围绕另一个点旋转一个点 (2D) [英] Rotating a point about another point (2D)

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本文介绍了围绕另一个点旋转一个点 (2D)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

我正在尝试制作一种纸牌散开的纸牌游戏.现在使用具有功能的 Allegro API 来显示它:

I'm trying to make a card game where the cards fan out. Right now to display it Im using the Allegro API which has a function:

al_draw_rotated_bitmap(OBJECT_TO_ROTATE,CENTER_X,CENTER_Y,X
        ,Y,DEGREES_TO_ROTATE_IN_RADIANS);

所以有了这个,我可以轻松地制作我的粉丝效果.问题是知道哪张卡在鼠标下面.为此,我想到了进行多边形碰撞测试.我只是不确定如何旋转卡片上的 4 个点来组成多边形.我基本上需要做和Allegro一样的操作.

so with this I can make my fan effect easily. The problem is then knowing which card is under the mouse. To do this I thought of doing a polygon collision test. I'm just not sure how to rotate the 4 points on the card to make up the polygon. I basically need to do the same operation as Allegro.

比如卡片的4点是:

card.x

card.y

card.x + card.width

card.y + card.height

我需要一个类似的功能:

I would need a function like:

POINT rotate_point(float cx,float cy,float angle,POINT p)
{
}

谢谢

推荐答案

先减去枢轴点(cx,cy),然后旋转,再添加点.

First subtract the pivot point (cx,cy), then rotate it, then add the point again.

未经测试:

POINT rotate_point(float cx,float cy,float angle,POINT p)
{
  float s = sin(angle);
  float c = cos(angle);

  // translate point back to origin:
  p.x -= cx;
  p.y -= cy;

  // rotate point
  float xnew = p.x * c - p.y * s;
  float ynew = p.x * s + p.y * c;

  // translate point back:
  p.x = xnew + cx;
  p.y = ynew + cy;
  return p;
}

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