在获取旋转矩阵值时,它包含public static boolean getRotationMatrix (float[] R, float[] I, float[] gravity, float[] geomagnetic)
我在这里如何计算float[] gravity
?
我找到了一个代码示例,它使用orientation
和Accelerometer
来计算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));
}
我的问题是:
要获得旋转矩阵,我使用此代码
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);
}
但问题是这个getRotationMatrixFromVector
对传感器来说是未定义的。任何想法?
答案 0 :(得分:4)
方向不是旋转矩阵,因为它只提供与磁北相关的角度。您可以获得旋转矩阵(方向余弦矩阵),它将帮助您以这种方式将坐标从设备框架转换为地球框架:
与
=方位角(弧度)
=音高(弧度)
= roll(弧度)
答案 1 :(得分:1)
我知道这是一个旧线程但是如果它有帮助,对于Android我认为3x3旋转矩阵实际上是由批准答案的变体给出的。具体来说,在Android中,旋转矩阵是
(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
,其中
φ = azimuth θ = pitch ψ = roll
对应于3x3 Android旋转矩阵R [0]到R [8](问题中的矩阵R)通过
R[0] R[1] R[2] R[3] R[4] R[5] R[6] R[7] R[8]