Java：基于距离绘制点 [英] Java: plot points based on distances

问题描述

我需要根据距离绘制一组点。我有三个未知点X，Y和Z.然后我得到另一个未知点（A）及其与原件（AX，AY，AZ）的距离。我会继续得到点和距离（B，BX，BY，BZ; C，CX，CY，CZ）等。

I need to plot a group of points based on distances. I have three unknown points X, Y, and Z. I then get another unknown point (A) and its distances from the originals (AX, AY, AZ). I will continue getting points and distances (B, BX, BY, BZ; C, CX, CY, CZ) etc.

我的问题是，的点。如果是这样，我需要多少点才能得到精确的绘图图？

My question is whether its possible to plot all of the points. If so, how many points would I need for an exact plot map? What about an approximate map?

这类似于这个问题，但我得到一组不同的距离，并不限于原点数。

This is similar to this question but I get a different set of distances and am not limited to the original number of points.

另外，如果它会帮助我可以添加更多的点到X，Y，Z组，这将给我更多的距离A，B等。我不知道，直到它是以某种方式计算的任何距离XY ，XZ，YZ，AB，AC等。

Also, if it would help I could add more points to the X, Y, Z group which would give me more distances for A, B, etc.What I don't know until it's been somehow calculated are any of the Distances XY, XZ, YZ, AB, AC, etc.

推荐答案

1. 2D空间

1. I assume you use 2D space

• 如果是1D，则2分就足够了（不一样!!!）


• 2D ，然后 3个距离就够了，但所使用的点必须不在同一行！

• if it is 1D then 2 points are enough (not identical !!!)
• if 2D then 3 distances is enough but the points used must not lay on the same line !!!

情节的位置/方向

• 知道1点消除偏移


• li>知道2点消除旋转


• 知道3点消除镜像

• for relative plot are above conditions enough
• if you want also the exact orientation and position
• then you need to know exact position of first 3 points
• otherwise your plot will look the same but can be offseted,rotated and mirrored to original geometry
• knowing 1 point eliminates offset
• knowing 2 point eliminates rotation
• knowing 3 point eliminates mirroring

[notes]

• 您需要 n + 1 nD 坐标系的点

• you need n+1 points for n-D coordinate system

/ strong>

[edit1] equations

• 原始问题文字未包含任何方程式，但需要注释，因此这里有一些：


• 您将需要2D球体，3D球体中的两个超球体之间的交叉点
  
  
  
  
  • ...
  
  
  • look here： circle-circle intersection
  
  
  • 从每个点作为中心投射圆形，其半径等于从该点开始的距离
  
  
  • 找到所有圆形组合之间相同的交叉点（0,1）， ，2），（1,2）
    
  
  
  • 黄色交集在所有3种组合中是相同的，因此是下一个点
  
  
  • original question text did not contain any equations need but comments require it so here are some:
  • you will need intersection point between two hyperspheres
    • in 2D circles, in 3D spheres,...
    • look here: circle-circle intersection
    • cast circle from each point as center with radius equal to the distance from that point
    • find out intersection point that is the same between all combinations of circles (0,1),(0,2),(1,2)
    • Yellow intersection is the same in all 3 combinations so that is the next point
    
    
    • （x-x0）^ 2 +（y-y0）^ 2 = l0 ^ 2
    
    
    • （x-x1）^ 2 +（y-y1）^ 2 = l1 ^ 2
    
    
    • （x-x2）^ 2 +（y-y2）^ 2 = l2 ^ 2
    
    
    
    
    • 
      
    • xi，yi是圆的中心
    
    
    • li是与该点的距离
    / li>
    
    • 第一个选项应该更简单和更准确，如果正确
    
    
    • 但需要一些关于向量和三角学数学的知识
    
    
    • 您需要在向量
    
    
    • 上添加旋转或计算，并在2D V（x，y） - > V0（+ y，-x），V1 -y，+ x）
    
    
    • 其中V0，V1垂直于V
    
    
    • (x-x0)^2+(y-y0)^2=l0^2
    • (x-x1)^2+(y-y1)^2=l1^2
    • (x-x2)^2+(y-y2)^2=l2^2
    • where x,y is the intersection point
    • xi,yi is center of circle
    • li is distance from that point

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