Android magnetometer global coordinates

Question

I have an app that reads the values returned by the magnetometer.
The way it does it at the moment is in the phone's coordinates system.

I want to be able to always read the magnetic field vector in a global coordinates system (x,y,z - east, north, sky)

The code I tried (inspired from this question) will result in a value of 0 for the x axis component, and variable y,z depending on the way I tilt the phone. As far I understood, the rotation matrix transform should make my coordinates system global, but it doesn't seem that way.

The expected behaviour would be to have the same value on the x,y,z components as long as I hold the phone in one place, no matter its tilt.

private final float alpha = (float) 0.8;
private float gravity[] = new float[3];
private float magnetic[] = new float[3];

public void onSensorChanged(SensorEvent event) {

  Sensor sensor = event.sensor;
  if (sensor.getType() == Sensor.TYPE_ACCELEROMETER) {
        // Isolate the force of gravity with the low-pass filter.
          gravity[0] = alpha * gravity[0] + (1 - alpha) * event.values[0];
          gravity[1] = alpha * gravity[1] + (1 - alpha) * event.values[1];
          gravity[2] = alpha * gravity[2] + (1 - alpha) * event.values[2];
  } else if (sensor.getType() == Sensor.TYPE_MAGNETIC_FIELD) {

        magnetic[0] = event.values[0];
        magnetic[1] = event.values[1];
        magnetic[2] = event.values[2];

        float[] R = new float[9];  //rotation matrix
        float[] I = new float[9];  //inclination
        SensorManager.getRotationMatrix(R, I, gravity, magnetic);
        float [] A_D = event.values.clone();  // device coordinates
        float [] A_W = new float[3];          // global coodinates
        A_W[0] = R[0] * A_D[0] + R[1] * A_D[1] + R[2] * A_D[2];
        A_W[1] = R[3] * A_D[0] + R[4] * A_D[1] + R[5] * A_D[2];
        A_W[2] = R[6] * A_D[0] + R[7] * A_D[1] + R[8] * A_D[2];

        Log.d("Field","\nX :"+A_W[0]+"\nY :"+A_W[1]+"\nZ :"+A_W[2]);
    }

score 0 · Answer 1 · answered Mar 06 '16 at 18:02

The code is correct but there is always fluctuation as gravity is only an estimate. Actually, if the device is still, the accelerometer in the world coordinate should theoretically be the same independent of position since the only force acting on it is minus gravity and thus accelerometer should be gravity when the device is still. You can only expect that the East and North coordinates are very small when the device is still.

Android magnetometer global coordinates

1 Answers1