旋转插值 [英] Rotation Interpolation

问题描述

注意:为了简单起见，我将用度数提出这个问题，弧度，度数，不同的零轴承，问题本质上是相同的.

NB: I'll present this question in degrees purely for simplicity, radians, degrees, different zero-bearing, the problem is essentially the same.

有人对旋转插值背后的代码有任何想法吗?给定一个线性插值函数:Lerp(from, to, amount)，其中 amount 为 0...1，它按数量返回 from 和 to 之间的值.如何将相同的函数应用于 0 到 360 度之间的旋转插值?鉴于度数不应返回 0 和 360 度之外.

Does anyone have any ideas on the code behind rotational interpolation? Given a linear interpolation function: Lerp(from, to, amount), where amount is 0...1 which returns a value between from and to, by amount. How could I apply this same function to a rotational interpolation between 0 and 360 degrees? Given that degrees should not be returned outside 0 and 360.

给定度数的单位圆:

从 = 45 到 = 315，算法应该采用最短路径到达角度，即它应该经过零，到 360，然后到 315 - 而不是一直绕到 90、180、270 到 315.

where from = 45 and to = 315, the algorithm should take the shortest path to the angle, i.e. it should go through zero, to 360 and then to 315 - and not all the way round 90, 180, 270 to 315.

有没有很好的方法来实现这一点?还是只是 if() 块的可怕混乱?我错过了一些很好理解的标准方法吗?任何帮助将不胜感激.

Is there a nice way to achieve this? Or is it going to just be a horrid mess of if() blocks? Am I missing some well understood standard way of doing this? Any help would be appreciated.

推荐答案

我知道这已经有 2 年历史了，但我最近一直在寻找同样的问题，如果没有发布 ifs，我看不到一个优雅的解决方案来了，就这样吧:

I know this is 2 years old, but I've recently been looking around for the same problem and I don't see an elegant solution without ifs posted in here, so here it goes:

    shortest_angle=((((end - start) % 360) + 540) % 360) - 180;
    return shortest_angle * amount;

就是这样

ps:当然，% 表示取模，shortest_angle 是保存整个插值角度的变量

ps: of course, % is meaning modulo and shortest_angle is the variable that holds the whole interpolation angle

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