我正在尝试将使用旋转向量传感器类型滚动,方位角和俯仰时返回的5个值。
我用来这样做的代码是以下内容。
@Override
public void onSensorChanged(SensorEvent event) {
double[] g = convertFloatsToDoubles(event.values.clone());
double norm = Math.sqrt(g[0] * g[0] + g[1] * g[1] + g[2] * g[2] + g[3] * g[3]);
g[0] /= norm;
g[1] /= norm;
g[2] /= norm;
g[3] /= norm;
double xAng = (2 * Math.acos(g[0])) * (180 / Math.PI);
double yAng = (2 * Math.acos(g[1])) * (180 / Math.PI);
double zAng = (2 * Math.acos(g[2])) * (180 / Math.PI);
}
private double[] convertFloatsToDoubles(float[] input)
{
if (input == null)
return null;
double[] output = new double[input.length];
for (int i = 0; i < input.length; i++)
output[i] = input[i];
return output;
}
问题是变量xAng
返回的值,yAng
似乎仅限于80-280。
至于zAng
(我认为是方位角(,它像指南针一样工作,但是当它返回0时,它似乎在磁性南方大约12度。
我认为我对所使用的数学做了一些问题,但我不确定到底是什么。
Sensor.TYPE_ROTATION_VECTOR
的值在这里定义为:
值[0]:x*sin(θ/2(
值[1]:y*sin(θ/2(
值[2]:z*sin(θ/2(
值[3]:cos(θ/2(
值[4]:估计的标题准确性(在弧度中((-1,如果不可用(
,以防任何人都希望完成相同的任务。数学的处理完全错误地处理。
onSensorChanged
下面已更新,因此它以度为单位。
@Override
public void onSensorChanged(SensorEvent event) {
//Get Rotation Vector Sensor Values
double[] g = convertFloatsToDoubles(event.values.clone());
//Normalise
double norm = Math.sqrt(g[0] * g[0] + g[1] * g[1] + g[2] * g[2] + g[3] * g[3]);
g[0] /= norm;
g[1] /= norm;
g[2] /= norm;
g[3] /= norm;
//Set values to commonly known quaternion letter representatives
double x = g[0];
double y = g[1];
double z = g[2];
double w = g[3];
//Calculate Pitch in degrees (-180 to 180)
double sinP = 2.0 * (w * x + y * z);
double cosP = 1.0 - 2.0 * (x * x + y * y);
double pitch = Math.atan2(sinP, cosP) * (180 / Math.PI);
//Calculate Tilt in degrees (-90 to 90)
double tilt;
double sinT = 2.0 * (w * y - z * x);
if (Math.abs(sinT) >= 1)
tilt = Math.copySign(Math.PI / 2, sinT) * (180 / Math.PI);
else
tilt = Math.asin(sinT) * (180 / Math.PI);
//Calculate Azimuth in degrees (0 to 360; 0 = North, 90 = East, 180 = South, 270 = West)
double sinA = 2.0 * (w * z + x * y);
double cosA = 1.0 - 2.0 * (y * y + z * z);
double azimuth = Math.atan2(sinA, cosA) * (180 / Math.PI);
}