我读了从旋转的矩形计算边界框坐标 https://stackoverflow.com/questions/622140/calculate-bounding-box-coordinates-from-a-rotated-rectangle了解如何从旋转的矩形计算边界框坐标。但特殊情况如下图：

如果知道了边界框大小、坐标和旋转度数，如何获取旋转后的矩形大小？

我尝试用 javascript 编写代码

//assume w=123,h=98,deg=35 and get calculate box size
var deg = 35;
var bw = 156.9661922099485;
var bh = 150.82680201149986;

//calculate w and h
var xMax = bw / 2;
var yMax = bh / 2;
var radian = (deg / 180) * Math.PI;
var cosine = Math.cos(radian);
var sine = Math.sin(radian);
var cx = (xMax * cosine) + (yMax * sine)   / (cosine * cosine + sine * sine);
var cy =  -(-(xMax * sine)  - (yMax * cosine) / (cosine * cosine + sine * sine));
var w = (cx * 2 - bw)*2;
var h = (cy * 2 - bh)*2;

但是...答案是 w 和 h 不匹配

Solution

给定边界框尺寸bx by by and t是矩形尺寸的逆时针旋转x by y:

x = (1/(cos(t)^2-sin(t)^2)) * (  bx * cos(t) - by * sin(t))
y = (1/(cos(t)^2-sin(t)^2)) * (- bx * sin(t) + by * cos(t))

推导

为什么是这样？

首先，考虑长度bx被切成两块，a and b，在矩形的角处。用三角函数来表达bx按照x, y, and theta:

bx = b          + a
bx = x * cos(t) + y * sin(t)            [1]

类似地对于by:

by = c          + d
by = x * sin(t) + y * cos(t)            [2]

1 https://i.stack.imgur.com/Oq85E.jpg and 2 http://www.mathwords.com/i/inverse_of_a_matrix.htm可以用矩阵形式表示为：

[ bx ] = [ cos(t)  sin(t) ] * [ x ]     [3]
[ by ]   [ sin(t)  cos(t) ]   [ y ]

请注意，矩阵是nearly旋转矩阵（但不完全是 - 它有一个负号。）

将矩阵左除两边，得到：

[ x ] = inverse ( [ cos(t)  sin(t) ]    * [ bx ]                        [4]
[ y ]             [ sin(t)  cos(t) ] )    [ by ]

矩阵的逆矩阵是易于计算 2x2 矩阵 http://www.mathwords.com/i/inverse_of_a_matrix.htm并扩展到：

[ x ] = (1/(cos(t)^2-sin(t)^2)) * [ cos(t) -sin(t) ] * [ bx ]           [5]
[ y ]                             [-sin(t)  cos(t) ]   [ by ]

[5]给出了两个公式：

x = (1/(cos(t)^2-sin(t)^2)) * (  bx * cos(t) - by * sin(t))             [6]
y = (1/(cos(t)^2-sin(t)^2)) * (- bx * sin(t) + by * cos(t))

易如反掌！