首页 > 后端开发 > C++ > 正文

如何直接计算两个向量之间的顺时针角度?

Mary-Kate Olsen
发布: 2024-11-14 16:19:02
原创
256 人浏览过

How do you calculate clockwise angles between two vectors directly?

Calculating Clockwise Angles Directly

Calculating the clockwise angle between two vectors is often addressed using the dot product, which determines the inner angle (0-180 degrees). However, if you prefer a direct method, here are the steps to consider:

2D Case

Just like the dot product measures the cosine of the angle, the determinant provides the sine of the angle. The clockwise angle can be computed as:

dot = x1*x2 + y1*y2  # Dot product between [x1, y1] and [x2, y2]
det = x1*y2 - y1*x2  # Determinant
angle = atan2(det, dot)  # atan2(y, x) or atan2(sin, cos)
登录后复制

The angle's orientation aligns with the coordinate system, with positive signs indicating clockwise angles. Swapping the inputs changes the orientation and hence the sign.

3D Case

For 3D vectors, the two vectors define an axis of rotation perpendicular to both. Since this axis has no fixed orientation, the angle of rotation's direction cannot be uniquely determined. A common convention involves orienting the axis to produce positive angles. In this scenario, the dot product of normalized vectors suffices:

dot = x1*x2 + y1*y2 + z1*z2  # Between [x1, y1, z1] and [x2, y2, z2]
lenSq1 = x1*x1 + y1*y1 + z1*z1
lenSq2 = x2*x2 + y2*y2 + z2*z2
angle = acos(dot/sqrt(lenSq1 * lenSq2))
登录后复制

Planes in 3D

If the vectors lie within a plane with a known normal vector n, their rotation axis lies along n. Adapting the 2D computation while incorporating n provides the clockwise angle:

dot = x1*x2 + y1*y2 + z1*z2
det = x1*y2*zn + x2*yn*z1 + xn*y1*z2 - z1*y2*xn - z2*yn*x1 - zn*y1*x2
angle = atan2(det, dot)
登录后复制

Ensure that n is normalized for this computation.

0-360 Degree Range

Many atan2 implementations return angles in the range [-180°, 180°]. To obtain positive angles in the range [0°, 360°], add 2π to any negative result.

以上是如何直接计算两个向量之间的顺时针角度?的详细内容。更多信息请关注PHP中文网其他相关文章!

来源:php.cn
本站声明
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系admin@php.cn
作者最新文章
热门教程
更多>
最新下载
更多>
网站特效
网站源码
网站素材
前端模板