2014-09-04 23 views
2

我查了各地關於如何使用設備的硬件(羅盤,加速度計)使用加速度計和獲得穩定的方位

下面的指南是驚人的旋轉3D對象: http://ibuzzlog.blogspot.co.uk/2012/08/how-to-use-android-sensors.html?showComment=1409868331799#c665801926070707020

然而方位在多種設備上是令人難以置信的跳躍。 (想象一下這是綁在你的臉,看着左和右)

@Override 
    public void onSensorChanged(SensorEvent event) { 

     int type = event.sensor.getType(); 

     //Log.i("TAG", "Sensor " + type); 

     if(event.sensor.getType() == Sensor.TYPE_MAGNETIC_FIELD) 
     { 
      MagneticFieldValues_last[0] = event.values[0]; 
      MagneticFieldValues_last[1] = event.values[1]; 
      MagneticFieldValues_last[2] = event.values[2]; 

      bHaveMagneticField = true; 
     } 
     if(event.sensor.getType() == Sensor.TYPE_ACCELEROMETER) 
     { 
      AccelerometerValues_last[0] = event.values[0]; 
      AccelerometerValues_last[1] = event.values[1]; 
      AccelerometerValues_last[2] = event.values[2]; 

      bHaveAccelerometer = true; 
     } 
     if(bHaveMagneticField && bHaveAccelerometer) 
     { 
      if(SensorManager.getRotationMatrix(R, null, AccelerometerValues_last, MagneticFieldValues_last)) 
      { 
       SensorManager.remapCoordinateSystem(R, SensorManager.AXIS_Y, SensorManager.AXIS_MINUS_X, remapR); 
       SensorManager.getOrientation(remapR, orientationValues); 

       Matrix.multiplyMV(orientationVector, 0, remapR, 0, sZVector, 0); 
       pitch2 = (float) (-Math.atan2(orientationVector[1], orientationVector[2]) * RADIANS_TO_DEGREES); 

       Matrix.multiplyMV(orientationVector, 0, remapR, 0, sZVector, 0); 
       orientation = (float) (-Math.atan2(orientationVector[0], orientationVector[1]) * RADIANS_TO_DEGREES); 

       Matrix.invertM(remapR_inv, 0, remapR, 0); 
       Matrix.multiplyMV(azimuthVector, 0, remapR_inv, 0, sZVector, 0); 
       azimuth = (float) (180 + Math.atan2(azimuthVector[0], azimuthVector[1]) * RADIANS_TO_DEGREES); 
      } 
     } 
    } 

UPDATE

發現這一點,似乎這樣的伎倆 http://blog.thomnichols.org/2011/08/smoothing-sensor-data-with-a-low-pass-filter

更新2

可悲的是,它並沒有做到這一點,仍然需要一種平滑價值的方法。其他人如何做到這一點?

回答