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
最后一步就是插值。我们将着重讨论其他两个步骤。
给定纬度和经度的单位向量为（+z指向北极，+x指向本初子午线）：
x = cos(lat)*cos(lon)
y = cos(lat)*sin(lon)
z = sin(lat)
for循环）。我还将立方体图像分成了6个单独的图像，因为我正在构建的应用程序中立方体图像是以这种格式呈现的。此外，代码中没有错误检查，并且假定所有立方体图像都具有相同的大小（n x n）。这也假设图像是以RGB格式呈现的。如果您想为单色图像执行此操作，只需注释掉需要访问多个通道的那些代码行即可。我们开始吧！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公开了代码，您可以在这里找到它。包括主要的转换脚本、一个用来展示其使用的测试脚本以及从Paul Bourke的网站上获取的一组6个立方图像样本。希望这对您有用！
所以，我找到了一个解决方案，混合使用维基百科上的球面坐标this article和OpenGL 4.1规范的第3.8.10节（加上一些技巧使其正常工作）。因此，假设立方体图像的高度为h_o，宽度为w_o，则等距投影将具有高度h = w_o / 3和宽度w = 2 * h。现在，对于等距投影中的每个像素(x, y) 0 <= x <= w, 0 <= y <= h，我们想要找到相应的立方体投影像素，我使用以下Python代码解决了这个问题（希望我在从C语言翻译时没有犯错）。
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])
find_corresponding_pixel函数。我认为从你在Python中的算法来看，你可能在计算theta和phi时颠倒了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的网站这里
theta = x pi phi = y pi / 2
在你的代码中，你在theta计算中使用了y，在phi计算中使用了x。
如果我错了，请纠正我。