随机GPS数据（地图坐标）

随机GPS数据（地图坐标）

昨天又接了个小任务，还是GPS的，要做一个随机生成GPS坐标数据的工具，而且要求生成的数据不能太"离谱"，也就是说要在一定范围之内，因为是模拟车行，所以可能有个限速的要求吧，不能一辆汽车每小时1w公里的速度行驶。。那就~~！

快下班接的任务~~！然后。。马上关机回家。。吃饭，看了会电视~~！11点了。。才想起来有任务。打开电脑~~！

既然是不能太离谱。。那就要在某个范围内移动。。唉，百度了下经纬度的2点距离公式~~！

double lat1,long1;//经度,纬度1

double lat2,long2;//经度,纬度2

const float pi = float(3.1415926);//pi

const float EarthRadiu = float(6378.245);//地球半径 

float Wei_Km = ( pi * EarthRadiu )/180 ; //每度纬度多少公里

double lat0; //比较纬度，取两点小纬度的那个

if(lat1<lat2) lat0 = lat1;

else lat0 = lat2;

Jing_Km = (pi * (EarthRadiu * float( cos(Lat*pi/180 ))))/180;//在lat0上，每度经度间的距离

double cx, cy, r;//x方向距离，y方向距离，两点距离；

cx = abs(lat2-lat1) * Wei_Km;//根据纬度差计算距离

cy = abs(long2-long1)*Jing_Km;//根据经度差计算距离

r = sqrt(cx*cx + cy *cy);

唉看了一大堆，三角函数。。。弧度。。。。公里。。英尺。。弄的头晕了。

以前算这个都是口算啊，唉，大学几年以后。。什么都忘记了~~！

还好我们有google，baidu，还有"病"。。信息检索这么发达的社会，还愁什么呢。唉~~！不会玩检索的人怎么冲浪呢~~！

检索了半天中文网页~~！没发现一个有用的，唉，直接google英文吧。Random longitude latitude 第一条记录俺就发现了想要的东西。。

http://www.geomidpoint.com/random/calculation.html

Calculation Methods - Random Points

详细内容自己看去吧，呵呵看不懂有google翻译。

意思大概就是2种方法，一个圆形区域一个矩形区域。详细内容只能意会不能言转，因为感觉把英语翻译成汉语是件十分恶心的事情，所以就不多说了，

1. Circular region
2. B. Rectangular region

Circular region calculation detail

1. Convert all latitudes and longitudes to radians.
2. rand1 and rand2 are unique random numbers generated in the range 0 to 1.0.
3. Given the initial values startlat, startlon and maxdist. (maxdist is in miles or kms).
4. For the mean radius of the earth use:
   radiusEarth = 3960.056052 miles or radiusEarth = 6372.796924 km
5. Convert maximum distance to radians.
   maxdist=maxdist/radiusEarth
6. Compute a random distance from 0 to maxdist scaled so that points on larger circles have a greater probability of being chosen than points on smaller circles as described earlier.
   dist = acos(rand1*(cos(maxdist) - 1) + 1)
7. Compute a random bearing from 0 to 2*PI radians (0 to 360 degrees), with all bearings having an equal probability of being chosen.
   brg = 2*PI*rand2
8. Use the starting point, random distance and random bearing to calculate the coordinates of the final random point.
   lat = asin(sin(startlat)*cos(dist) + cos(startlat)*sin(dist)*cos(brg))
   lon = startlon + atan2(sin(brg)*sin(dist)*cos(startlat), cos(dist)-sin(startlat)*sin(lat))
9. If lon is less than -PI then:
   lon = lon + 2*PI
   If lon is greater than PI then:
   lon = lon - 2*PI

Circular region calculation detail

1. Convert all latitudes and longitudes to radians.
2. rand1 and rand2 are unique random numbers generated in the range 0 to 1.0.
3. Given the initial values startlat, startlon and maxdist. (maxdist is in miles or kms).
4. For the mean radius of the earth use:
   radiusEarth = 3960.056052 miles or radiusEarth = 6372.796924 km
5. Convert maximum distance to radians.
   maxdist=maxdist/radiusEarth
6. Compute a random distance from 0 to maxdist scaled so that points on larger circles have a greater probability of being chosen than points on smaller circles as described earlier.
   dist = acos(rand1*(cos(maxdist) - 1) + 1)
7. Compute a random bearing from 0 to 2*PI radians (0 to 360 degrees), with all bearings having an equal probability of being chosen.
   brg = 2*PI*rand2
8. Use the starting point, random distance and random bearing to calculate the coordinates of the final random point.
   lat = asin(sin(startlat)*cos(dist) + cos(startlat)*sin(dist)*cos(brg))
   lon = startlon + atan2(sin(brg)*sin(dist)*cos(startlat), cos(dist)-sin(startlat)*sin(lat))
9. If lon is less than -PI then:
   lon = lon + 2*PI
   If lon is greater than PI then:
   lon = lon - 2*PI

大概讲述了计算过程，正准备还是用c#进行解析上面的内容。。。看了一下首页的图标突然发现了很多功能~~！唉，有的时候还是要多点击一下的~~！

Random point Generator哈哈这不是我正需要的么？

爱因斯坦都不记自己家的电话，咱何必费那劲用c#解释那2段方法呢？

打开http://www.geomidpoint.com/random/

右键查看源码Ctrl+f 输入.js只有一个结果

<script type="text/javascript" src="jFunc.js">

下载吧呵呵，虽然js方面是个小菜鸟，但是学习的精神是挡不住的~~！

点了跟烟，开始研究js~~！…………

主要算法calculate()

其实就是用JavaScript解释了一下上面那2段英文

好，c#转码~~！…………15分钟搞定，测试一下

~~!怎么随机出来的都是一样的。。。好吧线程休息一下

这次就对了，距离在设定的500Km以内，没什么问题。

睡了~~！

话说第二天~~！

来弄数据库的东西，根据数据库的设备生成"合理"数据

直接上图

根据最大速度，在间隔时间内生成一个点，然后继续以最后的点为中心继续移动~~！

主要方法:

 

var cosdif = Math.Cos(maxdist) - 1;

var sinstartlat = Math.Sin(startlat);

var cosstartlat = Math.Cos(startlat);

double dist = 0;

var rad360 = 2 * Math.PI;//圆周率

for (var i = 0; i < p; i++)//点数

{

//Thread.Sleep(100);

dist = Math.Acos((new Random().NextDouble() * cosdif + 1));//随机数

brg[0] = rad360 * new Random().NextDouble();

lat = Math.Asin(sinstartlat * Math.Cos(dist) + cosstartlat * Math.Sin(dist) * Math.Cos(brg[0]));

lon = deg(normalizeLongitude(startlon * 1 + Math.Atan2(Math.Sin(brg[0]) * Math.Sin(dist) * cosstartlat, Math.Cos(dist) - sinstartlat * Math.Sin(lat))));

lat = deg(lat);

dist = Math.Round(dist * radiusEarth * 10000) / 10000;

brg[0] = Math.Round(deg(brg[0]) * 1000) / 1000;//随机距离

 

rtnList.Add(new GPSData() { Latitude = padZeroRight(lat), longtitude = padZeroRight(lon), distance = dist.ToString(), bearing = brg[j].ToString() });

}

 

就写到这里吧，也算个记录了~~！