从一个坐标计算新坐标x米和y度

问题描述

如果我有一个坐标，想要得到一个新的坐标x数米之遥，在一些方向。如何做到这一点？

我正在寻找类似于

- （ CLLocationCoordinate2D）translateCoordinate：（CLLocationCoordinate2D）coordinate
translateMeters：（int）meters
translateDegrees：（double）degrees;

Thanks！

解决方案

不幸的是，API中没有提供这样的功能，所以您必须自己编写。 / p>

显示了更精确的计算距离大于地球四分之一的距离。

另外请注意，上面的代码接受以km为单位的距离，因此请务必在通过之前将米除以1000.0。

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

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

I am looking for something like

-(CLLocationCoordinate2D) translateCoordinate:(CLLocationCoordinate2D)coordinate translateMeters:(int)meters translateDegrees:(double)degrees;

Thanks!

解决方案

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

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.

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;
}

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.

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

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