C＃中 - 如何移动点给定的距离d（并得到一个新的坐标） [英] c# - how to move point a given distance d (and get a new coordinates)
问题描述
喜 我想知道是否有任何efficent的方式来计算点的坐标（这是移动距离d从它的原始位置）。
Hi I was wondering if there is any efficent way to calculating coordinates of point (which was moved distance d from it's original location).
让我们说我有一个点P（0.3,0.5），我需要移动这一点上任意方向与距离d。
Let's say I have a point P(0.3,0.5) and I need to move that point random direction with distance d.
到目前为止，我做到了随机挑选新的X和Y坐标，我被检查，如果旧的和新的点之间的距离等于Ð。我当然知道，岂不等于太eficient办法做到这一点。 你会怎么做呢？？
So far I did it by random picking new x and y coordinates and I was checking if distance between old and new point equals d. I do realize that is't too eficient way to do that. How would You do it ??
推荐答案
给定一个点（X1，Y1），我们希望找到一个随机点（x2，y2）在离它的距离 D 。
Given a point (x1, y1), we want to find a "random" point (x2, y2) at a distance d from it.
选择一个随机的角度 THETA 。然后：
Pick a random angle theta. Then:
x2 = x1 + d * cos(theta)
y2 = y1 + d * sin(theta)
这将是对半径 D 集中于（X1，Y1）的圆一个随机点
证明
Distance between (x1, y1) and (x2, y2)
= sqrt ( (x2 - x1) ^ 2 + (y2 - y1) ^ 2)
= sqrt ( d^2 * (sin^2 (theta) + cos^2 (theta) ) )
= d
您可能想看看：
- Polar coordinate system
- Distance Formula
- Pythagorean trigonometric identity
这篇关于C＃中 - 如何移动点给定的距离d（并得到一个新的坐标）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IT屋！
