Sun's position in Swift - r
I am trying to implement this solution for calculating the sun's position in Swift3. I then wrap this in another function that simply cycles through a day from midnight stepping every 10 minutes until 23:50.
I do not really understand R and there are some details of the answer I do not fully comprehend, notably what appears to be some sort of if/clamp function with the square brackets. I did my best, comparing with the Python version when I got confused. Otherwise the only differences are due to the use of NSDate, which simplified some of the code at the top.
Some of the values I get back seem correct and I can see the basis of a curve when I plot the results. However, the result from one call, say 7AM, and then the next, 7:10, are wildly different.
I strongly suspect I did something wrong with the clamping, and that minor changes in the inputs get mod/trunced in different ways and swing the output. But I can't spot it. Can anyone who understands this algo help?
Here's a sample of the output I'm getting:
2017-06-21 00:10:00 +0000 -16.0713262209521 31.7135341633943
2017-06-21 00:20:00 +0000 61.9971433936385 129.193513530349
2017-06-21 00:30:00 +0000 22.5263575559266 78.5445189561018
2017-06-21 00:40:00 +0000 29.5973897349096 275.081637736092
2017-06-21 00:50:00 +0000 41.9552795956374 262.989819486864
As you can see, it swings wildly between iterations. The Earth does not turn that way! My code follows, this version simply sends the results to the log:
class func julianDayFromDate(_ date: Date) -> Double {
let ti = date.timeIntervalSince1970
return ((ti / 86400.0) + 2440587)
}
class func sunPath(lat: Double, lon: Double, size: CGSize) -> UIImage {
var utzCal = Calendar(identifier: .gregorian)
utzCal.timeZone = TimeZone(secondsFromGMT: 0)!
let year = utzCal.component(.year, from: Date())
let june = DateComponents(calendar: utzCal, year: year, month: 6, day: 21).date!
// now we loop for every 10 minutes (2 degrees) and plot those points
for time in stride(from:0, to:(24 * 60), by: 10) {
let calcdate = june.addingTimeInterval(Double(time) * 60.0)
let (alt, az) = sun(date: calcdate, lat: lat, lon: lon)
print(calcdate, alt, az)
}
class func sun(date: Date, lat: Double, lon: Double) -> (altitude: Double, azimuth: Double) {
// these come in handy
let twopi = Double.pi * 2
let deg2rad = Double.pi / 180.0
// latitude to radians
let lat_radians = lat * deg2rad
// the Astronomer's Almanac method used here is based on Epoch 2000, so we need to
// convert the date into that format. We start by calculating "n", the number of
// days since 1 January 2000
let n = julianDayFromDate(date) - 2451545.0
// it continues by calculating the position in ecliptic coordinates,
// starting with the mean longitude of the sun in degrees, corrected for aberation
var meanlong_degrees = 280.460 + (0.9856474 * n)
meanlong_degrees = meanlong_degrees.truncatingRemainder(dividingBy: 360.0)
// and the mean anomaly in degrees
var meananomaly_degrees = 357.528 + (0.9856003 * n)
meananomaly_degrees = meananomaly_degrees.truncatingRemainder(dividingBy: 360.0)
let meananomaly_radians = meananomaly_degrees * deg2rad
// and finally, the eliptic longitude in degrees
var elipticlong_degrees = meanlong_degrees + (1.915 * sin(meananomaly_radians)) + (0.020 * sin(2 * meananomaly_radians))
elipticlong_degrees = elipticlong_degrees.truncatingRemainder(dividingBy: 360.0)
let elipticlong_radians = elipticlong_degrees * deg2rad
// now we want to convert that to equatorial coordinates
let obliquity_degrees = 23.439 - (0.0000004 * n)
let obliquity_radians = obliquity_degrees * deg2rad
// right ascention in radians
let num = cos(obliquity_radians) * sin(elipticlong_radians)
let den = cos(elipticlong_radians)
var ra_radians = atan(num / den)
ra_radians = ra_radians.truncatingRemainder(dividingBy: Double.pi)
if den < 0 {
ra_radians = ra_radians + Double.pi
} else if num < 0 {
ra_radians = ra_radians + twopi
}
// declination is simpler...
let dec_radians = asin(sin(obliquity_radians) * sin(elipticlong_radians))
// and from there, to local coordinates
// start with the UTZ sidereal time
let cal = Calendar.current
let h = Double(cal.component(.hour, from: date))
let m = Double(cal.component(.minute, from: date))
let f: Double
if h == 0 && m == 0 {
f = 0.0
} else if h == 0 {
f = 60.0 / m
} else if h == 0 {
f = 24.0 / h
} else {
f = (24.0 / h) + (60.0 / m)
}
var utz_sidereal_time = 6.697375 + 0.0657098242 * n + f
utz_sidereal_time = utz_sidereal_time.truncatingRemainder(dividingBy: 24.0)
// then convert that to local sidereal time
var localtime = utz_sidereal_time + lon / 15.0
localtime = localtime.truncatingRemainder(dividingBy: 24.0)
var localtime_radians = localtime * 15.0 * deg2rad
localtime_radians = localtime.truncatingRemainder(dividingBy: Double.pi)
// hour angle in radians
var hourangle_radians = localtime_radians - ra_radians
hourangle_radians = hourangle_radians.truncatingRemainder(dividingBy: twopi)
// get elevation in degrees
let elevation_radians = (asin(sin(dec_radians) * sin(lat_radians) + cos(dec_radians) * cos(lat_radians) * cos(hourangle_radians)))
let elevation_degrees = elevation_radians / deg2rad
// and azimuth
let azimuth_radians = asin( -cos(dec_radians) * sin(hourangle_radians) / cos(elevation_radians))
// now clamp the output
let azimuth_degrees: Double
if (sin(dec_radians) - sin(elevation_radians) * sin(lat_radians) < 0) {
azimuth_degrees = (Double.pi - azimuth_radians) / deg2rad
} else if (sin(azimuth_radians) < 0) {
azimuth_degrees = (azimuth_radians + twopi) / deg2rad
} else {
azimuth_degrees = azimuth_radians / deg2rad
}
return (elevation_degrees, azimuth_degrees)
}
Ok, after downloading an R interpreter for OSX, finding that it had no debugger, discovering that there are multiple ways to do a print all with their own caveats, etc etc, I found the problem I was looking for. It was indeed clamping one of the values incorrectly. Here is a working Swift3 version that should be easy to convert to any C-like language and easier to read than the originals. You will have to provide your own versions of the first two functions that work with the date format of your target platform. And the truncatingRemainer is someone's ferbile idea that there shouldn't be a % operator on Double, it's a normal MOD.
// convinience method to return a unit-epoch data from a julian date
class func dateFromJulianDay(_ julianDay: Double) -> Date {
let unixTime = (julianDay - 2440587) * 86400.0
return Date(timeIntervalSince1970: unixTime)
}
class func julianDayFromDate(_ date: Date) -> Double {
//==let JD = Integer(365.25 * (Y + 4716)) + Integer(30.6001 * (M +1)) +
let ti = date.timeIntervalSince1970
return ((ti / 86400.0) + 2440587.5)
}
// calculate the elevation and azimuth of the sun for a given date and location
class func sun(date: Date, lat: Double, lon: Double) -> (altitude: Double, azimuth: Double) {
// these come in handy
let twopi = Double.pi * 2
let deg2rad = Double.pi / 180.0
// latitude to radians
let lat_radians = lat * deg2rad
// the Astronomer's Almanac method used here is based on Epoch 2000, so we need to
// convert the date into that format. We start by calculating "n", the number of
// days since 1 January 2000. So if your date format is 1970-based, convert that
// a pure julian date and pass that in. If your date is 2000-based, then
// just let n = date
let n = julianDayFromDate(date) - 2451545.0
// it continues by calculating the position in ecliptic coordinates,
// starting with the mean longitude of the sun in degrees, corrected for aberation
var meanlong_degrees = 280.460 + (0.9856474 * n)
meanlong_degrees = meanlong_degrees.truncatingRemainder(dividingBy: 360.0)
// and the mean anomaly in degrees
var meananomaly_degrees = 357.528 + (0.9856003 * n)
meananomaly_degrees = meananomaly_degrees.truncatingRemainder(dividingBy: 360.0)
let meananomaly_radians = meananomaly_degrees * deg2rad
// and finally, the eliptic longitude in degrees
var elipticlong_degrees = meanlong_degrees + (1.915 * sin(meananomaly_radians)) + (0.020 * sin(2 * meananomaly_radians))
elipticlong_degrees = elipticlong_degrees.truncatingRemainder(dividingBy: 360.0)
let elipticlong_radians = elipticlong_degrees * deg2rad
// now we want to convert that to equatorial coordinates
let obliquity_degrees = 23.439 - (0.0000004 * n)
let obliquity_radians = obliquity_degrees * deg2rad
// right ascention in radians
let num = cos(obliquity_radians) * sin(elipticlong_radians)
let den = cos(elipticlong_radians)
var ra_radians = atan(num / den)
ra_radians = ra_radians.truncatingRemainder(dividingBy: Double.pi)
if den < 0 {
ra_radians = ra_radians + Double.pi
} else if num < 0 {
ra_radians = ra_radians + twopi
}
// declination is simpler...
let dec_radians = asin(sin(obliquity_radians) * sin(elipticlong_radians))
// and from there, to local coordinates
// start with the UTZ sidereal time, which is probably a lot easier in non-Swift languages
var utzCal = Calendar(identifier: .gregorian)
utzCal.timeZone = TimeZone(secondsFromGMT: 0)!
let h = Double(utzCal.component(.hour, from: date))
let m = Double(utzCal.component(.minute, from: date))
let f: Double // universal time in hours and decimals (not days!)
if h == 0 && m == 0 {
f = 0.0
} else if h == 0 {
f = m / 60.0
} else if m == 0 {
f = h
} else {
f = h + (m / 60.0)
}
var utz_sidereal_time = 6.697375 + 0.0657098242 * n + f
utz_sidereal_time = utz_sidereal_time.truncatingRemainder(dividingBy: 24.0)
// then convert that to local sidereal time
var localtime = utz_sidereal_time + lon / 15.0
localtime = localtime.truncatingRemainder(dividingBy: 24.0)
let localtime_radians = localtime * 15.0 * deg2rad
// hour angle in radians
var hourangle_radians = localtime_radians - ra_radians
hourangle_radians = hourangle_radians.truncatingRemainder(dividingBy: twopi)
// get elevation in degrees
let elevation_radians = (asin(sin(dec_radians) * sin(lat_radians) + cos(dec_radians) * cos(lat_radians) * cos(hourangle_radians)))
let elevation_degrees = elevation_radians / deg2rad
// and azimuth
let azimuth_radians = asin( -cos(dec_radians) * sin(hourangle_radians) / cos(elevation_radians))
// now clamp the output
let azimuth_degrees: Double
if (sin(dec_radians) - sin(elevation_radians) * sin(lat_radians) < 0) {
azimuth_degrees = (Double.pi - azimuth_radians) / deg2rad
} else if (sin(azimuth_radians) < 0) {
azimuth_degrees = (azimuth_radians + twopi) / deg2rad
} else {
azimuth_degrees = azimuth_radians / deg2rad
}
// all done!
return (elevation_degrees, azimuth_degrees)
}
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radian atan2 is showing opposite values then what they need to be
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How to make an animated wave in threejs
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Normalised compass bearing from start to end point as lat, lng pairs
From the net: Since atan2 returns values in the range -π ... +π (that is, -180° ... +180°), to normalise the result to a compass bearing (in the range 0° ... 360°, with -ve values transformed into the range 180° ... 360°), convert to degrees and then use (θ+360) % 360, where % is modulo. I'm using VB6 and my code is like this brng = (θ + 360) Mod 360 where θ = -68 I'm getting the answer 292°,but expected answer is 248°. am i committing a mistake..? or am I missing something.? Please help me. Update: I'll further explain my question, Dim b As Double Dim x As Double Dim y As Double Dim Dlat As Double Dim DLon As Double Private Const pi As Double = 3.14159265358979 Public Function Atn2(ByVal y As Double, ByVal x As Double) As Double On Error GoTo DivideError Atn2 = Atn(y / x) If (x < 0) Then If (y < 0) Then Atn2 = Atn2 - vbPI Else Atn2 = Atn2 + vbPI End If Exit Function DivideError: If Abs(y) > Abs(x) Then 'Must be an overflow If y > 0 Then Atn2 = vbPI / 2 Else Atn2 = -vbPI / 2 Else Atn2 = 0 'Must be an underflow End If Resume Next End Function Public Sub AFAMP() lat1 = Val(Text1.Text) * pi / 180 'conveting to radians lat2 = Val(Text2.Text) * pi / 180 Long1 = Val(Text3.Text) * pi / 180 Long2 = Val(Text4.Text) * pi / 180 Dlat = (lat1 - lat2) DLon = (Long1 - Long2) y = Math.Sin(DLon) * Math.Cos(lat2) x = Math.Cos(lat1) * Math.Sin(lat2) - Math.Sin(lat1) * Math.Cos(lat2) * Math.Cos(DLon) b = Atn2(y, x) * (180 / pi) Text5.Text = (b +360) mod 360 End Sub coordinates are as below in decimal degrees lat1 = Val(7.337361) * pi / 180 lat2 = Val(7.000667) * pi / 180 Long1 = Val(81.626198) * pi / 180 Long2 = Val(80.773737) * pi / 180 I have checked with an online coordinate calculator, the answer is 248. In my case it is 292 degrees. Can anyone please help me? I'm stuck.
As Deanna said in a comment, 292 is the correct answer. -68 + 360 = 292
This code is JavaScript because I can't remember VB syntax. Convert to degrees. Don't repeat this code, make it a method on Number. Or a first class function with two parameters, whatever your target language supports. Number.prototype.toDegrees = function (unit) { const u = (unit || "d").toLowerCase().charAt(0); switch (u) { case 'r': //radians return this * 180 / Math.PI; case 'g': //grads return this * 200 / Math.PI; case 'd': //degrees return this; default: //radians return this * 180 / Math.PI; } }; Normalise the range. Number.prototype.toNormalisedDegrees = function () { return (this - 360 * (Math.trunc(this / 360) - 1)) % 360; } Rotate 90 degrees and reverse the direction. Number.prototype.toCompassBearing = function (unit) { const degrees = unit && unit.toLowerCase().charAt(0) === "r" ? this.toDegrees() : this; return (90 - degrees).toNormalisedDegrees(); } Test with some known values Number.prototype.toDegrees = function (unit) { const u = (unit || "d").toLowerCase().charAt(0); switch (u) { case 'r': return this * 180 / Math.PI; case 'g': return this * 200 / Math.PI; default: return this; } }; Number.prototype.toNormalisedDegrees = function () { return (this - 360 * (Math.trunc(this / 360) - 1)) % 360; } Number.prototype.toCompassBearing = function (unit) { return (90 - this.toDegrees(unit)).toNormalisedDegrees(); } console.log(`0 cartesian (E) => ${(0).toCompassBearing("d")} compass bearing`); console.log(`90 cartesian (N) => ${(90).toCompassBearing("d")} compass bearing`); console.log(`90-22.5 cartesian (NNE) => ${(90-22.5).toCompassBearing("d")} compass bearing`); console.log(`180 cartesian (W) => ${(180).toCompassBearing("d")} compass bearing`); console.log(`-68 cartesian => ${(-68).toCompassBearing("d")} compass bearing`); console.log(`-68 compass bearing => ${(-68).toNormalisedDegrees()} normalised degrees`); console.log(`270 cartesian (S) => ${(270).toCompassBearing("d")} compass bearing`); console.log(`0 cartesian (E) => ${(0).toCompassBearing("r")} compass bearing`); console.log(`PI/2 cartesian (N) => ${(0.5 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`PI cartesian (W) => ${Math.PI.toCompassBearing("r")} compass bearing`); console.log(`1.5 PI cartesian (S) => ${(1.5 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`2 PI cartesian (E) => ${(2 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`20 PI cartesian (E) => ${(20 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`-2 PI cartesian (E) => ${(-2 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`-2.5 PI cartesian (S) => ${(-2.5 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`-3 PI cartesian (W) => ${(-3 * Math.PI).toCompassBearing("r")} compass bearing`); console.log(`-3.5 PI cartesian (N) => ${(-3.5 * Math.PI).toCompassBearing("r")} compass bearing`); If 292 is right, what's this 248 business? You asked about converting a compass heading into normalised degrees, but the sample code you provide and the result you think is correct both suggest that what you actually want to do is determine the heading defined by start and end points expressed as Lat,Lng pairs, and express this in normalised degrees. There's an article here suggesting a method and your code looks like a botched attempt to implement this. This is what it looks like if you get it right in JavaScript. Number.prototype.toDegrees = function(unit) { const u = (unit || "d").toLowerCase().charAt(0); switch (u) { case 'r': return this * 180 / Math.PI; case 'g': return this * 200 / Math.PI; default: return this; } }; Number.prototype.toRadians = function() { return this * Math.PI / 180; } Number.prototype.toNormalisedDegrees = function() { return (this - 360 * (Math.trunc(this / 360) - 1)) % 360; } function bearing(lat1, lng1, lat2, lng2) { const X = Math.cos(lat2) * Math.sin(lng2 - lng1); const Y = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(lng2 - lng1); return Math.atan2(X, Y); } const lat1 = (7.337361).toRadians(); const lng1 = (81.626198).toRadians(); const lat2 = (7.000667).toRadians(); const lng2 = (80.773737).toRadians(); const bearingRadians = bearing(lat1, lng1, lat2, lng2); const bearingDegrees = bearingRadians.toDegrees("radians"); console.log(`bearingRadians = ${bearingRadians}`); console.log(`bearingDegrees = ${bearingDegrees}`); console.log(`normalisedDegrees = ${bearingDegrees.toNormalisedDegrees()}`); There are a couple of things to note. Latitude and Longitude are in the compass frame of reference and do not require conversion from the cartesian frame of reference. This code is written so that the result is in the compass frame of reference. The result is in radians and then converted to degrees. The degrees are then normalised to the 0-360 range. Using your inputs I get your expected output.
Calculate distance between 2 GPS coordinates
How do I calculate distance between two GPS coordinates (using latitude and longitude)?
Calculate the distance between two coordinates by latitude and longitude, including a Javascript implementation. West and South locations are negative. Remember minutes and seconds are out of 60 so S31 30' is -31.50 degrees. Don't forget to convert degrees to radians. Many languages have this function. Or its a simple calculation: radians = degrees * PI / 180. function degreesToRadians(degrees) { return degrees * Math.PI / 180; } function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) { var earthRadiusKm = 6371; var dLat = degreesToRadians(lat2-lat1); var dLon = degreesToRadians(lon2-lon1); lat1 = degreesToRadians(lat1); lat2 = degreesToRadians(lat2); var a = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2); var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); return earthRadiusKm * c; } Here are some examples of usage: distanceInKmBetweenEarthCoordinates(0,0,0,0) // Distance between same // points should be 0 0 distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London // to Arlington 5918.185064088764
Look for haversine with Google; here is my solution: #include <math.h> #include "haversine.h" #define d2r (M_PI / 180.0) //calculate haversine distance for linear distance double haversine_km(double lat1, double long1, double lat2, double long2) { double dlong = (long2 - long1) * d2r; double dlat = (lat2 - lat1) * d2r; double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2); double c = 2 * atan2(sqrt(a), sqrt(1-a)); double d = 6367 * c; return d; } double haversine_mi(double lat1, double long1, double lat2, double long2) { double dlong = (long2 - long1) * d2r; double dlat = (lat2 - lat1) * d2r; double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2); double c = 2 * atan2(sqrt(a), sqrt(1-a)); double d = 3956 * c; return d; }
C# Version of Haversine double _eQuatorialEarthRadius = 6378.1370D; double _d2r = (Math.PI / 180D); private int HaversineInM(double lat1, double long1, double lat2, double long2) { return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2)); } private double HaversineInKM(double lat1, double long1, double lat2, double long2) { double dlong = (long2 - long1) * _d2r; double dlat = (lat2 - lat1) * _d2r; double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D); double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a)); double d = _eQuatorialEarthRadius * c; return d; } Here's a .NET Fiddle of this, so you can test it out with your own Lat/Longs.
Java Version of Haversine Algorithm based on Roman Makarov`s reply to this thread public class HaversineAlgorithm { static final double _eQuatorialEarthRadius = 6378.1370D; static final double _d2r = (Math.PI / 180D); public static int HaversineInM(double lat1, double long1, double lat2, double long2) { return (int) (1000D * HaversineInKM(lat1, long1, lat2, long2)); } public static double HaversineInKM(double lat1, double long1, double lat2, double long2) { double dlong = (long2 - long1) * _d2r; double dlat = (lat2 - lat1) * _d2r; double a = Math.pow(Math.sin(dlat / 2D), 2D) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r) * Math.pow(Math.sin(dlong / 2D), 2D); double c = 2D * Math.atan2(Math.sqrt(a), Math.sqrt(1D - a)); double d = _eQuatorialEarthRadius * c; return d; } }
This is very easy to do with geography type in SQL Server 2008. SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326)) -- computes distance in meters using eliptical model, accurate to the mm 4326 is SRID for WGS84 elipsoidal Earth model
Here's a Haversine function in Python that I use: from math import pi,sqrt,sin,cos,atan2 def haversine(pos1, pos2): lat1 = float(pos1['lat']) long1 = float(pos1['long']) lat2 = float(pos2['lat']) long2 = float(pos2['long']) degree_to_rad = float(pi / 180.0) d_lat = (lat2 - lat1) * degree_to_rad d_long = (long2 - long1) * degree_to_rad a = pow(sin(d_lat / 2), 2) + cos(lat1 * degree_to_rad) * cos(lat2 * degree_to_rad) * pow(sin(d_long / 2), 2) c = 2 * atan2(sqrt(a), sqrt(1 - a)) km = 6367 * c mi = 3956 * c return {"km":km, "miles":mi}
I needed to calculate a lot of distances between the points for my project, so I went ahead and tried to optimize the code, I have found here. On average in different browsers my new implementation runs 2 times faster than the most upvoted answer. function distance(lat1, lon1, lat2, lon2) { var p = 0.017453292519943295; // Math.PI / 180 var c = Math.cos; var a = 0.5 - c((lat2 - lat1) * p)/2 + c(lat1 * p) * c(lat2 * p) * (1 - c((lon2 - lon1) * p))/2; return 12742 * Math.asin(Math.sqrt(a)); // 2 * R; R = 6371 km } You can play with my jsPerf and see the results here. Recently I needed to do the same in python, so here is a python implementation: from math import cos, asin, sqrt def distance(lat1, lon1, lat2, lon2): p = 0.017453292519943295 a = 0.5 - cos((lat2 - lat1) * p)/2 + cos(lat1 * p) * cos(lat2 * p) * (1 - cos((lon2 - lon1) * p)) / 2 return 12742 * asin(sqrt(a)) And for the sake of completeness: Haversine on wiki.
It depends on how accurate you need it to be. If you need pinpoint accuracy, it is best to look at an algorithm which uses an ellipsoid, rather than a sphere, such as Vincenty's algorithm, which is accurate to the mm.
Here it is in C# (lat and long in radians): double CalculateGreatCircleDistance(double lat1, double long1, double lat2, double long2, double radius) { return radius * Math.Acos( Math.Sin(lat1) * Math.Sin(lat2) + Math.Cos(lat1) * Math.Cos(lat2) * Math.Cos(long2 - long1)); } If your lat and long are in degrees then divide by 180/PI to convert to radians.
PHP version: (Remove all deg2rad() if your coordinates are already in radians.) $R = 6371; // km $dLat = deg2rad($lat2-$lat1); $dLon = deg2rad($lon2-$lon1); $lat1 = deg2rad($lat1); $lat2 = deg2rad($lat2); $a = sin($dLat/2) * sin($dLat/2) + sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2); $c = 2 * atan2(sqrt($a), sqrt(1-$a)); $d = $R * $c;
A T-SQL function, that I use to select records by distance for a center Create Function [dbo].[DistanceInMiles] ( #fromLatitude float , #fromLongitude float , #toLatitude float, #toLongitude float ) returns float AS BEGIN declare #distance float select #distance = cast((3963 * ACOS(round(COS(RADIANS(90-#fromLatitude))*COS(RADIANS(90-#toLatitude))+ SIN(RADIANS(90-#fromLatitude))*SIN(RADIANS(90-#toLatitude))*COS(RADIANS(#fromLongitude-#toLongitude)),15)) )as float) return round(#distance,1) END
I. Regarding "Breadcrumbs" method Earth radius is different on different Lat. This must be taken into consideration in Haversine algorithm. Consider Bearing change, which turns straight lines to arches (which are longer) Taking Speed change into account will turn arches to spirals (which are longer or shorter than arches) Altitude change will turn flat spirals to 3D spirals (which are longer again). This is very important for hilly areas. Below see the function in C which takes #1 and #2 into account: double calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1, double rLat2, double rLon2, double rHeading2){ double rDLatRad = 0.0; double rDLonRad = 0.0; double rLat1Rad = 0.0; double rLat2Rad = 0.0; double a = 0.0; double c = 0.0; double rResult = 0.0; double rEarthRadius = 0.0; double rDHeading = 0.0; double rDHeadingRad = 0.0; if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0) || (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0) || (rLon2 > 180.0)) { return -1; }; rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS; rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS; rLat1Rad = rLat1 * DEGREE_TO_RADIANS; rLat2Rad = rLat2 * DEGREE_TO_RADIANS; a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin( rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad); if (a == 0.0) { return 0.0; } c = 2 * atan2(sqrt(a), sqrt(1 - a)); rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2)) / 2.0)); rResult = rEarthRadius * c; // Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0) && (rHeading2 < 360.0)) { rDHeading = fabs(rHeading1 - rHeading2); if (rDHeading > 180.0) { rDHeading -= 180.0; } rDHeadingRad = rDHeading * DEGREE_TO_RADIANS; if (rDHeading > 5.0) { rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2))); } else { rResult = rResult / cos(rDHeadingRad); } } return rResult; } II. There is an easier way which gives pretty good results. By Average Speed. Trip_distance = Trip_average_speed * Trip_time Since GPS Speed is detected by Doppler effect and is not directly related to [Lon,Lat] it can be at least considered as secondary (backup or correction) if not as main distance calculation method.
If you need something more accurate then have a look at this. Vincenty's formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid, developed by Thaddeus Vincenty (1975a) They are based on the assumption that the figure of the Earth is an oblate spheroid, and hence are more accurate than methods such as great-circle distance which assume a spherical Earth. The first (direct) method computes the location of a point which is a given distance and azimuth (direction) from another point. The second (inverse) method computes the geographical distance and azimuth between two given points. They have been widely used in geodesy because they are accurate to within 0.5 mm (0.020″) on the Earth ellipsoid.
If you're using .NET don't reivent the wheel. See System.Device.Location. Credit to fnx in the comments in another answer. using System.Device.Location; double lat1 = 45.421527862548828D; double long1 = -75.697189331054688D; double lat2 = 53.64135D; double long2 = -113.59273D; GeoCoordinate geo1 = new GeoCoordinate(lat1, long1); GeoCoordinate geo2 = new GeoCoordinate(lat2, long2); double distance = geo1.GetDistanceTo(geo2);
here is the Swift implementation from the answer func degreesToRadians(degrees: Double) -> Double { return degrees * Double.pi / 180 } func distanceInKmBetweenEarthCoordinates(lat1: Double, lon1: Double, lat2: Double, lon2: Double) -> Double { let earthRadiusKm: Double = 6371 let dLat = degreesToRadians(degrees: lat2 - lat1) let dLon = degreesToRadians(degrees: lon2 - lon1) let lat1 = degreesToRadians(degrees: lat1) let lat2 = degreesToRadians(degrees: lat2) let a = sin(dLat/2) * sin(dLat/2) + sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2) let c = 2 * atan2(sqrt(a), sqrt(1 - a)) return earthRadiusKm * c }
This is version from "Henry Vilinskiy" adapted for MySQL and Kilometers: CREATE FUNCTION `CalculateDistanceInKm`( fromLatitude float, fromLongitude float, toLatitude float, toLongitude float ) RETURNS float BEGIN declare distance float; select 6367 * ACOS( round( COS(RADIANS(90-fromLatitude)) * COS(RADIANS(90-toLatitude)) + SIN(RADIANS(90-fromLatitude)) * SIN(RADIANS(90-toLatitude)) * COS(RADIANS(fromLongitude-toLongitude)) ,15) ) into distance; return round(distance,3); END;
This Lua code is adapted from stuff found on Wikipedia and in Robert Lipe's GPSbabel tool: local EARTH_RAD = 6378137.0 -- earth's radius in meters (official geoid datum, not 20,000km / pi) local radmiles = EARTH_RAD*100.0/2.54/12.0/5280.0; -- earth's radius in miles local multipliers = { radians = 1, miles = radmiles, mi = radmiles, feet = radmiles * 5280, meters = EARTH_RAD, m = EARTH_RAD, km = EARTH_RAD / 1000, degrees = 360 / (2 * math.pi), min = 60 * 360 / (2 * math.pi) } function gcdist(pt1, pt2, units) -- return distance in radians or given units --- this formula works best for points close together or antipodal --- rounding error strikes when distance is one-quarter Earth's circumference --- (ref: wikipedia Great-circle distance) if not pt1.radians then pt1 = rad(pt1) end if not pt2.radians then pt2 = rad(pt2) end local sdlat = sin((pt1.lat - pt2.lat) / 2.0); local sdlon = sin((pt1.lon - pt2.lon) / 2.0); local res = sqrt(sdlat * sdlat + cos(pt1.lat) * cos(pt2.lat) * sdlon * sdlon); res = res > 1 and 1 or res < -1 and -1 or res res = 2 * asin(res); if units then return res * assert(multipliers[units]) else return res end end
private double deg2rad(double deg) { return (deg * Math.PI / 180.0); } private double rad2deg(double rad) { return (rad / Math.PI * 180.0); } private double GetDistance(double lat1, double lon1, double lat2, double lon2) { //code for Distance in Kilo Meter double theta = lon1 - lon2; double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta)); dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0)); return (dist); } private double GetDirection(double lat1, double lon1, double lat2, double lon2) { //code for Direction in Degrees double dlat = deg2rad(lat1) - deg2rad(lat2); double dlon = deg2rad(lon1) - deg2rad(lon2); double y = Math.Sin(dlon) * Math.Cos(lat2); double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon); double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0); if (direct < 0) direct = direct + 360; return (direct); } private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime) { //code for speed in Kilo Meter/Hour TimeSpan TimeDifference = CurTime.Subtract(PrevTime); double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0); double theta = lon1 - lon2; double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta)); dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344; double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0)); return (Speed); } private double GetDuration(DateTime CurTime, DateTime PrevTime) { //code for speed in Kilo Meter/Hour TimeSpan TimeDifference = CurTime.Subtract(PrevTime); double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0)); return (TimeDifferenceInSeconds); }
i took the top answer and used it in a Scala program import java.lang.Math.{atan2, cos, sin, sqrt} def latLonDistance(lat1: Double, lon1: Double)(lat2: Double, lon2: Double): Double = { val earthRadiusKm = 6371 val dLat = (lat2 - lat1).toRadians val dLon = (lon2 - lon1).toRadians val latRad1 = lat1.toRadians val latRad2 = lat2.toRadians val a = sin(dLat / 2) * sin(dLat / 2) + sin(dLon / 2) * sin(dLon / 2) * cos(latRad1) * cos(latRad2) val c = 2 * atan2(sqrt(a), sqrt(1 - a)) earthRadiusKm * c } i curried the function in order to be able to easily produce functions that have one of the two locations fixed and require only a pair of lat/lon to produce distance.
Here's a Kotlin variation: import kotlin.math.* class HaversineAlgorithm { companion object { private const val MEAN_EARTH_RADIUS = 6371.008 private const val D2R = Math.PI / 180.0 } private fun haversineInKm(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double { val lonDiff = (lon2 - lon1) * D2R val latDiff = (lat2 - lat1) * D2R val latSin = sin(latDiff / 2.0) val lonSin = sin(lonDiff / 2.0) val a = latSin * latSin + (cos(lat1 * D2R) * cos(lat2 * D2R) * lonSin * lonSin) val c = 2.0 * atan2(sqrt(a), sqrt(1.0 - a)) return MEAN_EARTH_RADIUS * c } }
you can find a implementation of this (with some good explanation) in F# on fssnip here are the important parts: let GreatCircleDistance<[<Measure>] 'u> (R : float<'u>) (p1 : Location) (p2 : Location) = let degToRad (x : float<deg>) = System.Math.PI * x / 180.0<deg/rad> let sq x = x * x // take the sin of the half and square the result let sinSqHf (a : float<rad>) = (System.Math.Sin >> sq) (a / 2.0<rad>) let cos (a : float<deg>) = System.Math.Cos (degToRad a / 1.0<rad>) let dLat = (p2.Latitude - p1.Latitude) |> degToRad let dLon = (p2.Longitude - p1.Longitude) |> degToRad let a = sinSqHf dLat + cos p1.Latitude * cos p2.Latitude * sinSqHf dLon let c = 2.0 * System.Math.Atan2(System.Math.Sqrt(a), System.Math.Sqrt(1.0-a)) R * c
I needed to implement this in PowerShell, hope it can help someone else. Some notes about this method Don't split any of the lines or the calculation will be wrong To calculate in KM remove the * 1000 in the calculation of $distance Change $earthsRadius = 3963.19059 and remove * 1000 in the calculation of $distance the to calulate the distance in miles I'm using Haversine, as other posts have pointed out Vincenty's formulae is much more accurate Function MetresDistanceBetweenTwoGPSCoordinates($latitude1, $longitude1, $latitude2, $longitude2) { $Rad = ([math]::PI / 180); $earthsRadius = 6378.1370 # Earth's Radius in KM $dLat = ($latitude2 - $latitude1) * $Rad $dLon = ($longitude2 - $longitude1) * $Rad $latitude1 = $latitude1 * $Rad $latitude2 = $latitude2 * $Rad $a = [math]::Sin($dLat / 2) * [math]::Sin($dLat / 2) + [math]::Sin($dLon / 2) * [math]::Sin($dLon / 2) * [math]::Cos($latitude1) * [math]::Cos($latitude2) $c = 2 * [math]::ATan2([math]::Sqrt($a), [math]::Sqrt(1-$a)) $distance = [math]::Round($earthsRadius * $c * 1000, 0) #Multiple by 1000 to get metres Return $distance }
Scala version def deg2rad(deg: Double) = deg * Math.PI / 180.0 def rad2deg(rad: Double) = rad / Math.PI * 180.0 def getDistanceMeters(lat1: Double, lon1: Double, lat2: Double, lon2: Double) = { val theta = lon1 - lon2 val dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta)) Math.abs( Math.round( rad2deg(Math.acos(dist)) * 60 * 1.1515 * 1.609344 * 1000) ) }
Here's my implementation in Elixir defmodule Geo do #earth_radius_km 6371 #earth_radius_sm 3958.748 #earth_radius_nm 3440.065 #feet_per_sm 5280 #d2r :math.pi / 180 def deg_to_rad(deg), do: deg * #d2r def great_circle_distance(p1, p2, :km), do: haversine(p1, p2) * #earth_radius_km def great_circle_distance(p1, p2, :sm), do: haversine(p1, p2) * #earth_radius_sm def great_circle_distance(p1, p2, :nm), do: haversine(p1, p2) * #earth_radius_nm def great_circle_distance(p1, p2, :m), do: great_circle_distance(p1, p2, :km) * 1000 def great_circle_distance(p1, p2, :ft), do: great_circle_distance(p1, p2, :sm) * #feet_per_sm #doc """ Calculate the [Haversine](https://en.wikipedia.org/wiki/Haversine_formula) distance between two coordinates. Result is in radians. This result can be multiplied by the sphere's radius in any unit to get the distance in that unit. For example, multiple the result of this function by the Earth's radius in kilometres and you get the distance between the two given points in kilometres. """ def haversine({lat1, lon1}, {lat2, lon2}) do dlat = deg_to_rad(lat2 - lat1) dlon = deg_to_rad(lon2 - lon1) radlat1 = deg_to_rad(lat1) radlat2 = deg_to_rad(lat2) a = :math.pow(:math.sin(dlat / 2), 2) + :math.pow(:math.sin(dlon / 2), 2) * :math.cos(radlat1) * :math.cos(radlat2) 2 * :math.atan2(:math.sqrt(a), :math.sqrt(1 - a)) end end
Dart Version Haversine Algorithm. import 'dart:math'; class GeoUtils { static double _degreesToRadians(degrees) { return degrees * pi / 180; } static double distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) { var earthRadiusKm = 6371; var dLat = _degreesToRadians(lat2-lat1); var dLon = _degreesToRadians(lon2-lon1); lat1 = _degreesToRadians(lat1); lat2 = _degreesToRadians(lat2); var a = sin(dLat/2) * sin(dLat/2) + sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2); var c = 2 * atan2(sqrt(a), sqrt(1-a)); return earthRadiusKm * c; } }
In Python, you can use the geopy library to compute the geodesic distance using the WGS84 ellipsoid: from geopy.distance import geodesic newport_ri = (41.49008, -71.312796) cleveland_oh = (41.499498, -81.695391) print(geodesic(newport_ri, cleveland_oh).km)
TypeScript Version export const degreeToRadian = (degree: number) => { return degree * Math.PI / 180; } export const distanceBetweenEarthCoordinatesInKm = (lat1: number, lon1: number, lat2: number, lon2: number) => { const earthRadiusInKm = 6371; const dLat = degreeToRadian(lat2 - lat1); const dLon = degreeToRadian(lon2 - lon1); lat1 = degreeToRadian(lat1); lat2 = degreeToRadian(lat2); const a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2); const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a)); return earthRadiusInKm * c; }
I think a version of the algorithm in R is still missing: gpsdistance<-function(lat1,lon1,lat2,lon2){ # internal function to change deg to rad degreesToRadians<- function (degrees) { return (degrees * pi / 180) } R<-6371e3 #radius of Earth in meters phi1<-degreesToRadians(lat1) # latitude 1 phi2<-degreesToRadians(lat2) # latitude 2 lambda1<-degreesToRadians(lon1) # longitude 1 lambda2<-degreesToRadians(lon2) # longitude 2 delta_phi<-phi1-phi2 # latitude-distance delta_lambda<-lambda1-lambda2 # longitude-distance a<-sin(delta_phi/2)*sin(delta_phi/2)+ cos(phi1)*cos(phi2)*sin(delta_lambda/2)* sin(delta_lambda/2) cc<-2*atan2(sqrt(a),sqrt(1-a)) distance<- R * cc return(distance) # in meters }
For java public static double degreesToRadians(double degrees) { return degrees * Math.PI / 180; } public static double distanceInKmBetweenEarthCoordinates(Location location1, Location location2) { double earthRadiusKm = 6371; double dLat = degreesToRadians(location2.getLatitude()-location1.getLatitude()); double dLon = degreesToRadians(location2.getLongitude()-location1.getLongitude()); double lat1 = degreesToRadians(location1.getLatitude()); double lat2 = degreesToRadians(location2.getLatitude()); double a = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2); double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); return earthRadiusKm * c; }
For anyone searching for a Delphi/Pascal version: function GreatCircleDistance(const Lat1, Long1, Lat2, Long2: Double): Double; var Lat1Rad, Long1Rad, Lat2Rad, Long2Rad: Double; const EARTH_RADIUS_KM = 6378; begin Lat1Rad := DegToRad(Lat1); Long1Rad := DegToRad(Long1); Lat2Rad := DegToRad(Lat2); Long2Rad := DegToRad(Long2); Result := EARTH_RADIUS_KM * ArcCos(Cos(Lat1Rad) * Cos(Lat2Rad) * Cos(Long1Rad - Long2Rad) + Sin(Lat1Rad) * Sin(Lat2Rad)); end; I take no credit for this code, I originally found it posted by Gary William on a public forum.