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到目前为止，我已经在互联网上搜索过。我正在尝试开发一个问题，我需要从手机起点旋转角度。要知道用户是否向上或向下移动手机，我正在使用加速度计，它在开始时有点不稳定，但我设法让它稳定。我现在需要围绕它自己的旋转度数。就像不推荐使用的方向传感器一样。然后我尝试使用 Magnometer，但我很不稳定。然后我确定我想尝试使用陀螺仪，当我使用示例代码时：

 // This timestep's delta rotation to be multiplied by the current rotation
 // after computing it from the gyro sample data.
 if (timestamp != 0) {
     final float dT = (event.timestamp - timestamp) * NS2S;
     // Axis of the rotation sample, not normalized yet.
     float axisX = event.values[0];
     float axisY = event.values[1];
     float axisZ = event.values[2];

     // Calculate the angular speed of the sample
     float omegaMagnitude = (float) Math.sqrt(axisX*axisX + axisY*axisY + axisZ*axisZ);

     // Normalize the rotation vector if it's big enough to get the axis
     if (omegaMagnitude > EPSILON) {
         axisX /= omegaMagnitude;
         axisY /= omegaMagnitude;
         axisZ /= omegaMagnitude;
     }

     // Integrate around this axis with the angular speed by the timestep
     // in order to get a delta rotation from this sample over the timestep
     // We will convert this axis-angle representation of the delta rotation
     // into a quaternion before turning it into the rotation matrix.
     float thetaOverTwo = omegaMagnitude * dT / 2.0f;
     float sinThetaOverTwo = (float) Math.sin(thetaOverTwo);
     float cosThetaOverTwo = (float) Math.cos(thetaOverTwo);
     deltaRotationVector[0] = sinThetaOverTwo * axisX;
     deltaRotationVector[1] = sinThetaOverTwo * axisY;
     deltaRotationVector[2] = sinThetaOverTwo * axisZ;
     deltaRotationVector[3] = cosThetaOverTwo;
 }
 timestamp = event.timestamp;
 float[] deltaRotationMatrix = new float[9];
 SensorManager.getRotationMatrixFromVector(deltaRotationMatrix, deltaRotationVector);
 // User code should concatenate the delta rotation we computed with the current rotation
 // in order to get the updated rotation.
 // rotationCurrent = rotationCurrent * deltaRotationMatrix;

此代码取自他们的文档，无论如何我可以将其转换为 360 度吗？或者我可以获得诸如手机从起点转动多少度的值？

提前致谢。

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1 回答 1

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使用getRotationMatrix()在两个感兴趣的时间点R1获取3x3 旋转矩阵。您想知道与对齐所带来的旋转。R2angleRR1R2

首先计算R：

R = R1 * transpose(R2)

然后计算这个旋转的角度：

angle = acos((trace(R)-1)/2)

就这些。

于 2012-11-12T21:22:56.087 回答