position prediction fails for close-to-the-dateline

[emilburzo]

It appears that the position predictor has difficulties near the dateline meridian.

You can see in the output below that it's a positive value instead of a negative one.

Testcase:

public void test180Meridian() {
	int distanceKm = 10;
	Position from = Position.create(-16.796686666666666, -179.99983333333333);
	Position north = from.predict(distanceKm, 0);
	Position south = from.predict(distanceKm, 180);
	Position east = from.predict(distanceKm, 90);
	Position west = from.predict(distanceKm, 270);

	System.out.println(north);
	System.out.println(south);
	System.out.println(east);
	System.out.println(west);

	Geometries.rectangle(west.getLon(), south.getLat(), east.getLon(), north.getLat());
}

Output:

[-16.706754647233197,-179.99983333333333]
[-16.886618686100135,-179.99983333333333]
[-16.79666536204166,-179.90589349094225]
[-16.79666536204166,179.9062268242756]

Stacktrace:

java.lang.IllegalArgumentException
	at com.github.davidmoten.guavamini.Preconditions.checkArgument(Preconditions.java:25)
	at com.github.davidmoten.rtree.geometry.internal.RectangleDouble.<init>(RectangleDouble.java:14)
	at com.github.davidmoten.rtree.geometry.internal.RectangleDouble.create(RectangleDouble.java:23)
	at com.github.davidmoten.rtree.geometry.Geometries.rectangleDouble(Geometries.java:58)
	at com.github.davidmoten.rtree.geometry.Geometries.rectangle(Geometries.java:36)

[davidmoten]

I think there are some normalization offerings with other methods like Geometries.rectangleGeographic. You may still have to break your desired search region into two rectangles (one left of the 180 longitude line and the other right of it).

[emilburzo]

Ok, I just thought it might make sense for the position predictor to be capped at the dateline, but I admit I am not sure of the implications.

Thanks for the Geometries.rectangleGeographic tip!