如何正確旋轉矢量以匹配另一個？ （OPENGL）

Question

我一直在努力解決以下問題：我有一個程序允許用戶繪制各種長度的Bezier曲線（第一個為4個點，所有其他曲線為3個，只是一個接一個地連接曲線）。 我需要在曲線上放置小的矩形火車軌道，讓用戶騎在他們制作的軌道上。 我已經定義了頂點，並制作了一種方法將它們正確放置在線上，但旋轉已被證明是棘手的。 對於平滑的曲線，我當前的實現工作正常，但尖銳的角落導致軌道不再對齊，並且經過該角落的所有軌道具有完全相同的旋轉，完全打破它。 所有相關代碼如下：

使用點填充曲線std :: vector的代碼：

p0 = glm::vec2(pointVertexData.at(0), pointVertexData.at(1));
p1 = glm::vec2(pointVertexData.at(3), pointVertexData.at(4));
p2 = glm::vec2(pointVertexData.at(6), pointVertexData.at(7));
p3 = glm::vec2(pointVertexData.at(9), pointVertexData.at(10));
curveVertexData = Subdivide(0.0f, 1.0f, 0.05f, curveVertexData);
for (int i = 0; i < timesToLoop; i++)
{
    p0 = p3;
    p1 = glm::vec2(pointVertexData.at(n), pointVertexData.at(n+1));
    p2 = glm::vec2(pointVertexData.at(n+3), pointVertexData.at(n+4));
    p3 = glm::vec2(pointVertexData.at(n+6), pointVertexData.at(n+7));
    std::vector<GLfloat> tempVec = Subdivide(0.0f, 1.0f, 0.05f, tempVec);
    curveVertexData.insert(curveVertexData.end(), tempVec.begin()+3, tempVec.end());
}

細分代碼：

std::vector<GLfloat> Subdivide(GLfloat u0, GLfloat u1, GLfloat maxLineLength, std::vector<GLfloat> recurVertices)
{
    GLfloat umid = (u0 + u1) / 2.0;
    glm::vec2 x0 = Interpolate(p0, p1, p2, p3, u0, pFinal);
    glm::vec2 x1 = Interpolate(p0, p1, p2, p3, u1, pFinal);
    GLfloat length = sqrt(pow((x1.x - x0.x), 2) + pow((x1.y - x0.y), 2));
    if (length > maxLineLength)
    {
        std::vector<GLfloat> firstVertices = Subdivide(u0, umid, maxLineLength, firstVertices);
        std::vector<GLfloat> secondVertices = Subdivide(umid, u1, maxLineLength, secondVertices);
        secondVertices.insert(secondVertices.begin(), firstVertices.begin(), firstVertices.end()-3);
        recurVertices = secondVertices;
        return recurVertices;
    }
    else
    {
        recurVertices.push_back(x0.x);
        recurVertices.push_back(x0.y);
        recurVertices.push_back(0.1f);
        recurVertices.push_back(x1.x);
        recurVertices.push_back(x1.y);
        recurVertices.push_back(0.1f);
        numberOfVertices += 6;
        return recurVertices;
    }
}

使用軌道頂點設置std :: vector的代碼：

    std::vector<GLfloat> tempVertices;
    numberOfTrackVertices = 0;
    for (int i = 0; i < curveVertexData.size() - 2; i+=3)
    {
        std::cout << "Now calculating point # " << i << " : ";
        if(i != 0 && i < curveVertexData.size() - 5)
            shiftVertices(trainVertices, curveVertexData[i], curveVertexData[i + 1], curveVertexData[i + 2], curveVertexData[i + 3], curveVertexData[i + 4], curveVertexData[i + 5], curveVertexData[i - 3], curveVertexData[i - 2], curveVertexData[i - 1], &tempVertices);
        else if (i == 0)
            shiftVertices(trainVertices, curveVertexData[i], curveVertexData[i + 1], curveVertexData[i + 2], curveVertexData[i + 3], curveVertexData[i + 4], curveVertexData[i + 5], curveVertexData[i], curveVertexData[i + 1], curveVertexData[i + 2], &tempVertices);
        else
            shiftVertices(trainVertices, curveVertexData[i], curveVertexData[i + 1], curveVertexData[i + 2], curveVertexData[i], curveVertexData[i + 1], curveVertexData[i + 2], curveVertexData[i - 3], curveVertexData[i - 2], curveVertexData[i - 1], &tempVertices);
    }

最后，我認為代碼最可能是罪魁禍首，代碼用於改變軌道的旋轉。 我當前的算法如下:(注意，“currentOrientation”被設置為等於彼此相減的頂點的前兩個元素是因為它們代表矩形的后下角，當相互減去時，給出一個向量，表示方框的方向）

    void shiftVertices(GLfloat inVertices[], GLfloat x, GLfloat y, GLfloat z, GLfloat rx, GLfloat ry, GLfloat rz, GLfloat qx, GLfloat qy, GLfloat qz, std::vector<GLfloat> *container)
{
    glm::vec3 tempVectors[36];
    glm::vec3 moveVector = glm::vec3(x, y, z);
    glm::vec3 rotateVector = glm::normalize(glm::vec3(rx - qx, ry - qy, rz - qz));
    rotateVector = glm::normalize(glm::cross(rotateVector, UP));
    bool unFilled = true;
    int i = 0;
    int n = 0;
    while(unFilled)
    {
        tempVectors[n].x = inVertices[i];
        i++;
        tempVectors[n].y = inVertices[i];
        i++;
        tempVectors[n].z = inVertices[i];
        i++;
        n++;
        if (n >= 36)
            unFilled = false;
    }
    glm::vec3 currentOrientation = glm::normalize(tempVectors[0] - tempVectors[1]);
    GLfloat angleToRotate = glm::acos(glm::dot(currentOrientation, rotateVector));
    angleToRotate = (180.0f * angleToRotate) / PI;
    std::cout << angleToRotate << "\n";
    glm::mat4 rotationMatrix;
    rotationMatrix = glm::rotate(rotationMatrix, angleToRotate, UP);
    for (int u = 0; u < 36; u++)
    {
        tempVectors[u] = glm::vec3(rotationMatrix * glm::vec4(tempVectors[u], 1.0));
        tempVectors[u] = tempVectors[u] + moveVector;
    }
    i = 0;
    n = 0;
    unFilled = true;
    while (unFilled)
    {
        container->push_back(tempVectors[n].x);
        container->push_back(tempVectors[n].y);
        container->push_back(tempVectors[n].z);
        numberOfTrackVertices++;
        n++;
        if (n >= 36)
            unFilled = false;
    }
}

這個實現給出了以下結果： http ： //imgur.com/a/8OI2E （除了最后一張圖片以外的所有圖片。抱歉，它不會讓我嵌入圖片）

我為此尋找了很多資源，收效甚微。 一個實現是Jur van den Berg對這個問題的回答： https ： //math.stackexchange.com/questions/180418/calculate-rotation-matrix-to-align-vector-a-to-vector-b-in-3d注意我將其稱為圖像中的“偏斜對稱方法”。 我的算法實現如下，並且如前所述，它不能正常工作:(注意，這段代碼替換了最后一個樣本中的代碼的中間部分，其中循環在保持相同之前和之后）

        glm::vec3 crossVector = glm::cross(currentOrientation, rotateVector);
        GLfloat sineAngle = crossVector.length();
        GLfloat cosAngle = glm::dot(currentOrientation, rotateVector);
        glm::mat3 experimentalRMatrix;
        glm::mat3 skewSymmetric = { 0, (-1.0f * crossVector.z), crossVector.y,
            crossVector.z, 0, (-1.0f *crossVector.x),
            (-1.0f * crossVector.y), crossVector.x, 0 };
        glm::mat3 skewSecond = skewSymmetric * skewSymmetric;
        skewSecond = skewSecond * ((1.0f - cosAngle) / (sineAngle * sineAngle));
        experimentalRMatrix = glm::mat3() + skewSymmetric + skewSecond;
        testVector = experimentalRMatrix * currentOrientation;
        rotationMatrix = glm::mat4(experimentalRMatrix);

有了這一切，我希望分析為什么我的解決問題的嘗試失敗了，和/或一個能正確旋轉頂點的解決方案。

非常感謝你。

Answer 1

看到評論......切線空間和四元數似乎已經完成了