将 2:1 等距柱状图全景图转换为立方体贴图 [英] Convert 2:1 equirectangular panorama to cube map
问题描述
我目前正在为一个网站开发一个简单的 3D 全景查看器.出于移动性能的原因，我使用了 three.js 我们得到这似乎使铺路中的大部分线条都正确.
我们得到了一些图像伪影.这是因为没有 1 比 1 的像素图.我们需要做的是使用逆变换.我们不是遍历源中的每个像素并找到目标中的对应像素，而是遍历目标图像并找到最接近的对应源像素.
导入系统从 PIL 导入图像从数学导入 pi,sin,cos,tan,atan2,hypot,floor从 numpy 导入剪辑# 从图像像素坐标中获取 x,y,z 坐标# i,j 是像素坐标# face 是人脸编号# edge 是边长def outImgToXYZ(i,j,face,edge):a = 2.0*float(i)/边b = 2.0*float(j)/边if face==0: # 返回(x,y,z) = (-1.0, 1.0-a, 3.0 - b)elif face==1: # 左(x,y,z) = (a-3.0, -1.0, 3.0 - b)elif face==2: # 前面(x,y,z) = (1.0, a - 5.0, 3.0 - b)elif face==3: # 对(x,y,z) = (7.0-a, 1.0, 3.0 - b)elif face==4: # 顶部(x,y,z) = (b-1.0, a -5.0, 1.0)elif face==5: # 底部(x,y,z) = (5.0-b, a-5.0, -1.0)返回 (x,y,z)# 使用逆变换进行转换def convertBack(imgIn,imgOut):inSize = imgIn.sizeoutSize = imgOut.sizeinPix = imgIn.load()outPix = imgOut.load()edge = inSize[0]/4 # 每条边的长度，以像素为单位对于 i in xrange(outSize[0]):face = int(i/edge) # 0 - 后，1 - 左 2 - 前，3 - 右如果脸==2:rng = xrange(0,edge*3)别的:rng = xrange(边，边*2)对于 rng 中的 j:如果 j<边缘:face2 = 4 # 顶部elif j>=2*边:face2 = 5 # 底部别的:face2 = 脸(x,y,z) = outImgToXYZ(i,j,face2,edge)theta = atan2(y,x) # 范围 -pi 到 pir = 假设(x，y)phi = atan2(z,r) # 范围 -pi/2 到 pi/2# 源 img 坐标uf = ( 2.0*edge*(theta + pi)/pi )vf = ( 2.0*edge * (pi/2 - phi)/pi)# 在四个周围像素之间使用双线性插值ui = floor(uf) # 像素到左下角的坐标vi = 地板(vf)u2 = ui+1 # 像素到右上角的坐标v2 = vi+1mu = uf-ui # 像素的比例nu = vf-vi# 四个角的像素值A = inPix[ui % inSize[0],clip(vi,0,inSize[1]-1)]B = inPix[u2 % inSize[0],clip(vi,0,inSize[1]-1)]C = inPix[ui % inSize[0],clip(v2,0,inSize[1]-1)]D = inPix[u2 % inSize[0],clip(v2,0,inSize[1]-1)]# 插值(r,g,b) = (A[0]*(1-mu)*(1-nu) + B[0]*(mu)*(1-nu) + C[0]*(1-mu)*nu+D[0]*穆*努，A[1]*(1-mu)*(1-nu) + B[1]*(mu)*(1-nu) + C[1]*(1-mu)*nu+D[1]*穆*努，A[2]*(1-mu)*(1-nu) + B[2]*(mu)*(1-nu) + C[2]*(1-mu)*nu+D[2]*穆*努 )outPix[i,j] = (int(round(r)),int(round(g)),int(round(b)))imgIn = Image.open(sys.argv[1])inSize = imgIn.sizeimgOut = Image.new(RGB",(inSize[0],inSize[0]*3/4),黑色")convertBack(imgIn,imgOut)imgOut.save(sys.argv[1].split('.')[0]+Out2.png")imgOut.show()这样的结果是
如果有人想反过来看JS Fiddle 页面
I'm currently working on a simple 3D panorama viewer for a website. For mobile performance reasons I'm using the three.js CSS3 renderer. This requires a cube map, split up into 6 single images.
I'm recording the images on the iPhone with the Google Photosphere app, or similar apps that create 2:1 equirectangular panoramas. I then resize and convert these to a cubemap with this website: http://gonchar.me/panorama/ (Flash)
Preferrably, I'd like to do the conversion myself, either on the fly in three.js, if that's possible, or in Photoshop. I found Andrew Hazelden's Photoshop actions, and they seem kind of close, but no direct conversion is available. Is there a mathematical way to convert these, or some sort of script that does it? I'd like to avoid going through a 3D app like Blender, if possible.
Maybe this is a long shot, but I thought I'd ask. I have okay experience with javascript, but I'm pretty new to three.js. I'm also hesitant to rely on the WebGL functionality, since it seems either slow or buggy on mobile devices. Support is also still spotty.
If you want to do it server side there are many options. http://www.imagemagick.org/ has a bunch of command line tools which could slice your image into pieces. You could put the command to do this into a script and just run that each time you have a new image.
Its hard to tell quite what algorithm is used in the program. We can try and reverse engineer quite what is happening by feeding a square grid into the program. I've used a grid from wikipedia
Which gives This gives us a clue as to how the box is constructed.
Imaging sphere with lines of latitude and longitude one it, and a cube surrounding it. Now project from the point at center of the sphere produces a distorted grid on the cube.
Mathematically take polar coordinates r, θ, ø, for the sphere r=1, 0 < θ < π, -π/4 < ø < 7π/4
- x= r sin θ cos ø
- y= r sin θ sin ø
- z= r cos θ
centrally project these to the cube. First we divide into four regions by the latitude -π/4 < ø < π/4, π/4 < ø < 3π/4, 3π/4 < ø < 5π/4, 5π/4 < ø < 7π/4. These will either project to one of the four sides the top or the bottom.
Assume we are in the first side -π/4 < ø < π/4. The central projection of (sin θ cos ø, sin θ sin ø, cos θ) will be (a sin θ cos ø, a sin θ sin ø, a cos θ) which hits the x=1 plane when
- a sin θ cos ø = 1
so
- a = 1 / (sin θ cos ø)
and the projected point is
- (1, tan ø, cot θ / cos ø)
If | cot θ / cos ø | < 1 this will be on the front face. Otherwise, it will be projected on the top or bottom and you will need a different projection for that. A better test for the top uses the fact that the minimum value of cos ø will be cos π/4 = 1/√2, so the projected point is always on the top if cot θ / (1/√2) > 1 or tan θ < 1/√2. This works out as θ < 35º or 0.615 radians.
Put this together in python
import sys
from PIL import Image
from math import pi,sin,cos,tan
def cot(angle):
return 1/tan(angle)
# Project polar coordinates onto a surrounding cube
# assume ranges theta is [0,pi] with 0 the north poll, pi south poll
# phi is in range [0,2pi]
def projection(theta,phi):
if theta<0.615:
return projectTop(theta,phi)
elif theta>2.527:
return projectBottom(theta,phi)
elif phi <= pi/4 or phi > 7*pi/4:
return projectLeft(theta,phi)
elif phi > pi/4 and phi <= 3*pi/4:
return projectFront(theta,phi)
elif phi > 3*pi/4 and phi <= 5*pi/4:
return projectRight(theta,phi)
elif phi > 5*pi/4 and phi <= 7*pi/4:
return projectBack(theta,phi)
def projectLeft(theta,phi):
x = 1
y = tan(phi)
z = cot(theta) / cos(phi)
if z < -1:
return projectBottom(theta,phi)
if z > 1:
return projectTop(theta,phi)
return ("Left",x,y,z)
def projectFront(theta,phi):
x = tan(phi-pi/2)
y = 1
z = cot(theta) / cos(phi-pi/2)
if z < -1:
return projectBottom(theta,phi)
if z > 1:
return projectTop(theta,phi)
return ("Front",x,y,z)
def projectRight(theta,phi):
x = -1
y = tan(phi)
z = -cot(theta) / cos(phi)
if z < -1:
return projectBottom(theta,phi)
if z > 1:
return projectTop(theta,phi)
return ("Right",x,-y,z)
def projectBack(theta,phi):
x = tan(phi-3*pi/2)
y = -1
z = cot(theta) / cos(phi-3*pi/2)
if z < -1:
return projectBottom(theta,phi)
if z > 1:
return projectTop(theta,phi)
return ("Back",-x,y,z)
def projectTop(theta,phi):
# (a sin θ cos ø, a sin θ sin ø, a cos θ) = (x,y,1)
a = 1 / cos(theta)
x = tan(theta) * cos(phi)
y = tan(theta) * sin(phi)
z = 1
return ("Top",x,y,z)
def projectBottom(theta,phi):
# (a sin θ cos ø, a sin θ sin ø, a cos θ) = (x,y,-1)
a = -1 / cos(theta)
x = -tan(theta) * cos(phi)
y = -tan(theta) * sin(phi)
z = -1
return ("Bottom",x,y,z)
# Convert coords in cube to image coords
# coords is a tuple with the side and x,y,z coords
# edge is the length of an edge of the cube in pixels
def cubeToImg(coords,edge):
if coords[0]=="Left":
(x,y) = (int(edge*(coords[2]+1)/2), int(edge*(3-coords[3])/2) )
elif coords[0]=="Front":
(x,y) = (int(edge*(coords[1]+3)/2), int(edge*(3-coords[3])/2) )
elif coords[0]=="Right":
(x,y) = (int(edge*(5-coords[2])/2), int(edge*(3-coords[3])/2) )
elif coords[0]=="Back":
(x,y) = (int(edge*(7-coords[1])/2), int(edge*(3-coords[3])/2) )
elif coords[0]=="Top":
(x,y) = (int(edge*(3-coords[1])/2), int(edge*(1+coords[2])/2) )
elif coords[0]=="Bottom":
(x,y) = (int(edge*(3-coords[1])/2), int(edge*(5-coords[2])/2) )
return (x,y)
# convert the in image to out image
def convert(imgIn,imgOut):
inSize = imgIn.size
outSize = imgOut.size
inPix = imgIn.load()
outPix = imgOut.load()
edge = inSize[0]/4 # the length of each edge in pixels
for i in xrange(inSize[0]):
for j in xrange(inSize[1]):
pixel = inPix[i,j]
phi = i * 2 * pi / inSize[0]
theta = j * pi / inSize[1]
res = projection(theta,phi)
(x,y) = cubeToImg(res,edge)
#if i % 100 == 0 and j % 100 == 0:
# print i,j,phi,theta,res,x,y
if x >= outSize[0]:
#print "x out of range ",x,res
x=outSize[0]-1
if y >= outSize[1]:
#print "y out of range ",y,res
y=outSize[1]-1
outPix[x,y] = pixel
imgIn = Image.open(sys.argv[1])
inSize = imgIn.size
imgOut = Image.new("RGB",(inSize[0],inSize[0]*3/4),"black")
convert(imgIn,imgOut)
imgOut.show()
The projection function takes the theta and phi values and returns coordinates in a cube from -1 to 1 in each direction. The cubeToImg takes the (x,y,z) coords and translates them to the output image coords.
The above algorithm seems to get the geometry right using an image of buckingham palace we get This seems to get most of the lines in the paving right.
We are getting a few image artefacts. This is due to not having a 1 to 1 map of pixels. What we need to do is use a inverse transformation. Rather than loop through each pixel in the source and find the corresponding pixel in the target we loop through the target images and find the closest corresponding source pixel.
import sys
from PIL import Image
from math import pi,sin,cos,tan,atan2,hypot,floor
from numpy import clip
# get x,y,z coords from out image pixels coords
# i,j are pixel coords
# face is face number
# edge is edge length
def outImgToXYZ(i,j,face,edge):
a = 2.0*float(i)/edge
b = 2.0*float(j)/edge
if face==0: # back
(x,y,z) = (-1.0, 1.0-a, 3.0 - b)
elif face==1: # left
(x,y,z) = (a-3.0, -1.0, 3.0 - b)
elif face==2: # front
(x,y,z) = (1.0, a - 5.0, 3.0 - b)
elif face==3: # right
(x,y,z) = (7.0-a, 1.0, 3.0 - b)
elif face==4: # top
(x,y,z) = (b-1.0, a -5.0, 1.0)
elif face==5: # bottom
(x,y,z) = (5.0-b, a-5.0, -1.0)
return (x,y,z)
# convert using an inverse transformation
def convertBack(imgIn,imgOut):
inSize = imgIn.size
outSize = imgOut.size
inPix = imgIn.load()
outPix = imgOut.load()
edge = inSize[0]/4 # the length of each edge in pixels
for i in xrange(outSize[0]):
face = int(i/edge) # 0 - back, 1 - left 2 - front, 3 - right
if face==2:
rng = xrange(0,edge*3)
else:
rng = xrange(edge,edge*2)
for j in rng:
if j<edge:
face2 = 4 # top
elif j>=2*edge:
face2 = 5 # bottom
else:
face2 = face
(x,y,z) = outImgToXYZ(i,j,face2,edge)
theta = atan2(y,x) # range -pi to pi
r = hypot(x,y)
phi = atan2(z,r) # range -pi/2 to pi/2
# source img coords
uf = ( 2.0*edge*(theta + pi)/pi )
vf = ( 2.0*edge * (pi/2 - phi)/pi)
# Use bilinear interpolation between the four surrounding pixels
ui = floor(uf) # coord of pixel to bottom left
vi = floor(vf)
u2 = ui+1 # coords of pixel to top right
v2 = vi+1
mu = uf-ui # fraction of way across pixel
nu = vf-vi
# Pixel values of four corners
A = inPix[ui % inSize[0],clip(vi,0,inSize[1]-1)]
B = inPix[u2 % inSize[0],clip(vi,0,inSize[1]-1)]
C = inPix[ui % inSize[0],clip(v2,0,inSize[1]-1)]
D = inPix[u2 % inSize[0],clip(v2,0,inSize[1]-1)]
# interpolate
(r,g,b) = (
A[0]*(1-mu)*(1-nu) + B[0]*(mu)*(1-nu) + C[0]*(1-mu)*nu+D[0]*mu*nu,
A[1]*(1-mu)*(1-nu) + B[1]*(mu)*(1-nu) + C[1]*(1-mu)*nu+D[1]*mu*nu,
A[2]*(1-mu)*(1-nu) + B[2]*(mu)*(1-nu) + C[2]*(1-mu)*nu+D[2]*mu*nu )
outPix[i,j] = (int(round(r)),int(round(g)),int(round(b)))
imgIn = Image.open(sys.argv[1])
inSize = imgIn.size
imgOut = Image.new("RGB",(inSize[0],inSize[0]*3/4),"black")
convertBack(imgIn,imgOut)
imgOut.save(sys.argv[1].split('.')[0]+"Out2.png")
imgOut.show()
The results of this are
If anyone want to go in the reverse see JS Fiddle page
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