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

我有一组包含(理论上)共面曲线的点.我的问题是平面是任意的，每次我收集数据时都可以在它们之间移动(这些点是从相机收集的).我想知道你们是否可以帮我弄清楚如何:

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. 找到离这些点共面的平面最近的平面
2. 以这样的方式投影该平面上的点，从而为我提供二维曲线

我相信我知道如何做第 2 点，我真正在努力解决的主要是第 1 点，但我也不介意在第二点上提供帮助.

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.

非常感谢！

推荐答案

1. 在您的数据中找出 3 个点 A,B,C

它们不能在一条线上，并且应该尽可能远离以提高准确性.

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

构造U,V基向量

Construct U,V basis vectors

 U = B-A
 V = C-A

标准化

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

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)

将您的数据转换为U,V平面

Convert your data to U,V plane

简单地用于数据计算中的任何点 P=(x,y,z):

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

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

如果您还需要反向转换:

in case you need also the reverse conversion:

 P = x'*U + y'*V

如果你想要/有一个原点A，转换将是:

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

这会将 A 映射到您的 2D 坐标中的 (0,0).

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

如果您不知道矢量数学，请看这里:

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

• 了解 4x4 同构变换矩阵

在底部您会找到向量运算的方程式.希望有帮助...

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

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