如何创建任意共面3d曲线的2d图 [英] How to create 2d plot of arbitrary, coplanar 3d curve

问题描述

我有一组点，这些点组成(理论上)共面曲线.我的问题是飞机是任意的，并且每次我收集数据时都可以移动(这些点是从相机收集的).我想知道你们是否可以帮助我弄清楚如何:

1. 找到最接近这些点共面的平面
2. 以这样的方式将这些点投影到该平面上，使我获得二维曲线

我相信我知道该怎么做的第二点，实际上我主要是在努力的第一点，但是我也不会介意在第二点上提供帮助.

一吨！

解决方案

1. 在数据中找到3个点A,B,C

   它们不得位于同一行上，并且应尽可能地彼此隔开以提高准确性.

2. 构造U,V基向量

   U = B-A
   V = C-A
   

   规范化

   U /= |U|
   V /= |U|
   

   使U,V垂直

   W = cross(U,V) // this will be near zero if A,B,C are on single line
   U = cross(V,W)
   

3. 将您的数据转换为U,V平面

   仅针对数据计算中的任意点P=(x,y,z):

   x' = dot(U,P)
   y' = dot(V,P)
   

   如果您还需要反向转换:

   P = x'*U + y'*V
   

   如果您要/有一个原点A，则转换为:

   x' = dot(U,P-A)
   y' = dot(V,P-A)
   P = A + x'*U + y'*V
   

   这将在您的2D坐标中将A映射到(0,0).

如果您不知道向量数学，请看这里:

• 了解4x4均匀变换矩阵

在底部，您将找到矢量运算的方程式.希望对您有帮助...

I have a set of points which comprise a (in theory) co-planar curve. My problem is that the plane is arbitrary and can move between each time I collect the data (these points are being collected from a camera). I was wondering if you guys could help me figure out how to:

1. find the plane which is closest to the one which these points are co-planar on
2. project the points on this plane in such a way that gives me a 2-d curve

I believe that I know how to do point 2, it is really mainly point 1 that i'm struggling with, but I wouldn't mind help on the second point as well.

Thanks a ton!

解决方案

1. find 3 points A,B,C in your data

   They must not be on single line and should be as far from each other as possible to improve accuracy.

2. construct U,V basis vectors

   U = B-A
   V = C-A
   

   normalize

   U /= |U|
   V /= |U|
   

   make U,V perpendicular

   W = cross(U,V) // this will be near zero if A,B,C are on single line
   U = cross(V,W)
   

3. convert your data to U,V plane

   simply for any point P=(x,y,z) in your data compute:

   x' = dot(U,P)
   y' = dot(V,P)
   

   in case you need also the reverse conversion:

   P = x'*U + y'*V
   

   In case you want/have an origin point A the conversions would be:

   x' = dot(U,P-A)
   y' = dot(V,P-A)
   P = A + x'*U + y'*V
   

   That will map A to (0,0) in your 2D coordinates.

In case you do not know your vector math look here:

• Understanding 4x4 homogenous transform matrices

at the bottom you will find the equation for vector operations. Hope that helps ...

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