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

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    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;    }    

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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

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