Matrix4x4 camera::GetVIEwMat(){ Matrix4x4 oRotate,oVIEw; oVIEw.SetIDentity(); Vector3 lookAtDir = m_targetposition - m_camposition; Vector3 lookAtHorizontal = Vector3(lookAtDir.GetX(),0.0f,lookAtDir.GetZ()); lookAtHorizontal.normalize(); float angle = acosf(Vector3(0.0f,-1.0f).Dot(lookAtHorizontal)); Quaternions horizontalOrIEnt(angle,Vector3(0.0f,1.0f,0.0f)); ori = horizontalOrIEnt; ori.Conjugate(); oRotate = ori.ToMatrix(); Vector3 inverseTranslate = Vector3(-m_camposition.GetX(),-m_camposition.GetY(),-m_camposition.GetZ()); oRotate.transform(inverseTranslate); oRotate.Set(0,3,inverseTranslate.GetX()); oRotate.Set(1,inverseTranslate.GetY()); oRotate.Set(2,inverseTranslate.GetZ()); oVIEw = oRotate; return oVIEw;}解决方法 正如所承诺的那样,一些代码显示了我让摄像机始终在空间中的特定点观察的方式.
首先,我们需要一种从角度和轴构造四元数的方法,我碰巧在pastebin上有这个,角度输入是弧度:
http://pastebin.com/vLcx4Qqh
确保你没有输入轴(0,0),这没有任何意义.
现在是实际的更新方法,我们需要让四元数从默认方向旋转相机指向目标点.请注意我刚刚写了这篇文章,它可能需要一些调试,可能需要一些优化,但这至少应该给你一个正确的方向.
voID camera::update(){ // First get the direction from the camera's position to the target point vec3 lookAtDir = m_targetPoint - m_position; // I'm going to divIDe the vector into two 'components',the Y axis rotation // and the Up/Down rotation,like a regular camera would work. // First to calculate the rotation around the Y axis,so we zero out the y // component: vec3 lookAtHorizontal = vec3(lookAtDir.x,lookAtDir.z).normalize(); // Get the quaternion from 'default' direction to the horizontal direction // In this case,'default' direction is along the -z axis,like most OpenGL // programs. Make sure the projection matrix works according to this. float angle = acos(vec3(0.0f,-1.0f).dot(lookAtHorizontal)); quaternion horizontalOrIEnt(angle,vec3(0.0f,0.0f)); // Since we already stripped the Y component,we can simply get the up/down // rotation from it as well. angle = acos(lookAtDir.normalize().dot(lookAtHorizontal)); if(angle) horizontalOrIEnt *= quaternion(angle,lookAtDir.cross(lookAtHorizontal)); // ... m_orIEntation = horizontalOrIEnt;}
现在实际上采用m_orIEntation和m_position来获取世界 – >相机矩阵
// First inverse each element (-position and inverse the quaternion),// the position is rotated since the position within a matrix is 'added' last// to the output vector,so it needs to account for rotation of the space.mat3 rotationMatrix = m_orIEntation.inverse().toMatrix();vec3 inverseTranslate = rotationMatrix * -m_position; // Note the minusmat4 matrix = mat3; // just means the matrix is expanded,the last entry (bottom right of the matrix) is a 1.0f like an IDentity matrix would be.// This bit is row-major in my case,you just need to set the translation of the matrix.matrix[3] = inverseTranslate.x;matrix[7] = inverseTranslate.y;matrix[11] = inverseTranslate.z;
编辑我认为它应该是显而易见的,但只是为了完整性,.dot()采用向量的点积,.cross()采用叉积,执行方法的对象是向量A,方法的参数是向量B.
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