计算“线"与“线"之间的最小距离.和一个“点" [英] Calculate min distance between a "line" and one "point"

问题描述

我有一个线串"(带有初始化点和结束点)和一个点"(两个坐标).

I have a "linestring" (with init and end points) and a single "point" (two coordinates).

并且我实现了以下ActionSctipt代码，以使用"haversine公式"来计算两个点之间的距离(每个点具有x和y坐标)；此函数可以以"kms"，"meters"，"feets"或"miles"返回距离":

And I have implemented the following ActionSctipt code to use "haversine formula" to calculate the distance between two points (each point has x & y coordinates); this function can return the "distance" in "kms", "meters", "feets" or "miles":

private function distanceBetweenCoordinates(lat1:Number, lon1:Number, lat2:Number, lon2:Number, units:String = "miles"):Number {
    var R:int = RADIUS_OF_EARTH_IN_MILES;

    if (units == "km") {
        R = RADIUS_OF_EARTH_IN_KM;
    }

    if (units == "meters") {
        R = RADIUS_OF_EARTH_IN_M;
    }

    if (units == "feet") {
        R = RADIUS_OF_EARTH_IN_FEET;
    }

    var dLat:Number = (lat2 - lat1) * Math.PI / 180;

    var dLon:Number = (lon2 - lon1) * Math.PI / 180;

    var lat1inRadians:Number = lat1 * Math.PI / 180;

    var lat2inRadians:Number = lat2 * Math.PI / 180;

    var a:Number = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1inRadians) * Math.cos(lat2inRadians);

    var c:Number = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));

    var d:Number = R * c;

    return d;
}

此代码运行良好.但是我需要改进此代码，以允许计算单点"和一个线串"(有2个点)之间的最小距离. 我该怎么办?

This code is functioning well. But I need to improving this code to allow calculate the minimum distance between a "single point" and one "linestring" (with 2 points). How can I do?

我认为这种解决方案: *对每个点(初始和结束)除以划线" ...，并为每个点计算到单点"的距离...在得到两个距离"后，返回最小距离. 下图说明了该解决方案的更好之处:

I thought this solution: * Divide the "linesting" for each point (Init and end)... and for each of these calculate the distance to the "single point"... and after I getting both "distances" return the minimum distance. This solution is not the better, this is explained in the following image:

"d1"和"d2"距离无效...因为只有"d0"是有效距离.

"d1" and "d2" distances are invalid... because only "d0" is the valid distance.

请！帮我！！！如何改进Haversine公式以计算线与单点之间的距离(以公里为单位)?

Please! help me!!! How can I improve the haversine formula to calculate the distance between a line and a single point in kilometres?

谢谢!!!!

推荐答案

在您的情况下，d0距离是三角形的高度.是Hb=2*A/b，其中A-区域& b-基本边的长度(您的线串).

In your case d0 distance is a height of triangle. It's Hb=2*A/b where A- Area & b-length of the base side (your linestring).

如果给定3个点，则可以计算它们之间的距离(三角形的边a，b，c).它将允许您计算三角形面积:A=sqrt(p*(p-a)*(p-b)*(p-c))，其中p是半周长:p=(a+b+c)/2.因此，现在您已经拥有计算距离Hb (您的"d0")所需的所有变量.

If given 3 points you can calculate the the distances between them (sides a, b, c of triangle). It will allow you to calculate triangle Area: A=sqrt(p*(p-a)*(p-b)*(p-c)) where p is half perimeter: p=(a+b+c)/2. So, now u have all variables u need to calculate the distance Hb (your "d0").

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