Question

我正在尝试开发一个使用iPhone的GPS和指南针的应用程序，以便指向某个特定位置的指针（就像指南针总是指向北方）。位置是固定的，无论用户位于何处，我总是需要指针指向该特定位置。我有这个位置的Lat / Long坐标，但不知道如何使用指南针和GPS指向该位置...就像http://www.youtube.com/watch?v=iC0Xn8hY80w此链接1:20'

我写了一些代码，然而，它无法正确旋转。

-(float) angleToRadians:(double) a {
    return ((a/180)*M_PI);
}

-(void)updateArrow {    
    double alon=[longi doubleValue];//source
    double alat=[lati doubleValue];//source
    double blon=[pointlongi doubleValue];//destination
    double blat=[pointlati doubleValue];//destination

    float fLat = [self angleToRadians:alat];
    float fLng = [self angleToRadians:alon];
    float tLat = [self angleToRadians:blat];
    float tLng = [self angleToRadians:blon];

    float temp = atan2(sin(tLng-fLng)*cos(tLat), 
        cos(fLat)*sin(tLat)-sin(fLat)*cos(tLat)*cos(tLng-fLng)); 
    double temp2= previousHeading;

    double temp1=temp-[self angleToRadians:temp2];

    /*I using this,but it can't rotate by :point even i change the coordinate
      in CGPointMake */
    Compass2.layer.anchorPoint=CGPointMake(0, 0.5);

    [Compass2 setTransform:CGAffineTransformMakeRotation(temp1)];
    /* Compass2 is a UIImageView like below picture I want to rotate it around 
     : point in image

        ^
        |
        | 
        |
        :
        |
    */

Answer 1

您可以使用标准的“航向”或“方位”等式 - 如果您在lat1，lon1，并且您感兴趣的点是lat2，lon2，则等式为：

heading = atan2( sin(lon2-lon1)*cos(lat2), cos(lat1)*sin(lat2) - sin(lat1)*cos(lat2)*cos(lon2-lon1))

这为您提供弧度方位，您可以通过乘以180 /π转换为度数。然后该值介于-180和180度之间，因此要获得标准罗盘轴承，请将360添加到任何否定答案。

atan2是与arctan相关的标准功能，它可以为您的目的地点所处的四个可能象限做正确的事情。

Answer 2

1）获取您当前的位置（来自GPS）

2）获得纬度和经度的差异

3）使用atan2方法获得角度

即。（警告：未经测试的代码）

CLLocation *targetLocation = [CLLocation alloc] initWithLatitude:1 longitude:2];
CLLocation *sourceLocation = <get from GPS>

double dx = [targetLocation coordinate].latitude - [sourceLocation coordinate].latitude;
double dy = [targetLocation coordinate].longitude - [sourceLocation coordinate].longitude;

double angle = atan2(dx, dy);

你可能不得不调整它来编译，但想法就在那里！

Answer 3

我之前做过这个，这里有两个不同的实现。第一个类似于你的方法，第二个是没有trig数学。第一个是我在我的应用程序中使用的，但第二个似乎也工作，虽然看起来并不干净。您还需要记住在UI中基于北方来抵消此轴承。

- (double) toRadian: (double) val
{
    return val * (M_PI / 180);
}

// Convert to degrees from radians
- (double) toDegrees: (double) val
{
    return val * 180 / M_PI;
}

// convert from a radian to a 360 degree format.
- (double) toBearing: (double) val
{
    return ( (int)([self toDegrees: val]) + 360 ) % 360;        // use mod to get the degrees
}

// Calculate the bearing based off of the passed coordinates and destination.  
//
- (double) calcBearingWithLatitude:(CLLocationDegrees)latSource 
                             latitude:(CLLocationDegrees)latDest 
                            longitude:(CLLocationDegrees)lonSrc 
                            longitude:(CLLocationDegrees)lonDest
{
    double lat1 = [self toRadian:latSource];
    double lat2 = [self toRadian:latDest];
    double dLon = [self toRadian:(lonDest - lonSrc)];

    double y = sin(dLon) * cos(lat2);
    double x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);
    return [self toBearing:atan2(y, x)];
}

第二个。

// got this code from some forums and modified it, thanks for posting it coullis!  Mostly here for reference on how to do this without sin and cos.
- (CLLocationDegrees) altCalcBearingWithLatitude:(CLLocationDegrees)latSource 
                                        latitude:(CLLocationDegrees)latDest 
                                       longitude:(CLLocationDegrees)lonSrc 
                                       longitude:(CLLocationDegrees)lonDest

{
    CLLocationDegrees result;


// First You calculate Delta distances.
float dx = lonSrc - latSource;
float dy = lonDest - latDest;

// If x part is 0 we could get into division by zero problems, but in that case result can only be 90 or 270:
if (dx==0)
{
    if (dy > 0)
        result = 90;
    else
        result = 270;
}
else
{
    result = [self toDegrees: atan(dy/dx)];
}

// This is only valid for two quadrants (for right side of the coordinate system) so modify result if necessary...
if (dx < 0) 
    result = result + 180;

// looks better if all numbers are positive (0 to 360 range)
if (result < 0)
    result = result + 360;

// return our result.
return result;

}

Answer 4

使用它。您必须从getHeadingForDirection的结果中减去实际的罗盘标题，以确定正确的相对标题。返回值以弧度为标题。

-(float) angleToRadians:(float) a {
    return ((a/180)*M_PI);
}


- (float) getHeadingForDirectionFromCoordinate:(CLLocationCoordinate2D)fromLoc toCoordinate:(CLLocationCoordinate2D)toLoc
{

    float fLat = [self angleToRadians:fromLoc.latitude];
    float fLng = [self angleToRadians:fromLoc.longitude];
    float tLat = [self angleToRadians:toLoc.latitude];
    float tLng = [self angleToRadians:toLoc.longitude];

    return atan2(sin(tLng-fLng)*cos(tLat), cos(fLat)*sin(tLat)-sin(fLat)*cos(tLat)*cos(tLng-fLng));         
}

iPhone指南针GPS方向

4 个答案: