Android中SensorManager.getRotationMatrix函数原理解释

SensorManager是Android中的一个类，其有一个函数getRotationMatrix，可以计算出旋转矩阵，进而通过getOrientation求得设备的方向（航向角、俯仰角、横滚角）。函数getRotationMatrix的源码如下所示，源码中虽然对该函数整体进行了解释，但是对代码中各个参数的计算没有说明，如为什么加速度的数值要和磁力计的数值做差乘。在网上各种搜索后，找到一段老外对这个问题的英文解释，很好的回答了上述问题。大意翻译（包括自己的理解）如下：加速度数值和磁力计数值均是向量，手机水平放置时，加速度读数实际上就是重力向量，方向是竖直朝下的；磁力计表示本地的磁场，不考虑环境影响及磁偏角的话，认为磁场方向是水平南北朝向的。因此，代码中首先对加速度和磁力计数据做了一个差乘，得出一个水平东西方向的向量（差乘的定义）。经过这个运算，本来只有一个平面的向量，变成了三个三维立体平面的向量，从而可以用来计算设备的方向。源码中后面又做了一次差乘，是用计算出的水平东西方向的向量和重力向量做的差乘，这次运算重新得出一个水平南北方向的向量，最后旋转矩阵中用这三个向量（两个计算出的水平向量、一个重力向量）构成。

1、SensorManager.getRotationMatrix

    public static boolean getRotationMatrix(float[] R, float[] I,
            float[] gravity, float[] geomagnetic) {
        // TODO: move this to native code for efficiency
        float Ax = gravity[0];
        float Ay = gravity[1];
        float Az = gravity[2];
        final float Ex = geomagnetic[0];
        final float Ey = geomagnetic[1];
        final float Ez = geomagnetic[2];
        float Hx = Ey*Az - Ez*Ay;
        float Hy = Ez*Ax - Ex*Az;
        float Hz = Ex*Ay - Ey*Ax;
        final float normH = (float)Math.sqrt(Hx*Hx + Hy*Hy + Hz*Hz);
        if (normH < 0.1f) {
            // device is close to free fall (or in space?), or close to
            // magnetic north pole. Typical values are  > 100.
            return false;
        }
        final float invH = 1.0f / normH;
        Hx *= invH;
        Hy *= invH;
        Hz *= invH;
        final float invA = 1.0f / (float)Math.sqrt(Ax*Ax + Ay*Ay + Az*Az);
        Ax *= invA;
        Ay *= invA;
        Az *= invA;
        final float Mx = Ay*Hz - Az*Hy;
        final float My = Az*Hx - Ax*Hz;
        final float Mz = Ax*Hy - Ay*Hx;
        if (R != null) {
            if (R.length == 9) {
                R[0] = Hx;     R[1] = Hy;     R[2] = Hz;
                R[3] = Mx;     R[4] = My;     R[5] = Mz;
                R[6] = Ax;     R[7] = Ay;     R[8] = Az;
            } else if (R.length == 16) {
                R[0]  = Hx;    R[1]  = Hy;    R[2]  = Hz;   R[3]  = 0;
                R[4]  = Mx;    R[5]  = My;    R[6]  = Mz;   R[7]  = 0;
                R[8]  = Ax;    R[9]  = Ay;    R[10] = Az;   R[11] = 0;
                R[12] = 0;     R[13] = 0;     R[14] = 0;    R[15] = 1;
            }
        }
        if (I != null) {
            // compute the inclination matrix by projecting the geomagnetic
            // vector onto the Z (gravity) and X (horizontal component
            // of geomagnetic vector) axes.
            final float invE = 1.0f / (float)Math.sqrt(Ex*Ex + Ey*Ey + Ez*Ez);
            final float c = (Ex*Mx + Ey*My + Ez*Mz) * invE;
            final float s = (Ex*Ax + Ey*Ay + Ez*Az) * invE;
            if (I.length == 9) {
                I[0] = 1;     I[1] = 0;     I[2] = 0;
                I[3] = 0;     I[4] = c;     I[5] = s;
                I[6] = 0;     I[7] =-s;     I[8] = c;
            } else if (I.length == 16) {
                I[0] = 1;     I[1] = 0;     I[2] = 0;
                I[4] = 0;     I[5] = c;     I[6] = s;
                I[8] = 0;     I[9] =-s;     I[10]= c;
                I[3] = I[7] = I[11] = I[12] = I[13] = I[14] = 0;
                I[15] = 1;
            }
        }
        return true;
    }
    

2、英文原文解释

If I understand your problem correctly you want your phone to know its 3D orientation. You need at least two vectors to do that. The xyz accelerometer produces a gravity vector that goes straight down. An xyz magnetometer can provide a second vector that is horizontally towards magnetic north and vertically downward in the northen hemisphere. The cross-product of these two vectors will be horizontal the magnetic east-west directions and the cross-product between the east-west vector and the gravity vector will be horizontal in the magnetic north-south directions. Formulas exist for converting magnetic to geographical north although I don't know them offhand. An xyz accelerometer by itself produces just the gravity vector, which could allow you to use your cell phone as an electronic bi-directional level (lateral and axial to the orientation of the accelerometer).

Read more: https://www.physicsforums.com