What is the inverse of the function
math.atan2
I use this in Lua where I can get the inverse of math.atan by math.tan.
But I am lost here.
EDIT
OK, let me give you more details.
I needed to calculate angle between 2 points (x1,y1) and (x2,y2) and I did,
local dy = y1-y2
local dx = x1-x2
local angle = atan2(dy,dx)* 180 / pi
Now if I have the angle, is it possible to get back dy and dx?
Given only the angle you can only derive a unit vector pointing to (dx, dy). To get the original (dx, dy) you also need to know the length of the vector (dx, dy), which I'll call len. You also have to convert the angle you derived from degrees back to radians and then use the trig equations mentioned elsewhere in this post. That is you have:
local dy = y1-y2
local dx = x1-x2
local angle = atan2(dy,dx) * 180 / pi
local len = sqrt(dx*dx + dy*dy)
Given angle (in degrees) and the vector length, len, you can derive dx and dy by:
local theta = angle * pi / 180
local dx = len * cos(theta)
local dy = len * sin(theta)
Apparently, something like this will help:
x = cos(theta)
y = sin(theta)
Simple Google search threw this up, and the guy who asked the question said it solved it.
You'll probably get the wrong numbers if you use:
local dy = y1-y2
local dx = x1-x2
local angle = atan2(dy,dx) * 180 / pi
If you are using the coordinate system where y gets bigger going down the screen and x gets bigger going to the right then you should use:
local dy = y1 - y2
local dx = x2 - x1
local angle = math.deg(math.atan2(dy, dx))
if (angle < 0) then
angle = 360 + angle
end
The reason you want to use this is because atan2 in lua will give you a number between -180 and 180. It will be correct until you hit 180 then as it should go beyond 180 (i.e. 187) it will invert it to a negative number going down from -180 to 0 as you get closer to 360. To correct for this we check to see if the angle is less than 0 and if it is we add 360 to give us the correct angle.
According this reference:
Returns the arc tangent of y/x (in radians), but uses the signs of
both parameters to find the quadrant of the result. (It also handles
correctly the case of x being zero.)
So I guess you can use math.tan to invert it also.
As atan2 works as tan-1, so the inverse could be tan, taking into consideration conversion between radian and degree
I'm using google calculator for an equation. The number sixty is a degree.
If I put it in like this, I get:
80 * cos(60) = -76.1930384
But if I put the word 'degrees' I get this:
80 * cos(60degrees) = 40
Which one is the right answer?
If you do not specify it to be actual degrees, it will assume it is in PI. Once around the sphere equals 2* PI, which means that your 60 equals 19.09 PI, meaning you got the cosinus for 1.09 PI, or 196 degrees.
Both answers are correct - first one calculates cosinus of 60 radians and the second of 60 degrees.
By default the 60 alone will be considered 60 radians, so that's why you get a different result
80 * cos(60) uses Radians
80 * cos(60degrees) uses Degrees
Since your value (60) is in Degrees, add the keyword 'degrees'
The first calculation is in Radians, the second is (obviously) in Degrees.
I am just stuck in the strange calculation.
I am trying to find distance by latitude and longitude by using this formula mentioned in this site.
http://www.movable-type.co.uk/scripts/latlong.html
every thing is Ok there when I do it by calculator but when I started doing it by programmatically there is a difference in result.
here is my part of code
lat = Convert.ToDouble( (s3[a]));
longt =Convert.ToDouble( (s4[a]));
Avg_lat = pc._Latitude - lat;
Avg_longt = pc._Longitude - longt;
a_hold =( Math.Sin(Avg_lat / 2) * Math.Sin(Avg_lat / 2) )+ (Math.Cos(pc._Latitude) * Math.Cos(lat)) * (Math.Sin(Avg_longt / 2) * Math.Sin(Avg_longt / 2));
c_hold = 2 * Math.Atan2(Math.Sqrt(a_hold), Math.Sqrt(1 - a_hold));
// c_hold = (c_hold * 180) / 3.12;
distance = 6371 * c_hold;
the problem is Math funcation here calculate the things in radian I want it in degree. so I just found a formula to convert radian into degree but that formula doesn't convert the calculation correctly.
I am just confused what to do bcoz there isn't any other option to convert radian into degree..
Are you converting correctly and in all places necessary? The correct radians-to-degrees formula is to multiply by (180.0 / Math.PI) and the degrees-to-radians formula is to multiply by (Math.PI / 180.0).
You need to make a degrees-to-radians conversion on the arguments to Math.Sin and Math.Cos. You would need to make a radians-to-degrees conversion on the return value of Math.Atan2 if you are trying to simulate the output of a scientific calculator in degrees mode, but the formula from the web page you linked to says to not do so (only convert the inputs).
You can find a Java implementation of a function that computes the distance between two latitude-longitude pairs in LinkedIn's datafu project.
Your calculator will have a radians option, so you verify your conversion.
Radians = Degrees * PI / 180
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?
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.
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
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.
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
}
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
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).
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
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.
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.
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;
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}.
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.