Default values on Google Maps

Mon, 20 Sep 2010 17:41:06 GMT

Hello,

I am having a problem with the auto calculation of the mid_lat and mid_long calculated by the Google Maps code example.

In my case I want to show the big picture of the map and not only one part of it (calculated by the coordinates that i have.)

Any idea how to create a default view of the map ? I supposed that i have to change something in the var_mid_lat and var_mid_long variables to make it work.

This is my code:



// Variables required for calculating map
// No need to change these
var_pi180= '=pi()/180';
var_lat_offset= '0';
var_mc2= '=256*pow(2,$(var_zoom))';
var_mc1= '=256*pow(2,($(var_zoom)-1))';
var_mid_lat= '=min(latitude)+(1+var_lat_offset)*((max(latitude)-min(latitude))/2)';
var_mid_long= '=min(longitude)+(max(longitude)-min(longitude))/2';
var_zoom= '=max(aggr(if(max( round(256*pow(2,(_zoom_level -1)))+( longitude *((256*pow(2,_zoom_level ))/360)) )-min( round(256*pow(2,(_zoom_level -1)))+( longitude *((256*pow(2,_zoom_level ))/360)) ) <map_size_x AND max((256*pow(2,(_zoom_level-1)))+((0.5*log((1+(sin((latitude)*pi()/180)))/(1-(sin((latitude)*pi()/180)))))*((-256*pow(2,_zoom_level))/(2*pi()))))-min((256*pow(2,(_zoom_level-1)))+((0.5*log((1+(sin((latitude)*pi()/180)))/(1-(sin((latitude)*pi()/180)))))*((-256*pow(2,_zoom_level))/(2*pi()))))<map_size_y,_zoom_level,1),_zoom_level))';
var_maptype= '=if(isnull(only(maptype)),fieldvalue( '&chr(39)&'maptype'&chr(39)&', 1 ),maptype)';
map_size_x= '400'; //640
map_size_y= '500'; //400

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Default values on Google Maps