BLE(iBeacons)三边测量 [英] BLE (iBeacons) Trilateration

问题描述

我是德国富特旺根大学的学生.

I'm a student at the University of Furtwangen in Germany.

我即将结束学期，现在正在写论文.我对iBeacon及其背后的技术非常感兴趣.我当前的项目是将信标技术与其他技术(例如GPS，无线定位，GSM和NFC)进行比较.对于我的论文，我将创建不同的用例并比较结果.

I am in my final term and I am writing my thesis now. I'm very interested in iBeacons and the technology behind them. My current project is to compare the beacon technology with other technologies like GPS, Wireless-location, GSM, and NFC. For my thesis, I will create different use-cases and compare the results.

在过去的几天里，我试图确定我在房间中的位置.我使用距三个信标的相对距离(准确性)，并在我的房间中给每个信标一个固定的位置. 我得到三个圆并计算6个交点. 当弧度(精度)太低时，我会人为地增加此值.然后，我查看6个点(交叉点)中哪一个最近. (三个最近的点) 有了这些点，我得到了一个三角形，并且有了这个三角形，我计算出了中间点.

Over the last few days I've tried to determine my position in a room. I use the relative distance (accuracy) from three beacons and gave every beacon a fixed position in my room. I get three circles and calculate 6 intersections. When a radian (accuracy) is too low I increase this value artificially. Then I look which of the 6 points (intersections) are the nearest. (The three nearest points) With those points I get an triangle, and with this I calculate the middle point.

我的问题是结果并不是最好的.

My problem is that the result is not really the best.

我在这里找到了更好的解决方案:

I found a better solution here:

> https://gis.stackexchange.com/questions/40660 /trilateration-algorithm-for-n-points-points

但是我在目标C中实现它遇到了麻烦. 但我知道解决方案.如何在Objective C中导入或获取它. 我找到了一些库(C，C ++)，但我不确定这些库中哪一个最好.

but I have am having trouble implementing this in Objective C. But I understand the solution. How can I import or get this in Objective C. I found some libs (C, C++) but I'm not really sure which of these libs is the best.

对我来说最好的解决方案是一个Objectice C数学库，该库可以使用这些点(x1，x2，x3，-，y1，y2，y3，-，r1，r2，r3)进行计算.

The best solution for me will be a Objectice C math library which can calculate with these points (x1,x2,x3, -- ,y1,y2,y3, --- ,r1,r2,r3).

我现在的计算图

推荐答案

我在同样的问题上苦苦挣扎，然后我发现了这个

I was struggling with the same problem, then I found this solution, written in python. I tried porting the code into objective-c and using the same case for testing and the result is accurate. I modified the code so it can accept 2-dimension vector as well.

测试用例是:

P1 = (3,0) r1 = 6.4031
P2 = (9,0) r2 = 4.1231
P3 = (4,8) r3 = 5.6568

我通过代码运行了这些数据:

I ran this data through the code :

//P1,P2,P3 is the point and 2-dimension vector
NSMutableArray *P1 = [[NSMutableArray alloc] initWithCapacity:0];
[P1 addObject:[NSNumber numberWithDouble:3]];
[P1 addObject:[NSNumber numberWithDouble:0]];


NSMutableArray *P2 = [[NSMutableArray alloc] initWithCapacity:0];
[P2 addObject:[NSNumber numberWithDouble:9]];
[P2 addObject:[NSNumber numberWithDouble:0]];

NSMutableArray *P3 = [[NSMutableArray alloc] initWithCapacity:0];
[P3 addObject:[NSNumber numberWithDouble:4]];
[P3 addObject:[NSNumber numberWithDouble:8]];

//this is the distance between all the points and the unknown point
double DistA = 6.4031;
double DistB = 4.1231;
double DistC = 5.6568;

// ex = (P2 - P1)/(numpy.linalg.norm(P2 - P1))
NSMutableArray *ex = [[NSMutableArray alloc] initWithCapacity:0];
double temp = 0;
for (int i = 0; i < [P1 count]; i++) {
    double t1 = [[P2 objectAtIndex:i] doubleValue];
    double t2 = [[P1 objectAtIndex:i] doubleValue];
    double t = t1 - t2;
    temp += (t*t);
}
for (int i = 0; i < [P1 count]; i++) {
    double t1 = [[P2 objectAtIndex:i] doubleValue];
    double t2 = [[P1 objectAtIndex:i] doubleValue];
    double exx = (t1 - t2)/sqrt(temp);
    [ex addObject:[NSNumber numberWithDouble:exx]];
}

// i = dot(ex, P3 - P1)
NSMutableArray *p3p1 = [[NSMutableArray alloc] initWithCapacity:0];
for (int i = 0; i < [P3 count]; i++) {
    double t1 = [[P3 objectAtIndex:i] doubleValue];
    double t2 = [[P1 objectAtIndex:i] doubleValue];
    double t3 = t1 - t2;
    [p3p1 addObject:[NSNumber numberWithDouble:t3]];
}

double ival = 0;
for (int i = 0; i < [ex count]; i++) {
    double t1 = [[ex objectAtIndex:i] doubleValue];
    double t2 = [[p3p1 objectAtIndex:i] doubleValue];
    ival += (t1*t2);
}

// ey = (P3 - P1 - i*ex)/(numpy.linalg.norm(P3 - P1 - i*ex))
NSMutableArray *ey = [[NSMutableArray alloc] initWithCapacity:0];
double p3p1i = 0;
for (int  i = 0; i < [P3 count]; i++) {
    double t1 = [[P3 objectAtIndex:i] doubleValue];
    double t2 = [[P1 objectAtIndex:i] doubleValue];
    double t3 = [[ex objectAtIndex:i] doubleValue] * ival;
    double t = t1 - t2 -t3;
    p3p1i += (t*t);
}
for (int i = 0; i < [P3 count]; i++) {
    double t1 = [[P3 objectAtIndex:i] doubleValue];
    double t2 = [[P1 objectAtIndex:i] doubleValue];
    double t3 = [[ex objectAtIndex:i] doubleValue] * ival;
    double eyy = (t1 - t2 - t3)/sqrt(p3p1i);
    [ey addObject:[NSNumber numberWithDouble:eyy]];
}


// ez = numpy.cross(ex,ey)
// if 2-dimensional vector then ez = 0
NSMutableArray *ez = [[NSMutableArray alloc] initWithCapacity:0];
double ezx;
double ezy;
double ezz;
if ([P1 count] !=3){
    ezx = 0;
    ezy = 0;
    ezz = 0;

}else{
    ezx = ([[ex objectAtIndex:1] doubleValue]*[[ey objectAtIndex:2]doubleValue]) - ([[ex objectAtIndex:2]doubleValue]*[[ey objectAtIndex:1]doubleValue]);
    ezy = ([[ex objectAtIndex:2] doubleValue]*[[ey objectAtIndex:0]doubleValue]) - ([[ex objectAtIndex:0]doubleValue]*[[ey objectAtIndex:2]doubleValue]);
    ezz = ([[ex objectAtIndex:0] doubleValue]*[[ey objectAtIndex:1]doubleValue]) - ([[ex objectAtIndex:1]doubleValue]*[[ey objectAtIndex:0]doubleValue]);

}

[ez addObject:[NSNumber numberWithDouble:ezx]];
[ez addObject:[NSNumber numberWithDouble:ezy]];
[ez addObject:[NSNumber numberWithDouble:ezz]];


// d = numpy.linalg.norm(P2 - P1)
double d = sqrt(temp);

// j = dot(ey, P3 - P1)
double jval = 0;
for (int i = 0; i < [ey count]; i++) {
    double t1 = [[ey objectAtIndex:i] doubleValue];
    double t2 = [[p3p1 objectAtIndex:i] doubleValue];
    jval += (t1*t2);
}

// x = (pow(DistA,2) - pow(DistB,2) + pow(d,2))/(2*d)
double xval = (pow(DistA,2) - pow(DistB,2) + pow(d,2))/(2*d);

// y = ((pow(DistA,2) - pow(DistC,2) + pow(i,2) + pow(j,2))/(2*j)) - ((i/j)*x)
double yval = ((pow(DistA,2) - pow(DistC,2) + pow(ival,2) + pow(jval,2))/(2*jval)) - ((ival/jval)*xval);

// z = sqrt(pow(DistA,2) - pow(x,2) - pow(y,2))
// if 2-dimensional vector then z = 0
double zval;
if ([P1 count] !=3){
    zval = 0;
}else{
    zval = sqrt(pow(DistA,2) - pow(xval,2) - pow(yval,2));
}

// triPt = P1 + x*ex + y*ey + z*ez
NSMutableArray *triPt = [[NSMutableArray alloc] initWithCapacity:0];
for (int i = 0; i < [P1 count]; i++) {
    double t1 = [[P1 objectAtIndex:i] doubleValue];
    double t2 = [[ex objectAtIndex:i] doubleValue] * xval;
    double t3 = [[ey objectAtIndex:i] doubleValue] * yval;
    double t4 = [[ez objectAtIndex:i] doubleValue] * zval;
    double triptx = t1+t2+t3+t4;
    [triPt addObject:[NSNumber numberWithDouble:triptx]];
}

NSLog(@"ex %@",ex);
NSLog(@"i %f",ival);
NSLog(@"ey %@",ey);
NSLog(@"d %f",d);
NSLog(@"j %f",jval);
NSLog(@"x %f",xval);
NSLog(@"y %f",yval);
NSLog(@"y %f",yval);
NSLog(@"final result %@",triPt);

我已经使用上面的测试用例数据在笛卡尔图上进行了测试，得出的结果是未知点位于(8,4)，然后使用上面的代码进行测试并得到了结果(7.999978,4.000021710625001 ).

I've tested by drawing on cartesian diagram using the test case data above, and got the result that the unknown point is located at (8,4), then testing using the code above and got the result (7.999978,4.000021710625001).

然后我使用数据进行了第二次测试:

Then I did a second test using data :

P1 = (2,0) r1 = 5.831
P2 = (8,0) r2 = 5.831
P3 = (8,10) r3 = 5.831

手动结果为(5,5)，使用代码的结果为(5,5). 因此，我认为代码是正确的.

The manual result is (5,5), and the result using the code is (5,5). So, I believe the code is correct.

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