OpenGL 是否将屏幕定位在相机前面?
Does OpenGL position the screen in front of the camera?
我正在尝试将鼠标点击转换为全局位置。尽管世界空间是 3d,但我目前正在尝试在 z = 0 处找到点击的 x 和 y 坐标。
我写了一些转换代码,几乎是正确的。但是似乎有一个错误随着与(0,0)的距离而增长。它似乎增长得太快而不是一个舍入误差,所以我认为屏幕没有设置到准确的相机位置。
对于此测试,我主要关注 x 坐标。
变量
screenWidth = 512;
screenHeight = 512;
cameraPos = glm::vec3(0.0f, 0.0f, 4.0f);
转换逻辑
std::cout << "mouse float x: " << inMouseX << ", mouse float y: " << inMouseY << std::endl;
float mouseX = (inMouseX / ((float)screenWidth * 0.5f)) - 1.0f;
float mouseY = (inMouseY / ((float)screenHeight * 0.5f)) - 1.0f;
std::cout << "mouse adjusted screen width x: " << (float)screenWidth * 0.5f << ", mouse adjusted screen height y: " << (float)screenHeight * 0.5f << std::endl;
std::cout << "mouse 2nd adjusted x: " << ((inMouseX / (float)screenWidth * 0.5f)) << ", mouse 2nd adjusted y: " << ((inMouseY / (float)screenHeight * 0.5f)) << std::endl;
std::cout << "mouse converted x: " << mouseX << ", mouse converted y: " << mouseY << std::endl;
glm::mat4 invVP = glm::inverse(*projection * *view);
glm::vec4 mouseClickVector = invVP * glm::vec4(mouseX, -mouseY, 1.0f, 1.0f);
glm::vec3 norm = glm::normalize(glm::vec3(mouseClickVector));
float t = (0 - (*cameraPos)[2]) / norm[2];
glm::vec3 trueWorldPos = *cameraPos + (norm * t);
return trueWorldPos;
回复示例
------------case 1------------
expected:
mouse world position x: 0
actual:
mouse float x: 256, mouse float y: 255
mouse adjusted screen width x: 256, mouse adjusted screen height y: 256
mouse 2nd adjusted x: 0.25, mouse 2nd adjusted y: 0.249023
mouse converted x: 0, mouse converted y: -0.00390625
mouse world position x: 0, mouse world y: 0.0162761
------------case 2------------
expected:
mouse world position x: 5
actual:
mouse float x: 189, mouse float y: 251
mouse adjusted screen width x: 256, mouse adjusted screen height y: 256
mouse 2nd adjusted x: 0.18457, mouse 2nd adjusted y: 0.245117
mouse converted x: -0.261719, mouse converted y: -0.0195313
mouse world position x: 5.23121, mouse world y: 0.0813803
------------case 3------------
expected:
mouse world position x: 7
actual:
mouse float x: 82, mouse float y: 254
mouse adjusted screen width x: 256, mouse adjusted screen height y: 256
mouse 2nd adjusted x: 0.0800781, mouse 2nd adjusted y: 0.248047
mouse converted x: -0.679688, mouse converted y: -0.0078125
mouse world position x: 7.25648, mouse world y: 0.0325521
由于我对鼠标的使用不精确,可能会有一些小的舍入问题,我预计会有一些波动,但它比我预期的要大得多。
假设视图矩阵有一些标准约定,这没有意义:
glm::vec4 mouseClickVector = invVP * glm::vec4(mouseX, -mouseY, 1.0f, 1.0f);
glm::vec3 norm = glm::normalize(glm::vec3(mouseClickVector));
通常,视图矩阵包含相机的方向和位置。您似乎将 mouseClickVector
视为只是一个方向。但是,如果您的相机不在世界 space 中心,则此规范化将产生完全错误的结果。
要获得未投影点的正确世界 space 位置,您不能再忽略 w 分量:
glm::vec3 posWorld = (1.0f / mouseClickVector.w) * glm::vec3(mouseClickVector)
这只会在背板上产生未投影的点(当您将 NDC z 设置为 1.0 时)。您仍然需要计算从相机位置到这一点的视线,以便您可以计算与某个世界 space 平面的交点。
我正在尝试将鼠标点击转换为全局位置。尽管世界空间是 3d,但我目前正在尝试在 z = 0 处找到点击的 x 和 y 坐标。
我写了一些转换代码,几乎是正确的。但是似乎有一个错误随着与(0,0)的距离而增长。它似乎增长得太快而不是一个舍入误差,所以我认为屏幕没有设置到准确的相机位置。
对于此测试,我主要关注 x 坐标。
变量
screenWidth = 512;
screenHeight = 512;
cameraPos = glm::vec3(0.0f, 0.0f, 4.0f);
转换逻辑
std::cout << "mouse float x: " << inMouseX << ", mouse float y: " << inMouseY << std::endl;
float mouseX = (inMouseX / ((float)screenWidth * 0.5f)) - 1.0f;
float mouseY = (inMouseY / ((float)screenHeight * 0.5f)) - 1.0f;
std::cout << "mouse adjusted screen width x: " << (float)screenWidth * 0.5f << ", mouse adjusted screen height y: " << (float)screenHeight * 0.5f << std::endl;
std::cout << "mouse 2nd adjusted x: " << ((inMouseX / (float)screenWidth * 0.5f)) << ", mouse 2nd adjusted y: " << ((inMouseY / (float)screenHeight * 0.5f)) << std::endl;
std::cout << "mouse converted x: " << mouseX << ", mouse converted y: " << mouseY << std::endl;
glm::mat4 invVP = glm::inverse(*projection * *view);
glm::vec4 mouseClickVector = invVP * glm::vec4(mouseX, -mouseY, 1.0f, 1.0f);
glm::vec3 norm = glm::normalize(glm::vec3(mouseClickVector));
float t = (0 - (*cameraPos)[2]) / norm[2];
glm::vec3 trueWorldPos = *cameraPos + (norm * t);
return trueWorldPos;
回复示例
------------case 1------------
expected:
mouse world position x: 0
actual:
mouse float x: 256, mouse float y: 255
mouse adjusted screen width x: 256, mouse adjusted screen height y: 256
mouse 2nd adjusted x: 0.25, mouse 2nd adjusted y: 0.249023
mouse converted x: 0, mouse converted y: -0.00390625
mouse world position x: 0, mouse world y: 0.0162761
------------case 2------------
expected:
mouse world position x: 5
actual:
mouse float x: 189, mouse float y: 251
mouse adjusted screen width x: 256, mouse adjusted screen height y: 256
mouse 2nd adjusted x: 0.18457, mouse 2nd adjusted y: 0.245117
mouse converted x: -0.261719, mouse converted y: -0.0195313
mouse world position x: 5.23121, mouse world y: 0.0813803
------------case 3------------
expected:
mouse world position x: 7
actual:
mouse float x: 82, mouse float y: 254
mouse adjusted screen width x: 256, mouse adjusted screen height y: 256
mouse 2nd adjusted x: 0.0800781, mouse 2nd adjusted y: 0.248047
mouse converted x: -0.679688, mouse converted y: -0.0078125
mouse world position x: 7.25648, mouse world y: 0.0325521
由于我对鼠标的使用不精确,可能会有一些小的舍入问题,我预计会有一些波动,但它比我预期的要大得多。
假设视图矩阵有一些标准约定,这没有意义:
glm::vec4 mouseClickVector = invVP * glm::vec4(mouseX, -mouseY, 1.0f, 1.0f); glm::vec3 norm = glm::normalize(glm::vec3(mouseClickVector));
通常,视图矩阵包含相机的方向和位置。您似乎将 mouseClickVector
视为只是一个方向。但是,如果您的相机不在世界 space 中心,则此规范化将产生完全错误的结果。
要获得未投影点的正确世界 space 位置,您不能再忽略 w 分量:
glm::vec3 posWorld = (1.0f / mouseClickVector.w) * glm::vec3(mouseClickVector)
这只会在背板上产生未投影的点(当您将 NDC z 设置为 1.0 时)。您仍然需要计算从相机位置到这一点的视线,以便您可以计算与某个世界 space 平面的交点。