that would have occurred doing that before rotating the center point away from the pole. Showing an explicit example of that would take a lot more space.

Witticism

Answer 13

Why limiting to one degree?

The formula is based on the proportion:

distance[m] : distance[deg] = max circumference[m] : 360[deg]

Lets say you are given an angle for a latitude and one for longitude both in degrees: (longitude[deg], latitude[deg])

For the latitude, the max circumference is always the one passing for the poles. In a spherical model, with radius R (in meters) the max circumference is 2 * pi * R and the proportions resolves to:

latitude[m] = ( 2 * pi * R[m] * latitude[deg] ) / 360[deg]

(note that deg and deg simplifies, and what remains is meters on both sides).

For the longitude the max circumference is proportional to the cosine of the latitude (as you can imagine running in circle the north pole is shorter than running in circle around the equator), so it is 2 * pi * R * cos(latitude[rad]). Therefore

longitude distance[m] = ( 2 * pi * R[m] * cos(latitude[rad]) * longitude[deg] ) / 360[deg]

Note that you will have to convert the latitude from deg to rad before computing the cos.

---

Omitting details for who is just looking for the formula:

lat_in_m = 111132.954 * lat_in_degree / 360
lon_in_m = 111132.954 * cos(lat_in_radians) * lon_in_deg ) / 360

Chough

I wrote a more detailed explanation on my blog: geospatial.netlify.app/posts/gds-2024-01-10-haversine-dist – Chough

Answer 14

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.

Faculty

No, that does not work! The x distance in m is different for different values of latitude. At the equator you might get away with it, but the closer you get to the poles the extremer your ellipsoids will get. – Myxoma

While your comment is reasonable, it doesn't answer the user's question about converting the lat/lng degree difference to meters. – Tough

Answer 15

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
  }

Timbal

Answer 16

To directly convert the distance between two latitudes/longitudes to the x y meters between them, you could use the Earth Radius equations from Wikipedia. In C, these are:

N = a/sqrt(1-e2)*(sin(Lat1)*sin(Lat1))));
latlen = ((1-e2)/(a*a))*N*N*N*(Lat1-Lat2)*PI/180;
lonlen = (N*cos(Lat1))*(Lon1-Lon2)*PI/180;

Where a is the radius of the Earth at the equator and e2 is the Earth's eccentricity squared. The equations get the radii between the two points and PI/180 gets the circumferences of those radii.

Lawler

Answer 17

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}.

Penitence

it is assuming the earth is a circle ^^ Some strange people do this nowadays ... but what you mean is probably rather it is assuming the earth is a sphere ;) – Prim

Answer 18

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.

Nathanson

Answer 19

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;

Unpolite

finally, a straightforward answer :) – Aliaalias

what if latitude is -179 and the other is 179, the x distance should be 2 degrees instead of 358 – Cheng

Do not use this answer (for some reason, it's upvoted). There is not a single scaling between longitude and distance; the Earth is not flat. – Yama

I believe it is 111.1 – Gyratory

Note that one degree of longitude is 111 km at the equator, but less for other latitudes. There's a simple approximative formula to find the length in km of 1° of longitude in function of latitude : 1° of longitude = 40000 km * cos (latitude) / 360 (and of course it gives 111 km for latitude = 90°). Also remark that 1° of longitude is almost always a different distance than 1° of latitude. – Sherlock