从已知的边界框坐标计算旋转的矩形大小

Question

我从旋转的矩形中读取了" 计算边界"框的坐标,以了解如何从旋转的矩形计算边界框坐标.但在特殊情况下如下图:

如果获得边界框大小,坐标和旋转度,如何获得旋转的矩形大小？

我尝试在javascript中编写代码

//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......答案与w和h不匹配

Answer 1

解

给定边界框尺寸bx,by并且t是矩形的逆时针旋转,其大小x为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))

求导

为什么是这样？

首先,考虑将长度bx切成两块,a并b在矩形的一角切割.使用三角来表达bx来讲x,y和theta:

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

同样适用于by:

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

1和2可以矩阵形式表示为:

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

请注意,矩阵几乎是一个旋转矩阵(但不完全 - 它是一个减号.)

左侧划分矩阵,给出:

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

对于2x2矩阵,矩阵求逆很容易评估,并扩展为:

[ 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))

非常简单!