test.cpp

GPS的数据处理

#include<iostream>
#include<math.h>
using namespace std;
const double c = 298200.0;

void getResult(double x1,double y1, double t1,double x2,double y2,double t2,int flag);

int main()
{
	cout<<"x坐标"<<'	'<<"y坐标"<<endl;
	//计算x和y的坐标值，一共9组数据
	getResult(30.0, 0.0, 51.1, 0.0, -45.0, 96.5, 1);
	getResult(30.0, 0.0, 29.3, 0.0, -45.0, 93.1, 1);
	getResult(30.0, 0.0, 44.4, 0.0, -45.0, 73.2, 1);
	getResult(-30.0, 0.0, 49.1, -30.0, -45.0, 123.4, 2);
	getResult(-30.0, 0.0, 23.0, -30.0, -45.0, 106.4, 2);
	getResult(-30.0, 0.0, 48.7, -30.0, -45.0, 105.2, 2);
	getResult(0.0, 45.0, 101.0, 30.0, 45.0, 129.0, 3);
	getResult(0.0, 45.0, 74.6, 30.0, 45.0, 101.5, 3);
	getResult(0.0, 45.0, 88.3, 30.0, 45.0, 104.9, 3);
	
	return 0;
}
/*
(x1,y1),(x2,y2)已知定点的坐标，flag为显示时间为0的定点的序号，最后，根据flag的不同值，进行坐标交换，换回以s3为坐标原点的坐标值，显示声源的x，y坐标
*/
void getResult(double x1,double y1, double t1,double x2,double y2,double t2,int flag)
{
	double d1 = c*t1 / 1000000.0;
	double d2 = c*t2 / 1000000.0;
	double var1 = x1*x1 + y1*y1 - d1*d1;
	double var2 = x2*x2 + y2*y2 - d2*d2;
	double A = x2*var1 - x1*var2;
	double B = y2*var1 - y1*var2;
	double D = d1*var2 - d2*var1;

	double Fi = atan(B/A);
	double mid_R = acos(D/sqrt(A*A+B*B));
	double r;
	
/*
根据flag的不同值，进行坐标变换，换回以s3为坐标原点的坐标系，显示声源的x，y坐标
*/
	if(flag == 2)
	{
		r = var1 / (2.0 * (x1*cos(Fi-mid_R) + y1*sin(Fi-mid_R) + d1));
		cout<<30.0+r*cos(Fi-mid_R)<<' '<<r*sin(Fi-mid_R)+45.0<<endl;
	}
	else if(flag == 3)
	{
		r = var1 / (2.0 * (x1*cos(Fi+mid_R) + y1*sin(Fi+mid_R) + d1));
		cout<< r*cos(Fi+mid_R)<<' '<<r*sin(Fi+mid_R)<<endl;
	}
	else
	{
		r = var1 / (2.0 * (x1*cos(Fi-mid_R) + y1*sin(Fi-mid_R) + d1));
		cout<< r*cos(Fi-mid_R)<<' '<<r*sin(Fi-mid_R)+45.0 <<endl;
	}
}

//运行结果如下:
/*
x坐标   y坐标
6.14286 37.3404
8.81791 35.6321
7.1493 36.4197
22.9282 39.5571
18.6228 40.0254
24.0142 35.3447
3.26622 7.15617
4.76157 11.0523
8.33742 6.02512

*/

/*
误差分析：
由于是操作板上的刻度有磨损，导致坐标记录时会产生微小误差。又由于机器的电源不太稳定，因而又有系统误差，致使测量结果有微小误差
*/