OpenGL: Get lookAt center from position and orientation with quaternion










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I'm working with OpenGL ES 2.0 on an Android augmented reality application. I would like to position the openGL camera at the exact same location as my headset (in terms of position and orientation in space). I have access to the position and orientation of the headset in real-time, but the orientation is based on a quaternion.



What I have tried so far: I converted the quaternion to Euler angles to have yaw, pitch and roll angles (which seems fine), and then computed a direction vector :



 float ray = 1.0f;
directionX = ray * (float) (Math.cos(rollRadian) * Math.sin(yawRadian));
directionY = ray * (float) (Math.cos(rollRadian) * Math.cos(yawRadian));
directionZ = ray * (float) Math.sin(rollRadian);


Then, I update my LookAt values :



 // LookAt = translation + direction
mLookAt[0] = headSetX + directionX;
mLookAt[1] = headSetY + directionY;
mLookAt[2] = headSetZ + directionZ;


However, it seems that the direction values are too weak compared to the headset translation, resulting in an incorrect display (ex: I can see my virtual object even when I'm not looking in its direction). I suspect that a part of the issue is related to the ray parameter in the formula above, which was from this question in StackExchange



Questions: Did I miss something with this implementation ? Is there another way to get a LookAt vector (or directly the point where I'm looking to) from a position in world-space and a orientation quaternion ?










share|improve this question

















  • 2





    Don't bother with Euler angles. Simply convert the quaternion + translation directly to a matrix and use its inverse as the view matrix.

    – Nico Schertler
    Nov 15 '18 at 17:10











  • Additionally, roll should only affect the side vectors, not the direction vector.

    – meowgoesthedog
    Nov 15 '18 at 23:19















0















I'm working with OpenGL ES 2.0 on an Android augmented reality application. I would like to position the openGL camera at the exact same location as my headset (in terms of position and orientation in space). I have access to the position and orientation of the headset in real-time, but the orientation is based on a quaternion.



What I have tried so far: I converted the quaternion to Euler angles to have yaw, pitch and roll angles (which seems fine), and then computed a direction vector :



 float ray = 1.0f;
directionX = ray * (float) (Math.cos(rollRadian) * Math.sin(yawRadian));
directionY = ray * (float) (Math.cos(rollRadian) * Math.cos(yawRadian));
directionZ = ray * (float) Math.sin(rollRadian);


Then, I update my LookAt values :



 // LookAt = translation + direction
mLookAt[0] = headSetX + directionX;
mLookAt[1] = headSetY + directionY;
mLookAt[2] = headSetZ + directionZ;


However, it seems that the direction values are too weak compared to the headset translation, resulting in an incorrect display (ex: I can see my virtual object even when I'm not looking in its direction). I suspect that a part of the issue is related to the ray parameter in the formula above, which was from this question in StackExchange



Questions: Did I miss something with this implementation ? Is there another way to get a LookAt vector (or directly the point where I'm looking to) from a position in world-space and a orientation quaternion ?










share|improve this question

















  • 2





    Don't bother with Euler angles. Simply convert the quaternion + translation directly to a matrix and use its inverse as the view matrix.

    – Nico Schertler
    Nov 15 '18 at 17:10











  • Additionally, roll should only affect the side vectors, not the direction vector.

    – meowgoesthedog
    Nov 15 '18 at 23:19













0












0








0








I'm working with OpenGL ES 2.0 on an Android augmented reality application. I would like to position the openGL camera at the exact same location as my headset (in terms of position and orientation in space). I have access to the position and orientation of the headset in real-time, but the orientation is based on a quaternion.



What I have tried so far: I converted the quaternion to Euler angles to have yaw, pitch and roll angles (which seems fine), and then computed a direction vector :



 float ray = 1.0f;
directionX = ray * (float) (Math.cos(rollRadian) * Math.sin(yawRadian));
directionY = ray * (float) (Math.cos(rollRadian) * Math.cos(yawRadian));
directionZ = ray * (float) Math.sin(rollRadian);


Then, I update my LookAt values :



 // LookAt = translation + direction
mLookAt[0] = headSetX + directionX;
mLookAt[1] = headSetY + directionY;
mLookAt[2] = headSetZ + directionZ;


However, it seems that the direction values are too weak compared to the headset translation, resulting in an incorrect display (ex: I can see my virtual object even when I'm not looking in its direction). I suspect that a part of the issue is related to the ray parameter in the formula above, which was from this question in StackExchange



Questions: Did I miss something with this implementation ? Is there another way to get a LookAt vector (or directly the point where I'm looking to) from a position in world-space and a orientation quaternion ?










share|improve this question














I'm working with OpenGL ES 2.0 on an Android augmented reality application. I would like to position the openGL camera at the exact same location as my headset (in terms of position and orientation in space). I have access to the position and orientation of the headset in real-time, but the orientation is based on a quaternion.



What I have tried so far: I converted the quaternion to Euler angles to have yaw, pitch and roll angles (which seems fine), and then computed a direction vector :



 float ray = 1.0f;
directionX = ray * (float) (Math.cos(rollRadian) * Math.sin(yawRadian));
directionY = ray * (float) (Math.cos(rollRadian) * Math.cos(yawRadian));
directionZ = ray * (float) Math.sin(rollRadian);


Then, I update my LookAt values :



 // LookAt = translation + direction
mLookAt[0] = headSetX + directionX;
mLookAt[1] = headSetY + directionY;
mLookAt[2] = headSetZ + directionZ;


However, it seems that the direction values are too weak compared to the headset translation, resulting in an incorrect display (ex: I can see my virtual object even when I'm not looking in its direction). I suspect that a part of the issue is related to the ray parameter in the formula above, which was from this question in StackExchange



Questions: Did I miss something with this implementation ? Is there another way to get a LookAt vector (or directly the point where I'm looking to) from a position in world-space and a orientation quaternion ?







math opengl-es quaternions






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asked Nov 15 '18 at 16:10









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  • 2





    Don't bother with Euler angles. Simply convert the quaternion + translation directly to a matrix and use its inverse as the view matrix.

    – Nico Schertler
    Nov 15 '18 at 17:10











  • Additionally, roll should only affect the side vectors, not the direction vector.

    – meowgoesthedog
    Nov 15 '18 at 23:19












  • 2





    Don't bother with Euler angles. Simply convert the quaternion + translation directly to a matrix and use its inverse as the view matrix.

    – Nico Schertler
    Nov 15 '18 at 17:10











  • Additionally, roll should only affect the side vectors, not the direction vector.

    – meowgoesthedog
    Nov 15 '18 at 23:19







2




2





Don't bother with Euler angles. Simply convert the quaternion + translation directly to a matrix and use its inverse as the view matrix.

– Nico Schertler
Nov 15 '18 at 17:10





Don't bother with Euler angles. Simply convert the quaternion + translation directly to a matrix and use its inverse as the view matrix.

– Nico Schertler
Nov 15 '18 at 17:10













Additionally, roll should only affect the side vectors, not the direction vector.

– meowgoesthedog
Nov 15 '18 at 23:19





Additionally, roll should only affect the side vectors, not the direction vector.

– meowgoesthedog
Nov 15 '18 at 23:19












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