How to convert latitude or longitude to meters?

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.

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.

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.

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.

    '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

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 ;

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

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
  }

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.

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}.