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摘要:GPS Term ProjectGPS条款项目全球定位系统(GPS)项项目 GPS Term Project Mathematical Approach- Satellite Satellite will receive standard input from pipe line, then we will make 2 key data structure, whose name are matrix and satellite. Proces......

GPS Term Project  GPS条款项目  全球定位系统(GPS)项项目

GPS Term Project
Mathematical Approach-
Satellite
Satellite will receive standard input from pipe line, then we will make 2 key data structure, whose name are matrix and satellite. Process matrix and satellite by x position with function named by HigherThanHorizon.
Finally, we will send out the result to standard output.
1st, Approach towards our satellite, the programs were written in order to convert angles from degrees, minutes, and seconds to radians, and vice versa. 2nd, we used the formula derived from exercise #4 from equations; (19), (18), (17), (14), and (13) in the project homework assignment. This gives us a formula from converting position and general time t as given in [ tV  Ød  Øm  Øs  NS  ëd  ëm  ës  EW  h (10)] into Cartesian coordinates. Since the main purpose for each of our programs is to take in standard input, and out standard output, we used the information gained in exercise 6 to convert the data from Cartesian coordinates into form (10). In order to complete this task we used and added some additional parameters to (20).

全球定位系统(GPS)项项目
 
数学方法,
 
卫星
 
卫星将从管道接收标准输入,然后我们将制作2个关键数据结构,分别为矩阵和卫星。过程矩阵和卫星由x位置,函数由HigherThanHorizon命名。
 
最后,我们将把结果发送到标准输出。
 
第一,接近我们的卫星时,编写的程序是为了将角度、分、秒转换为弧度,反之亦然。第二,我们使用了从公式中推导出的练习#4的公式;(19)、(18)、(17)、(14)、(13)。这给了我们一个公式从转换位置和时间t在[电视ØdØmØs NS ed em es电子战h(10)]到笛卡尔坐标系。由于我们每个程序的主要目的是接收标准输入,输出标准输出,因此我们使用练习6中获得的信息将数据从笛卡尔坐标转换为形式(10)。为了完成这个任务,我们使用并向(20)添加了一些额外的参数。
h = sqrt(x2 + y2 +z2) – R
 
{arctan (z/sqrt(x2 + y2))
Ø = {π/2
{-π/2
For x2 + y2 ≠ 0
For x = y = 0, z > 0
For x = y = 0, z < 0
{ð/2
{-ð/2
ë =    {arctan (y/x)
{ð + arctan(y/x)
{2ð + arctan(y/x)
For x = 0, y > 0
For x = 0, y < 0
For x > 0, y > 0
For x < 0
For x > 0, y < 0

3rd, We were then compelled to find a condition set to assure us the satellites used for our data were above the horizon. Given in Cartesian coordinates of the following equation (31), we figure x being a point on earth and s, in space, this tells us whether s viewed from x is in fact higher than the horizon.

第三,我们被迫找到一个条件,以确保我们用于数据的卫星在地平线之上。在笛卡尔坐标系下,用下面的方程(31)表示,我们可以看出x是地球上的一点,s是空间中的一点,这告诉我们,从x角度看,s是否高于视界。

t0 = tV - ǁxs(tV)-xVǁ2/c

4th,  is to compute our ts and xs, being time and position of the satellite respectively. We continued following the iteration from exercise 9 and solving t0 in the form (40).

第四,计算我们的ts和xs,分别是卫星的时间和位置。我们继续遵循练习9中的迭代,并以(40)的形式求解t0。


Receiver

Receiver will receive standard input from pipe line. Then it will 2 geography, one is a temperature geography, one is to save the final result.
Loop and get inputs, process and check to find the final result.
As mentioned in the project assignment, we have 4 unknowns with coordinates of location and time, which leaves us a range within c(tS – tV), where we do not know tV. For our mathematical approach in solving our system of 4 nonlinear equations, we chose to follow exercise 14, and solving thru Newton’s method, the nonlinear system obtained by the Least Squares approach. Due to the possibility of failure in solving this non-linear system, we chose our initial points as such: for the receiver, denote geo0 and geo1 be last two geographic coordinates of vehicle's solution. Let d_geo be the difference of geo1 and geo0 and geo = d_geo + geo1.  Then init_t = t_s[1]+.08 and init_x = geography2cartesian(init_t, geo). We used .08s for the addition to t_s[1], because it is experimentally the average time that the signal takes to go to the vehicle. Also, let it be known that the init_x is approximated by assuming the velocity of the vehicle, constant over a short period.

接收机
 
接收器将接收来自管道的标准输入。然后它会有两个地理位置,一个是温度地理位置,一个是保存最终结果。
 
循环和获取输入,处理和检查,以找到最终的结果。
 
正如在项目作业中提到的,我们有4个未知数,位置和时间的坐标,这给我们留下了一个c(tS - tV)的范围,我们不知道电视。对于我们求解4个非线性方程组的数学方法,我们选择了练习14,并通过牛顿法求解,通过最小二乘法得到的非线性方程组。由于求解该非线性系统存在失败的可能性,我们的初始点如下:对于接收器,表示geo0, geo1为车辆解的最后两个地理坐标。设d_geo为geo1和geo0的差值,geo = d_geo + geo1。然后init_t = t_s[1]+。和init_x = geography2cartesian(init_t, geo)我们用0。08来表示t_s[1]的加法,因为从实验上看,它是信号到达飞行器的平均时间。同样,让我们知道,init_x是通过假设飞行器的速度来近似的,在短时间内恒定。

How to use our program?
Our programming language consisted of using C++.
1.we can run the program by command:make run
2.we can compile the program by command: make
3. we can clear all binary files by command: make clean
4.we can run the program annual by command : cat bm.dat | java vehicle | ./satellite | ./receiver

如何使用我们的程序?
 
我们的编程语言由使用c++组成。
 
1.我们可以通过命令来运行程序:make run
 
2.我们可以通过命令:make来编译程序
 
3.我们可以通过命令清除所有二进制文件:make clean
 
4.我们可以通过以下命令每年运行该程序:cat bm.dat | java车辆| ./卫星| ./接收器

Satellite takes geographic data from standard input [ tV  Ød  Øm  Øs  NS  ëd  ëm  ës  EW  h (10)] and generates data from satellites above the horizon in the form [iS tS xS] (26). The receiver takes data in form (26), and prints to standard output in (10). Both the satellite and receiver use data.dat parameters. These programs can be used by giving a command like:
cat bm.dat | java vehicle | ./satellite | ./receiver
What we can learn?
We noticed in derivatives that there could have been problems in diving the Ni = ǁxS - xǁ2, since Ni is large, we avoided dividing by zero. From this, we learned that we should divide by larger numbers, but if in another project Ni was small, we would know to modify the problem or equation to divide by a reasonably large number to avoid having to divide by zero.
 
卫星将地理数据从标准输入(电视ØdØmØs NS ed em es电子战h(10)]和生成数据从卫星在地平线形式是tS xS(26)。接收器以(26)形式接收数据,并以(10)形式打印到标准输出。卫星和接收器都使用数据。dat参数。这些程序可以使用的命令,如:
 
cat bm.dat | java车辆| ./卫星| ./接收机
 
我们能学到什么?
 
我们注意到在衍生品可能是问题在潜水倪=ǁxS - xǁ2,因为倪很大,我们避免了除以零。从这里,我们知道我们应该除以更大的数,但如果在另一个项目中,Ni很小,我们就会知道修改问题或方程,除以一个相当大的数,以避免必须除以零。

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