Principle
Three-point positioning, as the name implies, requires three points. These three points are generally base stations. Whose position is being determined? Generally, the position of the client or terminal. By measuring the distance between the terminal and the base station (there may be errors in this measurement process, so let’s assume there are no errors), use it as the radius of the three circles, draw a picture, and finally find the intersection of the three circles. The intersection is the terminal position, achieving the positioning effect. As shown in the figure. Point O in the figure is what we are looking for.
Next, consider the case of measurement error. The following situation is easy to occur in the actual measurement process. In this case, the intersection is an area. As shown in the figure, the intersection of the three circles is the BCE area.
How to solve this problem? Let’s first consider the case of two circles. Here we first find the coordinates of point C.
Two circles (center is the base station location, radius is our measurement data, i.e. the distance from the terminal to the base station) intersect at points A and B, connect AB, PQ, AQ, AP, AB and PQ intersect at point C, the distance PQ of the center of the circle is known after the base station is arranged, the center of the circle and the radius are known, the coordinates of points A and B can be obtained by simultaneous equations, and then PA and AQ can be obtained according to the Euclidean distance. According to the Pythagorean theorem:
Obtaining coordinates based on proportional relationships
Calculating the three circles in pairs can get three points. Then we can take the average of the coordinates of these three points as the coordinates of the terminal.
In addition, there may be a situation where the following circles do not intersect.
How to solve this problem? Let’s first consider the case of two circles. As shown in the figure. Now we find the point O.
The simple method is to calculate directly based on the proportional radius.
Similarly, we will eventually get three points, and take the mean of the coordinates of these three points as the coordinates of the terminal.
Code
struct Point
{
int x; //x coordinate
int y; //y coordinate
Point() :x(0), y(0) {};
};
//Three-point positioning method
//dis: radius
//points: center of circle
Point threePoints(float *dis, Point *ps)
{
Point p;
if (dis == NULL || ps== NULL)
return p;
for (int i = 0; i < 3; ++i)
{
//Check if there is a problem with the distance
if (dis[i] < 0)
return Point();
for (int j = i + 1; j < 3; ++j)
{
//Distance of center of circle PQ
float p2p = (float)sqrt((ps[i].x – ps[j].x)*(ps[i].x – ps[j].x) +
(ps[i].y – ps[j].y)*(ps[i].y – ps[j].y));
//Judge whether the two circles intersect
if (dis[i] + dis[j] <= p2p)
{
//If they do not intersect, calculate according to the proportion
p.x += ps[i].x + (ps[j].x – ps[i].x)*dis[i] / (dis[i] + dis[j]);
p.y += ps[i].y + (ps[j].y – ps[i].y)*dis[i] / (dis[i] + dis[j]);
}
else
{
//If they intersect, apply the formula
//PC
float dr = p2p / 2 + (dis[i] * dis[i] – dis[j] * dis[j]) / (2 * p2p);
//x = xp + (xq-xp) * PC / PQ
p.x += ps[i].x + (ps[j].x – ps[i].x)*dr / p2p;
//y = yp + (yq-yp) * PC / PQ
p.y += ps[i].y + (ps[j].y – ps[i].y)*dr / p2p;
}
}
}
//Three circles are divided into two groups, and finally three points are obtained. Find their average
p.x /= 3;
p.y /= 3;
return p;
}
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