所以我推出了一个方法,它返回给定距离的一组点。
/// <summary>
/// Gets all the points between a two vectors at given distance, including the starting and the ending point.
/// </summary>
public static List<Vector2> GetPointsAlongTwoVectors(Vector2 startingPoint, Vector2 endingPoint, float distance)
{
Vector2 direction = (endingPoint - startingPoint).normalized;
float totalDistance = (endingPoint - startingPoint).magnitude;
float increasingDistance = 0.0f;
List<Vector2> points = new List<Vector2>();
points.Add(startingPoint);
if (totalDistance > distance)
{
do
{
increasingDistance += distance;
points.Add(startingPoint + increasingDistance * direction);
} while (increasingDistance + distance < totalDistance);
}
points.Add(endingPoint);
return points;
}
该方法有效,但我最终想做的是将这些点均匀地分布在给定的向量上。这让我认为距离最终将变成近似距离,因为可能不可能均匀分布具有完全精确距离的点,但是只要该方法返回起点、终点以及它们之间均匀分布的点就可以了。有人可以帮助我吗?
最佳答案
也许添加此代码:
...
float totalDistance = (endingPoint - startingPoint).magnitude;
float sectionsCount = (float)Math.Round(totalDistance / distance, MidpointRounding.AwayFromZero);
distance = totalDistance / sectionsCount;
...
请务必检查 sectionsCount
为 0 的情况。
关于c# - 沿着矢量线以大约定义的距离获得均匀分布的点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26127763/