将法线向量给定的平面坐标映射到 XY 平面 [英] Mapping coordinates from plane given by normal vector to XY plane

问题描述

所以，我有这个算法来计算 3D 形状的横截面，平面由法线向量给出.

So, I have this algorithm to calculate cross-section of 3D shape with plane given with normal vector.

但是，我目前的问题是，横截面是一组 3D 点(都位于给定平面上)，为了显示它，我需要将此坐标映射到 XY 平面.

However, my current problem is, that the cross-section is set of 3D points (all lying on that given plane) and to display it I need to map this coordinates to XY plane.

如果平面法线类似于 (0,0,c)，这将非常有效 - 我只是复制 x 和 y 坐标而丢弃 z.

This works perfect if the plane normal is something like (0,0,c) - I just copy x and y coordinates discarding z.

这是我的问题:由于我不知道如何转换任何其他纯文本，任何人都可以给我任何暗示我现在应该做什么吗?

And here is my question: Since I have no idea how to convert any other plain could anybody give me any hint as to what should I do now?

推荐答案

您的窗格由法线向量定义

Your pane is defined by a normal vector

n=(xn,yn,zn)

对于协调变换，我们需要 2 个基向量和窗格的零点

For coordination transformation we need 2 base vectors and a zero point for the pane

基向量

我们选择了那些自然地"适合 x/y 窗格的(见后面的边缘情况):

We chose those "naturally" fitting to the x/y pane (see later for edge case):

b1=(1,0,zb1)
b2=(0,1,zb2)

我们想要

b1 x b2 = n*c(c 常量标量)

确保这两个是真正的基础

to make sure these two are really bases

现在解决这个问题:

b1 x b2= (0*zb2-zb1*1,zb1*0-1*zb2,1*1-0*0) = (zb1,zb2,1)
zb1*c=xn
zb2*c=yn
1*c=zn

c=zn,
zb2=yn/c=yn/zn
zb1=xn/c=xn/zn

b1=(1,0,yn/zn)
b2=(0,1,xn/zn)

并将其标准化

bv1=(1,0,yn/zn)*sqrt(1+(yn/zn*yn/zn))
bv2=(0,1,yn/zn)*sqrt(1+(xn/zn*xn/zn))

一个边缘情况是，当 zn=0 时:在这种情况下，法线向量平行于 x/y 窗格并且不存在自然基向量，在这种情况下，您必须通过审美 POV 选择基 b1 和 b2 向量并通过相同的解决方案过程或只是选择 bv1 和 bv2.

An edge case is, when zn=0: In this case the normal vector is parallel to the x/y pane and no natural base vectors exist, ind this case you have to chose base b1 and b2 vectors by an esthetic POV and go through the same solution process or just chose bv1 and bv2.

零点

您提到在 OQ 中您的窗格没有锚点，但有必要将您的窗格与无限的平行窗格系列区分开来.

you spoke of no anchor point for your pane in the OQ, but it is necessary to differentiate your pane from the infinite family of parallel panes.

如果您的锚点是 (0,0,0)，这是坐标转换的完美锚点，并且您的窗格具有

If your anchor point is (0,0,0) this is a perfect anchor point for the coordinate transformation and your pane has

x*xn+y*yn+z*zn=0,
(y0,y0,z0)=(0,0,0)

如果没有，我假设你有一个锚点 (xa,ya,za) 并且你的窗格有

If not, I assume you have an anchor point of (xa,ya,za) and your pane has

x*xn+y*yn+z*zn=d

带有 d const 标量.自然拟合将是窗格的点，由原始零点在窗格上的法线投影定义:

with d const scalar. A natural fit would be the point of the pane, that is defined by normal projection of the original zero point onto the pane:

P0=(x0,y0,z0)

与

(x0, y0, z0) = c * (xn,yn,zn)

解决这个问题

x*xn+y*yn+z*zn=d

给予

c*xn*xn+c*yn*yn+c*zn*zn=d

和

c=d/(xn*xn+yn*yn+zn*zn)

因此

P0=(x0,y0,z0)=c*(xn,yn,zn)

找到了.

最终转化

通过将窗格的每个点(即您要显示的点)表示为

is achieved by representing every point of your pane (i.e. those points you want to show) as

P0+x'*bv1+y'*bv2

x' 和 y' 是新坐标.因为我们知道 P0、bv1 和 bv2，所以这很简单.如果我们不在边缘情况下，我们在 bv1.y 和 bv2.x 中有零，进一步减少了问题.

with x' and y' being the new coordinates. Since we know P0, bv1 and bv2 this is quite trivial. If we are not on the edge case, we have zeroes in bv1.y and bv2.x further reducing the problem.

x' 和 y' 是你想要的新坐标.

x' and y' are the new coordinates you want.

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