OpenGL: Get lookAt center from position and orientation with quaternion
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 ?
math opengl-es quaternions
asked Nov 15 at 16:10
C.IHM
114
add a comment |
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 ?
math opengl-es quaternions
asked Nov 15 at 16:10
C.IHM
114
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,rollshould only affect the side vectors, not the direction vector.
– meowgoesthedog
Nov 15 at 23:19
add a comment |
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 ?
math opengl-es quaternions
asked Nov 15 at 16:10
C.IHM
114
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
math opengl-es quaternions
asked Nov 15 at 16:10
C.IHM
114
asked Nov 15 at 16:10
C.IHM
114
asked Nov 15 at 16:10
C.IHM
114
asked Nov 15 at 16:10
C.IHM
114
asked Nov 15 at 16:10
C.IHM
114
114
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,rollshould only affect the side vectors, not the direction vector.
– meowgoesthedog
Nov 15 at 23:19
add a comment |
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,rollshould 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
add a comment |
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53323520%2fopengl-get-lookat-center-from-position-and-orientation-with-quaternion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53323520%2fopengl-get-lookat-center-from-position-and-orientation-with-quaternion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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,
rollshould only affect the side vectors, not the direction vector.– meowgoesthedog
Nov 15 at 23:19