Figuring out distance and course between two coord

问题:

I have 2 coordinates and would like to do something seemingly straightforward. I want to figure out, given:

1) Coordinate A 2) Course provided by Core Location 3) Coordinate B

the following:

1) Distance between A and B (can currently be done using distanceFromLocation) so ok on that one. 2) The course that should be taken to get from A to B (different from course currently traveling)

Is there a simple way to accomplish this, any third party or built in API?

Apple doesn't seem to provide this but I could be wrong.

Thanks, ~Arash

EDIT:

Thanks for the fast responses, I believe there may have been some confusion, I am looking to get the course (bearing from point a to point b in degrees so that 0 degrees = north, 90 degrees = east, similar to the course value return by CLLocation. Not trying to compute actual turn by turn directions.

回答1:

See How to get angle between two POI?

回答2:

I have some code on github that does that. Take a look at headingInRadians here. It is based on the Spherical Law of Cosines. I derived the code from the algorithm on this page.

/*-------------------------------------------------------------------------
* Given two lat/lon points on earth, calculates the heading
* from lat1/lon1 to lat2/lon2.
*
* lat/lon params in radians
* result in radians
*-------------------------------------------------------------------------*/
double headingInRadians(double lat1, double lon1, double lat2, double lon2)
{
    //-------------------------------------------------------------------------
    // Algorithm found at http://www.movable-type.co.uk/scripts/latlong.html
    //
    // Spherical Law of Cosines
    //
    // Formula: θ = atan2( sin(Δlong) * cos(lat2),
    // cos(lat1) * sin(lat2) − sin(lat1) * cos(lat2) * cos(Δlong) )
    // JavaScript:
    //
    // 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).toDeg();
    //-------------------------------------------------------------------------
    double dLon = lon2 - lon1;
    double y = sin(dLon) * cos(lat2);
    double x = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(dLon);

    return atan2(y, x);
}

回答3:

Depending on how much work you want to put in this one, I would suggest looking at Tree Traversal Algorithms (check the column on the right), things like A* alpha star, that you can use to find your find from one point to another, even if obstacles are in-between.

回答4:

If I understand you correctly, you have the current location and you have some other location. You want to find the distance (as the crow flies) between the two points, and to find a walking path between the points.

To answer your first question, distanceFromLocation will find the distance across the earth's surface between 2 points, that is it follows the curvature of the earth, but it will give you the distance as the crow flies. So I think you're right about that.

The second question is a much harder. What you want to do is something called path-finding. Path finding, require's not only a search algorithm that will decide on the path, but you also need data about the possible paths. That is to say, if you want to find a path through the streets, the computer has to know how the streets are connected to each other. Furthermore, if you're trying to make a pathfinder that takes account for traffic and the time differences between taking two different possible paths, you will need a whole lot more data. It is for this reason that we usually leave these kinds of tasks up to big companies, with lots of resources, like Google, and Yahoo.

However, If you're still interested in doing it, check this out http://www.youtube.com/watch?v=DoamZwkEDK0