由网友(縂究是莪太傻)分享简介：我想开发自己的增强现实引擎。I'm trying to develop my own augmented reality engine.在搜索在互联网上，我发现这个有用教程。读它，我看到，重要的是用户的位置，点位置和北部之间的轴承。Searching on internet, I've found this us...

我想开发自己的增强现实引擎。

I'm trying to develop my own augmented reality engine.

在搜索在互联网上，我发现这个有用教程。读它，我看到，重要的是用户的位置，点位置和北部之间的轴承。

Searching on internet, I've found this useful tutorial. Reading it I see that the important thing is bearing between user location, point location and north.

下面的图片是从该教程。

The following picture is from that tutorial.

这之后，我写了一个Objective-C的方法来获得测试版：

Following it, I wrote an Objective-C method to obtain beta:

+ (float) calculateBetaFrom:(CLLocationCoordinate2D)user to:(CLLocationCoordinate2D)destination
{
    double beta = 0;
    double a, b = 0;

    a = destination.latitude - user.latitude;
    b = destination.longitude - user.longitude;

    beta = atan2(a, b) * 180.0 / M_PI;
    if (beta < 0.0)
        beta += 360.0;
    else if (beta > 360.0)
        beta -= 360;

    return beta;
}

但是，当我尝试它，它不能很好地工作。

But, when I try it, it doesn't work very well.

所以，我检查了iPhone AR工具包，看看它是如何工作（我一直在使用这个工具，但它是如此之大对我来说）。

So, I checked iPhone AR Toolkit, to see how it works (I've been working with this toolkit, but it is so big for me).

和，在ARGeoCoordinate.m还有另一种实施如何获得的β

And, in ARGeoCoordinate.m there is another implementation of how to obtain beta:

- (float)angleFromCoordinate:(CLLocationCoordinate2D)first toCoordinate:(CLLocationCoordinate2D)second {

    float longitudinalDifference    = second.longitude - first.longitude;
    float latitudinalDifference     = second.latitude  - first.latitude;
    float possibleAzimuth           = (M_PI * .5f) - atan(latitudinalDifference / longitudinalDifference);

    if (longitudinalDifference > 0) 
        return possibleAzimuth;
    else if (longitudinalDifference < 0) 
        return possibleAzimuth + M_PI;
    else if (latitudinalDifference < 0) 
        return M_PI;

    return 0.0f;
}

它使用这个公式：

It uses this formula:

float possibleAzimuth = (M_PI * .5f) - atan(latitudinalDifference / longitudinalDifference);

为什么（M_PI * .5f）在这个公式？我不明白。

Why is (M_PI * .5f) in this formula? I don't understand it.

和继续搜索，我发现了另一个页面谈到如何计算距离和轴承的2个地点。在该页面中还有另外一种实现：

And continue searching, I've found another page talking about how to calculate distance and bearing of 2 locations. In this page there is another implementation:

/**
 * Returns the (initial) bearing from this point to the supplied point, in degrees
 *   see http://williams.best.vwh.net/avform.htm#Crs
 *
 * @param   {LatLon} point: Latitude/longitude of destination point
 * @returns {Number} Initial bearing in degrees from North
 */
LatLon.prototype.bearingTo = function(point) {
  var lat1 = this._lat.toRad(), lat2 = point._lat.toRad();
  var dLon = (point._lon-this._lon).toRad();

  var y = Math.sin(dLon) * Math.cos(lat2);
  var x = Math.cos(lat1)*Math.sin(lat2) -
          Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
  var brng = Math.atan2(y, x);

  return (brng.toDeg()+360) % 360;
}

哪一个是正确的？

Which one is the right one?

推荐答案

轴承计算

//Source
JSONObject source = step.getJSONObject("start_location");
double lat1 = Double.parseDouble(source.getString("lat"));
double lng1 = Double.parseDouble(source.getString("lng"));

// destination
JSONObject destination = step.getJSONObject("end_location");
double lat2 = Double.parseDouble(destination.getString("lat"));
double lng2 = Double.parseDouble(destination.getString("lng"));

double dLon = (lng2-lng1);
double y = Math.sin(dLon) * Math.cos(lat2);
double x = Math.cos(lat1)*Math.sin(lat2) - Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
double brng = Math.toDegrees((Math.atan2(y, x)));
brng = (360 - ((brng + 360) % 360));

转换度到弧度

Radians = Degrees * PI / 180

弧度转换成度

Degrees = Radians * 180 / PI