如何计算多个经度和纬度之间的距离?

时间:2019-09-30 10:17:44

标签: java android google-maps google-maps-api-3 google-maps-markers

我想计算包含经度和纬度的每个点之间的总距离,这些点都存储在本地数据库中,所以方案是我要计算从点a到b&b到c&c到d的距离使用Google Place API enter image description here

将点(纬度和经度)存储到数据库中

所有点都从数据库中获取。

4 个答案:

答案 0 :(得分:0)

您可以计算两个位置点之间的距离,而无需使用places api

    private double distance(double lat1, double lon1, double lat2, double lon2) {
        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.acos(dist);
        dist = rad2deg(dist);
        dist = dist * 60 * 1.1515;
        return (dist);
    }

    private double deg2rad(double deg) {
        return (deg * Math.PI / 180.0);
    }

    private double rad2deg(double rad) {


return (rad * 180.0 / Math.PI);
}

答案 1 :(得分:0)

我不确定我是否理解此问题的范围。您是否需要考虑地球的曲率?还是只有2个随机点?

如果是第一个:由于无法帮助您,我可以向您显示此链接:

https://www.movable-type.co.uk/scripts/latlong.html

如果是后者:

假设1是纬度,2是经度,那么a1是点A的纬度。

Calculating distance from 2 points

在平面中计算2个点就像这样:

距离= from pymongo import MongoClient import gridfs db = MongoClient('mongodb://localhost:27017/').myDB fs = gridfs.GridFS( db ) fileID = fs.put( open(('Test.pdf') )) out = fs.get(fileID)

因此,要计算从D到E的距离,例如:sqrt((a1−b1)+(a2−b2))

答案 2 :(得分:0)

class DistanceCalculator
{
    public static void main (String[] args) throws java.lang.Exception
    {
        System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, "M") + " Miles\n");
        System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, "K") + " Kilometers\n");
        System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, "N") + " Nautical Miles\n");
    }

    private static double distance(double lat1, double lon1, double lat2, double lon2, String unit) {
        if ((lat1 == lat2) && (lon1 == lon2)) {
            return 0;
        }
        else {
            double theta = lon1 - lon2;
            double dist = Math.sin(Math.toRadians(lat1)) * Math.sin(Math.toRadians(lat2)) + Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) * Math.cos(Math.toRadians(theta));
            dist = Math.acos(dist);
            dist = Math.toDegrees(dist);
            dist = dist * 60 * 1.1515;
            if (unit == "K") {
                dist = dist * 1.609344;
            } else if (unit == "N") {
                dist = dist * 0.8684;
            }
            return (dist);
        }
    }
}

答案 3 :(得分:0)

float[] distanceWidth = new float[2];
Location.distanceBetween(
                startLatitude,
                startLongitude,
                endLatitude,
                endLongitude,
                distanceWidth);

//You will get float[] of results after Location.distanceBetween().

float distance = distanceHeight[0];
float distanceInKm = distance/1000;