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我正在尝试在纵向放置设备时进行俯仰和滚动。 轴看起来像这样，它将分别关于 x 和 z 轴。现在我正在使用 SensorManager API 来获取设备平放的俯仰、滚动和偏航，这是默认设置。

当我尝试将平面设备的旋转值转换为垂直方向时，我遇到了其他 SO 用户所说的万向节锁定，这是欧拉角工作方式固有的问题。问题是我已经尝试实现旋转矩阵，因为其他用户必须解决类似的问题，但即使我仍然遇到相同的万向节锁定问题。我已经包含了我的 onSensorChanged 方法，希望有人可以帮助找出问题所在。

public void onSensorChanged(SensorEvent se) {

    float degPitch = 0;
    float degRoll = 0;
    float degYaw = 0;
    float radPitch = 0;
    float radRoll = 0;
    float radYaw = 0;

    if (se.sensor.getType() == Sensor.TYPE_ACCELEROMETER) {
        mAccelerometerResult = se.values;
        Log.d("onSensorChanged", "Accelerometer: " + mAccelerometerResult.length);
    }

    if (se.sensor.getType() == Sensor.TYPE_MAGNETIC_FIELD) {
        mMagneticFieldResult = se.values;
        Log.d("onSensorChanged", "Magnetic Field: " + mMagneticFieldResult.length);
    }

    if (mAccelerometerResult != null && mMagneticFieldResult != null) {
        float[] rotation = new float[9];
        float[] inclination = new float[9];

        boolean rotationMatrixCheck = mSensorManager.getRotationMatrix(rotation, inclination, mAccelerometerResult, mMagneticFieldResult);

        if (rotationMatrixCheck) {

            float[] orientation = new float[3];
            mSensorManager.getOrientation(rotation, orientation);


            radYaw = orientation[0];      //Yaw = Z axis
            radPitch = orientation[1];      //Pitch = X axis
            radRoll = orientation[2];      //Roll = Y axis

            degYaw = round((float) Math.toDegrees(radYaw), 2);
            degPitch = round((float)Math.toDegrees(radPitch), 2);
            degRoll = round((float)Math.toDegrees(radRoll), 2);


            if ((counter % 10) == 0) {
                //mYawTextView.setText(degYaw + "°");
                mPitchTextView.setText(degPitch + "°");
                mRollTextView.setText(degRoll + "°");
                counter = 0;
            } else {
                counter++;
            }
        }
    }

此外，我什至不确定我是否了解我正在寻找的旋转值，如果我能在纵向轴上获得良好的旋转。如果我想要纵向的设备滚动（大约从我的原始图像的 z 轴），那仍然是平放的设备的滚动（大约来自平轴图像的 y）？

任何可以在这里分享的见解将不胜感激。

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1 回答 1

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我找到了问题的解决方案。如原始问题中所述，获取 3x3 网格中的旋转矩阵会返回欧拉角的旋转矩阵。在 4x4 网格中获取矩阵会返回旋转矩阵的四元数表示（根据 SO 用户 Stochastically 的答案在这里找到）。

此外，必须重新映射旋转矩阵坐标系才能在纵向模式下使用。注释更改如下所示。

public void onSensorChanged(SensorEvent se) {

float degPitch = 0;
float degRoll = 0;
float degYaw = 0;
float radPitch = 0;
float radRoll = 0;
float radYaw = 0;

if (se.sensor.getType() == Sensor.TYPE_ACCELEROMETER) {
    mAccelerometerResult = se.values;
    Log.d("onSensorChanged", "Accelerometer: " + mAccelerometerResult.length);
}

if (se.sensor.getType() == Sensor.TYPE_MAGNETIC_FIELD) {
    mMagneticFieldResult = se.values;
    Log.d("onSensorChanged", "Magnetic Field: " + mMagneticFieldResult.length);
}

if (mAccelerometerResult != null && mMagneticFieldResult != null) {

//~~~This is where the 3x3 matrix has changed to a 4x4.
    float[] rotation = new float[16];
    float[] inclination = new float[16];

    boolean rotationMatrixCheck = mSensorManager.getRotationMatrix(rotation, inclination, mAccelerometerResult, mMagneticFieldResult);

    if (rotationMatrixCheck) {

        float[] orientation = new float[3];

//~~~This is where the rotational matrix is remapped to be used in portrait mode.
        float[] remappedRotation = new float[16];
        SensorManager.remapCoordinateSystem(rotation, SensorManager.AXIS_X, SensorManager.AXIS_Z, remappedRotation);

        mSensorManager.getOrientation(rotation, orientation);


        radYaw = orientation[0];      //Yaw = Z axis
        radPitch = orientation[1];      //Pitch = X axis
        radRoll = orientation[2];      //Roll = Y axis

        degYaw = round((float) Math.toDegrees(radYaw), 2);
        degPitch = round((float)Math.toDegrees(radPitch), 2);
        degRoll = round((float)Math.toDegrees(radRoll), 2);


        if ((counter % 10) == 0) {
            //mYawTextView.setText(degYaw + "°");
            mPitchTextView.setText(degPitch + "°");
            mRollTextView.setText(degRoll + "°");
            counter = 0;
        } else {
            counter++;
        }
    }
}
于 2016-06-14T21:16:47.050 回答