计算 3D 平面的斜率 [英] compute slope of a 3D plane

问题描述

我有一组 (X,Y,Z) 点代表不同的平面特征.我需要使用法向量计算每个平面的斜率.我认为斜率由每个平面的法向量 (NV) 与假想水平面的 NV 之间的角度给出.假设，我使用的平面方程是；Ax+By+c=z.然后我猜我的平面的法向量是 (a,b, -1).对于我的平面方程，假想水平面的方程应该是什么?我认为水平面的方程是 z=c.因此，法向量是 (0,0,-1).这样对吗?那么我的平面与水平面的夹角是；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 ))〗

这样对吗?请评论我并给我正确的方程式.

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

推荐答案

是的，大部分是正确的，但是您在用表达式替换角度时犯了一些小错误.角度为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})

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})

这篇关于计算 3D 平面的斜率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！