Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
219 views
in Technique[技术] by (71.8m points)

android - TileProvider method getTile - need to translate x and y to lat/long

I’m porting an iOS app to Android, and using Google Maps Android API v2. The application needs to draw a heatmap overlay onto the map.

So far, it looks like the best option is to use a TileOverlay and implement a custom TileProvider. In the method getTile, my method is given x, y, and zoom, and needs to return a bitmap in the form of a Tile. So far, so good.

I have an array of heatmap items that I will use to draw radial gradients onto the bitmap, each with a lat/long. I am having trouble with the following two tasks:

  1. How do I determine if the tile represented by x, y, and zoom contains the lat/long of the heatmap item?
  2. How do I translate the lat/long of the heatmap item to x/y coordinates of the bitmap.

Thank you for your help!

UPDATE

Thanks to MaciejGórski's answer below, and marcin's implementation I was able to get the 1st half of my question answered, but I still need help with the 2nd part. To clarify, I need a function to return the x/y coordinates of the tile for a specified lat/long. I've tried reversing the calculations of MaciejGórski's and marcin's answer with no luck.

public static Point fromLatLng(LatLng latlng, int zoom){
    int noTiles = (1 << zoom);
    double longitudeSpan = 360.0 / noTiles;
    double mercator = fromLatitude(latlng.latitude);
    int y = ((int)(mercator / 360 * noTiles)) + 180;
    int x = (int)(latlng.longitude / longitudeSpan) + 180;
    return new Point(x, y);
}

Any help is appreciated!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This worked for me:

double n = Math.pow(2, zoom);
double longitudeMin = x/n * 360 -180;
double lat_rad = Math.atan(Math.sinh(Math.PI * (1 - 2 * y/n)));
double latitudeMin = lat_rad * 180/Math.PI;

double longitudeMax = (x + 1)/n * 360 -180;
lat_rad = Math.atan(Math.sinh(Math.PI * (1 - 2 * (y + 1)/n)));
double latitudeMax = lat_rad * 180/Math.PI;

References: http://wiki.openstreetmap.org/wiki/Slippy_map_tilenames


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...