How to calculate the bounding box for a given lat lng location

Question

I have given a location defined by latitude and longitude  Now i want to calculate a bounding box within e g  10 kilometers of that point   The bounding box should be defined as latmin  lngmin and latmax  lngmax   I need this stuff in order to use the panoramio API   Does someone know the formula of how to get thos points   Edit  Guys i am looking for a formula function which takes lat  amp  lng as input and returns a bounding box as latmin  amp  lngmin and latmax  amp  latmin  Mysql  php  c   javascript is fine but also pseudocode should be okay   Edit  I am not looking for a solution which shows me the distance of 2 points

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Since I needed a very rough estimate  so to filter out some needless documents in an elasticsearch query  I employed the below formula     Min lat   Given Lat -  0 009 x N  Max lat   Given Lat    0 009 x N  Min lon   Given lon -  0 009 x N  Max lon   Given lon    0 009 x N    N   kms required form the given location  For your case N 10  Not accurate but handy

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Thanks  Fedrico A  for the Phyton implementation  I have ported it into a Objective C category class  Here is    import  LocationService Bounds h     Semi-axes of WGS-84 geoidal reference const double WGS84 a   6378137 0    Major semiaxis  m  const double WGS84 b   6356752 3    Minor semiaxis  m    implementation LocationService  Bounds   struct BoundsLocation       double maxLatitude      double minLatitude      double maxLongitude      double minLongitude         struct BoundsLocation locationBoundsWithLatitude  double aLatitude longitude  double aLongitude maxDistanceKm  NSInteger aMaxKmDistance       return  self boundingBoxWithLatitude aLatitude longitude aLongitude halfDistanceKm aMaxKmDistance 2       pragma mark - Algorithm      struct BoundsLocation boundingBoxWithLatitude  double aLatitude longitude  double aLongitude halfDistanceKm  double aDistanceKm       double radianLatitude    self degreesToRadians aLatitude       double radianLongitude    self degreesToRadians aLongitude       double halfDistanceMeters   aDistanceKm 1000        double earthRadius    self earthRadiusAtLatitude radianLatitude       double parallelRadius   earthRadius cosl radianLatitude        double radianMinLatitude   radianLatitude - halfDistanceMeters earthRadius      double radianMaxLatitude   radianLatitude   halfDistanceMeters earthRadius      double radianMinLongitude   radianLongitude - halfDistanceMeters parallelRadius      double radianMaxLongitude   radianLongitude   halfDistanceMeters parallelRadius       struct BoundsLocation bounds      bounds minLatitude    self radiansToDegrees radianMinLatitude       bounds maxLatitude    self radiansToDegrees radianMaxLatitude       bounds minLongitude    self radiansToDegrees radianMinLongitude       bounds maxLongitude    self radiansToDegrees radianMaxLongitude        return bounds        double earthRadiusAtLatitude  double aRadianLatitude       double An   WGS84 a   WGS84 a   cosl aRadianLatitude       double Bn   WGS84 b   WGS84 b   sinl aRadianLatitude       double Ad   WGS84 a   cosl aRadianLatitude       double Bd   WGS84 b   sinl aRadianLatitude       return sqrtl    An   An     Bn   Bn     Ad   Ad     Bd   Bd            double degreesToRadians  double aDegrees       return M PI aDegrees 180 0        double radiansToDegrees  double aRadians       return 180 0 aRadians M PI        end   I have tested it and seems be working nice  Struct BoundsLocation should be replaced by a class  I have used it just to share it here

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Here is an simple implementation using javascript which is based on the conversion of latitude degree to kms where 1 degree latitude   111 2 km  I am calculating bounds of the map from a given latitude  longitude and radius in kilometers  function getBoundsFromLatLng lat  lng  radiusInKm        var lat change   radiusInKm 111 2       var lon change   Math abs Math cos lat  Math PI 180          var bounds               lat min   lat - lat change           lon min   lng - lon change           lat max   lat   lat change           lon max   lng   lon change              return bounds

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You re looking for an ellipsoid formula     The best place I ve found to start coding is based on the Geo  Ellipsoid library from CPAN   It gives you a baseline to create your tests off of and to compare your results with its results   I used it as the basis for a similar library for PHP at my previous employer   Geo  Ellipsoid  Take a look at the location method   Call it twice and you ve got your bbox   You didn t post what language you were using   There may already be a geocoding library available for you   Oh  and if you haven t figured it out by now  Google maps uses the WGS84 ellipsoid

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Here I have converted Federico A  Ramponi s answer to PHP if anybody is interested    lt  php   deg2rad and rad2deg are already within PHP    Semi-axes of WGS-84 geoidal reference  WGS84 a   6378137 0     Major semiaxis  m   WGS84 b   6356752 3     Minor semiaxis  m     Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  function WGS84EarthRadius  lat        global  WGS84 a   WGS84 b        an    WGS84 a    WGS84 a   cos  lat        bn    WGS84 b    WGS84 b   sin  lat        ad    WGS84 a   cos  lat        bd    WGS84 b   sin  lat        return sqrt   an  an    bn  bn    ad  ad    bd  bd         Bounding box surrounding the point at given coordinates    assuming local approximation of Earth surface as a sphere   of radius given by WGS84 function boundingBox  latitudeInDegrees   longitudeInDegrees   halfSideInKm         lat   deg2rad  latitudeInDegrees        lon   deg2rad  longitudeInDegrees        halfSide   1000    halfSideInKm         Radius of Earth at given latitude      radius   WGS84EarthRadius  lat         Radius of the parallel at given latitude      pradius    radius cos  lat         latMin    lat -  halfSide    radius       latMax    lat    halfSide    radius       lonMin    lon -  halfSide    pradius       lonMax    lon    halfSide    pradius       return array rad2deg  latMin   rad2deg  lonMin   rad2deg  latMax   rad2deg  lonMax        gt

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I adapted a PHP script I found to do just this   You can use it to find the corners of a box around a point  say  20 km out    My specific example is for Google Maps API   http   www richardpeacock com blog 2011 11 draw-box-around-coordinate-google-maps-based-miles-or-kilometers

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I suggest to approximate locally the Earth surface as a sphere with radius given by the WGS84 ellipsoid at the given latitude  I suspect that the exact computation of latMin and latMax would require elliptic functions and would not yield an appreciable increase in accuracy  WGS84 is itself an approximation    My implementation follows  It s written in Python  I have not tested it        degrees to radians def deg2rad degrees       return math pi degrees 180 0   radians to degrees def rad2deg radians       return 180 0 radians math pi    Semi-axes of WGS-84 geoidal reference WGS84 a   6378137 0    Major semiaxis  m  WGS84 b   6356752 3    Minor semiaxis  m     Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  def WGS84EarthRadius lat         http   en wikipedia org wiki Earth radius     An   WGS84 a WGS84 a   math cos lat      Bn   WGS84 b WGS84 b   math sin lat      Ad   WGS84 a   math cos lat      Bd   WGS84 b   math sin lat      return math sqrt   An An   Bn Bn   Ad Ad   Bd Bd       Bounding box surrounding the point at given coordinates    assuming local approximation of Earth surface as a sphere   of radius given by WGS84 def boundingBox latitudeInDegrees  longitudeInDegrees  halfSideInKm       lat   deg2rad latitudeInDegrees      lon   deg2rad longitudeInDegrees      halfSide   1000 halfSideInKm        Radius of Earth at given latitude     radius   WGS84EarthRadius lat        Radius of the parallel at given latitude     pradius   radius math cos lat       latMin   lat - halfSide radius     latMax   lat   halfSide radius     lonMin   lon - halfSide pradius     lonMax   lon   halfSide pradius      return  rad2deg latMin   rad2deg lonMin   rad2deg latMax   rad2deg lonMax     EDIT  The following code converts  degrees  primes  seconds  to degrees   fractions of a degree  and vice versa  not tested    def dps2deg degrees  primes  seconds       return degrees   primes 60 0   seconds 3600 0  def deg2dps degrees       intdeg   math floor degrees      primes    degrees - intdeg  60 0     intpri   math floor primes      seconds    primes - intpri  60 0     intsec   round seconds      return  int intdeg   int intpri   int intsec

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Here I have converted Federico A  Ramponi s answer to C  for anybody interested   public class MapPoint       public double Longitude   get  set       In Degrees     public double Latitude   get  set       In Degrees    public class BoundingBox       public MapPoint MinPoint   get  set        public MapPoint MaxPoint   get  set                  Semi-axes of WGS-84 geoidal reference private const double WGS84 a   6378137 0     Major semiaxis  m  private const double WGS84 b   6356752 3     Minor semiaxis  m       halfSideInKm  is the half length of the bounding box you want in kilometers  public static BoundingBox GetBoundingBox MapPoint point  double halfSideInKm                       Bounding box surrounding the point at given coordinates         assuming local approximation of Earth surface as a sphere        of radius given by WGS84     var lat   Deg2rad point Latitude       var lon   Deg2rad point Longitude       var halfSide   1000   halfSideInKm          Radius of Earth at given latitude     var radius   WGS84EarthRadius lat          Radius of the parallel at given latitude     var pradius   radius   Math Cos lat        var latMin   lat - halfSide   radius      var latMax   lat   halfSide   radius      var lonMin   lon - halfSide   pradius      var lonMax   lon   halfSide   pradius       return new BoundingBox            MinPoint   new MapPoint   Latitude   Rad2deg latMin   Longitude   Rad2deg lonMin             MaxPoint   new MapPoint   Latitude   Rad2deg latMax   Longitude   Rad2deg lonMax                             degrees to radians private static double Deg2rad double degrees        return Math PI   degrees   180 0        radians to degrees private static double Rad2deg double radians        return 180 0   radians   Math PI        Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  private static double WGS84EarthRadius double lat           http   en wikipedia org wiki Earth radius     var An   WGS84 a   WGS84 a   Math Cos lat       var Bn   WGS84 b   WGS84 b   Math Sin lat       var Ad   WGS84 a   Math Cos lat       var Bd   WGS84 b   Math Sin lat       return Math Sqrt  An An   Bn Bn     Ad Ad   Bd Bd

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You re looking for an ellipsoid formula     The best place I ve found to start coding is based on the Geo  Ellipsoid library from CPAN   It gives you a baseline to create your tests off of and to compare your results with its results   I used it as the basis for a similar library for PHP at my previous employer   Geo  Ellipsoid  Take a look at the location method   Call it twice and you ve got your bbox   You didn t post what language you were using   There may already be a geocoding library available for you   Oh  and if you haven t figured it out by now  Google maps uses the WGS84 ellipsoid

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Illustration of  Jan Philip Matuschek excellent explanation  Please up-vote his answer  not this  I am adding this as I took a little time in understanding the original answer   The bounding box technique of optimizing of finding nearest neighbors would need to derive the minimum and maximum latitude longitude pairs  for a point P at distance d   All points that fall outside these are definitely at a distance  greater than d from the point  One thing to note here is the calculation of latitude of intersection as is highlighted in Jan Philip Matuschek explanation  The latitude of intersection is not at the latitude of point P but slightly offset from it  This is a often missed but important  part in determining the correct minimum and maximum bounding longitude for point P for the distance d This is also useful in verification    The haversine distance between  latitude of intersection longitude high  to  latitude longitude  of P is equal to distance d   Python gist here https   gist github com alexcpn f95ae83a7ee0293a5225

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I adapted a PHP script I found to do just this   You can use it to find the corners of a box around a point  say  20 km out    My specific example is for Google Maps API   http   www richardpeacock com blog 2011 11 draw-box-around-coordinate-google-maps-based-miles-or-kilometers

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You re looking for an ellipsoid formula     The best place I ve found to start coding is based on the Geo  Ellipsoid library from CPAN   It gives you a baseline to create your tests off of and to compare your results with its results   I used it as the basis for a similar library for PHP at my previous employer   Geo  Ellipsoid  Take a look at the location method   Call it twice and you ve got your bbox   You didn t post what language you were using   There may already be a geocoding library available for you   Oh  and if you haven t figured it out by now  Google maps uses the WGS84 ellipsoid

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Here is Federico Ramponi s answer in Go   Note   no error-checking     import        math        Semi-axes of WGS-84 geoidal reference const          Major semiaxis  meters      WGS84A   6378137 0        Minor semiaxis  meters      WGS84B   6356752 3       BoundingBox represents the geo-polygon that encompasses the given point and radius type BoundingBox struct       LatMin float64     LatMax float64     LonMin float64     LonMax float64       Convert a degree value to radians func deg2Rad deg float64  float64       return math Pi   deg   180 0       Convert a radian value to degrees func rad2Deg rad float64  float64       return 180 0   rad   math Pi       Get the Earth s radius in meters at a given latitude based on the WGS84 ellipsoid func getWgs84EarthRadius lat float64  float64       an    WGS84A   WGS84A   math Cos lat      bn    WGS84B   WGS84B   math Sin lat       ad    WGS84A   math Cos lat      bd    WGS84B   math Sin lat       return math Sqrt  an an   bn bn     ad ad   bd bd         GetBoundingBox returns a BoundingBox encompassing the given lat long point and radius func GetBoundingBox latDeg float64  longDeg float64  radiusKm float64  BoundingBox       lat    deg2Rad latDeg      lon    deg2Rad longDeg      halfSide    1000   radiusKm         Radius of Earth at given latitude     radius    getWgs84EarthRadius lat       pradius    radius   math Cos lat       latMin    lat - halfSide radius     latMax    lat   halfSide radius     lonMin    lon - halfSide pradius     lonMax    lon   halfSide pradius      return BoundingBox          LatMin  rad2Deg latMin           LatMax  rad2Deg latMax           LonMin  rad2Deg lonMin           LonMax  rad2Deg lonMax

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I suggest to approximate locally the Earth surface as a sphere with radius given by the WGS84 ellipsoid at the given latitude  I suspect that the exact computation of latMin and latMax would require elliptic functions and would not yield an appreciable increase in accuracy  WGS84 is itself an approximation    My implementation follows  It s written in Python  I have not tested it        degrees to radians def deg2rad degrees       return math pi degrees 180 0   radians to degrees def rad2deg radians       return 180 0 radians math pi    Semi-axes of WGS-84 geoidal reference WGS84 a   6378137 0    Major semiaxis  m  WGS84 b   6356752 3    Minor semiaxis  m     Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  def WGS84EarthRadius lat         http   en wikipedia org wiki Earth radius     An   WGS84 a WGS84 a   math cos lat      Bn   WGS84 b WGS84 b   math sin lat      Ad   WGS84 a   math cos lat      Bd   WGS84 b   math sin lat      return math sqrt   An An   Bn Bn   Ad Ad   Bd Bd       Bounding box surrounding the point at given coordinates    assuming local approximation of Earth surface as a sphere   of radius given by WGS84 def boundingBox latitudeInDegrees  longitudeInDegrees  halfSideInKm       lat   deg2rad latitudeInDegrees      lon   deg2rad longitudeInDegrees      halfSide   1000 halfSideInKm        Radius of Earth at given latitude     radius   WGS84EarthRadius lat        Radius of the parallel at given latitude     pradius   radius math cos lat       latMin   lat - halfSide radius     latMax   lat   halfSide radius     lonMin   lon - halfSide pradius     lonMax   lon   halfSide pradius      return  rad2deg latMin   rad2deg lonMin   rad2deg latMax   rad2deg lonMax     EDIT  The following code converts  degrees  primes  seconds  to degrees   fractions of a degree  and vice versa  not tested    def dps2deg degrees  primes  seconds       return degrees   primes 60 0   seconds 3600 0  def deg2dps degrees       intdeg   math floor degrees      primes    degrees - intdeg  60 0     intpri   math floor primes      seconds    primes - intpri  60 0     intsec   round seconds      return  int intdeg   int intpri   int intsec

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Here I have converted Federico A  Ramponi s answer to C  for anybody interested   public class MapPoint       public double Longitude   get  set       In Degrees     public double Latitude   get  set       In Degrees    public class BoundingBox       public MapPoint MinPoint   get  set        public MapPoint MaxPoint   get  set                  Semi-axes of WGS-84 geoidal reference private const double WGS84 a   6378137 0     Major semiaxis  m  private const double WGS84 b   6356752 3     Minor semiaxis  m       halfSideInKm  is the half length of the bounding box you want in kilometers  public static BoundingBox GetBoundingBox MapPoint point  double halfSideInKm                       Bounding box surrounding the point at given coordinates         assuming local approximation of Earth surface as a sphere        of radius given by WGS84     var lat   Deg2rad point Latitude       var lon   Deg2rad point Longitude       var halfSide   1000   halfSideInKm          Radius of Earth at given latitude     var radius   WGS84EarthRadius lat          Radius of the parallel at given latitude     var pradius   radius   Math Cos lat        var latMin   lat - halfSide   radius      var latMax   lat   halfSide   radius      var lonMin   lon - halfSide   pradius      var lonMax   lon   halfSide   pradius       return new BoundingBox            MinPoint   new MapPoint   Latitude   Rad2deg latMin   Longitude   Rad2deg lonMin             MaxPoint   new MapPoint   Latitude   Rad2deg latMax   Longitude   Rad2deg lonMax                             degrees to radians private static double Deg2rad double degrees        return Math PI   degrees   180 0        radians to degrees private static double Rad2deg double radians        return 180 0   radians   Math PI        Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  private static double WGS84EarthRadius double lat           http   en wikipedia org wiki Earth radius     var An   WGS84 a   WGS84 a   Math Cos lat       var Bn   WGS84 b   WGS84 b   Math Sin lat       var Ad   WGS84 a   Math Cos lat       var Bd   WGS84 b   Math Sin lat       return Math Sqrt  An An   Bn Bn     Ad Ad   Bd Bd

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All of the above answer are only partially correct  Specially in region like Australia  they always include pole and calculate a very large rectangle even for 10kms    Specially the algorithm by Jan Philip Matuschek at http   janmatuschek de LatitudeLongitudeBoundingCoordinates UsingIndex included a very large rectangle from  -37  -90  -180  180  for almost every point in Australia  This hits a large users in database and distance have to be calculated for all of the users in almost half the country   I found that the Drupal API Earth Algorithm by Rochester Institute of Technology works better around pole as well as elsewhere and is much easier to implement   https   www rit edu drupal api drupal sites 21all 21modules 21location 21earth inc 7 54  Use earth latitude range and earth longitude range from the above algorithm for calculating bounding rectangle   And use the distance calculation formula documented by google maps to calculate distance  https   developers google com maps solutions store-locator clothing-store-locator outputting-data-as-xml-using-php  To search by kilometers instead of miles  replace 3959 with 6371  For  Lat  Lng     37  -122  and a Markers table with columns lat and lng  the formula is   SELECT id    3959   acos  cos  radians 37      cos  radians  lat       cos  radians  lng   - radians -122      sin  radians 37      sin  radians  lat         AS distance FROM markers HAVING distance  lt  25 ORDER BY distance LIMIT 0   20    Read my detailed answer at https   stackoverflow com a 45950426 5076414

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I suggest to approximate locally the Earth surface as a sphere with radius given by the WGS84 ellipsoid at the given latitude  I suspect that the exact computation of latMin and latMax would require elliptic functions and would not yield an appreciable increase in accuracy  WGS84 is itself an approximation    My implementation follows  It s written in Python  I have not tested it        degrees to radians def deg2rad degrees       return math pi degrees 180 0   radians to degrees def rad2deg radians       return 180 0 radians math pi    Semi-axes of WGS-84 geoidal reference WGS84 a   6378137 0    Major semiaxis  m  WGS84 b   6356752 3    Minor semiaxis  m     Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  def WGS84EarthRadius lat         http   en wikipedia org wiki Earth radius     An   WGS84 a WGS84 a   math cos lat      Bn   WGS84 b WGS84 b   math sin lat      Ad   WGS84 a   math cos lat      Bd   WGS84 b   math sin lat      return math sqrt   An An   Bn Bn   Ad Ad   Bd Bd       Bounding box surrounding the point at given coordinates    assuming local approximation of Earth surface as a sphere   of radius given by WGS84 def boundingBox latitudeInDegrees  longitudeInDegrees  halfSideInKm       lat   deg2rad latitudeInDegrees      lon   deg2rad longitudeInDegrees      halfSide   1000 halfSideInKm        Radius of Earth at given latitude     radius   WGS84EarthRadius lat        Radius of the parallel at given latitude     pradius   radius math cos lat       latMin   lat - halfSide radius     latMax   lat   halfSide radius     lonMin   lon - halfSide pradius     lonMax   lon   halfSide pradius      return  rad2deg latMin   rad2deg lonMin   rad2deg latMax   rad2deg lonMax     EDIT  The following code converts  degrees  primes  seconds  to degrees   fractions of a degree  and vice versa  not tested    def dps2deg degrees  primes  seconds       return degrees   primes 60 0   seconds 3600 0  def deg2dps degrees       intdeg   math floor degrees      primes    degrees - intdeg  60 0     intpri   math floor primes      seconds    primes - intpri  60 0     intsec   round seconds      return  int intdeg   int intpri   int intsec

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Here is an simple implementation using javascript which is based on the conversion of latitude degree to kms where 1 degree latitude   111 2 km  I am calculating bounds of the map from a given latitude  longitude and radius in kilometers  function getBoundsFromLatLng lat  lng  radiusInKm        var lat change   radiusInKm 111 2       var lon change   Math abs Math cos lat  Math PI 180          var bounds               lat min   lat - lat change           lon min   lng - lon change           lat max   lat   lat change           lon max   lng   lon change              return bounds

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I wrote a JavaScript function that returns the four coordinates of a square bounding box  given a distance and a pair of coordinates    use strict            param  number  distance - distance  km  from the point represented by centerPoint     param  array  centerPoint - two-dimensional array containing center coords  latitude  longitude      description      Computes the bounding coordinates of all points on the surface of a sphere      that has a great circle distance to the point represented by the centerPoint      argument that is less or equal to the distance argument       Technique from  Jan Matuschek  lt http   JanMatuschek de LatitudeLongitudeBoundingCoordinates gt      author Alex Salisbury     getBoundingBox   function  centerPoint  distance      var MIN LAT  MAX LAT  MIN LON  MAX LON  R  radDist  degLat  degLon  radLat  radLon  minLat  maxLat  minLon  maxLon  deltaLon    if  distance  lt  0        return  Illegal arguments            helper functions  degrees lt     gt radians    Number prototype degToRad   function          return this    Math PI   180          Number prototype radToDeg   function          return  180   this    Math PI            coordinate limits   MIN LAT    -90  degToRad      MAX LAT    90  degToRad      MIN LON    -180  degToRad      MAX LON    180  degToRad         Earth s radius  km    R   6378 1       angular distance in radians on a great circle   radDist   distance   R       center point coordinates  deg    degLat   centerPoint 0     degLon   centerPoint 1        center point coordinates  rad    radLat   degLat degToRad      radLon   degLon degToRad         minimum and maximum latitudes for given distance   minLat   radLat - radDist    maxLat   radLat   radDist       minimum and maximum longitudes for given distance   minLon   void 0    maxLon   void 0       define deltaLon to help determine min and max longitudes   deltaLon   Math asin Math sin radDist    Math cos radLat      if  minLat  gt  MIN LAT  amp  amp  maxLat  lt  MAX LAT        minLon   radLon - deltaLon      maxLon   radLon   deltaLon      if  minLon  lt  MIN LON          minLon   minLon   2   Math PI            if  maxLon  gt  MAX LON          maxLon   maxLon - 2   Math PI                 a pole is within the given distance   else       minLat   Math max minLat  MIN LAT       maxLat   Math min maxLat  MAX LAT       minLon   MIN LON      maxLon   MAX LON        return       minLon radToDeg        minLat radToDeg        maxLon radToDeg        maxLat radToDeg

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I wrote an article about finding the bounding coordinates   http   JanMatuschek de LatitudeLongitudeBoundingCoordinates  The article explains the formulae and also provides a Java implementation   It also shows why Federico s formula for the min max longitude is inaccurate

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Here is Federico Ramponi s answer in Go   Note   no error-checking     import        math        Semi-axes of WGS-84 geoidal reference const          Major semiaxis  meters      WGS84A   6378137 0        Minor semiaxis  meters      WGS84B   6356752 3       BoundingBox represents the geo-polygon that encompasses the given point and radius type BoundingBox struct       LatMin float64     LatMax float64     LonMin float64     LonMax float64       Convert a degree value to radians func deg2Rad deg float64  float64       return math Pi   deg   180 0       Convert a radian value to degrees func rad2Deg rad float64  float64       return 180 0   rad   math Pi       Get the Earth s radius in meters at a given latitude based on the WGS84 ellipsoid func getWgs84EarthRadius lat float64  float64       an    WGS84A   WGS84A   math Cos lat      bn    WGS84B   WGS84B   math Sin lat       ad    WGS84A   math Cos lat      bd    WGS84B   math Sin lat       return math Sqrt  an an   bn bn     ad ad   bd bd         GetBoundingBox returns a BoundingBox encompassing the given lat long point and radius func GetBoundingBox latDeg float64  longDeg float64  radiusKm float64  BoundingBox       lat    deg2Rad latDeg      lon    deg2Rad longDeg      halfSide    1000   radiusKm         Radius of Earth at given latitude     radius    getWgs84EarthRadius lat       pradius    radius   math Cos lat       latMin    lat - halfSide radius     latMax    lat   halfSide radius     lonMin    lon - halfSide pradius     lonMax    lon   halfSide pradius      return BoundingBox          LatMin  rad2Deg latMin           LatMax  rad2Deg latMax           LonMin  rad2Deg lonMin           LonMax  rad2Deg lonMax

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All of the above answer are only partially correct  Specially in region like Australia  they always include pole and calculate a very large rectangle even for 10kms    Specially the algorithm by Jan Philip Matuschek at http   janmatuschek de LatitudeLongitudeBoundingCoordinates UsingIndex included a very large rectangle from  -37  -90  -180  180  for almost every point in Australia  This hits a large users in database and distance have to be calculated for all of the users in almost half the country   I found that the Drupal API Earth Algorithm by Rochester Institute of Technology works better around pole as well as elsewhere and is much easier to implement   https   www rit edu drupal api drupal sites 21all 21modules 21location 21earth inc 7 54  Use earth latitude range and earth longitude range from the above algorithm for calculating bounding rectangle   And use the distance calculation formula documented by google maps to calculate distance  https   developers google com maps solutions store-locator clothing-store-locator outputting-data-as-xml-using-php  To search by kilometers instead of miles  replace 3959 with 6371  For  Lat  Lng     37  -122  and a Markers table with columns lat and lng  the formula is   SELECT id    3959   acos  cos  radians 37      cos  radians  lat       cos  radians  lng   - radians -122      sin  radians 37      sin  radians  lat         AS distance FROM markers HAVING distance  lt  25 ORDER BY distance LIMIT 0   20    Read my detailed answer at https   stackoverflow com a 45950426 5076414

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Thanks  Fedrico A  for the Phyton implementation  I have ported it into a Objective C category class  Here is    import  LocationService Bounds h     Semi-axes of WGS-84 geoidal reference const double WGS84 a   6378137 0    Major semiaxis  m  const double WGS84 b   6356752 3    Minor semiaxis  m    implementation LocationService  Bounds   struct BoundsLocation       double maxLatitude      double minLatitude      double maxLongitude      double minLongitude         struct BoundsLocation locationBoundsWithLatitude  double aLatitude longitude  double aLongitude maxDistanceKm  NSInteger aMaxKmDistance       return  self boundingBoxWithLatitude aLatitude longitude aLongitude halfDistanceKm aMaxKmDistance 2       pragma mark - Algorithm      struct BoundsLocation boundingBoxWithLatitude  double aLatitude longitude  double aLongitude halfDistanceKm  double aDistanceKm       double radianLatitude    self degreesToRadians aLatitude       double radianLongitude    self degreesToRadians aLongitude       double halfDistanceMeters   aDistanceKm 1000        double earthRadius    self earthRadiusAtLatitude radianLatitude       double parallelRadius   earthRadius cosl radianLatitude        double radianMinLatitude   radianLatitude - halfDistanceMeters earthRadius      double radianMaxLatitude   radianLatitude   halfDistanceMeters earthRadius      double radianMinLongitude   radianLongitude - halfDistanceMeters parallelRadius      double radianMaxLongitude   radianLongitude   halfDistanceMeters parallelRadius       struct BoundsLocation bounds      bounds minLatitude    self radiansToDegrees radianMinLatitude       bounds maxLatitude    self radiansToDegrees radianMaxLatitude       bounds minLongitude    self radiansToDegrees radianMinLongitude       bounds maxLongitude    self radiansToDegrees radianMaxLongitude        return bounds        double earthRadiusAtLatitude  double aRadianLatitude       double An   WGS84 a   WGS84 a   cosl aRadianLatitude       double Bn   WGS84 b   WGS84 b   sinl aRadianLatitude       double Ad   WGS84 a   cosl aRadianLatitude       double Bd   WGS84 b   sinl aRadianLatitude       return sqrtl    An   An     Bn   Bn     Ad   Ad     Bd   Bd            double degreesToRadians  double aDegrees       return M PI aDegrees 180 0        double radiansToDegrees  double aRadians       return 180 0 aRadians M PI        end   I have tested it and seems be working nice  Struct BoundsLocation should be replaced by a class  I have used it just to share it here

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It is very simple just go to panoramio website and then open World Map from panoramio website Then go to specified location whichs latitude and longitude required   Then you found latitude and longitude in address bar for example in this address   http   www panoramio com map lt 32 739485 amp ln 70 491211 amp z 9 amp k 1 amp a 1 amp tab 1 amp pl all  lt 32 739485   latitude ln 70 491211   longitude  this Panoramio JavaScript API widget create a bounding box around a lat long pair and then returning all photos with in those bounds   Another type of Panoramio JavaScript API widget in which you can also change background color with example and code is here   It does not show in composing mood It show after publishing    lt div dir  ltr  style  text-align  center   trbidi  on  gt   lt script src  https   ssl panoramio com wapi wapi js v 1 amp amp hl en  gt  lt  script gt   lt div id  wapiblock  style  float  right  margin  10px 15px  gt  lt  div gt   lt script type  text javascript  gt  var myRequest        tag    kahna      rect     sw     lat   -30   lng   10 5    ne     lat   50 5   lng   30        var myOptions        width   300     height   200    var wapiblock   document getElementById  wapiblock    var photo widget   new panoramio PhotoWidget  wapiblock   myRequest  myOptions   photo widget setPosition 0    lt  script gt   lt  div gt

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I was working on the bounding box problem as a side issue to finding all the points within SrcRad radius of a static LAT  LONG point   There have been quite a few calculations that use  maxLon    lon   rad2deg  rad  R cos deg2rad  lat     minLon    lon - rad2deg  rad  R cos deg2rad  lat       to calculate the longitude bounds  but I found this to not give all the answers that were needed   Because what you really want to do is    SrcRad RadEarth  cos deg2rad lat     I know  I know the answer should be the same  but I found that it wasn t  It appeared that by not making sure I was doing the  SRCrad RadEarth  First and then dividing by the Cos part I was leaving out some location points     After you get all your bounding box points  if you have a function that calculates the Point to Point Distance given lat  long it is easy to only get those points that are a certain distance radius from the fixed point   Here is what I did  I know it took a few extra steps but it helped me  -- GLOBAL Constants gc pi CONSTANT REAL    3 14159265359   -- Pi  -- Conversion Factor Constants gc rad to degs          CONSTANT NUMBER    180 gc pi  -- Conversion for Radians to Degrees 180 pi gc deg to rads          CONSTANT NUMBER    gc pi 180  --Conversion of Degrees to Radians  lv stat lat    -- The static latitude point that I am searching from  lv stat long   -- The static longitude point that I am searching from   -- Angular radius ratio in radians lv ang radius    lv search radius   lv earth radius  lv bb maxlat    lv stat lat    gc rad to deg   lv ang radius   lv bb minlat    lv stat lat -  gc rad to deg   lv ang radius    --Here s the tricky part  accounting for the Longitude getting smaller as we move up the latitiude scale -- I seperated the parts of the equation to make it easier to debug and understand -- I may not be a smart man but I know what the right answer is     -   lv int calc    gc deg to rads   lv stat lat  lv int calc    COS lv int calc   lv int calc    lv ang radius lv int calc  lv int calc    gc rad to degs lv int calc   lv bb maxlong    lv stat long   lv int calc  lv bb minlong    lv stat long - lv int calc   -- Now select the values from your location datatable  SELECT    FROM   SELECT cityaliasname  city  state  zipcode  latitude  longitude   -- The actual distance in miles spherecos pnttopntdist lv stat lat  lv stat long  latitude  longitude   M   as miles dist     FROM Location Table  WHERE latitude between lv bb minlat AND lv bb maxlat AND   longitude between lv bb minlong and lv bb maxlong  WHERE miles dist  lt   lv limit distance miles order by miles dist

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It is very simple just go to panoramio website and then open World Map from panoramio website Then go to specified location whichs latitude and longitude required   Then you found latitude and longitude in address bar for example in this address   http   www panoramio com map lt 32 739485 amp ln 70 491211 amp z 9 amp k 1 amp a 1 amp tab 1 amp pl all  lt 32 739485   latitude ln 70 491211   longitude  this Panoramio JavaScript API widget create a bounding box around a lat long pair and then returning all photos with in those bounds   Another type of Panoramio JavaScript API widget in which you can also change background color with example and code is here   It does not show in composing mood It show after publishing    lt div dir  ltr  style  text-align  center   trbidi  on  gt   lt script src  https   ssl panoramio com wapi wapi js v 1 amp amp hl en  gt  lt  script gt   lt div id  wapiblock  style  float  right  margin  10px 15px  gt  lt  div gt   lt script type  text javascript  gt  var myRequest        tag    kahna      rect     sw     lat   -30   lng   10 5    ne     lat   50 5   lng   30        var myOptions        width   300     height   200    var wapiblock   document getElementById  wapiblock    var photo widget   new panoramio PhotoWidget  wapiblock   myRequest  myOptions   photo widget setPosition 0    lt  script gt   lt  div gt

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I was working on the bounding box problem as a side issue to finding all the points within SrcRad radius of a static LAT  LONG point   There have been quite a few calculations that use  maxLon    lon   rad2deg  rad  R cos deg2rad  lat     minLon    lon - rad2deg  rad  R cos deg2rad  lat       to calculate the longitude bounds  but I found this to not give all the answers that were needed   Because what you really want to do is    SrcRad RadEarth  cos deg2rad lat     I know  I know the answer should be the same  but I found that it wasn t  It appeared that by not making sure I was doing the  SRCrad RadEarth  First and then dividing by the Cos part I was leaving out some location points     After you get all your bounding box points  if you have a function that calculates the Point to Point Distance given lat  long it is easy to only get those points that are a certain distance radius from the fixed point   Here is what I did  I know it took a few extra steps but it helped me  -- GLOBAL Constants gc pi CONSTANT REAL    3 14159265359   -- Pi  -- Conversion Factor Constants gc rad to degs          CONSTANT NUMBER    180 gc pi  -- Conversion for Radians to Degrees 180 pi gc deg to rads          CONSTANT NUMBER    gc pi 180  --Conversion of Degrees to Radians  lv stat lat    -- The static latitude point that I am searching from  lv stat long   -- The static longitude point that I am searching from   -- Angular radius ratio in radians lv ang radius    lv search radius   lv earth radius  lv bb maxlat    lv stat lat    gc rad to deg   lv ang radius   lv bb minlat    lv stat lat -  gc rad to deg   lv ang radius    --Here s the tricky part  accounting for the Longitude getting smaller as we move up the latitiude scale -- I seperated the parts of the equation to make it easier to debug and understand -- I may not be a smart man but I know what the right answer is     -   lv int calc    gc deg to rads   lv stat lat  lv int calc    COS lv int calc   lv int calc    lv ang radius lv int calc  lv int calc    gc rad to degs lv int calc   lv bb maxlong    lv stat long   lv int calc  lv bb minlong    lv stat long - lv int calc   -- Now select the values from your location datatable  SELECT    FROM   SELECT cityaliasname  city  state  zipcode  latitude  longitude   -- The actual distance in miles spherecos pnttopntdist lv stat lat  lv stat long  latitude  longitude   M   as miles dist     FROM Location Table  WHERE latitude between lv bb minlat AND lv bb maxlat AND   longitude between lv bb minlong and lv bb maxlong  WHERE miles dist  lt   lv limit distance miles order by miles dist

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I suggest to approximate locally the Earth surface as a sphere with radius given by the WGS84 ellipsoid at the given latitude  I suspect that the exact computation of latMin and latMax would require elliptic functions and would not yield an appreciable increase in accuracy  WGS84 is itself an approximation    My implementation follows  It s written in Python  I have not tested it        degrees to radians def deg2rad degrees       return math pi degrees 180 0   radians to degrees def rad2deg radians       return 180 0 radians math pi    Semi-axes of WGS-84 geoidal reference WGS84 a   6378137 0    Major semiaxis  m  WGS84 b   6356752 3    Minor semiaxis  m     Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  def WGS84EarthRadius lat         http   en wikipedia org wiki Earth radius     An   WGS84 a WGS84 a   math cos lat      Bn   WGS84 b WGS84 b   math sin lat      Ad   WGS84 a   math cos lat      Bd   WGS84 b   math sin lat      return math sqrt   An An   Bn Bn   Ad Ad   Bd Bd       Bounding box surrounding the point at given coordinates    assuming local approximation of Earth surface as a sphere   of radius given by WGS84 def boundingBox latitudeInDegrees  longitudeInDegrees  halfSideInKm       lat   deg2rad latitudeInDegrees      lon   deg2rad longitudeInDegrees      halfSide   1000 halfSideInKm        Radius of Earth at given latitude     radius   WGS84EarthRadius lat        Radius of the parallel at given latitude     pradius   radius math cos lat       latMin   lat - halfSide radius     latMax   lat   halfSide radius     lonMin   lon - halfSide pradius     lonMax   lon   halfSide pradius      return  rad2deg latMin   rad2deg lonMin   rad2deg latMax   rad2deg lonMax     EDIT  The following code converts  degrees  primes  seconds  to degrees   fractions of a degree  and vice versa  not tested    def dps2deg degrees  primes  seconds       return degrees   primes 60 0   seconds 3600 0  def deg2dps degrees       intdeg   math floor degrees      primes    degrees - intdeg  60 0     intpri   math floor primes      seconds    primes - intpri  60 0     intsec   round seconds      return  int intdeg   int intpri   int intsec

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You re looking for an ellipsoid formula     The best place I ve found to start coding is based on the Geo  Ellipsoid library from CPAN   It gives you a baseline to create your tests off of and to compare your results with its results   I used it as the basis for a similar library for PHP at my previous employer   Geo  Ellipsoid  Take a look at the location method   Call it twice and you ve got your bbox   You didn t post what language you were using   There may already be a geocoding library available for you   Oh  and if you haven t figured it out by now  Google maps uses the WGS84 ellipsoid

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Illustration of  Jan Philip Matuschek excellent explanation  Please up-vote his answer  not this  I am adding this as I took a little time in understanding the original answer   The bounding box technique of optimizing of finding nearest neighbors would need to derive the minimum and maximum latitude longitude pairs  for a point P at distance d   All points that fall outside these are definitely at a distance  greater than d from the point  One thing to note here is the calculation of latitude of intersection as is highlighted in Jan Philip Matuschek explanation  The latitude of intersection is not at the latitude of point P but slightly offset from it  This is a often missed but important  part in determining the correct minimum and maximum bounding longitude for point P for the distance d This is also useful in verification    The haversine distance between  latitude of intersection longitude high  to  latitude longitude  of P is equal to distance d   Python gist here https   gist github com alexcpn f95ae83a7ee0293a5225

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I wrote a JavaScript function that returns the four coordinates of a square bounding box  given a distance and a pair of coordinates    use strict            param  number  distance - distance  km  from the point represented by centerPoint     param  array  centerPoint - two-dimensional array containing center coords  latitude  longitude      description      Computes the bounding coordinates of all points on the surface of a sphere      that has a great circle distance to the point represented by the centerPoint      argument that is less or equal to the distance argument       Technique from  Jan Matuschek  lt http   JanMatuschek de LatitudeLongitudeBoundingCoordinates gt      author Alex Salisbury     getBoundingBox   function  centerPoint  distance      var MIN LAT  MAX LAT  MIN LON  MAX LON  R  radDist  degLat  degLon  radLat  radLon  minLat  maxLat  minLon  maxLon  deltaLon    if  distance  lt  0        return  Illegal arguments            helper functions  degrees lt     gt radians    Number prototype degToRad   function          return this    Math PI   180          Number prototype radToDeg   function          return  180   this    Math PI            coordinate limits   MIN LAT    -90  degToRad      MAX LAT    90  degToRad      MIN LON    -180  degToRad      MAX LON    180  degToRad         Earth s radius  km    R   6378 1       angular distance in radians on a great circle   radDist   distance   R       center point coordinates  deg    degLat   centerPoint 0     degLon   centerPoint 1        center point coordinates  rad    radLat   degLat degToRad      radLon   degLon degToRad         minimum and maximum latitudes for given distance   minLat   radLat - radDist    maxLat   radLat   radDist       minimum and maximum longitudes for given distance   minLon   void 0    maxLon   void 0       define deltaLon to help determine min and max longitudes   deltaLon   Math asin Math sin radDist    Math cos radLat      if  minLat  gt  MIN LAT  amp  amp  maxLat  lt  MAX LAT        minLon   radLon - deltaLon      maxLon   radLon   deltaLon      if  minLon  lt  MIN LON          minLon   minLon   2   Math PI            if  maxLon  gt  MAX LON          maxLon   maxLon - 2   Math PI                 a pole is within the given distance   else       minLat   Math max minLat  MIN LAT       maxLat   Math min maxLat  MAX LAT       minLon   MIN LON      maxLon   MAX LON        return       minLon radToDeg        minLat radToDeg        maxLon radToDeg        maxLat radToDeg

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Here I have converted Federico A  Ramponi s answer to PHP if anybody is interested    lt  php   deg2rad and rad2deg are already within PHP    Semi-axes of WGS-84 geoidal reference  WGS84 a   6378137 0     Major semiaxis  m   WGS84 b   6356752 3     Minor semiaxis  m     Earth radius at a given latitude  according to the WGS-84 ellipsoid  m  function WGS84EarthRadius  lat        global  WGS84 a   WGS84 b        an    WGS84 a    WGS84 a   cos  lat        bn    WGS84 b    WGS84 b   sin  lat        ad    WGS84 a   cos  lat        bd    WGS84 b   sin  lat        return sqrt   an  an    bn  bn    ad  ad    bd  bd         Bounding box surrounding the point at given coordinates    assuming local approximation of Earth surface as a sphere   of radius given by WGS84 function boundingBox  latitudeInDegrees   longitudeInDegrees   halfSideInKm         lat   deg2rad  latitudeInDegrees        lon   deg2rad  longitudeInDegrees        halfSide   1000    halfSideInKm         Radius of Earth at given latitude      radius   WGS84EarthRadius  lat         Radius of the parallel at given latitude      pradius    radius cos  lat         latMin    lat -  halfSide    radius       latMax    lat    halfSide    radius       lonMin    lon -  halfSide    pradius       lonMax    lon    halfSide    pradius       return array rad2deg  latMin   rad2deg  lonMin   rad2deg  latMax   rad2deg  lonMax        gt

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Since I needed a very rough estimate  so to filter out some needless documents in an elasticsearch query  I employed the below formula     Min lat   Given Lat -  0 009 x N  Max lat   Given Lat    0 009 x N  Min lon   Given lon -  0 009 x N  Max lon   Given lon    0 009 x N    N   kms required form the given location  For your case N 10  Not accurate but handy

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I wrote an article about finding the bounding coordinates   http   JanMatuschek de LatitudeLongitudeBoundingCoordinates  The article explains the formulae and also provides a Java implementation   It also shows why Federico s formula for the min max longitude is inaccurate

[geocoding] How to calculate the bounding box for a given lat/lng location?

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