本文介绍了围绕另一个任意点旋转任意点的问题的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我编写了以下方法,可以在持续时间内围绕任意点旋转任意点一个角度.现在,所讨论的问题不固定,但最终绕到了我认为是理想目的地的地方.我需要使它平稳地移动到一定角度.

I have written the following methods to rotate an arbitrary point around another arbitrary point by an angle over a duration. The point in question now moves erratically, but ends up around where I believe is the desired destination. I need to get it to move smoothly to the angle.

请注意,这些点与游戏对象无关.

从下图开始,我试图在给定的时间段内将贝塞尔曲线的一个点(用LineRenderer绘制)绕另一点移动一个角度.这些点都不与包含贝塞尔曲线的游戏对象的位置重合.

From the diagram below, I am trying to move one point of a bezier curve (drawn with a LineRenderer) around another point by an angle over a given period of time. None of these points coincide with the position of the game object which contains the bezier curve. 


IEnumerator runMovement()  {

    yield return new WaitForSeconds(2.0f);
    Vector2 pivot = points[3];

    StartCoroutine(RotateControlPointWithDuration(pivot, 2.0f, 90.0f));

}

   IEnumerator RotateControlPointWithDuration(Vector2 pivot, float duration, float angle)
{
    float currentTime = 0.0f;
    float ourTimeDelta = 0;
    Vector2 startPos = points[0];

    ourTimeDelta = Time.deltaTime;
    float angleDelta = angle / duration; //how many degress to rotate in one second

    while (currentTime < duration)
    {
        currentTime += Time.deltaTime;
        ourTimeDelta = Time.deltaTime;

        points[0] = new Vector2(Mathf.Cos(angleDelta * ourTimeDelta) * (startPos.x - pivot.x) - Mathf.Sin(angleDelta * ourTimeDelta) * (startPos.y - pivot.y) + pivot.x,
                                                        Mathf.Sin(angleDelta * ourTimeDelta) * (startPos.x - pivot.x) + Mathf.Cos(angleDelta * ourTimeDelta) * (startPos.y - pivot.y) + pivot.y);

        yield return null;
    }
}

推荐答案

您想要的模式只是

public IEnumerator HowToSmoothly()
    {
    // move "X" from value "A" to value "B"

    float duration = 2.5f;
    float delta = B - A;

    float startTime = Time.time;
    float finishTime = Time.time+duration;

    while(Time.time<finishTime)
        {

        float soFarTime = Time.time-startTime;
        float fractionThisFrame = soFarTime / duration;
        float valueThisFrame = A + delta * fractionThisFrame;

        X = valueThisFrame
        if (X > B) X = B;

        yield return 0;
        }

    X = B;
    yield break;
    }

这篇关于围绕另一个任意点旋转任意点的问题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!

06-18 22:54