问题描述
如果椭圆的长轴是垂直或水平的，那么计算边界框很容易，但是当椭圆被旋转时怎么样？唯一的方法我能想到的是计算周边的所有点并找出最大/最小x和y值。这似乎应该有一个更简单的方法。
如果有一个函数（在数学意义上）描述一个任意角度的椭圆，那么我可以使用它的导数找到斜率为零或未定义的点，但我似乎无法找到一个。
编辑：为了阐明，我需要轴对齐边界框，即它不应该与椭圆一起旋转，而是保持与x轴对齐，因此变换边界框将不起作用。 解决方案
您可以尝试使用以任意角度旋转的椭圆的参数化方程：
x = h + a * cos（t）* cos（phi）-b * sin（t）* sin（phi）[1]
y = k + b * sin（t）* cos（phi）+ a * cos（t）* sin（phi）[2]
...其中椭圆有中心点（h ，k）半长轴a和半短轴b，并旋转角度φ。
然后，您可以区分并求解gradient = 0：
0 = dx / dt = -a * sin（t）* cos（phi） - b * cos（t）* sin（phi）
=>
tan（t）= -b * tan（phi）/ a [3]
给你许多t的解决方案（你感兴趣的两个解决方案），把它插回到[1]以获得你的最大和最小x。
重复[2 ]：
$ b $ pre $ code> 0 = dy / dt = b * cos（t）* cos（phi） - a * sin（t） * sin（phi）
=>
tan（t）= b * cot（phi）/ a [4]
可以尝试一个例子：
考虑一个椭圆（0,0），a = 2，b = 1，由PI / 4旋转：
[1] =>
x = 2 * cos（t）* cos（PI / 4） - sin（t）* sin（PI / 4）
$
$ p $ tan $（t）= -tan（PI / 4）/ 2 = -1/2
=>
t = -0.4636 + n * PI
因此我们得到：
pre $ $
$ b code> x = 2 * cos（-0.4636）* cos（PI / 4） - sin（-0.4636）* sin（PI / 4）= 1.5811
$ b $ p
$ $ $ $ $ $ c $ x $ 2 $ cos $ -3.6052 * cos （PI / 4） - sin（-3.6052）* sin（PI / 4）= -1.5811
If the major axis of the ellipse is vertical or horizontal, it's easy to calculate the bounding box, but what about when the ellipse is rotated?
The only way I can think of so far is to calculate all the points around the perimeter and find the max/min x and y values. It seems like there should be a simpler way.
If there's a function (in the mathematical sense) that describes an ellipse at an arbitrary angle, then I could use its derivative to find points where the slope is zero or undefined, but I can't seem to find one.
Edit: to clarify, I need the axis-aligned bounding box, i.e. it should not be rotated with the ellipse, but stay aligned with the x axis so transforming the bounding box won't work.
You could try using the parametrized equations for an ellipse rotated at an arbitrary angle:
x = h + a*cos(t)*cos(phi) - b*sin(t)*sin(phi) [1]
y = k + b*sin(t)*cos(phi) + a*cos(t)*sin(phi) [2]
...where ellipse has centre (h,k) semimajor axis a and semiminor axis b, and is rotated through angle phi.
You can then differentiate and solve for gradient = 0:
0 = dx/dt = -a*sin(t)*cos(phi) - b*cos(t)*sin(phi)
=>
tan(t) = -b*tan(phi)/a [3]
Which should give you many solutions for t (two of which you are interested in), plug that back into [1] to get your max and min x.
Repeat for [2]:
0 = dy/dt = b*cos(t)*cos(phi) - a*sin(t)*sin(phi)
=>
tan(t) = b*cot(phi)/a [4]
Lets try an example:
Consider an ellipse at (0,0) with a=2, b=1, rotated by PI/4:
[1] =>
x = 2*cos(t)*cos(PI/4) - sin(t)*sin(PI/4)
[3] =>
tan(t) = -tan(PI/4)/2 = -1/2
=>
t = -0.4636 + n*PI
We are interested in t = -0.4636 and t = -3.6052
So we get:
x = 2*cos(-0.4636)*cos(PI/4) - sin(-0.4636)*sin(PI/4) = 1.5811
and
x = 2*cos(-3.6052)*cos(PI/4) - sin(-3.6052)*sin(PI/4) = -1.5811
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