python - 如何找到 OpenGL 顶点的 PyGame 窗口坐标？

Question

我试图找出使用 OpenGL 创建 3D 对象的 pygame 窗口中两个矩形顶点的坐标。

import pygame
from pygame.locals import *
import random
from OpenGL.GL import *
from OpenGL.GLU import *

rect1 = [(-5.125,0,-40),(-3.125,0,-40),(-3.125,5,-40),(-5.125,5,-40),]
rect2 = [(3.125,0,-40),(5.125,0,-40),(5.125,5,-40),(3.125,5,-40)]
edges = ((0,1),(1,2),(2,3),(3,0))

#This draws the rectangles edges
def Target():
    glBegin(GL_LINES)
    for edge in edges:
        for vertex in edge:
            glVertex3fv(rect1[vertex])
    glEnd()
    glBegin(GL_LINES)
    for edge in edges:
        for vertex in edge:
            glVertex3fv(rect2[vertex])
    glEnd()

def main():
    try:
        pygame.init()
        display = (320,240)
        pygame.display.set_mode(display, DOUBLEBUF|OPENGL)
        gluPerspective(45, (display[0]/display[1]), .1, 1000)

        while True:
            #iterates through events to check for quits
            for event in pygame.event.get():
                if event.type == pygame.QUIT:
                    pygame.quit()
                    quit()        
            Target()
            pygame.display.flip()
            pygame.time.wait(100)
            glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT)         
    except Exception as e:
        print (e)
main()

如何获取对象的 pygame 窗口(320,240)上的坐标？

Answer 1

投影矩阵描述了从场景的 3D 点到视口(viewport)的 2D 点的映射。它从眼睛空间转换到剪辑空间，并且剪辑空间中的坐标通过除以剪辑坐标的 w 分量而转换为标准化设备坐标 (NDC)。 NDC 的范围为 (-1,-1,-1) 到 (1,1,1)。

在透视投影中，投影矩阵描述了从针孔相机看到的世界上的 3D 点到视口(viewport)的 2D 点的映射。
相机视锥体(截棱锥体)中的眼睛空间坐标被映射到立方体(标准化设备坐标)。

透视投影矩阵:

r = right, l = left, b = bottom, t = top, n = near, f = far

2*n/(r-l)      0              0                0
0              2*n/(t-b)      0                0
(r+l)/(r-l)    (t+b)/(t-b)    -(f+n)/(f-n)    -1    
0              0              -2*f*n/(f-n)     0

哪里:

aspect = w / h
tanFov = tan( fov_y / 2 );

2 * n / (r-l) = 1 / (tanFov * aspect)
2 * n / (t-b) = 1 / tanFov

由于投影矩阵是由视场和纵横比定义的，因此可以通过视场和纵横比恢复视口(viewport)位置。前提是它是一个对称的透视投影，其中视野不会移位(如您的情况)。

首先，您必须将鼠标位置转换为标准化设备坐标:

w = with of the viewport
h = height of the viewport
x = X position of the mouse
y = Y position ot the mouse

ndc_x = 2.0 * x/w - 1.0;
ndc_y = 1.0 - 2.0 * y/h; // invert Y axis

然后你必须将标准化的设备坐标转换为 View 坐标:

z = z coodinate of the geometry in view space

viewPos.x = -z * ndc_x * aspect * tanFov;
viewPos.y = -z * ndc_y * tanFov;  

如果您想检查鼠标是否悬停在矩形上，则代码可能如下所示:

mpos = pygame.mouse.get_pos()
z = 40

ndc = [ 2.0 * mpos[0]/width - 1.0, 1.0 - 2.0 * mpos[1]/height ]
tanFov = math.tan( fov_y * 0.5 * math.pi / 180 )
aspect = width / height 
viewPos = [z * ndc[0] * aspect * tanFov, z * ndc[1] * tanFov ]

onRect1 = 1 if (viewPos[0]>=rect1[0][0] and viewPos[0]<=rect1[1][0] and viewPos[1]>=rect1[0][1] and viewPos[1]<=rect1[2][1] ) else 0
onRect2 = 1 if (viewPos[0]>=rect2[0][0] and viewPos[0]<=rect2[1][0] and viewPos[1]>=rect2[0][1] and viewPos[1]<=rect2[2][1] ) else 0

进一步查看:

• How to recover view space position given view space depth value and ndc xy
• Is it possble get which surface of cube will be click in OpenGL?
• OpenGL - Mouse coordinates to Space coordinates

下面我将算法添加到您的示例中。如果鼠标悬停在矩形上，则该矩形将显示为红色。

import pygame
from pygame.locals import *
import random
from OpenGL.GL import *
from OpenGL.GLU import *
import math

rect1 = [(-5.125,0,-40),(-3.125,0,-40),(-3.125,5,-40),(-5.125,5,-40),]
rect2 = [(3.125,0,-40),(5.125,0,-40),(5.125,5,-40),(3.125,5,-40)]
edges = ((0,1),(1,2),(2,3),(3,0))

fov_y = 45
width = 320
height = 200

#This draws the rectangles edges
def Target():
    mpos = pygame.mouse.get_pos()
    z = 40

    ndc = [ 2.0 * mpos[0]/width - 1.0, 1.0 - 2.0 * mpos[1]/height ]
    tanFov = math.tan( fov_y * 0.5 * math.pi / 180 )
    aspect = width / height 
    viewPos = [z * ndc[0] * aspect * tanFov, z * ndc[1] * tanFov ]

    onRect1 = 1 if (viewPos[0]>=rect1[0][0] and viewPos[0]<=rect1[1][0] and viewPos[1]>=rect1[0][1] and viewPos[1]<=rect1[2][1] ) else 0
    onRect2 = 1 if (viewPos[0]>=rect2[0][0] and viewPos[0]<=rect2[1][0] and viewPos[1]>=rect2[0][1] and viewPos[1]<=rect2[2][1] ) else 0

    glColor3f( 1, 1-onRect1, 1-onRect1 )
    glBegin(GL_LINES)
    for edge in edges:
        for vertex in edge:
            glVertex3fv(rect1[vertex])
    glEnd()

    glColor3f( 1, 1-onRect2, 1-onRect2 )
    glBegin(GL_LINES)
    for edge in edges:
        for vertex in edge:
            glVertex3fv(rect2[vertex])
    glEnd()

def main():
    try:
        pygame.init()
        display = (width,height)
        pygame.display.set_mode(display, DOUBLEBUF|OPENGL)
        glMatrixMode(GL_PROJECTION)
        gluPerspective(fov_y, (display[0]/display[1]), .1, 1000)
        glMatrixMode(GL_MODELVIEW)

        while True:
            #iterates through events to check for quits
            for event in pygame.event.get():
                if event.type == pygame.QUIT:
                    pygame.quit()
                    quit()        
            Target()
            pygame.display.flip()
            pygame.time.wait(100)
            glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT)         
    except Exception as e:
        print (e)
main()

答案的扩展

当然你也可以反过来做。您可以将矩形的角点转换为标准化设备坐标，并将它们与标准化设备坐标中的鼠标位置进行比较。

为此，您必须通过 glGetFloatv(GL_PROJECTION_MATRIX) 读取投影矩阵:

prjMat = (GLfloat * 16)()
glGetFloatv(GL_PROJECTION_MATRIX, prjMat)

并且您需要一个通过投影矩阵转换 3 维笛卡尔向量的函数。这是通过将向量乘以投影矩阵来完成的，这给出了齐次剪辑空间坐标。标准化设备坐标是通过将 x、y 和 z 分量除以 w 分量来计算的。

def TransformVec3(vecA,mat44):
    vecB = [0, 0, 0, 0]
    for i0 in range(0, 4):
        vecB[i0] = vecA[0] * mat44[0*4+i0] + vecA[1] * mat44[1*4+i0] + vecA[2] * mat44[2*4+i0] + mat44[3*4+i0]
    return [vecB[0]/vecB[3], vecB[1]/vecB[3], vecB[2]/vecB[3]]

以下函数测试鼠标位置是否位于由左下角和右上角点定义的矩形中(角点必须位于 View 空间坐标中):

def TestRec(prjMat, mpos, ll, tr):
    ll_ndc = TransformVec3(ll, prjMat)
    tr_ndc = TransformVec3(tr, prjMat)
    ndc = [ 2.0 * mpos[0]/width - 1.0, 1.0 - 2.0 * mpos[1]/height ]
    inRect = 1 if (ndc[0]>=ll_ndc[0] and ndc[0]<=tr_ndc[0] and ndc[1]>=ll_ndc[1] and ndc[1]<=tr_ndc[1] ) else 0
    return inRect

我再次将算法添加到您的示例中。如果鼠标悬停在矩形上，则该矩形将显示为红色。

import pygame
from pygame.locals import *
import random
from OpenGL.GL import *
from OpenGL.GLU import *
import math

rect1 = [(-5.125,0,-40),(-3.125,0,-40),(-3.125,5,-40),(-5.125,5,-40),]
rect2 = [(3.125,0,-40),(5.125,0,-40),(5.125,5,-40),(3.125,5,-40)]
edges = ((0,1),(1,2),(2,3),(3,0))

fov_y = 45
width = 320
height = 200

def TransformVec3(vecA,mat44):
    vecB = [0, 0, 0, 0]
    for i0 in range(0, 4):
        vecB[i0] = vecA[0] * mat44[0*4+i0] + vecA[1] * mat44[1*4+i0] + vecA[2] * mat44[2*4+i0] + mat44[3*4+i0]
    return [vecB[0]/vecB[3], vecB[1]/vecB[3], vecB[2]/vecB[3]]

def TestRec(prjMat, mpos, ll, tr):
    ll_ndc = TransformVec3(ll, prjMat)
    tr_ndc = TransformVec3(tr, prjMat)
    ndc = [ 2.0 * mpos[0]/width - 1.0, 1.0 - 2.0 * mpos[1]/height ]
    inRect = 1 if (ndc[0]>=ll_ndc[0] and ndc[0]<=tr_ndc[0] and ndc[1]>=ll_ndc[1] and ndc[1]<=tr_ndc[1] ) else 0
    return inRect


#This draws the rectangles edges
def Target():

    prjMat = (GLfloat * 16)()
    glGetFloatv(GL_PROJECTION_MATRIX, prjMat)

    mpos = pygame.mouse.get_pos()
    onRect1 = TestRec(prjMat, mpos, rect1[0], rect1[2])
    onRect2 = TestRec(prjMat, mpos, rect2[0], rect2[2])

    glColor3f( 1, 1-onRect1, 1-onRect1 )
    glBegin(GL_LINES)
    for edge in edges:
        for vertex in edge:
            glVertex3fv(rect1[vertex])
    glEnd()

    glColor3f( 1, 1-onRect2, 1-onRect2 )
    glBegin(GL_LINES)
    for edge in edges:
        for vertex in edge:
            glVertex3fv(rect2[vertex])
    glEnd()

def main():
    try:
        pygame.init()
        display = (width,height)
        pygame.display.set_mode(display, DOUBLEBUF|OPENGL)
        glMatrixMode(GL_PROJECTION)
        gluPerspective(fov_y, (display[0]/display[1]), .1, 1000)
        glMatrixMode(GL_MODELVIEW)

        while True:
            #iterates through events to check for quits
            for event in pygame.event.get():
                if event.type == pygame.QUIT:
                    pygame.quit()
                    quit()        
            Target()
            pygame.display.flip()
            pygame.time.wait(100)
            glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT)         
    except Exception as e:
        print (e)
main()