Hi

I am currently in the process with making an app that shall show the
loaction retrived from the GPS sensor and displying the position in UTM
format (WGS83 datum).

Is there a way to get the location info from the GPS as UTM?

No. Location is provides as lat/lon in WGS84. You have to convert that to
UTM.

I figured as much.

I hav found the forumla but Im not sure how to use the formula in code. I
mean do I just put the numbers and calculations in or do I need a special
syntax for things like sinus and cosinus?

Here is link to formula : Lat long to UTM<http://www.uwgb.edu/dutchs/usefuldata/utmformulas.htm>

Can someone pleas tell me how to get numbers in power ?

Like this (syntax used in Excel): 45^(3/2) ?

Math.pow(45.0, 1.5);

On Wednesday, October 10, 2012 1:03:48 PM UTC-5, Karl Kristian Markman
wrote:

--
Lew

Thanks Lew. I have figured it out (after reading the API docs ++.)

So here is the code I use to transform the Lat long to UTM ( in case anyone
else wouders..)

[code]

public void convert (){

                double a = 6378137;
        double b = 6356752.3142;

                e = Math.sqrt(1-Math.pow(b,2)/Math.pow(a, 2));  // a and b
are constants.
e2 = e*e/(1-(e*e));
n =(a-b)/(a+b);        
 A0 =  a*(1 - n + (5/4)*(Math.pow(n,2) - Math.pow(n,3)) +
(81/64)*(Math.pow(n,4) - Math.pow(n,5)));
  B0 =(3*a*n/2)*(1 - n - (7*n*n/8)*(1-n) + (55/64)*(Math.pow(n,
4)-Math.pow(n, 5)));
C0 =(15*a*n*n/16)*(1 - n +(3*n*n/4)*(1-n));
D0 =(35*a*Math.pow(n, 3)/48)*(1 - n + 11*n*n/16);
E0 =(315*a*Math.pow(n,4)/51)*(1-n);
 zone = (int) (31 + (lngdeg/6));
 double pi = 6* zone -183;
double pii = (lngdeg-pi)*Math.PI/180;
double rho1 = (1-(e*e) * (Math.sin(lat)*( Math.sin(lat))));
rho = a * (1-e*e)/ Math.pow(rho1, (3/2));
nu = a/(Math.pow((1-(e*e *(Math.sin(lat))*(Math.sin(lat)))), (1/2)));
S  = A0* lat - B0 * Math.sin(2*lat) + C0 * Math.sin(4*lat) - D0 *
Math.sin(6*lat) + E0 * Math.sin(8*lat);
double Ki = S * knu;
double Kii = knu * nu *Math.sin(lat)*Math.cos(lat)/2;
double Kiii = (knu * nu
*Math.sin(lat)*Math.pow(Math.cos(lat),3)/24)*(5-Math.pow(Math.tan(lat),2)+9 *Math.pow(e2,2)*Math.pow(Math.cos(lat),2)+4*Math.pow(e2
,2)*Math.pow(Math.cos(lat),4));
double Kiv = knu * nu *Math.cos(lat);
double Kv = knu *
Math.pow(Math.cos(lat),3)*(nu/6)*(1-Math.pow(Math.tan(lat),2)+e2*Math.pow(M ath.cos(lat),2));

double UTMni = (Ki+Kii*Math.pow(pii, 2)+ Kiii * Math.pow(pii,4));
double UTMei = 500000 + (Kiv*pii + Kv * Math.pow(pii, 3));
 UTMn = (int) UTMni;
UTMe = (int) UTMei;

--
Lew

Yes i agree I have not been good with comments. If its any constalation I
will update the code mow that its working....

Lew is this more what you are looking for??

Also I have a link to the forumla I used<http://www.uwgb.edu/dutchs/usefuldata/utmformulas.htm>

public void onLocationChanged(Location location) {

  lat = Math.toRadians(location.getLatitude()); // Latitude in radians
  lng = Math.toRadians(location.getLongitude()); // Longitude in radians
  lngdeg = location.getLongitude();  // Latitude in "normal" form. dd.dd
 convert();

     pos.setText(zone + "\t"+ UTMn +"\n \t\t\t" +UTMe );

  }

  public void convert(){ // konverter fra Lat /Long til UTM (WGS84 / NAD84)
  public final double knu = 0.9996;  // scale along central meridian of zone
  public final double a = 6378137;   //Equatorial radius in meters
  public final double b = 6356752.3142; // Polar radius in meters

e = Math.sqrt(1-(b*b)/(a*a)); // e = the eccenticity of the earth's
elliptical cross-section
e2 = e*e/(1-(e*e)); // The quantile e' only occurs in even powers
n =(a-b)/(a+b);
  zone = (int) (31 + (lngdeg/6)); // Calculating UTM zone using Longetude
in dd.dd form as supplied by the GPS
 double pi = 6* zone -183; // Central meridian of zone
double pii = (lngdeg-pi)*Math.PI/180; //Differance between Longitude and
central meridian of zone
 rho = a * (1-e*e)/ Math.pow((1-(e*e) * (Math.sin(lat)*( Math.sin(lat)))),
(3/2)); // The radius of the curvature of the earth in meridian plane
 nu = a/(Math.pow((1-(e*e *(Math.sin(lat))*(Math.sin(lat)))), (1/2)));
//The radius of the curvature of the earth perpendicular to the meridian
plane
 /*
 * A0 - E0 is used for calclulating the Meridinol arc through the given
point (lat long)
 * The distance from the earth's surface form the equator. All angles are
in radians
 *
 */
 A0 =  a*(1 - n + (5/4)*(Math.pow(n,2) - Math.pow(n,3)) +
(81/64)*(Math.pow(n,4) - Math.pow(n,5)));
  B0 =(3*a*n/2)*(1 - n - (7*n*n/8)*(1-n) + (55/64)*(Math.pow(n,
4)-Math.pow(n, 5)));
C0 =(15*a*n*n/16)*(1 - n +(3*n*n/4)*(1-n));
D0 =(35*a*Math.pow(n, 3)/48)*(1 - n + 11*n*n/16);
E0 =(315*a*Math.pow(n,4)/51)*(1-n);
 // Calculation of the Meridional Arc
S  = A0* lat - B0 * Math.sin(2*lat) + C0 * Math.sin(4*lat) - D0 *
Math.sin(6*lat) + E0 * Math.sin(8*lat);
 /*
 * y = northing = Ki+ Kiip^2 + Kiip^4
 *
 *
 */
 double Ki = S * knu;
double Kii = knu * nu *Math.sin(lat)*Math.cos(lat)/2;
double Kiii = (knu * nu
*Math.sin(lat)*Math.pow(Math.cos(lat),3)/24)*(5-Math.pow(Math.tan(lat),2)+9 *Math.pow(e2,2)*Math.pow(Math.cos(lat),2)+4*Math.pow(e2
,2)*Math.pow(Math.cos(lat),4));
 /*
 *  x = easting = Kivpii + Kvpii^3 +
 */
double Kiv = knu * nu *Math.cos(lat);
double Kv = knu *
Math.pow(Math.cos(lat),3)*(nu/6)*(1-Math.pow(Math.tan(lat),2)+e2*Math.pow(M ath.cos(lat),2));

double UTMni = (Ki+Kii*Math.pow(pii, 2)+ Kiii * Math.pow(pii,4));// Northing
double UTMei = 500000 + (Kiv*pii + Kv * Math.pow(pii, 3));  //Easting is
relative to the central meridain. Forconvetional UTM Easting add 5000000
meters to x
 UTMn = (int) UTMni; // Northing, rounded to closest integer
UTMe = (int) UTMei; // Easting, rounded to closest integer
  }

--
Lew

Sorry for the TAB usage but I just copy/past from eclipse..

I have one more problem here. The conversion sort of works but I miss the
target by about 200 meters West. I have figured it out to bee this line
that screws everything up.:

nu = a/(Math.pow((1-(e*e* Math.sin(lat)*Math.sin(lat))), (1/2));

As it is written in plain language:

nu = a/(1-e^2sin(lat)^2)^1/2

When I enter this in to excel I get 6394567 and from my app I get 6378922.
This follows through on every calculation done after.

I saw one place (in Java code though) where they used
Math.pow(Math.sin(lat),(1/2.0)) for setting odd powers.

It's not  java.lang.math that needs to have the decimal point but Java the
language:
1/2 is integer 1 divided by integer 2 result integer 0
1.0/2.0 is double 1.0 divided by double 2.0 result double 0.5