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作为概念证明,我想创建一个应用程序来检索当前坐标,计算朝向另一个点的方向,并使用指南针旋转箭头图像以指向空间中的那个点。

我知道如何检索当前坐标并通过 CGAffineTransformMakeRotation 旋转图像,但我还没有找到计算正确角度的公式。

有什么提示吗?

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2 回答 2

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首先,您需要计算方位角。此页面提供了一个简洁的公式:

http://www.movable-type.co.uk/scripts/latlong.html

然后,您可以做一些简单的算术运算来找出该方位与 iPhone 指向的航向之间的差异。通过该差异旋转您的图像。

于 2010-10-12T00:44:45.083 回答
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轴承是:

double bearingUsingStartCoordinate(CLLocation *start, CLLocation *end)
{
    double tc1;
    tc1 = 0.0;

    //dlat = lat2 - lat1
    //CLLocationDegrees dlat = end.coordinate.latitude - start.coordinate.latitude; 

    //dlon = lon2 - lon1
    CLLocationDegrees dlon = end.coordinate.longitude - start.coordinate.longitude;

    //y = sin(lon2-lon1)*cos(lat2)
    double y = sin(d2r(dlon)) * cos(d2r(end.coordinate.latitude));

    //x = cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1)
    double x = cos(d2r(start.coordinate.latitude))*sin(d2r(end.coordinate.latitude)) - sin(d2r(start.coordinate.latitude))*cos(d2r(end.coordinate.latitude))*cos(d2r(dlon)); 

    if (y > 0)
    {
        if (x > 0)
            tc1 = r2d(atan(y/x));

        if (x < 0)
            tc1 = 180 - r2d(atan(-y/x));

        if (x == 0)
            tc1 = 90;

    } else if (y < 0)
    {
        if (x > 0)
            tc1 = r2d(-atan(-y/x));

        if (x < 0)
            tc1 = r2d(atan(y/x)) - 180;

        if (x == 0)
            tc1 = 270;

    } else if (y == 0)
    {
        if (x > 0)
            tc1 = 0;

        if (x < 0)
            tc1 = 180;

        if (x == 0)
            tc1 = nan(0);
    }
    if (tc1 < 0)
        tc1 +=360.0;
        return tc1;
}

对于那些寻找两点之间距离的人:

double haversine_km(double lat1, double long1, double lat2, double long2)
{
    double dlong = d2r(long2 - long1);
    double dlat = d2r(lat2 - lat1);
    double a = pow(sin(dlat/2.0), 2) + cos(d2r(lat1)) * cos(d2r(lat2)) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 6367 * c;

    return d;
}

double haversine_mi(double lat1, double long1, double lat2, double long2)
{
    double dlong = d2r(long2 - long1);
    double dlat = d2r(lat2 - lat1);
    double a = pow(sin(dlat/2.0), 2) + cos(d2r(lat1)) * cos(d2r(lat2)) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 3956 * c; 

    return d;
}
于 2011-01-10T18:31:35.800 回答