计算3D平面坡度 [英] compute slope of a 3D plane

问题描述

我有一组重presenting不同的平面功能（X，Y，Z）点。我需要计算使用法线向量每个平面的斜率。 我认为斜率由每个平面的法向量（NV）和虚水平面的NV之间的角度给出。假设，我用的是平面方程; AX +到+ C = Z。那么我想我的飞机的法向量（A，B，-1）。对于我的平面方程，应该是什么假想水平面的方程？我认为水平面的方程为z = C。因此，正常的矢量为（0,0，-1）。它是否正确？ 然后我的面与水平面之间的夹角; COS ^（ - 1）⁡〖（A.0 + B.0 +（ - 1）.1）/（√（〖〗A1 ^ 2 +〖B1〗^ 2 +〖〗C1 ^ 2）.√（0 ^ 2 + 0 ^ 2 + 1 ^ 2））〗

I have a set of (X,Y,Z) points representing different planar features. I need to calculate the slope of each plane using normal vectors. i think slope is given by the angle between normal vector (NV) of each plane and NV of imaginary horizontal plane. Assume, the plane equation that I use is; Ax+By+c=z. Then i guess the normal vector of my plane is (a,b, -1). For my plane equation, what should be the equation of imaginary horizontal plane? I think equation of horizontal plane is z=c. Hence, the normal vector is (0,0,-1). Is this correct? Then the angle between my plane and the horizontal plane is; cos^(-1)⁡〖(a.0+b.0+(-1).1)/(√(〖a1〗^2+〖b1〗^2+〖c1〗^2 ).√(0^2+0^2+1^2 ))〗

这是否正确？请评论我，给我正确的公式。

Is that correct? please comment me and give me the correct equation.

推荐答案

是的，这主要是正确的，但你已经做了一些小的失误代入EX pression的角度。的角度为余弦^ { - 1} [（一个* 0 + B * 0 +（ - 1）*（ - 1）/（√{一个^ 2 + B ^ 2 +（ - 1）^ 2} *√{ 0 ^ 2 + 0 ^ 2 +（ - 1）^ 2]} = COS ^ { - 1}（1 /√{A ^ 2 + B ^ 2 + 1}）

Yes, that's mostly correct, but you've made some small mistakes substituting into the expression for the angle. The angle is cos^{-1} [(a * 0 + b * 0 + (-1) * (-1) / (√{a^2 + b^2 + (-1)^2} * √{0^2+0^2+(-1)^2}] = cos^{-1}(1/√{a^2 + b^2 + 1})

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