-57 -11 75%
-57 -52 89%
EDIT:
I wrote a java program so that anyone can calculate it.
[code]// USEAGE: java HotspotCalc LAT1 LON1 EFF1 LAT2 LON2 EFF2, etc
public class HotspotCalc {
public static final int nPointsMax = 99;
public static void main(String[] args) {
double[] lon = new double[nPointsMax];
double[] lat = new double[nPointsMax];
double[] eff = new double[nPointsMax];
double bestFit;
double bestFitLat;
double bestFitLong;
double thisFit;
double temp;
bestFit = 99999999;
bestFitLat = 0;
bestFitLong = 0;
int nPoints = args.length / 3;
int ii; int jj; int kk;
//Load the points from command line
for (ii = 0; ii < nPoints * 3; ii += 3) {
lon[ii/3] = Double.parseDouble(args[ii]);
lat[ii/3] = Double.parseDouble(args[ii+1]);
eff[ii/3] = Double.parseDouble(args[ii+2]);
System.out.println ("Point: ("+lon[ii/3]+","+lat[ii/3]+"), "+eff[ii/3]+"%");
}
//Scan through lat, long
for (ii = -90; ii <= 90; ii++ ) {
for (jj = -180; jj <= 180; jj++) {
thisFit = 0;
//Scan through poitns
for (kk = 0; kk < nPoints; kk++) {
//TEMP = sqrt(100 - 125*d(p,x)) -eff)
temp = 100.0 - 125.0*Math.acos(Math.sin(lat[kk]*Math.PI/180.0)*Math.sin(ii*Math.PI/180.0) + Math.cos(lat[kk]*Math.PI/180.0)*Math.cos(ii*Math.PI/180.0)*Math.cos(Math.PI*(lon[kk]-jj)/180.0))/Math.PI -eff[kk];
thisFit += temp * temp;
}
//If this fits the data better than the current best fit, store the lat, long
if (thisFit < bestFit) {
bestFit = thisFit;
bestFitLat = ii;
bestFitLong = jj;
}
}
}
System.out.println ("Best lat: "+bestFitLat);
System.out.println ("Best lon: "+bestFitLong);
System.out.println ("Fit Quality (closest to zero is better): "+(bestFit/nPoints) + " (" +(bestFit)+")");
}
}[/code]
According to it, the best Lat, Long is -31, -72.
Note:
Only include points with eff > 50%!