问题描述

我一定是在文档中遗漏了一些东西，我认为这应该很容易......

I must be missing somthing out in the docs, I thought this should be easy...

如果我有一个坐标并且想要在 x 米外的某个方向上获得一个新坐标.我该怎么做?

If I have one coordinate and want to get a new coordinate x meters away, in some direction. How do I do this?

我正在寻找类似的东西

-(CLLocationCoordinate2D) translateCoordinate:(CLLocationCoordinate2D)坐标translateMeters:(int) 米translateDegrees:(double)degrees;

谢谢！

推荐答案

很遗憾，API 中没有提供这样的功能，所以你必须自己编写.

Unfortunately, there's no such function provided in the API, so you'll have to write your own.

这个网站给出了几个涉及纬度/经度和样本的计算JavaScript 代码.具体来说，标题为给定距离起点的目标点和方位角"的部分显示了如何计算您的要求.

This site gives several calculations involving latitude/longitude and sample JavaScript code. Specifically, the section titled "Destination point given distance and bearing from start point" shows how to calculate what you're asking.

JavaScript 代码位于该页面的底部，这是将其转换为 Objective-C 的一种可能方法:

The JavaScript code is at the bottom of that page and here's one possible way to convert it to Objective-C:

- (double)radiansFromDegrees:(double)degrees
{
    return degrees * (M_PI/180.0);    
}

- (double)degreesFromRadians:(double)radians
{
    return radians * (180.0/M_PI);
}

- (CLLocationCoordinate2D)coordinateFromCoord:
        (CLLocationCoordinate2D)fromCoord 
        atDistanceKm:(double)distanceKm 
        atBearingDegrees:(double)bearingDegrees
{
    double distanceRadians = distanceKm / 6371.0;
      //6,371 = Earth's radius in km
    double bearingRadians = [self radiansFromDegrees:bearingDegrees];
    double fromLatRadians = [self radiansFromDegrees:fromCoord.latitude];
    double fromLonRadians = [self radiansFromDegrees:fromCoord.longitude];

    double toLatRadians = asin( sin(fromLatRadians) * cos(distanceRadians) 
        + cos(fromLatRadians) * sin(distanceRadians) * cos(bearingRadians) );

    double toLonRadians = fromLonRadians + atan2(sin(bearingRadians) 
        * sin(distanceRadians) * cos(fromLatRadians), cos(distanceRadians) 
        - sin(fromLatRadians) * sin(toLatRadians));

    // adjust toLonRadians to be in the range -180 to +180...
    toLonRadians = fmod((toLonRadians + 3*M_PI), (2*M_PI)) - M_PI;

    CLLocationCoordinate2D result;
    result.latitude = [self degreesFromRadians:toLatRadians];
    result.longitude = [self degreesFromRadians:toLonRadians];
    return result;
}

在 JS 代码中，它包含 这个链接 显示更准确的计算距离大于地球周长的 1/4.

In the JS code, it contains this link which shows a more accurate calculation for distances greater than 1/4 of the Earth's circumference.

另请注意，上述代码接受以公里为单位的距离，因此请务必在通过之前将米除以 1000.0.

Also note the above code accepts distance in km so be sure to divide meters by 1000.0 before passing.

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