确定点是否在半径范围内

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确定点X在点Y的100米半径内的最佳方法是什么？确定点是否在半径范围内

CLLocation是否有方法？

由于

2012-01-31 Laurent May

计算距离两个纬度/长之间以及与100比较，以便看看它是否小于100为您案件。 – samfisher 2012-01-31 08:40:25

CoreLocation框架包含一些方法，假设您正在处理纬度和经度坐标。查看这篇文章了解更多信息：http://stackoverflow.com/questions/9029445/how-to-find-the-nearest-100-meters-latitude-and-longitude-depending-upon-the-cur/9031836# 9031836 – 2012-01-31 08:41:42

看到

- (CLLocationDistance) distanceFromLocation:(const CLLocation *)location

documentation

它计算从另一CLLocation物体的距离。

来源

2012-01-31 08:40:32

您可以使用此方法：

// proximity distance calculation 
static const double kDegToRad = 0.017453292519943295769236907684886; 
static const double kEarthRadiusM = 6372797.560856; 

+ (double)distanceInMetersFromLoc:(CLLocation *)from toLoc:(CLLocation *)to 
{ 
    return kEarthRadiusM * [self radianArcFrom:from.coordinate to:to.coordinate]; 
} 

+ (double)radianArcFrom:(CLLocationCoordinate2D)from to:(CLLocationCoordinate2D)to 
{ 
    double latitudeArc = (from.latitude - to.latitude) * kDegToRad; 
    double longitudeArc = (from.longitude - to.longitude) * kDegToRad; 
    double latitudeHS = sin(latitudeArc * 0.5); 
    latitudeHS *= latitudeHS; 
    double lontitudeHS = sin(longitudeArc * 0.5); 
    lontitudeHS *= lontitudeHS; 
    double factor = cos(from.latitude * kDegToRad) * cos(to.latitude * kDegToRad); 
    return 2.0 * asin(sqrt(latitudeHS + factor * lontitudeHS)); 
}

比较距离

if([distanceInMetersFromLoc:location1 to:location2] < 100) 
{ 
    // your condition is satisfied. you can write your code here 
}

来源

2012-01-31 08:44:36 samfisher

确定点是否在半径范围内

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