绕任意轴做圆周旋转 [英] Circular rotation around an arbitrary axis

问题描述

我正在编写星际争霸 2 自定义地图，并在 3D 中遇到了一些数学问题.目前，我正在尝试围绕由 x、y 和 z 给定的任意轴创建和旋转一个点(xyz 向量已归一化).

I am programming Starcraft 2 custom maps and got some proglems with math in 3D. Currently I am trying to create and rotate a point around an arbitrary axis, given by x,y and z (the xyz vector is normalized).

我已经尝试了很多，并在互联网上阅读了很多东西，但我就是不知道它是如何正常工作的.我当前的脚本(您可能不了解该语言，但没什么特别的)是将所有内容破坏数小时的结果(无法正常工作):

I've been trying around a lot and read through a lot of stuff on the internet, but I just cant get how it works correctly. My current script (you probably dont know the language, but it's nothing special) is the result of breaking everything for hours (doesn't work correctly):

    point CP;
fixed AXY;
point D;
point DnoZ;
point DXY_Z;
fixed AZ;
fixed LXY;
missile[Missile].Angle = (missile[Missile].Angle + missile[Missile].Acceleration) % 360.0;
missile[Missile].Acceleration += missile[Missile].AirResistance;
if (missile[Missile].Parent > -1) {
    D = missile[missile[Missile].Parent].Direction;
    DnoZ = Point(PointGetX(D),0.0);
    DXY_Z = Normalize(Point(SquareRoot(PointDot(DnoZ,DnoZ)),PointGetHeight(D)));
    AZ = MaxF(ACos(PointGetX(DXY_Z)),ASin(PointGetY(DXY_Z)))+missile[Missile].Angle;
    DnoZ = Normalize(DnoZ);
    AXY = MaxF(ACos(PointGetX(DnoZ)),ASin(PointGetY(DnoZ)));
    CP = Point(Cos(AXY+90),Sin(AXY+90));
    LXY = SquareRoot(PointDot(CP,CP));
    if (LXY > 0) {
        CP = PointMult(CP,Cos(AZ)/LXY);
        PointSetHeight(CP,Sin(AZ));
    } else {
        CP = Point3(0.0,0.0,1.0);
    }
} else {
    CP = Point(Cos(missile[Missile].Angle),Sin(missile[Missile].Angle));
}
missile[Missile].Direction = Normalize(CP);
missile[Missile].Position = PointAdd(missile[Missile].Position,PointMult(missile[Missile].Direction,missile[Missile].Distance));

我就是想不通数学.如果你能用简单的术语来解释它是最好的解决方案，那么剪下的代码也会很好(但没有那么有用，因为我计划在未来做更多的 3D 东西).

I just cant get my mind around the math. If you can explain it in simple terms that would be the best solution, a code snipped would be good as well (but not quite as helpful, because I plan to do more 3D stuff in the future).

推荐答案

http://en.wikipedia.org/wiki/Rotation_matrix.查看来自轴和角度的旋转矩阵部分.为了您的方便，这里是您需要的矩阵.它有点毛.theta 是角度，ux、uy 和 uz 是归一化轴向量

http://en.wikipedia.org/wiki/Rotation_matrix. Look under the section Rotation matrix from axis and angle. For your convenience, here's the matrix you need. It's a bit hairy. theta is the angle, and ux, uy, and uz are the x, y, and z components of the normalized axis vector

如果您不懂矩阵和向量，请回帖，我会帮助您.

If you don't understand matrices and vectors, post back and I'll help you.

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