Iphone 根据起点、方位和距离计算新纬度和经度的计算？

有人计算过新的价格吗 基于起点、方位和经度的纬度和经度 距离

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我将非常感谢人们可能提供的任何帮助。

您将在

您需要的代码（在JavaScript中）位于标题“给定距离和方向的目标点”下。摘录：

var lat2 = Math.asin( Math.sin(lat1)*Math.cos(d/R) + 
          Math.cos(lat1)*Math.sin(d/R)*Math.cos(brng) );
var lon2 = lon1 + Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(lat1), 
                 Math.cos(d/R)-Math.sin(lat1)*Math.sin(lat2));

其中R=地球半径，d=距离（单位相同），lat/long以弧度为单位（因为这是

sin

函数所期望的）。你可以从度到弧度

radians = pi * degrees / 180;

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你应该可以从这里开始。请务必查看我提供的链接以获取更多信息。

您将在

您需要的代码（在JavaScript中）位于标题“给定距离和方向的目标点”下。摘录：

var lat2 = Math.asin( Math.sin(lat1)*Math.cos(d/R) + 
          Math.cos(lat1)*Math.sin(d/R)*Math.cos(brng) );
var lon2 = lon1 + Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(lat1), 
                 Math.cos(d/R)-Math.sin(lat1)*Math.sin(lat2));

其中R=地球半径，d=距离（单位相同），lat/long以弧度为单位（因为这是

sin

函数所期望的）。你可以从度到弧度

radians = pi * degrees / 180;

你应该可以从这里开始。查看我提供的链接了解更多信息。

我使用了以下代码：

或使用Swift：

extension CLLocationCoordinate2D {
    func adjusted(distance: Double, degrees: Double) -> CLLocationCoordinate2D {
        let distanceRadians = distance / 6_371  // 6,371 == Earth's radius in km
        let bearingRadians = degrees.radians
        let fromLatRadians = latitude.radians
        let fromLonRadians = longitude.radians

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

        var 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 * .pi), (2 * .pi)) - .pi

        return CLLocationCoordinate2D(latitude: toLatRadians.degrees, longitude: toLonRadians.degrees)
    }
}

extension CLLocationDegrees {
    var radians: Double { self * .pi / 180 }
}

extension Double {
    var degrees: CLLocationDegrees { self * 180 / .pi }
}

我使用了以下代码：

或使用Swift：

extension CLLocationCoordinate2D {
    func adjusted(distance: Double, degrees: Double) -> CLLocationCoordinate2D {
        let distanceRadians = distance / 6_371  // 6,371 == Earth's radius in km
        let bearingRadians = degrees.radians
        let fromLatRadians = latitude.radians
        let fromLonRadians = longitude.radians

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

        var 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 * .pi), (2 * .pi)) - .pi

        return CLLocationCoordinate2D(latitude: toLatRadians.degrees, longitude: toLonRadians.degrees)
    }
}

extension CLLocationDegrees {
    var radians: Double { self * .pi / 180 }
}

extension Double {
    var degrees: CLLocationDegrees { self * 180 / .pi }
}

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计算是否必须在大距离上精确？当你沿着一个大圈走的时候，“方位”会改变。你的意思是“起始方位”还是“恒定方位”？计算是否必须在大距离上精确？当你沿着一个大圈走的时候，“方位”会改变。你是说“起始方位”还是“持续方位”？