iphone - 将MKMapPoint距离转换为米 | 将MKMapPoint距离转换为米

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问题描述

假设我有一个由四个CLLocationCoordinate2D点组成的正方形，这些点位于lat，lon，我想要以米为单位查找正方形的面积。我将CLLocationCoordinate2D点转换为MKMapPoints，并在X-Y空间中找到该区域。但是，我发现的区域是MKMapPoint的单位，它不直接转换为米。 MapKit函数

 MKMetersBetweenMapPoints 
 如何将MKMapPoint空间中的这个区域转换回米？
 / code>使得这一点更容易。
 
 
 例如，如果您想获取当前显示区域的区域：
  MKMapPoint mpTopLeft = mapView.visibleMapRect.origin; 
 
 MKMapPoint mpTopRight = MKMapPointMake（
 mapView.visibleMapRect.origin.x + mapView.visibleMapRect.size.width，
 mapView.visibleMapRect.origin.y）; 
 
 MKMapPoint mpBottomRight = MKMapPointMake（
 mapView.visibleMapRect.origin.x + mapView.visibleMapRect.size.width，
 mapView.visibleMapRect.origin.y + mapView.visibleMapRect.size 。高度）; 
 
 CLLocationDistance hDist = MKMetersBetweenMapPoints（mpTopLeft，mpTopRight）; 
 CLLocationDistance vDist = MKMetersBetweenMapPoints（mpTopRight，mpBottomRight）; 
 
 double vmrArea = hDist * vDist; 
  
该文件指出该函数需要考虑地球的曲率。
 
Say I have a square which consists of four CLLocationCoordinate2D points, which are in lat, lon, and I want to find the area of the square in meters.  I convert the CLLocationCoordinate2D points into MKMapPoints, and I find the area in X-Y space.  However, the area I find is in the units of MKMapPoint, which don't directly translate to meters.  How can I translate this area in MKMapPoint-space back into meters?
 解决方案 
The MapKit function MKMetersBetweenMapPoints makes this easier.
For example, if you wanted to get the area of the currently displayed region:
MKMapPoint mpTopLeft = mapView.visibleMapRect.origin;

MKMapPoint mpTopRight = MKMapPointMake(
    mapView.visibleMapRect.origin.x + mapView.visibleMapRect.size.width,
    mapView.visibleMapRect.origin.y);

MKMapPoint mpBottomRight = MKMapPointMake(
    mapView.visibleMapRect.origin.x + mapView.visibleMapRect.size.width,
    mapView.visibleMapRect.origin.y + mapView.visibleMapRect.size.height);

CLLocationDistance hDist = MKMetersBetweenMapPoints(mpTopLeft, mpTopRight);
CLLocationDistance vDist = MKMetersBetweenMapPoints(mpTopRight, mpBottomRight);

double vmrArea = hDist * vDist;
The documentation states that the function takes "into account the curvature of the Earth."
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