cocos2d 计算给定起点、角度和距离的目的地点 [英] cocos2d calculate destination point given start point, angle and distance

问题描述

我想快点.Cocos2d 和 xcode 中的 2d 问题.

Quick one I think. 2d problem in Cocos2d and xcode.

我有

CGPoint currPoint;
float lineLength;
float angle;

现在，我需要找到 lineLength 与 currPoint 相距角度 Degrees 的点.

Now, I need to find the point that is lineLength away from currPoint at angle Degrees.

试图搜索，但我找到的答案并不完全符合我的要求.希望有人指出我忽略的(我认为)非常简单的数学.

Tried to search, but the answers i have found aren't quite what I was looking for. Would appreciate anyone pointing out the (I assume) pretty simple math I have overlooked.

推荐答案

从我的头顶:

CGPoint endPoint;
endPoint.x = sinf(CC_DEGREES_TO_RADIANS(angle)) * lineLength;
endPoint.y = cosf(CC_DEGREES_TO_RADIANS(angle)) * lineLength;
endPoint = ccpAdd(currPoint, endPoint);

不确定向量指向的位置，如果它可以旋转 90、180 或 270 度.如果是这样，只需从角度添加/减去该数量.

Not sure where the vector points to, if it may be rotated by 90, 180 or 270 degrees. If so, just add/subtract that amount from angle.

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