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trilateration

Trilateration with Beacons for indoor positioning

I have this project where I am trying to expand our app, that can transmit realtime location of the appuser by locating the position of the smartphone. I've spent several hours looking for the right kind of beacons, but I can't seem to understand which ones suits the best for my project. What I am trying to achieve is to put beacons all over my workplace to be able to see the realtime location of everyone who uses the app within that area by using trilateration. I need to cover a large area of around 4.000m². I am not looking for a pre-built software solution, but just the hardware to be able

2022-05-01 09:58:31    分类:问答    android   ios   ibeacon   beacon   trilateration

Trilateration and locating the point (x,y,z)

I want to find the coordinate of an unknown node which lie somewhere in the space which has its reference distance away from 3 or more nodes which all of them have known coordinate. This problem is exactly like Trilateration as described here Trilateration. However, I don't understand the part about "Preliminary and final computations" (refer to the wikipedia site). I don't get where I could find P1, P2 and P3 just so I can put to those equation? Thanks

2022-02-19 22:52:14    分类:问答    algorithm   localization   geometry   triangulation   trilateration

Localizing a point using distances to three other points in 3-D

Assume that we have 4 points in 3-D (P1, P2, P3, P4). If the coordinates of those points are given with their euclidian distances to a fifth point P5 (r1, r2, r3, r4), how to calculate the coordinates of P5? In this post, answer of Don Reba is perfect for 2-D. But how do I extend it to 3-D? Here is my code for 2D: static void localize(double[] P1, double[] P2, double[] P3, double r1, double r2, double r3) { double[] ex = normalize(difference(P2, P1)); double i = dotProduct(ex, difference(P3, P1)); double[] ey = normalize(difference(difference(P3, P1), scalarProduct(i, ex))); double d =

2022-02-08 19:01:28    分类:问答    math   geometry   trilateration

Java中的多点三边测量算法(Multi-point trilateration algorithm in Java)

问题 我正在尝试在我的 Android 应用程序中实施三边测量算法来确定用户的室内位置。 我正在使用超宽带信标来获取到固定点的距离。 我能够采用 Trilateration Method Android Java 中建议的方法如下: public LatLng getLocationByTrilateration( LatLng location1, double distance1, LatLng location2, double distance2, LatLng location3, double distance3){ //DECLARE VARIABLES double[] P1 = new double[2]; double[] P2 = new double[2]; double[] P3 = new double[2]; double[] ex = new double[2]; double[] ey = new double[2]; double[] p3p1 = new double[2]; double jval = 0; double temp = 0; double ival = 0; double p3p1i = 0; double triptx; double tripty; double xval; double yval; double t1

2021-12-21 09:14:49    分类:技术分享    java   android   gps   indoor-positioning-system   trilateration

Multi-point trilateration algorithm in Java

I'm trying to implement a trilateration algorithm into my Android app to determine a user's indoor location. I'm using ultra-wideband beacons to get the distances to fixed points. I was able to adapt the method suggested in Trilateration Method Android Java as follows: public LatLng getLocationByTrilateration( LatLng location1, double distance1, LatLng location2, double distance2, LatLng location3, double distance3){ //DECLARE VARIABLES double[] P1 = new double[2]; double[] P2 = new double[2]; double[] P3 = new double[2]; double[] ex = new double[2]; double[] ey = new double[2]; double[] p3p1

2021-12-12 08:16:12    分类:问答    java   android   gps   indoor-positioning-system   trilateration

为 TDOA 三边测量实施方算法的问题(Issues implementing the Fang Algorithm for TDOA Trilateration)

问题 我一直在关注这篇论文(特别是关于方的方法的部分),试图使用 TDOA 技术解决三边测量问题。 我希望在Fang/TDOA 方面有经验的人可以向我伸出援助之手。 出于某种原因,我的实现将错误的根返回到最终的二次方。 这是我到目前为止编写的代码: #include <stdio.h> #include <math.h> struct Point { double x; double y; }; inline double sqr(double n) { return n * n; } // r1 and r2 are the TDOA of the sound impulse to p1 and p2, respectively void fang(double r1, double r2) { // transmitter coords Point tx = {0.7, -0.1}; // receiver coordinates Point p0 = {0, 0}; Point p1 = {1.7320508075688772, 0}; Point p2 = {0.8660254037844388, 1.5}; // linear coefficients double g = ((r2 * (p1.x/r1)) - p2.x) / p2.y; double h = (sqr

2021-10-25 11:29:33    分类:技术分享    c   algebra   trilateration

三边测量公式(编程)(Trilateration Formula (Programming))

问题 我目前正在尝试开发一个三边测量应用程序来使用 3 部手机跟踪信标。 我将在 python 中找到的代码转换为 c#,但我无法让它工作。 这是我的代码: public double[] getPosition(double phoneADistance, double phoneBDistance, double phoneCDistance) { //meterToFeet is just a conversion method which takes the distance parameter and multiplies it by 3.28. double PhoneADist = meterToFeet(phoneADistance); double PhoneBDist = meterToFeet(phoneBDistance); double PhoneCDist = meterToFeet(phoneCDistance); //The phone's x and y coordinates are pre-set Vector<double> P1 = new DenseVector(new[] { PhoneA_x, PhoneA_y }); Vector<double> P2 = new DenseVector(new[] { PhoneB_x

2021-10-20 17:21:16    分类:技术分享    c#   algorithm   math   trilateration

Issues implementing the Fang Algorithm for TDOA Trilateration

I've been following this paper (notably the section on Fang's Method) in an attempt to achieve a solution to the problem of trilateration using the TDOA technique. I'm hoping that someone experienced in Fang / TDOA can lend me a helping hand. For some reason, my implementation is returning incorrect roots to the final quadratic. Here's the code I've written so far: #include <stdio.h> #include <math.h> struct Point { double x; double y; }; inline double sqr(double n) { return n * n; } // r1 and r2 are the TDOA of the sound impulse to p1 and p2, respectively void fang(double r1, double r2) { //

2021-10-20 12:28:35    分类:问答    c   algebra   trilateration