透视变形矩形的比例
computer-vision
image-processing
5
0

给定矩形的2d图片,其透视会变形:

在此处输入图片说明

我知道形状最初是一个矩形,但我不知道它的原始大小。

如果我知道这张图中拐角的像素坐标,如何计算原始比例,即矩形的商(宽度/高度)?

(背景:目标是使矩形文档的照片自动变形,边缘检测可能会通过霍夫变换完成)

更新:

关于是否有可能完全根据给出的信息确定宽高比的问题进行了讨论。我幼稚的想法是必须做到这一点,因为我无法想到将例如1:4矩形投影到上述四边形上的方法。该比率显然很接近1:1,因此应该有一种数学上确定比率的方法。但是,除了我的直觉猜测之外,我没有任何证据。

我还没有完全理解下面提出的论点,但是我认为必须存在一些隐含的假设,即我们在这里不见了,并且对此有不同的解释。

但是,经过数小时的搜索,我终于找到了一些与该问题有关的论文。到目前为止,我一直在努力了解其中使用的数学,但没有成功。尤其是第一篇论文似乎完全在讨论我想做的事,不幸的是,没有代码示例和非常密集的数学。

  • 张正友,何立伟,“白板扫描和图像增强” http://research.microsoft.com/zh-cn/um/people/zhang/papers/tr03-39.pdf第 11页

    “由于透视畸变,矩形的图像看起来像是一个四边形。但是,由于我们知道它是空间中的矩形,因此我们能够估计相机的焦距和矩形的纵横比。”

  • ROBERT M. HARALICK“从矩形的透视投影确定相机参数” http://portal.acm.org/citation.cfm?id=87146

    “我们展示了如何使用未知大小和位置在3D空间中的矩形的2D透视投影来确定相对于矩形平面的摄影机视角参数。”

参考资料:
Stack Overflow
收藏
评论
共 2 个回答
高赞 时间 活跃

更新资料

阅读您的更新并查看第一个参考资料(白板扫描和图像增强)后,我看到了缺失的地方。

问题的输入数据是一个四元(A,B,C,d),将投影图像的中心O。在本文中,它对应于假设u0 = v0 = 0。加上这一点,问题就变得足够受约束,无法获得矩形的纵横比。

然后按如下方式重述该问题:给定Z = 0平面中的四元组(A,B,C,D),找到眼睛位置E(0,0,h),h> 0和3D平面P,使得(A,B,C,D)在P上的投影是矩形。

请注意,P由E确定:要获得平行四边形,P必须包含与(EU)和(EV)的平行线,其中U =(AB)x(CD)和V =(AD)x(BC)。

从实验上看,似乎这个问题通常具有一个唯一的解决方案,对应于矩形的w / h比的唯一值。

替代文字替代文字

上一篇

不,您无法从投影中确定矩形比例。

在一般情况下,Z = 0平面的四个非共线点的四倍(A,B,C,D)是无限多个矩形的投影,具有无限多个宽度/高度比。

考虑两个消失点U,即(AB)和(CD)的交点和V,(AD)和(BC)的交点,以及点I,两个对角线(AC)和(BD)的交点。若要投影为ABCD,必须将中心I的平行四边形放置在一个包含与通过点I平行于(UV)的线的平面上。在这样的平面上,您可以找到许多投影到ABCD的矩形,所有矩形的w / h比率均不同。

观看使用Cabri 3D完成的这两个图像。在这两种情况下,ABCD不变(在灰色的Z = 0平面上),并且包含矩形的蓝色平面也未更改。部分隐藏的绿线是(UV)线,可见的绿线与之平行并包含I。

替代文字替代文字

收藏
评论

这是我阅读本文后试图回答我的问题的尝试

我在SAGE中处理了一段时间的方程式,然后想到了c风格的伪代码:


// in case it matters: licensed under GPLv2 or later
// legend:
// sqr(x)  = x*x
// sqrt(x) = square root of x

// let m1x,m1y ... m4x,m4y be the (x,y) pixel coordinates
// of the 4 corners of the detected quadrangle
// i.e. (m1x, m1y) are the cordinates of the first corner, 
// (m2x, m2y) of the second corner and so on.
// let u0, v0 be the pixel coordinates of the principal point of the image
// for a normal camera this will be the center of the image, 
// i.e. u0=IMAGEWIDTH/2; v0 =IMAGEHEIGHT/2
// This assumption does not hold if the image has been cropped asymmetrically

// first, transform the image so the principal point is at (0,0)
// this makes the following equations much easier
m1x = m1x - u0;
m1y = m1y - v0;
m2x = m2x - u0;
m2y = m2y - v0;
m3x = m3x - u0;
m3y = m3y - v0;
m4x = m4x - u0;
m4y = m4y - v0;


// temporary variables k2, k3
double k2 = ((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x) /
            ((m2y - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) ;

double k3 = ((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y - m1y*m4x) / 
            ((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) ;

// f_squared is the focal length of the camera, squared
// if k2==1 OR k3==1 then this equation is not solvable
// if the focal length is known, then this equation is not needed
// in that case assign f_squared= sqr(focal_length)
double f_squared = 
    -((k3*m3y - m1y)*(k2*m2y - m1y) + (k3*m3x - m1x)*(k2*m2x - m1x)) / 
                      ((k3 - 1)*(k2 - 1)) ;

//The width/height ratio of the original rectangle
double whRatio = sqrt( 
    (sqr(k2 - 1) + sqr(k2*m2y - m1y)/f_squared + sqr(k2*m2x - m1x)/f_squared) /
    (sqr(k3 - 1) + sqr(k3*m3y - m1y)/f_squared + sqr(k3*m3x - m1x)/f_squared) 
) ;

// if k2==1 AND k3==1, then the focal length equation is not solvable 
// but the focal length is not needed to calculate the ratio.
// I am still trying to figure out under which circumstances k2 and k3 become 1
// but it seems to be when the rectangle is not distorted by perspective, 
// i.e. viewed straight on. Then the equation is obvious:
if (k2==1 && k3==1) whRatio = sqrt( 
    (sqr(m2y-m1y) + sqr(m2x-m1x)) / 
    (sqr(m3y-m1y) + sqr(m3x-m1x))


// After testing, I found that the above equations 
// actually give the height/width ratio of the rectangle, 
// not the width/height ratio. 
// If someone can find the error that caused this, 
// I would be most grateful.
// until then:
whRatio = 1/whRatio;

更新:这是确定这些方程的方式:

以下是SAGE中的代码。可以从http://www.sagenb.org/home/pub/704/在线访问 。 (Sage在求解方程式中确实很有用,并且可以在任何浏览器中使用,请查看)

# CALCULATING THE ASPECT RATIO OF A RECTANGLE DISTORTED BY PERSPECTIVE

#
# BIBLIOGRAPHY:
# [zhang-single]: "Single-View Geometry of A Rectangle 
#  With Application to Whiteboard Image Rectification"
#  by Zhenggyou Zhang
#  http://research.microsoft.com/users/zhang/Papers/WhiteboardRectification.pdf

# pixel coordinates of the 4 corners of the quadrangle (m1, m2, m3, m4)
# see [zhang-single] figure 1
m1x = var('m1x')
m1y = var('m1y')
m2x = var('m2x')
m2y = var('m2y')
m3x = var('m3x')
m3y = var('m3y')
m4x = var('m4x')
m4y = var('m4y')

# pixel coordinates of the principal point of the image
# for a normal camera this will be the center of the image, 
# i.e. u0=IMAGEWIDTH/2; v0 =IMAGEHEIGHT/2
# This assumption does not hold if the image has been cropped asymmetrically
u0 = var('u0')
v0 = var('v0')

# pixel aspect ratio; for a normal camera pixels are square, so s=1
s = var('s')

# homogenous coordinates of the quadrangle
m1 = vector ([m1x,m1y,1])
m2 = vector ([m2x,m2y,1])
m3 = vector ([m3x,m3y,1])
m4 = vector ([m4x,m4y,1])


# the following equations are later used in calculating the the focal length 
# and the rectangle's aspect ratio.
# temporary variables: k2, k3, n2, n3

# see [zhang-single] Equation 11, 12
k2_ = m1.cross_product(m4).dot_product(m3) / m2.cross_product(m4).dot_product(m3)
k3_ = m1.cross_product(m4).dot_product(m2) / m3.cross_product(m4).dot_product(m2)
k2 = var('k2')
k3 = var('k3')

# see [zhang-single] Equation 14,16
n2 = k2 * m2 - m1
n3 = k3 * m3 - m1


# the focal length of the camera.
f = var('f')
# see [zhang-single] Equation 21
f_ = sqrt(
         -1 / (
          n2[2]*n3[2]*s^2
         ) * (
          (
           n2[0]*n3[0] - (n2[0]*n3[2]+n2[2]*n3[0])*u0 + n2[2]*n3[2]*u0^2
          )*s^2 + (
           n2[1]*n3[1] - (n2[1]*n3[2]+n2[2]*n3[1])*v0 + n2[2]*n3[2]*v0^2
          ) 
         ) 
        )


# standard pinhole camera matrix
# see [zhang-single] Equation 1
A = matrix([[f,0,u0],[0,s*f,v0],[0,0,1]])


#the width/height ratio of the original rectangle
# see [zhang-single] Equation 20
whRatio = sqrt (
               (n2*A.transpose()^(-1) * A^(-1)*n2.transpose()) / 
               (n3*A.transpose()^(-1) * A^(-1)*n3.transpose())
              ) 

C代码中的简化方程由

print "simplified equations, assuming u0=0, v0=0, s=1"
print "k2 := ", k2_
print "k3 := ", k3_
print "f  := ", f_(u0=0,v0=0,s=1)
print "whRatio := ", whRatio(u0=0,v0=0,s=1)

    simplified equations, assuming u0=0, v0=0, s=1
    k2 :=  ((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)/((m2y
    - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x)
    k3 :=  ((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)/((m3y
    - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x)
    f  :=  sqrt(-((k3*m3y - m1y)*(k2*m2y - m1y) + (k3*m3x - m1x)*(k2*m2x
    - m1x))/((k3 - 1)*(k2 - 1)))
    whRatio :=  sqrt(((k2 - 1)^2 + (k2*m2y - m1y)^2/f^2 + (k2*m2x -
    m1x)^2/f^2)/((k3 - 1)^2 + (k3*m3y - m1y)^2/f^2 + (k3*m3x -
    m1x)^2/f^2))

print "Everything in one equation:"
print "whRatio := ", whRatio(f=f_)(k2=k2_,k3=k3_)(u0=0,v0=0,s=1)

    Everything in one equation:
    whRatio :=  sqrt(((((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y -
    m1y*m4x)/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) -
    1)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)/((m2y -
    m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) - 1)*(((m1y -
    m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)*m2y/((m2y - m4y)*m3x
    - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) - m1y)^2/((((m1y - m4y)*m2x -
    (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)*m3y/((m3y - m4y)*m2x - (m3x -
    m4x)*m2y + m3x*m4y - m3y*m4x) - m1y)*(((m1y - m4y)*m3x - (m1x -
    m4x)*m3y + m1x*m4y - m1y*m4x)*m2y/((m2y - m4y)*m3x - (m2x - m4x)*m3y
    + m2x*m4y - m2y*m4x) - m1y) + (((m1y - m4y)*m2x - (m1x - m4x)*m2y +
    m1x*m4y - m1y*m4x)*m3x/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y
    - m3y*m4x) - m1x)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y -
    m1y*m4x)*m2x/((m2y - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x)
    - m1x)) + (((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y -
    m1y*m4x)/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) -
    1)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)/((m2y -
    m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) - 1)*(((m1y -
    m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)*m2x/((m2y - m4y)*m3x
    - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) - m1x)^2/((((m1y - m4y)*m2x -
    (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)*m3y/((m3y - m4y)*m2x - (m3x -
    m4x)*m2y + m3x*m4y - m3y*m4x) - m1y)*(((m1y - m4y)*m3x - (m1x -
    m4x)*m3y + m1x*m4y - m1y*m4x)*m2y/((m2y - m4y)*m3x - (m2x - m4x)*m3y
    + m2x*m4y - m2y*m4x) - m1y) + (((m1y - m4y)*m2x - (m1x - m4x)*m2y +
    m1x*m4y - m1y*m4x)*m3x/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y
    - m3y*m4x) - m1x)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y -
    m1y*m4x)*m2x/((m2y - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x)
    - m1x)) - (((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y -
    m1y*m4x)/((m2y - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) -
    1)^2)/((((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y -
    m1y*m4x)/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) -
    1)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)/((m2y -
    m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) - 1)*(((m1y -
    m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)*m3y/((m3y - m4y)*m2x
    - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) - m1y)^2/((((m1y - m4y)*m2x -
    (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)*m3y/((m3y - m4y)*m2x - (m3x -
    m4x)*m2y + m3x*m4y - m3y*m4x) - m1y)*(((m1y - m4y)*m3x - (m1x -
    m4x)*m3y + m1x*m4y - m1y*m4x)*m2y/((m2y - m4y)*m3x - (m2x - m4x)*m3y
    + m2x*m4y - m2y*m4x) - m1y) + (((m1y - m4y)*m2x - (m1x - m4x)*m2y +
    m1x*m4y - m1y*m4x)*m3x/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y
    - m3y*m4x) - m1x)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y -
    m1y*m4x)*m2x/((m2y - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x)
    - m1x)) + (((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y -
    m1y*m4x)/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) -
    1)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y - m1y*m4x)/((m2y -
    m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x) - 1)*(((m1y -
    m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)*m3x/((m3y - m4y)*m2x
    - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) - m1x)^2/((((m1y - m4y)*m2x -
    (m1x - m4x)*m2y + m1x*m4y - m1y*m4x)*m3y/((m3y - m4y)*m2x - (m3x -
    m4x)*m2y + m3x*m4y - m3y*m4x) - m1y)*(((m1y - m4y)*m3x - (m1x -
    m4x)*m3y + m1x*m4y - m1y*m4x)*m2y/((m2y - m4y)*m3x - (m2x - m4x)*m3y
    + m2x*m4y - m2y*m4x) - m1y) + (((m1y - m4y)*m2x - (m1x - m4x)*m2y +
    m1x*m4y - m1y*m4x)*m3x/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y
    - m3y*m4x) - m1x)*(((m1y - m4y)*m3x - (m1x - m4x)*m3y + m1x*m4y -
    m1y*m4x)*m2x/((m2y - m4y)*m3x - (m2x - m4x)*m3y + m2x*m4y - m2y*m4x)
    - m1x)) - (((m1y - m4y)*m2x - (m1x - m4x)*m2y + m1x*m4y -
    m1y*m4x)/((m3y - m4y)*m2x - (m3x - m4x)*m2y + m3x*m4y - m3y*m4x) -
    1)^2))


# some testing:
# - choose a random rectangle, 
# - project it onto a random plane,
# - insert the corners in the above equations,
# - check if the aspect ratio is correct.

from sage.plot.plot3d.transform import rotate_arbitrary

#redundandly random rotation matrix
rand_rotMatrix = \
           rotate_arbitrary((uniform(-5,5),uniform(-5,5),uniform(-5,5)),uniform(-5,5)) *\
           rotate_arbitrary((uniform(-5,5),uniform(-5,5),uniform(-5,5)),uniform(-5,5)) *\
           rotate_arbitrary((uniform(-5,5),uniform(-5,5),uniform(-5,5)),uniform(-5,5))

#random translation vector
rand_transVector = vector((uniform(-10,10),uniform(-10,10),uniform(-10,10))).transpose()

#random rectangle parameters
rand_width =uniform(0.1,10)
rand_height=uniform(0.1,10)
rand_left  =uniform(-10,10)
rand_top   =uniform(-10,10)

#random focal length and principal point
rand_f  = uniform(0.1,100)
rand_u0 = uniform(-100,100)
rand_v0 = uniform(-100,100)

# homogenous standard pinhole projection, see [zhang-single] Equation 1
hom_projection = A * rand_rotMatrix.augment(rand_transVector)

# construct a random rectangle in the plane z=0, then project it randomly 
rand_m1hom = hom_projection*vector((rand_left           ,rand_top            ,0,1)).transpose()
rand_m2hom = hom_projection*vector((rand_left           ,rand_top+rand_height,0,1)).transpose()
rand_m3hom = hom_projection*vector((rand_left+rand_width,rand_top            ,0,1)).transpose()
rand_m4hom = hom_projection*vector((rand_left+rand_width,rand_top+rand_height,0,1)).transpose()

#change type from 1x3 matrix to vector
rand_m1hom = rand_m1hom.column(0)
rand_m2hom = rand_m2hom.column(0)
rand_m3hom = rand_m3hom.column(0)
rand_m4hom = rand_m4hom.column(0)

#normalize
rand_m1hom = rand_m1hom/rand_m1hom[2]
rand_m2hom = rand_m2hom/rand_m2hom[2]
rand_m3hom = rand_m3hom/rand_m3hom[2]
rand_m4hom = rand_m4hom/rand_m4hom[2]

#substitute random values for f, u0, v0
rand_m1hom = rand_m1hom(f=rand_f,s=1,u0=rand_u0,v0=rand_v0)
rand_m2hom = rand_m2hom(f=rand_f,s=1,u0=rand_u0,v0=rand_v0)
rand_m3hom = rand_m3hom(f=rand_f,s=1,u0=rand_u0,v0=rand_v0)
rand_m4hom = rand_m4hom(f=rand_f,s=1,u0=rand_u0,v0=rand_v0)

# printing the randomly choosen values
print "ground truth: f=", rand_f, "; ratio=", rand_width/rand_height

# substitute all the variables in the equations:
print "calculated: f= ",\
f_(k2=k2_,k3=k3_)(s=1,u0=rand_u0,v0=rand_v0)(
  m1x=rand_m1hom[0],m1y=rand_m1hom[1],
  m2x=rand_m2hom[0],m2y=rand_m2hom[1],
  m3x=rand_m3hom[0],m3y=rand_m3hom[1],
  m4x=rand_m4hom[0],m4y=rand_m4hom[1],
),"; 1/ratio=", \
1/whRatio(f=f_)(k2=k2_,k3=k3_)(s=1,u0=rand_u0,v0=rand_v0)(
  m1x=rand_m1hom[0],m1y=rand_m1hom[1],
  m2x=rand_m2hom[0],m2y=rand_m2hom[1],
  m3x=rand_m3hom[0],m3y=rand_m3hom[1],
  m4x=rand_m4hom[0],m4y=rand_m4hom[1],
)

print "k2 = ", k2_(
  m1x=rand_m1hom[0],m1y=rand_m1hom[1],
  m2x=rand_m2hom[0],m2y=rand_m2hom[1],
  m3x=rand_m3hom[0],m3y=rand_m3hom[1],
  m4x=rand_m4hom[0],m4y=rand_m4hom[1],
), "; k3 = ", k3_(
  m1x=rand_m1hom[0],m1y=rand_m1hom[1],
  m2x=rand_m2hom[0],m2y=rand_m2hom[1],
  m3x=rand_m3hom[0],m3y=rand_m3hom[1],
  m4x=rand_m4hom[0],m4y=rand_m4hom[1],
)

# ATTENTION: testing revealed, that the whRatio 
# is actually the height/width ratio, 
# not the width/height ratio
# This contradicts [zhang-single]
# if anyone can find the error that caused this, I'd be grateful

    ground truth: f= 72.1045134124554 ; ratio= 3.46538779959142
    calculated: f=  72.1045134125 ; 1/ratio= 3.46538779959
    k2 =  0.99114614987 ; k3 =  1.57376280159
收藏
评论
新手导航
  • 社区规范
  • 提出问题
  • 进行投票
  • 个人资料
  • 优化问题
  • 回答问题

关于我们

常见问题

内容许可

联系我们

@2020 AskGo
京ICP备20001863号