计算两个位置（纬度，长）之间的轴承 [英] Calculate bearing between two locations (lat, long)

问题描述

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<一个href="https://github.com/nielswh/iPhone-AR-Toolkit/blob/master/ARKit/ARGeoCoordinate.m">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

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