Question

我有一个定义周围环境的立方体贴图纹理，但我需要将它传递给仅适用于纬度/经度贴图的程序。我真的迷失在这里如何翻译。这里有什么帮助吗？

换句话说，我需要来自这里：

enter image description here

对此（我认为图像在x轴上有-90°旋转）：

enter image description here

更新：我得到了预测的正式名称。顺便说一句，我找到了相反的投影here

Answer 1

投影光栅图像的一般步骤如下：

for each pixel of the destination image:
    calculate the corresponding unit vector in 3-dimensional space
    calculate the x,y coordinate for that vector in the source image
    sample the source image at that coordinate and assign the value to the destination pixel

最后一步是插值。我们将重点关注其他两个步骤。

给定纬度和经度的单位矢量是（+ z朝向北极，+ x朝向本初子午线）：

x = cos(lat)*cos(lon)
y = cos(lat)*sin(lon)
z = sin(lat)

假设立方体在原点周围为+/- 1单位（即2x2x2整体大小）。一旦我们得到单位向量，我们就可以通过查看具有最大绝对值的元素来找到它所在的立方体的面。例如，如果我们的单位矢量是<0.2099，-0.7289,0.6516>，那么y元素具有最大的绝对值。它是负面的，所以这一点将在立方体的-y面上找到。通过除以y幅度来标准化其他两个坐标以获得该面部内的位置。因此，该点将在-y面上的x = 0.2879，z = 0.8939。

Answer 2

项目名称更改为libcube2cyl。在C和C ++中，同样的优点，更好的工作示例。

现在也可以在C。

中找到

我碰巧解决了你所描述的完全相同的问题。

我写了这个名为“Cube2Cyl”的小型C ++库，你可以在这里找到算法的详细解释：Cube2Cyl

请从github找到源代码：Cube2Cyl

在MIT许可下发布，免费使用！

Answer 3

我想分享我对此转换的MATLAB实现。我还借用了OpenGL 4.1规范，第3.8.10章（found here）以及Paul Bourke的网站（found here）。请务必查看副标题：转换为6立方环境地图和球形地图。

我也使用Sambatyon的帖子作为灵感。它起初是从Python到MATLAB的端口，但我制作了代码，使其完全矢量化（即没有for循环）。我也拍摄立方图像并将其分成6个单独的图像，因为我正在构建的应用程序具有此格式的立方图像。此外，代码中没有错误检查，并且这假定所有立方图像具有相同的大小（n x n）。这也假设图像是RGB格式。如果您想为单色图像执行此操作，只需注释掉那些需要访问多个通道的代码行。我们走了！

function [out] = cubic2equi(top, bottom, left, right, front, back)

% Height and width of equirectangular image
height = size(top, 1);
width = 2*height;

% Flags to denote what side of the cube we are facing
% Z-axis is coming out towards you
% X-axis is going out to the right
% Y-axis is going upwards
% Assuming that the front of the cube is towards the
% negative X-axis
FACE_Z_POS = 1; % Left
FACE_Z_NEG = 2; % Right
FACE_Y_POS = 3; % Top
FACE_Y_NEG = 4; % Bottom
FACE_X_NEG = 5; % Front 
FACE_X_POS = 6; % Back

% Place in a cell array
stackedImages{FACE_Z_POS} = left;
stackedImages{FACE_Z_NEG} = right;
stackedImages{FACE_Y_POS} = top;
stackedImages{FACE_Y_NEG} = bottom;
stackedImages{FACE_X_NEG} = front;
stackedImages{FACE_X_POS} = back;

% Place in 3 3D matrices - Each matrix corresponds to a colour channel
imagesRed = uint8(zeros(height, height, 6));
imagesGreen = uint8(zeros(height, height, 6));
imagesBlue = uint8(zeros(height, height, 6));

% Place each channel into their corresponding matrices
for i = 1 : 6
    im = stackedImages{i};
    imagesRed(:,:,i) = im(:,:,1);
    imagesGreen(:,:,i) = im(:,:,2);
    imagesBlue(:,:,i) = im(:,:,3);
end

% For each co-ordinate in the normalized image...
[X, Y] = meshgrid(1:width, 1:height);

% Obtain the spherical co-ordinates
Y = 2*Y/height - 1;
X = 2*X/width - 1;
sphereTheta = X*pi;
spherePhi = (pi/2)*Y;

texX = cos(spherePhi).*cos(sphereTheta);
texY = sin(spherePhi);
texZ = cos(spherePhi).*sin(sphereTheta);

% Figure out which face we are facing for each co-ordinate
% First figure out the greatest absolute magnitude for each point
comp = cat(3, texX, texY, texZ);
[~,ind] = max(abs(comp), [], 3);
maxVal = zeros(size(ind));
% Copy those values - signs and all
maxVal(ind == 1) = texX(ind == 1);
maxVal(ind == 2) = texY(ind == 2);
maxVal(ind == 3) = texZ(ind == 3);

% Set each location in our equirectangular image, figure out which
% side we are facing
getFace = -1*ones(size(maxVal));

% Back
ind = abs(maxVal - texX) < 0.00001 & texX < 0;
getFace(ind) = FACE_X_POS;

% Front
ind = abs(maxVal - texX) < 0.00001 & texX >= 0;
getFace(ind) = FACE_X_NEG;

% Top
ind = abs(maxVal - texY) < 0.00001 & texY < 0;
getFace(ind) = FACE_Y_POS;

% Bottom
ind = abs(maxVal - texY) < 0.00001 & texY >= 0;
getFace(ind) = FACE_Y_NEG;

% Left
ind = abs(maxVal - texZ) < 0.00001 & texZ < 0;
getFace(ind) = FACE_Z_POS;

% Right
ind = abs(maxVal - texZ) < 0.00001 & texZ >= 0;
getFace(ind) = FACE_Z_NEG;

% Determine the co-ordinates along which image to sample
% based on which side we are facing
rawX = -1*ones(size(maxVal));
rawY = rawX;
rawZ = rawX;

% Back
ind = getFace == FACE_X_POS;
rawX(ind) = -texZ(ind);
rawY(ind) = texY(ind);
rawZ(ind) = texX(ind);

% Front
ind = getFace == FACE_X_NEG;
rawX(ind) = texZ(ind);
rawY(ind) = texY(ind);
rawZ(ind) = texX(ind);

% Top
ind = getFace == FACE_Y_POS;
rawX(ind) = texZ(ind);
rawY(ind) = texX(ind);
rawZ(ind) = texY(ind);

% Bottom
ind = getFace == FACE_Y_NEG;
rawX(ind) = texZ(ind);
rawY(ind) = -texX(ind);
rawZ(ind) = texY(ind);

% Left
ind = getFace == FACE_Z_POS;
rawX(ind) = texX(ind);
rawY(ind) = texY(ind);
rawZ(ind) = texZ(ind);

% Right
ind = getFace == FACE_Z_NEG;
rawX(ind) = -texX(ind);
rawY(ind) = texY(ind);
rawZ(ind) = texZ(ind);

% Concatenate all for later
rawCoords = cat(3, rawX, rawY, rawZ);

% Finally determine co-ordinates (normalized)
cubeCoordsX = ((rawCoords(:,:,1) ./ abs(rawCoords(:,:,3))) + 1) / 2;
cubeCoordsY = ((rawCoords(:,:,2) ./ abs(rawCoords(:,:,3))) + 1) / 2;
cubeCoords = cat(3, cubeCoordsX, cubeCoordsY);

% Now obtain where we need to sample the image
normalizedX = round(cubeCoords(:,:,1) * height);
normalizedY = round(cubeCoords(:,:,2) * height);

% Just in case.... cap between [1, height] to ensure
% no out of bounds behaviour
normalizedX(normalizedX < 1) = 1;
normalizedX(normalizedX > height) = height;
normalizedY(normalizedY < 1) = 1;
normalizedY(normalizedY > height) = height;

% Place into a stacked matrix
normalizedCoords = cat(3, normalizedX, normalizedY);

% Output image allocation
out = uint8(zeros([size(maxVal) 3]));

% Obtain column-major indices on where to sample from the
% input images
% getFace will contain which image we need to sample from
% based on the co-ordinates within the equirectangular image
ind = sub2ind([height height 6], normalizedCoords(:,:,2), ...
    normalizedCoords(:,:,1), getFace);

% Do this for each channel
out(:,:,1) = imagesRed(ind);
out(:,:,2) = imagesGreen(ind);
out(:,:,3) = imagesBlue(ind);

我还通过github和you can go here for it公开提供了代码。其中包括主转换脚本，显示其用途的测试脚本以及从Paul Bourke网站上提取的6个立方图像样本集。我希望这很有用！

Answer 4

所以，我找到了一个解决方案，混合this article来自维基百科的球面坐标，以及第3.8.10节来自OpenGL 4.1规范（加上一些黑客使其工作）。因此，假设立方图像的高度为h_o，宽度为w_o，则此要素将具有高度h = w_o / 3和宽度w = 2 * h。现在对于equirectangular投影中的每个像素(x, y) 0 <= x <= w, 0 <= y <= h，我们想在立方投影中找到相应的像素，我在python中使用以下代码解决它（我希望我从C中翻译它时没有犯错）

import math

# from wikipedia
def spherical_coordinates(x, y):
    return (math.pi*((y/h) - 0.5), 2*math.pi*x/(2*h), 1.0)

# from wikipedia
def texture_coordinates(theta, phi, rho):
    return (rho * math.sin(theta) * math.cos(phi),
            rho * math.sin(theta) * math.sin(phi),
            rho * math.cos(theta))

FACE_X_POS = 0
FACE_X_NEG = 1
FACE_Y_POS = 2
FACE_Y_NEG = 3
FACE_Z_POS = 4
FACE_Z_NEG = 5

# from opengl specification
def get_face(x, y, z):
    largest_magnitude = max(x, y, z)
    if largest_magnitude - abs(x) < 0.00001:
        return FACE_X_POS if x < 0 else FACE_X_NEG
    elif largest_magnitude - abs(y) < 0.00001:
        return FACE_Y_POS if y < 0 else FACE_Y_NEG
    elif largest_magnitude - abs(z) < 0.00001:
        return FACE_Z_POS if z < 0 else FACE_Z_NEG

# from opengl specification
def raw_face_coordinates(face, x, y, z):
    if face == FACE_X_POS:
        return (-z, -y, x)
    elif face == FACE_X_NEG:
        return (-z, y, -x)
    elif face == FACE_Y_POS:
        return (-x, -z, -y)
    elif face == FACE_Y_NEG:
        return (-x, z, -y)
    elif face == FACE_Z_POS:
        return (-x, y, -z)
    elif face == FACE_Z_NEG:
        return (-x, -y, z)

# computes the topmost leftmost coordinate of the face in the cube map
def face_origin_coordinates(face):
    if face == FACE_X_POS:
        return (2*h, h)
    elif face == FACE_X_NEG:
        return (0, 2*h)
    elif face == FACE_Y_POS:
        return (h, h)
    elif face == FACE_Y_NEG:
        return (h, 3*h)
    elif face == FACE_Z_POS:
        return (h, 0)
    elif face == FACE_Z_NEG:
        return (h, 2*h)

# from opengl specification
def raw_coordinates(xc, yc, ma):
    return ((xc/abs(ma) + 1) / 2, (yc/abs(ma) + 1) / 2)


def normalized_coordinates(face, x, y):
    face_coords = face_origin_coordinates(face)
    normalized_x = int(math.floor(x * h + 0.5))
    normalized_y = int(math.floor(y * h + 0.5))
    # eliminates black pixels
    if normalized_x == h:
      --normalized_x
    if normalized_y == h:
      --normalized_y
    return (face_coords[0] + normalized_x, face_coords[1] + normalized_y)

def find_corresponding_pixel(x, y):
    spherical = spherical_coordinates(x, y)
    texture_coords = texture_coordinates(spherical[0], spherical[1], spherical[2])
    face = get_face(texture_coords[0], texture_coords[1], texture_coords[2])

    raw_face_coords = raw_face_coordinates(face, texture_coords[0], texture_coords[1], texture_coords[2])
    cube_coords = raw_coordinates(raw_face_coords[0], raw_face_coords[1], raw_face_coords[2])
    # this fixes some faces being rotated 90°
    if face in [FACE_X_NEG, FACE_X_POS]:
      cube_coords = (cube_coords[1], cube_coords[0])
    return normalized_coordinates(face, cube_coords[0], cube_coords[1])

最后我们只是为equirectangular投影中的每个像素调用find_corresponding_pixel

Answer 5

我认为从你的Python算法中你可能在theta和phi的计算中反转了x和y。

def spherical_coordinates(x, y):
    return (math.pi*((y/h) - 0.5), 2*math.pi*x/(2*h), 1.0)

来自Paul Bourke的网站here

theta = x pi phi = y pi / 2

在您的代码中，您在theta计算中使用y，在phi计算中使用x。

如果我错了，请纠正我。

立方到equirectangular投影算法

5 个答案: