Mixing
How to rotate an orientation (Euler angles)
$begingroup$
If I have an orientation defined by Euler angles and I want to simulate a rotation of the coordinate system about the origin (doesn't matter to me how the rotation is specified), how would I get the new Euler angles? I understand how to transform a point via a rotation matrix, but I'm not sure how to approach this in terms of Euler angles.
For example, I'm thinking of an airplane's orientation in terms of pitch, yaw, and roll, and say I want to simulate a random rotation of the airplane about its center.
transformation coordinate-systems rotations orientation
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
$endgroup$
add a comment |
$begingroup$
If I have an orientation defined by Euler angles and I want to simulate a rotation of the coordinate system about the origin (doesn't matter to me how the rotation is specified), how would I get the new Euler angles? I understand how to transform a point via a rotation matrix, but I'm not sure how to approach this in terms of Euler angles.
For example, I'm thinking of an airplane's orientation in terms of pitch, yaw, and roll, and say I want to simulate a random rotation of the airplane about its center.
transformation coordinate-systems rotations orientation
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
$endgroup$
$begingroup$
The rotation matrix is given here: en.wikipedia.org/wiki/Euler_angles#Rotation_matrix
$endgroup$
– Yves Daoust
Mar 12 '15 at 15:50
$begingroup$
How would I apply the rotation matrix to my original Euler angles?
$endgroup$
– rmp251
Mar 12 '15 at 16:21
$begingroup$
You just compute it from the Euler angles !
$endgroup$
– Yves Daoust
Mar 12 '15 at 17:22
$begingroup$
Thanks for the comment but you are misunderstanding my question. How do I apply (not compute) such a rotation transformation to an orientation (as specified via Euler angles) as opposed to points in 3-space?
$endgroup$
– rmp251
Mar 12 '15 at 19:03
$begingroup$
I guess that the only way is to multiply the rotation matrices and inverse the process to retrieve the Euler angles corresponding to the combined rotations (see my answer about this).
$endgroup$
– Yves Daoust
Mar 12 '15 at 19:58
add a comment |
$begingroup$
If I have an orientation defined by Euler angles and I want to simulate a rotation of the coordinate system about the origin (doesn't matter to me how the rotation is specified), how would I get the new Euler angles? I understand how to transform a point via a rotation matrix, but I'm not sure how to approach this in terms of Euler angles.
For example, I'm thinking of an airplane's orientation in terms of pitch, yaw, and roll, and say I want to simulate a random rotation of the airplane about its center.
transformation coordinate-systems rotations orientation
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
$endgroup$
If I have an orientation defined by Euler angles and I want to simulate a rotation of the coordinate system about the origin (doesn't matter to me how the rotation is specified), how would I get the new Euler angles? I understand how to transform a point via a rotation matrix, but I'm not sure how to approach this in terms of Euler angles.
For example, I'm thinking of an airplane's orientation in terms of pitch, yaw, and roll, and say I want to simulate a random rotation of the airplane about its center.
transformation coordinate-systems rotations orientation
transformation coordinate-systems rotations orientation
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
asked Mar 12 '15 at 15:19
rmp251rmp251
1253
1253
$begingroup$
The rotation matrix is given here: en.wikipedia.org/wiki/Euler_angles#Rotation_matrix
$endgroup$
– Yves Daoust
Mar 12 '15 at 15:50
$begingroup$
How would I apply the rotation matrix to my original Euler angles?
$endgroup$
– rmp251
Mar 12 '15 at 16:21
$begingroup$
You just compute it from the Euler angles !
$endgroup$
– Yves Daoust
Mar 12 '15 at 17:22
$begingroup$
Thanks for the comment but you are misunderstanding my question. How do I apply (not compute) such a rotation transformation to an orientation (as specified via Euler angles) as opposed to points in 3-space?
$endgroup$
– rmp251
Mar 12 '15 at 19:03
$begingroup$
I guess that the only way is to multiply the rotation matrices and inverse the process to retrieve the Euler angles corresponding to the combined rotations (see my answer about this).
$endgroup$
– Yves Daoust
Mar 12 '15 at 19:58
add a comment |
$begingroup$
The rotation matrix is given here: en.wikipedia.org/wiki/Euler_angles#Rotation_matrix
$endgroup$
– Yves Daoust
Mar 12 '15 at 15:50
$begingroup$
How would I apply the rotation matrix to my original Euler angles?
$endgroup$
– rmp251
Mar 12 '15 at 16:21
$begingroup$
You just compute it from the Euler angles !
$endgroup$
– Yves Daoust
Mar 12 '15 at 17:22
$begingroup$
Thanks for the comment but you are misunderstanding my question. How do I apply (not compute) such a rotation transformation to an orientation (as specified via Euler angles) as opposed to points in 3-space?
$endgroup$
– rmp251
Mar 12 '15 at 19:03
$begingroup$
I guess that the only way is to multiply the rotation matrices and inverse the process to retrieve the Euler angles corresponding to the combined rotations (see my answer about this).
$endgroup$
– Yves Daoust
Mar 12 '15 at 19:58
$begingroup$
The rotation matrix is given here: en.wikipedia.org/wiki/Euler_angles#Rotation_matrix
$endgroup$
– Yves Daoust
Mar 12 '15 at 15:50
$begingroup$
The rotation matrix is given here: en.wikipedia.org/wiki/Euler_angles#Rotation_matrix
$endgroup$
– Yves Daoust
Mar 12 '15 at 15:50
$begingroup$
How would I apply the rotation matrix to my original Euler angles?
$endgroup$
– rmp251
Mar 12 '15 at 16:21
$begingroup$
How would I apply the rotation matrix to my original Euler angles?
$endgroup$
– rmp251
Mar 12 '15 at 16:21
$begingroup$
You just compute it from the Euler angles !
$endgroup$
– Yves Daoust
Mar 12 '15 at 17:22
$begingroup$
You just compute it from the Euler angles !
$endgroup$
– Yves Daoust
Mar 12 '15 at 17:22
$begingroup$
Thanks for the comment but you are misunderstanding my question. How do I apply (not compute) such a rotation transformation to an orientation (as specified via Euler angles) as opposed to points in 3-space?
$endgroup$
– rmp251
Mar 12 '15 at 19:03
$begingroup$
Thanks for the comment but you are misunderstanding my question. How do I apply (not compute) such a rotation transformation to an orientation (as specified via Euler angles) as opposed to points in 3-space?
$endgroup$
– rmp251
Mar 12 '15 at 19:03
$begingroup$
I guess that the only way is to multiply the rotation matrices and inverse the process to retrieve the Euler angles corresponding to the combined rotations (see my answer about this).
$endgroup$
– Yves Daoust
Mar 12 '15 at 19:58
$begingroup$
I guess that the only way is to multiply the rotation matrices and inverse the process to retrieve the Euler angles corresponding to the combined rotations (see my answer about this).
$endgroup$
– Yves Daoust
Mar 12 '15 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us assume that the rotation is known by a $3times3$ matrix, and let us focus on the one specified as $X_1Z_2X_3$ in the Wikipedia article.
You get $alpha$ from $tanalpha=dfrac{s_1s_2}{c_1s_2}$.
You get $beta$ from $tanbeta=dfrac{sqrt{(s_2c_3)^2+(s_2s_3)^2}}{c_2}$.
You get $gamma$ from $tangamma=dfrac{s_3s_2}{c_3s_2}$.
As expected, $beta=0$ creates an indeterminacy, but this case is easy to deal with (it reduces to a single rotation $alpha+gamma$).
You can probably improve the accuracy by a least-squares fitting of the matrix derived from the angles to the given matrix, but this will require the Levenberg-Marquardt algorithm as this is a non-linear problem.
I guess that you can also address the computation of the angles by using three well-chosen Givens rotations that will transform the input matrix to a unit one (backwards, $X_1Z_2X_3to X_1Z_2to X_1to I$, corresponding to angles $(alpha,beta,gamma)to(alpha,beta,0)to(alpha,0,0)to(0,0,0)$).
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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',
autoActivateHeartbeat: false,
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
},
noCode: 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%2fmath.stackexchange.com%2fquestions%2f1186923%2fhow-to-rotate-an-orientation-euler-angles%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us assume that the rotation is known by a $3times3$ matrix, and let us focus on the one specified as $X_1Z_2X_3$ in the Wikipedia article.
You get $alpha$ from $tanalpha=dfrac{s_1s_2}{c_1s_2}$.
You get $beta$ from $tanbeta=dfrac{sqrt{(s_2c_3)^2+(s_2s_3)^2}}{c_2}$.
You get $gamma$ from $tangamma=dfrac{s_3s_2}{c_3s_2}$.
As expected, $beta=0$ creates an indeterminacy, but this case is easy to deal with (it reduces to a single rotation $alpha+gamma$).
You can probably improve the accuracy by a least-squares fitting of the matrix derived from the angles to the given matrix, but this will require the Levenberg-Marquardt algorithm as this is a non-linear problem.
I guess that you can also address the computation of the angles by using three well-chosen Givens rotations that will transform the input matrix to a unit one (backwards, $X_1Z_2X_3to X_1Z_2to X_1to I$, corresponding to angles $(alpha,beta,gamma)to(alpha,beta,0)to(alpha,0,0)to(0,0,0)$).
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
add a comment |
$begingroup$
Let us assume that the rotation is known by a $3times3$ matrix, and let us focus on the one specified as $X_1Z_2X_3$ in the Wikipedia article.
You get $alpha$ from $tanalpha=dfrac{s_1s_2}{c_1s_2}$.
You get $beta$ from $tanbeta=dfrac{sqrt{(s_2c_3)^2+(s_2s_3)^2}}{c_2}$.
You get $gamma$ from $tangamma=dfrac{s_3s_2}{c_3s_2}$.
As expected, $beta=0$ creates an indeterminacy, but this case is easy to deal with (it reduces to a single rotation $alpha+gamma$).
You can probably improve the accuracy by a least-squares fitting of the matrix derived from the angles to the given matrix, but this will require the Levenberg-Marquardt algorithm as this is a non-linear problem.
I guess that you can also address the computation of the angles by using three well-chosen Givens rotations that will transform the input matrix to a unit one (backwards, $X_1Z_2X_3to X_1Z_2to X_1to I$, corresponding to angles $(alpha,beta,gamma)to(alpha,beta,0)to(alpha,0,0)to(0,0,0)$).
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
add a comment |
$begingroup$
Let us assume that the rotation is known by a $3times3$ matrix, and let us focus on the one specified as $X_1Z_2X_3$ in the Wikipedia article.
You get $alpha$ from $tanalpha=dfrac{s_1s_2}{c_1s_2}$.
You get $beta$ from $tanbeta=dfrac{sqrt{(s_2c_3)^2+(s_2s_3)^2}}{c_2}$.
You get $gamma$ from $tangamma=dfrac{s_3s_2}{c_3s_2}$.
As expected, $beta=0$ creates an indeterminacy, but this case is easy to deal with (it reduces to a single rotation $alpha+gamma$).
You can probably improve the accuracy by a least-squares fitting of the matrix derived from the angles to the given matrix, but this will require the Levenberg-Marquardt algorithm as this is a non-linear problem.
I guess that you can also address the computation of the angles by using three well-chosen Givens rotations that will transform the input matrix to a unit one (backwards, $X_1Z_2X_3to X_1Z_2to X_1to I$, corresponding to angles $(alpha,beta,gamma)to(alpha,beta,0)to(alpha,0,0)to(0,0,0)$).
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
$endgroup$
Let us assume that the rotation is known by a $3times3$ matrix, and let us focus on the one specified as $X_1Z_2X_3$ in the Wikipedia article.
You get $alpha$ from $tanalpha=dfrac{s_1s_2}{c_1s_2}$.
You get $beta$ from $tanbeta=dfrac{sqrt{(s_2c_3)^2+(s_2s_3)^2}}{c_2}$.
You get $gamma$ from $tangamma=dfrac{s_3s_2}{c_3s_2}$.
As expected, $beta=0$ creates an indeterminacy, but this case is easy to deal with (it reduces to a single rotation $alpha+gamma$).
You can probably improve the accuracy by a least-squares fitting of the matrix derived from the angles to the given matrix, but this will require the Levenberg-Marquardt algorithm as this is a non-linear problem.
I guess that you can also address the computation of the angles by using three well-chosen Givens rotations that will transform the input matrix to a unit one (backwards, $X_1Z_2X_3to X_1Z_2to X_1to I$, corresponding to angles $(alpha,beta,gamma)to(alpha,beta,0)to(alpha,0,0)to(0,0,0)$).
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
edited Mar 12 '15 at 19:59
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
answered Mar 12 '15 at 15:35
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
add a comment |
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
Thanks for your answer. I was hoping for something simpler but I'll have to digest it a little bit to see if it's what I was looking for.
$endgroup$
– rmp251
Mar 13 '15 at 13:09
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
$begingroup$
The other option is to modify the angles randomly.
$endgroup$
– Yves Daoust
Mar 13 '15 at 14:05
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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.
Use MathJax to format equations. MathJax reference.
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%2fmath.stackexchange.com%2fquestions%2f1186923%2fhow-to-rotate-an-orientation-euler-angles%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
$begingroup$
The rotation matrix is given here: en.wikipedia.org/wiki/Euler_angles#Rotation_matrix
$endgroup$
– Yves Daoust
Mar 12 '15 at 15:50
$begingroup$
How would I apply the rotation matrix to my original Euler angles?
$endgroup$
– rmp251
Mar 12 '15 at 16:21
$begingroup$
You just compute it from the Euler angles !
$endgroup$
– Yves Daoust
Mar 12 '15 at 17:22
$begingroup$
Thanks for the comment but you are misunderstanding my question. How do I apply (not compute) such a rotation transformation to an orientation (as specified via Euler angles) as opposed to points in 3-space?
$endgroup$
– rmp251
Mar 12 '15 at 19:03
$begingroup$
I guess that the only way is to multiply the rotation matrices and inverse the process to retrieve the Euler angles corresponding to the combined rotations (see my answer about this).
$endgroup$
– Yves Daoust
Mar 12 '15 at 19:58