沿直线段的路径匀速移动答案

【问题标题】：Moving along a path of straight line segments at constant speed沿直线段的路径匀速移动
【发布时间】：2017-11-30 14:40:10
【问题描述】：

我有一个向量点列表，它定义了一个对象要遵循的直线段路径。目前，我使用线性插值来为沿路径的运动设置动画，如下所示：

public class Demo
{
    public float speed = 1;
    private List<Vector3> points;

    private float t; // [0..1]

    private Vector3 Evaluate(float t)
    {
        // Find out in between which points we currently are
        int lastPointIndex = GetLast(t);
        int nextPointIndex = GetNext(t);

        // Obviously, I need to somehow transform parameter t
        // to adjust for the individual length of each segment.

        float segmentLength = GetLength(lastPointIndex, nextPointIndex);

        // But how would I do this?
        return Vector3.Lerp(points[lastPointIndex], points[nextPointIndex], t);
    }

    public void Update()
    {
        // Curve parameter t moves between 0 and 1 at constant speed.
        t = Mathf.PingPong(Time.time * speed, 1);

        // Then just get the evaluated position for the curve time, but
        // this gives variant speed if points are not evenly spaced.
        Vector3 position = Evaluate(t);
        SetObjectPosition(position);
    }
}

我意识到，要实现恒定速度，我需要重新调整参数 t 以考虑每个段的长度，但我似乎无法准确找出方法。

我也知道，我可以通过以我想要的速度移动到下一个点来近似路径，并且只有在我非常接近或跟踪 t 时才改变方向，并在它移过时改变方向下一段，但这似乎很hacky，当我实际上知道每个段的确切长度并且应该能够准确地插值时。

【问题讨论】：

标签： math line interpolation motion

【解决方案1】：

这实际上很容易。首先，为您的对象定义一个您想要的速度。例如，每秒 6 个单位。这意味着如果线段的长度为 6 个单位，那么您的对象从起点到终点将需要 1 秒。这也意味着，如果您的线段是该长度的一半（即 3 个单位），则对象需要 0.5 秒才能穿过它。因此，您需要做的是计算所有线段的长度，然后将其除以您想要行驶的速度(3/6 = 0.5 = scaleFactor)。然后不是在 0 和 1 之间进行插值，而是在 0 和 1*scaleFactor 之间进行插值。然后你的代码变成：

public class Demo
{
public float speed = 1;
private List<Vector3> points;

private float t; // [0..1]

private Vector3 Evaluate(float t)
{
    // Find out in between which points we currently are
    int lastPointIndex = GetLast(t);
    int nextPointIndex = GetNext(t);

    float segmentLength = GetLength(lastPointIndex, nextPointIndex);
    float scaleFactor = segmentLength/speed;

    // note that I divided t by scaleFactor instead of multiplication. 
    // That's because Lerp always takes an interval of [0..1]. So, we
    // adjust the curve parameter instead.
    return Vector3.Lerp(points[lastPointIndex], points[nextPointIndex], t/scaleFactor);
}

public void Update()
{
    // Curve parameter t moves between 0 and 1 at constant speed.
    t = Mathf.PingPong(Time.time * speed, 1);

    // Then just get the evaluated position for the curve time, but
    // this gives variant speed if points are not evenly spaced.
    Vector3 position = Evaluate(t);
    SetObjectPosition(position);
}
}

【讨论】：