从球坐标旋转体 [英] Rotating body from spherical coordinates

问题描述

是否可以旋转其顶点定义为球坐标的物体.目前，我正在VHDL中进行拼贴项目，并且涉及旋转十二面体并通过VGA进行呈现.

Is it possible to rotate body which has its vertices defined in spherical coordinates. Currently I am doing collage project in VHDL and is about rotating dodecahedron and presenting over VGA.

我应用了针孔相机模型方程，只需要两个sin/cos计算和两个乘法每个顶点.我只是在考虑绕3轴在两个角度上使用3步旋转，但是即使有可能，我也无法找出合适的方程式.

I applied pinhole camera model equations and that required just two sin/cos calculation and two multiplication per vertice. I was just thinking about rotating around 3rd axis using 3 steps over two angles but i am unable to figure out proper equations and even if this is possible.

修改

我想我明白了.

在计算与相机坐标相同的方向的第3轴上旋转仅是相机坐标的2D变换.这意味着，与在3个轴上旋转(2个轴和1个倾斜度)相比，您总共需要进行4次sin/cos计算和4次乘法.如果有人提出更好的建议，请随意发表答案.

Rotating over 3rd axis which is in same direction as camera is just 2D transform of camera coordinates once you you compute them. That mean than for rotating in 3 axes (ok two axis and one inclination) you need to apply total of 4 sin/cos computations and 4 multiplications. If somebody came up whit something better, fell free to post answer.

推荐答案

可以通过更改θ绕y轴旋转，通过更改φ绕z轴旋转.但是，绕x轴旋转会更困难.

You can rotate around the y-axis by changing θ, and rotate around the z-axis by changing φ. Rotating around the x-axis, though, is a bit tougher.

一种简单的方法是将所有内容都转换为直角坐标，执行旋转，然后再转换回去.

One simple way would be to convert everything to catesian coordinates, perform the rotation, and convert back.

(x，y，z)(球面到笛卡尔坐标系)的方程是

The equations for (x,y,z) (spherical-to-cartesian) are

x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ

绕x轴以角度α旋转(x，y，z)到新点(x'，y'，z')的方程为

The equations for rotating (x,y,z) to new points (x', y', z') around the x-axis by an angle α are

x' = x
   = r sin θ cos φ
y' = y cos α - z sin α
   = (r sin θ sin φ) cos α - (r cos θ) sin α
z' = y sin α + z cos α
   = (r sin θ sin φ) sin α + (r cos θ) cos α

(r，θ，φ)(笛卡尔对球面)的方程为

The equations for (r, θ, φ) (cartesian-to-spherical) are

r' = sqrt(x'2 + y'2 + z'2)
   = r
θ' = cos-1(z'/r')
   = cos-1(sin θ sin φ sin α + cos θ cos α)
φ' = tan-1(y'/x')
   = tan-1(tan φ cos α - cotan θ sin α sec φ)

我不知道是否有办法进一步降低它，但是应该可以.

I don't know if there is a way to reduce that any further, but it should work.

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