一个点A (x0 ,y0)绕原点旋转一个角度后的点 A' (x1,y1) 的坐标值计算方法是怎么推出来的 一个点(x,y)绕着原点旋转a角度到达(x1,y1点)，求用...

\u4e00\u4e2a\u70b9\u7ed5\u539f\u70b9\u65cb\u8f6c\u4e00\u4e2a\u89d2\u5ea6\u540e\u7684\u5750\u6807 \u70b9 A (x0 ,y0)\u4ee5\u539f\u70b9\u4e3a\u5706\u5fc3\u65cb\u8f6cn\u5ea6,\u6c42\u65cb\u8f6c\u540e A' (x1,y1) \u5750\u6807\u503c.

\u8bc1\uff1a\u8bbe\u70b9\uff08x0,y0\uff09\u5230\u70b9\uff08rx0,ry0\uff09\u7684\u8ddd\u79bb\u4e3aLa,\u70b9\uff08x,y\uff09\u5230\u70b9\uff08rx0,ry0\uff09\u7684\u8ddd\u79bb\u4e3aLb

\u5219\u7531\u56fe1\u53ef\u5f97\uff1a



2

( x0 - rx0 ) / La = cos\uff08a + b\uff09                    -    \u2460

( x - rx0 ) / Lb = cos\uff08b\uff09                            -    \u2461

La = Lb                                                        -    \u2462

( y0 - ry0 ) / La = sin\uff08a+b\uff09                       -    \u2463

( y - ry0 ) / Lb = sin\uff08b\uff09                             -    \u2464

\u5f53cos\uff08b\uff09\uff0ccos\uff08a + b\uff09\u4e0d\u4e3a\u96f6\u65f6,\u7531\u2460\u2461\u2462\u5f97\uff1a

     (x0- rx0)/ (x-rx0) = cos(a+b)/cos(b)

    (x0- rx0)/ (x-rx0) = (cos(a)cos(b)-sin(a)sin(b))/cos(b)

    (x0- rx0)/ (x-rx0) = cos(a) - sin(a)tan(b)

    (x0- rx0) = (cos(a) - sin(a)tan(b))(x - rx0)

    x0 = (x - rx0)cos(a) - sin(a)tan(b)(x - rx0) + rx0

    x0 = (x - rx0)cos(a) - (y - ry0)sin(a) + rx0                        -    A

\u5f53sin\uff08b\uff09\uff0csin\uff08a + b\uff09\u4e0d\u4e3a\u96f6\u65f6,\u7531\u2462\u2463\u2464\u5f97\uff1a

    (y0 - ry0)/(y - ry0) = sin(a+b)/sin(b)

    (y0 - ry0)/(y - ry0) = (sin(a)cos(b) + cos(a)sin(b))/sin(b)

    (y0 - ry0)/(y - ry0) = sin(a)cos(b)/sin(b) + cos(a)

    y0 = (y - ry0)sin(a)cos(b)/sin(b) + (y - ry0)cos(a) + ry0

    y0 = (y - ry0)sin(a)(x - rx0)/(y - ry0) + (y - ry0)cos(a) + ry0

    y0 = (x - rx0)sin(a) + (y - ry0)cos(a) + ry0                        -    B

\u2234\u5f53cos\uff08b\uff09\uff0ccos\uff08a + b\uff09\u4e0d\u4e3a\u96f6\u65f6A\u3001B\u5f0f\u6210\u7acb

3

 \u5f53cos\uff08a+b\uff09= 0\u65f6\uff0c\u5373x0 = rx0\uff0ca+b = \u03c0/2+k\u03c0(k>=0\u7684\u81ea\u7136\u6570)\u5982\u56fe2\uff1a



\u2235cos\uff08a+b\uff09= 0

    cos(a)cos(b) - sin(a)sin(b) = 0

    tan(a) = 1/tan(b)

    sin(a)/cos(a) = (x - rx0)/(y - ry0)

    (x - rx0)cos(a) = (y - ry0)sin(a)

\u5c06x0 = rx0\u5f0f\u4ee3\u5165A\u5f0f\u4e5f\u5f97

    (x - rx0)cos(a) = (y - ry0)sin(a)

\u2234\u5f53cos\uff08a+b\uff09= 0\u65f6A\u5f0f\u6210\u7acb\u3002

 

 \u2235tan(a) = (x - rx0)/(y - ry0) - \u2465

    La = Lb = y0 - ry0 - \u2466

\u7531\u2465\u5f97

    (y - ry0)sin(a)/cos(a) = (x - rx0)

   (y - ry0)sin²(a)/cos(a) = (x - rx0)sin(a)

   (y - ry0)(1-cos²(a))/cos(a) = (x - rx0)sin(a)

   (y - ry0)(1/cos(a)-cos(a)) = (x - rx0)sin(a)

   (y - ry0)/cos(a)-(y - ry0)cos(a)) = (x - rx0)sin(a)

 

\u987a\u65f6\u9488\u8fd8\u662f\u9006\u65f6\u9488

是逆时针还是顺时针旋转啊？给一个逆时针的图解给你：

这个是顺时针的吧。。。坐标系逆时针了坐标对应的应该是顺时针移动了

建立直角坐标三角系 分解直线A到 x y轴 不就好了？

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