Determining Midpoint Between 2 Cooridinates

问题:

I am trying to determine the midpoint between to locations in an MKMapView. I am following the method outlined here (and here) and rewrote it in Objective-C, but the map is being centered somewhere northeast of Baffin Island, which is no where near the two points.

My method based on the java method linked above:

+(CLLocationCoordinate2D)findCenterPoint:(CLLocationCoordinate2D)_lo1 :(CLLocationCoordinate2D)_loc2 {
    CLLocationCoordinate2D center;

    double lon1 = _lo1.longitude * M_PI / 180;
    double lon2 = _loc2.longitude * M_PI / 100;

    double lat1 = _lo1.latitude * M_PI / 180;
    double lat2 = _loc2.latitude * M_PI / 100;

    double dLon = lon2 - lon1;

    double x = cos(lat2) * cos(dLon);
    double y = cos(lat2) * sin(dLon);

    double lat3 = atan2( sin(lat1) + sin(lat2), sqrt((cos(lat1) + x) * (cos(lat1) + x) + y * y) );
    double lon3 = lon1 + atan2(y, cos(lat1) + x);

    center.latitude  = lat3 * 180 / M_PI;
    center.longitude = lon3 * 180 / M_PI;

    return center;
}

The 2 parameters have the following data:

_loc1:
    latitude = 45.4959839
    longitude = -73.67826455

_loc2:
    latitude = 45.482889
    longitude = -73.57522299

The above are correctly place on the map (in and around Montreal). I am trying to center the map in the midpoint between the 2, yet my method return the following:

latitude = 65.29055
longitude = -82.55425

which somewhere in the arctic, when it should be around 500 miles south.

回答1:

Just a hunch, but I noticed your lon2 and lat2 variables are being computed with M_PI/100 and not M_PI/180.

double lon1 = _lo1.longitude * M_PI / 180;
double lon2 = _loc2.longitude * M_PI / 100;

double lat1 = _lo1.latitude * M_PI / 180;
double lat2 = _loc2.latitude * M_PI / 100;

Changing those to 180 might help you out a bit.

回答2:

In case someone need code in Swift, I have written library function in Swift to calculate the midpoint between MULTIPLE coordinates:

//        /** Degrees to Radian **/
class func degreeToRadian(angle:CLLocationDegrees) -> CGFloat {
    return (  (CGFloat(angle)) / 180.0 * CGFloat(M_PI)  )
}

//        /** Radians to Degrees **/
class func radianToDegree(radian:CGFloat) -> CLLocationDegrees {
    return CLLocationDegrees(  radian * CGFloat(180.0 / M_PI)  )
}

class func middlePointOfListMarkers(listCoords: [CLLocationCoordinate2D]) -> CLLocationCoordinate2D {

    var x = 0.0 as CGFloat
    var y = 0.0 as CGFloat
    var z = 0.0 as CGFloat

    for coordinate in listCoords{
        var lat:CGFloat = degreeToRadian(coordinate.latitude)
        var lon:CGFloat = degreeToRadian(coordinate.longitude)
        x = x + cos(lat) * cos(lon)
        y = y + cos(lat) * sin(lon)
        z = z + sin(lat)
    }

    x = x/CGFloat(listCoords.count)
    y = y/CGFloat(listCoords.count)
    z = z/CGFloat(listCoords.count)

    var resultLon: CGFloat = atan2(y, x)
    var resultHyp: CGFloat = sqrt(x*x+y*y)
    var resultLat:CGFloat = atan2(z, resultHyp)

    var newLat = radianToDegree(resultLat)
    var newLon = radianToDegree(resultLon)
    var result:CLLocationCoordinate2D = CLLocationCoordinate2D(latitude: newLat, longitude: newLon)

    return result

}

Detailed answer can be found here

回答3:

For swift users, corrected variant as @dinjas suggest

import Foundation
import MapKit
extension CLLocationCoordinate2D {
    // MARK: CLLocationCoordinate2D+MidPoint
    func middleLocationWith(location:CLLocationCoordinate2D) -> CLLocationCoordinate2D {

        let lon1 = longitude * M_PI / 180
        let lon2 = location.longitude * M_PI / 180
        let lat1 = latitude * M_PI / 180
        let lat2 = location.latitude * M_PI / 180
        let dLon = lon2 - lon1
        let x = cos(lat2) * cos(dLon)
        let y = cos(lat2) * sin(dLon)

        let lat3 = atan2( sin(lat1) + sin(lat2), sqrt((cos(lat1) + x) * (cos(lat1) + x) + y * y) )
        let lon3 = lon1 + atan2(y, cos(lat1) + x)

        let center:CLLocationCoordinate2D = CLLocationCoordinate2DMake(lat3 * 180 / M_PI, lon3 * 180 / M_PI)
        return center
    }
}

回答4:

It's important to say that the formula the OP used to calculate geographic midpoint is based on this formula which explains the cos/sin/sqrt calculation.

This formula will give you the geographic midpoint for any long distance including the four quarters and the prime meridian.

But, if your calculation is for short-range around 1 Kilometer, using a simple average will produce the same midpoint results. i.e:

let firstPoint = CLLocation(....)
let secondPoint = CLLocation(....)

let midPointLat = (firstPoint.coordinate.latitude + secondPoint.coordinate.latitude) / 2
let midPointLong = (firstPoint.coordinate.longitude + secondPoint.coordinate.longitude) / 2

You can actually use it for 10km but expect a deviation - if you only need an estimation for a short range midpoint with a fast solution it will be sufficient.

回答5:

I think you are over thinking it a bit. Just do:

float lon3 = ((lon1 + lon2) / 2)
float lat3 = ((lat1 + lat2) / 2)

lat3 and lon3 will be the center point.