Javascript 将2:1等矩形全景转换为立方体贴图

Javascript 将2:1等矩形全景转换为立方体贴图,javascript,math,three.js,Javascript,Math,Three.js,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。这需要一个立方体贴图,分成6个单独的图像 我正在用谷歌Photosphere应用程序或类似的创建2:1等矩形全景的应用程序在iPhone上录制图像。然后,我用这个网站(Flash)调整大小并将其转换为立方体贴图 我更愿意自己进行转换,如果可能的话,可以在three.js中动态转换,也可以在Photoshop中转换。我发现Andrew Hazelden的Photoshop操作,它们看起来很接近,但没有直

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
。这需要一个立方体贴图,分成6个单独的图像

我正在用谷歌Photosphere应用程序或类似的创建2:1等矩形全景的应用程序在iPhone上录制图像。然后,我用这个网站(Flash)调整大小并将其转换为立方体贴图

我更愿意自己进行转换,如果可能的话,可以在three.js中动态转换,也可以在Photoshop中转换。我发现Andrew Hazelden的Photoshop操作,它们看起来很接近,但没有直接转换。有没有一种数学方法来转换这些,或者某种脚本来转换它们?如果可能的话,我想避免使用Blender这样的3D应用程序


也许这不太可能,但我想我应该问问。我在javascript方面有不错的经验,但我对
three.js
还是相当陌生的。我也不太愿意依赖WebGL功能,因为它在移动设备上看起来很慢或者有问题。支持也仍然参差不齐。

如果您想在服务器端实现,有很多选择。有一系列的命令行工具,可以将图像分割成碎片。您可以将执行此操作的命令放入脚本中,并在每次有新图像时运行该命令

很难说程序中使用了什么算法。我们可以尝试通过向程序中输入一个正方形网格来反向工程正在发生的事情。我用了一个

这给了我们一个关于盒子是如何构造的线索

用一条经纬线和一个立方体围绕球体成像。现在,从球体中心的点投影会在立方体上生成扭曲的栅格

数学上取球面r=1的极坐标r,θ,ø,0<θ<π,-π/4<ø<7π/4

  • x=r sinθcosø
  • y=r sinθsinø
  • z=r cosθ
集中地将这些项目投影到多维数据集。首先,我们根据纬度将其分为四个区域——π/4<248<π/4,π/4<248<3π/4,3π/4<248<5π/4,5π/4<248<7π/4。这些将投影到顶部或底部的四个侧面之一

假设我们在第一面-π/4<ø<π/4。中心投影 (sinθcosø,sinθsinø,cosθ)将是(sinθcosø,sinθsinø,cosθ),当

  • a sinθcosø=1
所以

  • a=1/(sinθcosø)
投影点是

  • (1,tanø,cotθ/cosø)
如果| cotθ/cos|<1,这将出现在正面。否则,它将被投影到顶部或底部,您将需要一个不同的投影。更好的测试方法是使用cosø的最小值为cosπ/4=1这一事实/√2,所以如果cotθ/(1),投影点总是在顶部/√2) 大于1或tanθ<1/√2.计算结果为θ<35º或0.615弧度

用python将其组合起来

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()

其结果是

环境地图有多种表示形式。这里是一个很好的概述

如果您使用Photosphere(或任何全景应用程序),您很可能已经拥有了水平表示。 然后,您可以简单地绘制一个带纹理的three.js。下面是一个关于如何渲染地球的教程


祝你好运:)。

我编写了一个脚本,将生成的立方体贴图剪切成单独的文件(posx.png、negx.png、posy.png、negy.png、posz.png和negz.png)。它还将把这6个文件打包成一个.zip文件

资料来源如下:

您可以修改阵列以设置图像文件:

name_map = [ \
 ["", "", "posy", ""],
 ["negz", "negx", "posz", "posx"],
 ["", "", "negy", ""]]
转换的文件包括:


找到了这个问题,尽管答案很好,但我认为还有一些地方没有找到,所以这是我的两分钱

第一:除非您真的必须自己转换图像(即,由于某些特定的软件要求),不要

原因是,尽管在等矩形投影和立方投影之间有一个非常简单的映射,但区域之间的映射并不简单:当通过基本计算在目标图像的特定点和源中的点之间建立对应关系时,当你将两个点通过舍入转换成像素时,你正在做一个<>强>非常/强>原始近似,它不考虑像素的大小,并且图像的质量必然是低的。 第二:即使您需要在运行时进行转换,您是否确实需要进行转换?除非有一些非常严格的性能问题,如果你只需要一个skybox,创建一个非常大的球体,在上面缝合等矩形纹理,然后就可以了。据我记忆所及,三个JS已经提供了球体;-)

第三:美国宇航局提供了一种工具,可以在所有可以想象的投影之间进行转换(我刚刚发现并测试了它,效果非常好)。你可以在这里找到它:

我觉得有理由认为这些人知道他们在做什么;-)

希望这有帮助

更新:事实证明,这些“家伙”在某种程度上知道他们在做什么:生成的立方体贴图有一个可怕的边框,这使得转换不那么容易

更新2:找到了等矩形到立方体映射转换的权威工具,它被称为
erect2cubic

这是一个小型实用程序,它生成一个脚本供hugin使用,如下所示:

$ erect2cubic --erect=input.png --ptofile=cube.pto
$ nona -o cube_prefix cube.pto 
(从页面中提取的信息)

并将生成所有6个立方体贴图面。我在我的项目中使用它,它的效果非常好

这种方法唯一的缺点是脚本
erect2cubit
不在标准的Ubuntu发行版中(这就是我正在使用的),我不得不求助于此链接中的说明:

$ erect2cubic --erect=input.png --ptofile=cube.pto
$ nona -o cube_prefix cube.pto 
// Define our six cube faces. 
// 0 - 3 are side faces, clockwise order
// 4 and 5 are top and bottom, respectively
float faceTransform[6][2] = 
{ 
    {0, 0},
    {M_PI / 2, 0},
    {M_PI, 0},
    {-M_PI / 2, 0},
    {0, -M_PI / 2},
    {0, M_PI / 2}
};

// Map a part of the equirectangular panorama (in) to a cube face
// (face). The ID of the face is given by faceId. The desired
// width and height are given by width and height. 
inline void createCubeMapFace(const Mat &in, Mat &face, 
        int faceId = 0, const int width = -1, 
        const int height = -1) {

    float inWidth = in.cols;
    float inHeight = in.rows;

    // Allocate map
    Mat mapx(height, width, CV_32F);
    Mat mapy(height, width, CV_32F);

    // Calculate adjacent (ak) and opposite (an) of the
    // triangle that is spanned from the sphere center 
    //to our cube face.
    const float an = sin(M_PI / 4);
    const float ak = cos(M_PI / 4);

    const float ftu = faceTransform[faceId][0];
    const float ftv = faceTransform[faceId][1];

    // For each point in the target image, 
    // calculate the corresponding source coordinates. 
    for(int y = 0; y < height; y++) {
        for(int x = 0; x < width; x++) {

            // Map face pixel coordinates to [-1, 1] on plane
            float nx = (float)y / (float)height - 0.5f;
            float ny = (float)x / (float)width - 0.5f;

            nx *= 2;
            ny *= 2;

            // Map [-1, 1] plane coords to [-an, an]
            // thats the coordinates in respect to a unit sphere 
            // that contains our box. 
            nx *= an; 
            ny *= an; 

            float u, v;

            // Project from plane to sphere surface.
            if(ftv == 0) {
                // Center faces
                u = atan2(nx, ak);
                v = atan2(ny * cos(u), ak);
                u += ftu; 
            } else if(ftv > 0) { 
                // Bottom face 
                float d = sqrt(nx * nx + ny * ny);
                v = M_PI / 2 - atan2(d, ak);
                u = atan2(ny, nx);
            } else {
                // Top face
                float d = sqrt(nx * nx + ny * ny);
                v = -M_PI / 2 + atan2(d, ak);
                u = atan2(-ny, nx);
            }

            // Map from angular coordinates to [-1, 1], respectively.
            u = u / (M_PI); 
            v = v / (M_PI / 2);

            // Warp around, if our coordinates are out of bounds. 
            while (v < -1) {
                v += 2;
                u += 1;
            } 
            while (v > 1) {
                v -= 2;
                u += 1;
            } 

            while(u < -1) {
                u += 2;
            }
            while(u > 1) {
                u -= 2;
            }

            // Map from [-1, 1] to in texture space
            u = u / 2.0f + 0.5f;
            v = v / 2.0f + 0.5f;

            u = u * (inWidth - 1);
            v = v * (inHeight - 1);

            // Save the result for this pixel in map
            mapx.at<float>(x, y) = u;
            mapy.at<float>(x, y) = v; 
        }
    }

    // Recreate output image if it has wrong size or type. 
    if(face.cols != width || face.rows != height || 
        face.type() != in.type()) {
        face = Mat(width, height, in.type());
    }

    // Do actual resampling using OpenCV's remap
    remap(in, face, mapx, mapy, 
         CV_INTER_LINEAR, BORDER_CONSTANT, Scalar(0, 0, 0));
}
#!/usr/bin/env python
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
# faceIdx is face number
# faceSize is edge length
def outImgToXYZ(i, j, faceIdx, faceSize):
    a = 2.0 * float(i) / faceSize
    b = 2.0 * float(j) / faceSize

    if faceIdx == 0: # back
        (x,y,z) = (-1.0, 1.0 - a, 1.0 - b)
    elif faceIdx == 1: # left
        (x,y,z) = (a - 1.0, -1.0, 1.0 - b)
    elif faceIdx == 2: # front
        (x,y,z) = (1.0, a - 1.0, 1.0 - b)
    elif faceIdx == 3: # right
        (x,y,z) = (1.0 - a, 1.0, 1.0 - b)
    elif faceIdx == 4: # top
        (x,y,z) = (b - 1.0, a - 1.0, 1.0)
    elif faceIdx == 5: # bottom
        (x,y,z) = (1.0 - b, a - 1.0, -1.0)

    return (x, y, z)

# convert using an inverse transformation
def convertFace(imgIn, imgOut, faceIdx):
    inSize = imgIn.size
    outSize = imgOut.size
    inPix = imgIn.load()
    outPix = imgOut.load()
    faceSize = outSize[0]

    for xOut in xrange(faceSize):
        for yOut in xrange(faceSize):
            (x,y,z) = outImgToXYZ(xOut, yOut, faceIdx, faceSize)
            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 = 0.5 * inSize[0] * (theta + pi) / pi
            vf = 0.5 * inSize[0] * (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[xOut, yOut] = (int(round(r)), int(round(g)), int(round(b)))

imgIn = Image.open(sys.argv[1])
inSize = imgIn.size
faceSize = inSize[0] / 4
components = sys.argv[1].rsplit('.', 2)

FACE_NAMES = {
  0: 'back',
  1: 'left',
  2: 'front',
  3: 'right',
  4: 'top',
  5: 'bottom'
}

for face in xrange(6):
  imgOut = Image.new("RGB", (faceSize, faceSize), "black")
  convertFace(imgIn, imgOut, face)
  imgOut.save(components[0] + "_" + FACE_NAMES[face] + "." + components[1])
        // convert using an inverse transformation
        function convertFace(imgIn, imgOut, faceIdx) {
            var inPix = shimImgData(imgIn),
                        outPix = shimImgData(imgOut),
                        faceSize = imgOut.width,
                        pi = Math.PI,
                        pi_2 = pi/2;

            for(var xOut=0;xOut<faceSize;xOut++) {
                    for(var yOut=0;yOut<faceSize;yOut++) {

                    var xyz = outImgToXYZ(xOut, yOut, faceIdx, faceSize);
                    var theta = Math.atan2(xyz.y, xyz.x); // range -pi to pi
                    var r = Math.hypot(xyz.x,xyz.y);
                    var phi = Math.atan2(xyz.z,r); // range -pi/2 to pi/2

                    // source img coords
                    var uf = 0.5 * imgIn.width * (theta + pi) / pi;
                    var vf = 0.5 * imgIn.width * (pi_2 - phi) / pi;

                    // Use bilinear interpolation between the four surrounding pixels
                    var ui = Math.floor(uf);  // coord of pixel to bottom left
                    var vi = Math.floor(vf);
                    var u2 = ui+1;       // coords of pixel to top right
                    var v2 = vi+1;
                    var mu = uf-ui;      // fraction of way across pixel
                    var nu = vf-vi;

                    // Pixel values of four corners
                    var A = inPix.getPx(ui % imgIn.width, clip(vi, 0, imgIn.height-1));
                    var B = inPix.getPx(u2 % imgIn.width, clip(vi, 0, imgIn.height-1));
                    var C = inPix.getPx(ui % imgIn.width, clip(v2, 0, imgIn.height-1));
                    var D = inPix.getPx(u2 % imgIn.width, clip(v2, 0, imgIn.height-1));

                    // interpolate
                    var rgb = {
                      r:A[0]*(1-mu)*(1-nu) + B[0]*(mu)*(1-nu) + C[0]*(1-mu)*nu+D[0]*mu*nu,
                      g:A[1]*(1-mu)*(1-nu) + B[1]*(mu)*(1-nu) + C[1]*(1-mu)*nu+D[1]*mu*nu,
                      b:A[2]*(1-mu)*(1-nu) + B[2]*(mu)*(1-nu) + C[2]*(1-mu)*nu+D[2]*mu*nu
                    };

                    rgb.r=Math.round(rgb.r);
                    rgb.g=Math.round(rgb.g);
                    rgb.b=Math.round(rgb.b);

                    outPix.setPx(xOut, yOut, rgb);

                } // for(var yOut=0;yOut<faceSize;yOut++) {...}
             } // for(var xOut=0;xOut<faceSize;xOut++) {...}
        } // function convertFace(imgIn, imgOut, faceIdx) {...}

        // get x,y,z coords from out image pixels coords
        // i,j are pixel coords
        // faceIdx is face number
        // faceSize is edge length
        function outImgToXYZ(i, j, faceIdx, faceSize) {
            var a = 2 * i / faceSize,
                    b = 2 * j / faceSize;

            switch(faceIdx) {
                case 0: // back
                return({x:-1, y:1-a, z:1-b});
            case 1: // left
                return({x:a-1, y:-1, z:1-b});
            case 2: // front
                return({x: 1, y:a-1, z:1-b});
            case 3: // right
                return({x:1-a, y:1, z:1-b});
            case 4: // top
                return({x:b-1, y:a-1, z:1});
            case 5: // bottom
                return({x:1-b, y:a-1, z:-1});

            }
        } // function outImgToXYZ(i, j, faceIdx, faceSize) {...}

        function clip(val, min, max) {
            return(val<min?min:(val>max?max:val));
        }

        function shimImgData(imgData) {
            var w=imgData.width*4,
                    d=imgData.data;

            return({
                getPx:function(x,y) {
                    x=x*4+y*w;
                    return([ d[x], d[x+1], d[x+2] ]);
                },
                setPx:function(x,y,rgb) {
                    x=x*4+y*w;
                    d[x]=rgb.r;
                    d[x+1]=rgb.g;
                    d[x+2]=rgb.b;
                    d[x+3]=255; // alpha
                }
            });
        } // function shimImgData(imgData) {...}
$ pip install convert360
$ convert360 -i ~/Pictures/Barcelona/sagrada-familia.jpg -o example.png -s 300 300