Compute rotation matrix using the magnetic field

问题

In get rotation matrix value it contains public static boolean getRotationMatrix (float[] R, float[] I, float[] gravity, float[] geomagnetic) Here how can i calculate the float[] gravity? I found a sample of code where it calculate the orientation using both Accelerometer and Magnetic field

boolean success = SensorManager.getRotationMatrix(
   matrixR,
   matrixI,
   valuesAccelerometer,
   valuesMagneticField);

if(success){
SensorManager.getOrientation(matrixR, matrixValues);

double azimuth = Math.toDegrees(matrixValues[0]);
double pitch = Math.toDegrees(matrixValues[1]);
double roll = Math.toDegrees(matrixValues[2]);

readingAzimuth.setText("Azimuth: " + String.valueOf(azimuth));
readingPitch.setText("Pitch: " + String.valueOf(pitch));
 readingRoll.setText("Roll: "+String.valueOf(roll));
}

My questions are :

Is orientation value is the rotation matrix value?
If no then how can i implement this code to get the rotation matrix value using magnetic? field?

To get the rotation matrix i use this code

 public void onSensorChanged(SensorEvent sensorEvent) {
    if (timestamp != 0) {
        final double dT = (sensorEvent.timestamp - timestamp) * NS2S;
            double magneticX = sensorEvent.values[0];
            double magneticY = sensorEvent.values[1];
            double magneticZ = sensorEvent.values[2];
                        double omegaMagnitude =Math.sqrt(magneticX*magneticX + magneticY*magneticY + magneticZ*magneticZ);

                        if (omegaMagnitude > EPSILON) {
                            magneticX /= omegaMagnitude;
                            magneticY /= omegaMagnitude;
                            magneticZ /= omegaMagnitude;
        }
                        double thetaOverTwo = omegaMagnitude * dT / 2.0f;
                        double sinThetaOverTwo =Math.sin(thetaOverTwo);
                        double cosThetaOverTwo = Math.cos(thetaOverTwo);
                        deltaRotationVector[0] = (double) (sinThetaOverTwo * magneticX);
                        deltaRotationVector[1] = (double) (sinThetaOverTwo * magneticY);
                        deltaRotationVector[2] = (double) (sinThetaOverTwo * magneticZ);
                        deltaRotationVector[3] = cosThetaOverTwo;


    }
     double[] deltaRotationMatrix = new double[9];
     SensorManager.getRotationMatrixFromVector(deltaRotationMatrix, deltaRotationVector);
}

But the problem is this getRotationMatrixFromVector is says undefine for sensor.Any idea?

回答1:

Orientation is not a rotation matrix as it only provides you angles related to magnetic North. You can obtain the rotation matrix (Direction Cosine Matrix) that will help you to transform coordinates from your device frame to the Earth's frame this way :

with

= azimuth (radians)

= pitch (radians)

= roll (radians)

回答2:

I know that this is an old thread but in case it helps, for Android I think the 3x3 rotation matrix is actually given by a variation of the approved answer. To be specific, in Android the rotation matrix is

     (cosφ cosψ - sinφ sinψ sinθ)     sinφ cosθ     ( cosφ sinψ + sinφ cosψ sinθ)
    -(sinφ cosψ + cosφ sinψ sinθ)     cosφ cosθ     (-sinφ sinψ + cosφ cosψ sinθ)
              -sinψ cosθ                 -sinθ                  cosφ cosθ

where

    φ = azimuth
    θ = pitch
    ψ = roll

which corresponds to the 3x3 Android rotation matrix R[0] to R[8] (matrixR in the question) via

    R[0] R[1] R[2]
    R[3] R[4] R[5]
    R[6] R[7] R[8]

来源：https://stackoverflow.com/questions/10698092/compute-rotation-matrix-using-the-magnetic-field

标签

android

accelerometer

sensor

rotational-matrices

magnetometer