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 ?










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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 at 17:10










  • Additionally, roll should only affect the side vectors, not the direction vector.
    – meowgoesthedog
    Nov 15 at 23:19















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0
down vote

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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 at 17:10










  • Additionally, roll should only affect the side vectors, not the direction vector.
    – meowgoesthedog
    Nov 15 at 23:19













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 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 at 17:10










  • Additionally, roll should only affect the side vectors, not the direction vector.
    – meowgoesthedog
    Nov 15 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 at 17:10










  • Additionally, roll should only affect the side vectors, not the direction vector.
    – meowgoesthedog
    Nov 15 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 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 at 17:10












Additionally, roll should only affect the side vectors, not the direction vector.
– meowgoesthedog
Nov 15 at 23:19




Additionally, roll should only affect the side vectors, not the direction vector.
– meowgoesthedog
Nov 15 at 23:19

















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