计算纬度/经度坐标与另一对纬度/经度坐标之间的特定距离 [英] Calculate lat/long coords a specific distance away from another pair of lat/long coords

问题描述

我的应用程序需要能够获取用户的起始位置，然后计算该点(在所有方向上)与纬度/经度的特定距离.有人知道我该怎么做吗?

My app needs to be able to get the user starting location and then calculate lat/long coords specific distances away from that spot (in all directions). Does anyone know how I might accomplish this?

推荐答案

距中心点固定距离的点集具有无限个成员，因此不可能列出每个点.但是，给定一个点，您可以使用distanceFromLocation:方法检查它是否在集合内.

The set of points a fixed distance from a central point has infinite members, so it's not possible to list out every one. However, given a single point you could check if it is within the set by using the distanceFromLocation: method.

在数学上，给定半径坐标(长，纬度)并假设地球是平坦的，则半径(距离)为r的圆的等式为

Mathematically, given a radius coordinate (long, lat) and assuming the Earth is flat, the equation of the circle with radius (distance) r would be

r ^ 2 =(x-long)^ 2 +(y-lat)^ 2

r^2 = (x - long)^2 + (y - lat)^2

通过使用一些魔术，这变成了

With the use of some magic, this becomes

x = rcosθ-长； y = rsinθ-lat

x = r cosθ - long; y = r sinθ - lat

因此，您现在可以按任意角度打孔，并且以已知的距离/半径在圆的边缘上获得一个点.用地图术语来说，角度0将为您提供圆心，该圆心位于中心正东.正角围绕中心逆时针旋转.

So now you can punch in any arbitrary angle and for a known distance/radius get a point on the edge of the circle. In map terms, the angle 0 will give you the point on the circle straight east of the center. Positive angles go counterclockwise around the center.

在代码中:

-(CLLocation*)locationForAngle:(float)angle fromCenterLocation:(CLLocation *)center withDistance:(float)distance {
    //angle must be in radians
    float longitude = distance * cosf(angle) - center.coordinate.longitude;
    float latitude = distance * sinf(angle) - center.coordinate.latitude;
    return [[CLLocation alloc] initWithLatitude:latitude longitude:longitude];
}

我不小心删除了几个词

这篇关于计算纬度/经度坐标与另一对纬度/经度坐标之间的特定距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！