GPS坐标以度数计算距离

问题描述

在iPhone上，我以十进制度得到用户的位置，例如：纬度39.470920，经度= -0.373192;这就是A点。

我需要创建一个带有另一个GPS坐标的线，也是十进制度，点B。然后，计算线之间的距离（垂直） A到B和另一个点C。

问题是我与度数值混淆。我想把结果以米计。需要什么转换？如何计算这个样子的最终公式如何？

解决方案

为什么不使用 s distanceFromLocation：方法？它会告诉你接收者和另一个CLLocation之间的精确距离。

  CLLocation * locationA = [[CLLocation alloc] initWithLatitude：12.123456经度：12.123456]; 
 CLLocation * locationB = [[CLLocation alloc] initWithLatitude：21.654321 longitude：21.654321]; 
 
 CLLocationDistance distanceInMeters = [locationA distanceFromLocation：locationB]; 
 
 // CLLocation is aka double 
 
 [locationA release]; 
 [locationB release]; 
  

这样很简单。

On the iPhone, I get the user's location in decimal degrees, for example: latitude 39.470920 and longitude = -0.373192; That's point A.

I need to create a line with another GPS coordinate, also in decimal degrees, point B. Then, calculate the distance (perpendicular) between the line from A to B and another point C.

The problem is I get confused with the values in degrees. I would like to have the result in meters. What's the conversion needed? How will the final formula to compute this look like?

解决方案

Why don't you use CLLocations distanceFromLocation: method? It will tell you the precise distance between the receiver and another CLLocation.

CLLocation *locationA = [[CLLocation alloc] initWithLatitude:12.123456 longitude:12.123456];
CLLocation *locationB = [[CLLocation alloc] initWithLatitude:21.654321 longitude:21.654321];

CLLocationDistance distanceInMeters = [locationA distanceFromLocation:locationB];

// CLLocation is aka double

[locationA release];
[locationB release];

It's as easy as that.

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