当我只有很少的变量时如何计算每个传感器的位置
假设我有 3 个传感器：sensor1、sensor2 和 sensor3。
我知道的唯一变量是：
Distance from sensor1 to origin is 36.05
Distance from sensor2 to origin is 62.00
Distance from sensor3 to origin is 63.19
Distance from sensor1 to sensor2 is 61.03
Distance from sensor1 to sensor3 is 90.07
Distance from sensor2 to sensor3 is 59.50
如果你有以下职位，情况会是这样的：
如何仅使用这些变量来计算每个点的位置？
这不是作业，只是好奇心。
Suppose I have 3 sensors: sensor1, sensor2 and sensor3.
The only variables I know are:
Distance from sensor1 to origin is 36.05
Distance from sensor2 to origin is 62.00
Distance from sensor3 to origin is 63.19
Distance from sensor1 to sensor2 is 61.03
Distance from sensor1 to sensor3 is 90.07
Distance from sensor2 to sensor3 is 59.50
This is how it would look like if you had the positions:
How can I calculate the position of every point using only those variables?
This is not homework, just curiosity.
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您无法准确找到点的位置，因为围绕原点的任何旋转以及对称性仍然给出相同的距离。
您想要找到所有可能结果的方法吗？
找到这些点非常简单，但是您是否需要该方法对噪声具有鲁棒性？
You cannot find the position of the points exactly, as any rotation around the origin, as well as symmetry still give the same distances.
Do you want a way to find all the possible results?
Finding the points is pretty straightforward, but do you need the method to be robust on noise?
此过程称为三边测量。正如其他人所指出的，如果没有更多信息，就不可能找到传感器的绝对、明确的位置 - 您将需要 3 个非重合、非共线传感器的 2D 位置、4 个非重合、非共面传感器的 3D 位置，解决所有旋转/反射的歧义。
无线传感器网络定位领域对此问题进行了大量研究——处理不完整、有噪声的距离测量、不可靠的通信和高度受限的资源使其变得有趣。
这可能是一个合适的方法 - 基本思想是建立一个系统分段定位节点 - 从具有明确定义的相对位置的 3 或 4 个节点的种子形成开始，并随着节点的位置相对于已定位的节点变得明确可计算，逐一添加节点。
如果可能的话，具有已知位置的锚节点可以用作系统增长的种子，或者在定位所有节点之后用于计算校正变换。
This process is called trilateration. As others have noted, finding absolute, unambiguous positions for the sensors is not possible without more information - you'll need the positions of three non-coincident, non-colinear sensors in 2D, 4 non-coincident, non-coplanar sensors in 3D, to resolve all rotation/reflection ambiguities.
There's been an enormous amount of research into this problem in the field of wireless sensor network localisation - dealing with incomplete, noisy range measurements, unreliable communication and highly constrained resources make it interesting.
This might be an apt approach - the basic idea is to build up a system of located nodes piecewise - start with a seed formation of 3 or 4 nodes with well-defined relative locations and add nodes one by one as their locations become unambiguously computable relative to already-located nodes.
The anchor nodes with known locations can be used as the seed for system growth if possible, or used to compute a corrective transform after all nodes have been located.
如果没有更多信息，所提出的问题是不可能的。如果您添加更多信息和一些噪音，那么这是可行的。请参阅查找最适合交集的点n spheres 讨论了如何解决此类问题。
The problem as posed is impossible without more information. If you add more information and some noise, then it is doable. See Finding a point that best fits the intersection of n spheres discusses how to solve that type of problem.
看看这些图像。
和
您将看到三角形可以自由旋转（因此不存在“固定”位置），并且一般情况下不需要第三个传感器间距离，因为它由其他两个距离确定。
Look at these images.
And
You will see that the triangle can rotate freely (so no "fixed" position exists), and also the third intersensor distance is not needed in the general case, as it is determined by the other two distances.