問題描述

我一定是在文檔中遺漏了一些東西，我認(rèn)為這應(yīng)該很容易......

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

如果我有一個(gè)坐標(biāo)并且想要在 x 米外的某個(gè)方向上獲得一個(gè)新坐標(biāo).我該怎么做?

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

我正在尋找類似的東西

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

謝謝！

推薦答案

很遺憾，API 中沒有提供這樣的功能，所以你必須自己編寫.

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

這個(gè)網(wǎng)站給出了幾個(gè)涉及緯度/經(jīng)度和樣本的計(jì)算JavaScript 代碼.具體來說，標(biāo)題為給定距離起點(diǎn)的目標(biāo)點(diǎn)和方位角"的部分顯示了如何計(jì)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 代碼位于該頁面的底部，這是將其轉(zhuǎn)換為 Objective-C 的一種可能方法:

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 代碼中，它包含 這個(gè)鏈接 顯示更準(zhǔn)確的計(jì)算距離大于地球周長的 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.

另請注意，上述代碼接受以公里為單位的距離，因此請務(wù)必在通過之前將米除以 1000.0.

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

這篇關(guān)于計(jì)算距離一個(gè)坐標(biāo)的新坐標(biāo) x 米和 y 度的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！