Mixing
Given Tait–Bryan angles $ x-y'-z^{''} $ (intrinsic), how can i get Tait–Bryan angles $ z-y'-x^{''} $...
$begingroup$
I've been reading https://en.wikipedia.org/wiki/Euler_angles for hours and cannot figure out how to solve it. https://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Euler_angles_.E2.86.94_quaternion gives formulas on conventions between ($ z-y′-x^{″} $ intrinsic) and Quaternion but I don't know how to calculate quaternion from $ x-y′-z^{″} $.
linear-transformations rotations
edited Jan 21 at 9:54
user549397
1,5061418
asked Jan 21 at 7:39
ZenZen
1155
$endgroup$
add a comment |
$begingroup$
I've been reading https://en.wikipedia.org/wiki/Euler_angles for hours and cannot figure out how to solve it. https://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Euler_angles_.E2.86.94_quaternion gives formulas on conventions between ($ z-y′-x^{″} $ intrinsic) and Quaternion but I don't know how to calculate quaternion from $ x-y′-z^{″} $.
linear-transformations rotations
edited Jan 21 at 9:54
user549397
1,5061418
asked Jan 21 at 7:39
ZenZen
1155
$endgroup$
add a comment |
$begingroup$
I've been reading https://en.wikipedia.org/wiki/Euler_angles for hours and cannot figure out how to solve it. https://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Euler_angles_.E2.86.94_quaternion gives formulas on conventions between ($ z-y′-x^{″} $ intrinsic) and Quaternion but I don't know how to calculate quaternion from $ x-y′-z^{″} $.
linear-transformations rotations
edited Jan 21 at 9:54
user549397
1,5061418
asked Jan 21 at 7:39
ZenZen
1155
$endgroup$
I've been reading https://en.wikipedia.org/wiki/Euler_angles for hours and cannot figure out how to solve it. https://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Euler_angles_.E2.86.94_quaternion gives formulas on conventions between ($ z-y′-x^{″} $ intrinsic) and Quaternion but I don't know how to calculate quaternion from $ x-y′-z^{″} $.
linear-transformations rotations
linear-transformations rotations
edited Jan 21 at 9:54
user549397
1,5061418
asked Jan 21 at 7:39
ZenZen
1155
edited Jan 21 at 9:54
user549397
1,5061418
asked Jan 21 at 7:39
ZenZen
1155
edited Jan 21 at 9:54
user549397
1,5061418
edited Jan 21 at 9:54
user549397
1,5061418
edited Jan 21 at 9:54
user549397
1,5061418
1,5061418
asked Jan 21 at 7:39
ZenZen
1155
asked Jan 21 at 7:39
ZenZen
1155
asked Jan 21 at 7:39
ZenZen
1155
1155
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With Tait-Bryan angles orientation representation with rotation order $xyz$, the rotation matrix is
begin{equation}label{eq:euler-model-matrix}
%hspace*{-6cm}
begin{split}
R &= R_Z(psi) R_Y(theta) R_X(phi) \
&=
left[
begin{array}{ccc}
cos{psi} & -sin{psi} & 0 \
sin{psi} & cos{psi} & 0 \
0 & 0 & 1 \
end{array} right]
left[
begin{array}{ccc}
cos{theta} & 0 & sin{theta} \
0 & 1 & 0 \
-sin{theta} & 0 & cos{theta} \
end{array} right]
left[
begin{array}{ccc}
1 & 0 & 0 \
0 & cos{phi} & -sin{phi} \
0 & sin{phi} & cos{phi} \
end{array} right]
\
&=
left[
begin{array}{ccc}
cos{theta}cos{psi} & cos{psi} sin{theta} sin{phi} - cos{phi} sin{psi} & cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} \
%
cos{theta}sin{psi} & cos{phi} cos{psi} + sin{theta} sin{phi} sin{psi} & cos{phi}sin{theta} sin{psi} - cos{psi} sin{phi} \
%
-sin{theta} & cos{theta}sin{phi} & cos{theta}cos{phi}
end{array}
right]
end{split}
end{equation}
With the $zyx$ rotation order, the same matrix is
$$
R =
left[
begin{array}{cccc}
cos{theta'}cos{phi'} & -cos{theta'}sin{phi'} & sin{theta'} \
cos{psi'}sin{phi'}+ cos{phi'} sin{theta'}sin{psi'} & cos{phi'}cos{psi'} - sin{theta'}sin{phi'}sin{psi'} & -cos{theta'}sin{psi'} \
sin{phi'}sin{psi'}- cos{phi'}cos{psi'}sin{theta'} & cos{psi'}sin{theta'}sin{phi'} + cos{phi'}sin{psi'} & cos{theta'}cos{psi'} \
end{array}
right]
$$
What you're asking is how to convert this to the $zyx$ rotation order. So let's assume that you have calculated the rotation matrix $R$ with the $xyz$ rotation order (the first method). We need to now interpret the angles as if you had created the matrix with the second method. It's relatively easy to extract the angles from the second matrix:
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-R_{01}}{R_{00}}} \
theta' =& arcsin{R_{02}} \
psi'=& arctan{frac{-R_{12}}{R_{22}}} \
end{array}
right.
$$
Note that the entries here are from the second $R$-matrix. I've added the prime symbol to these angles to denote that they are the angles of the $zyx$-rotation order. The non-prime angle symbols are from the $xyz$-rotation order. Now we can plug in the values from the first matrix to obtain
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-cos{psi} sin{theta} sin{phi} + cos{phi} sin{psi}}{cos{theta}cos{psi} }} \
theta' =& arcsin{left( cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} right)} \
psi'=& arctan{frac{-cos{phi}sin{theta} sin{psi} + cos{psi} sin{phi} }{cos{theta}cos{phi}}} \
end{array}
right.
$$
You might want to double-check these results yourself, as there are quite many terms.
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
$endgroup$
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With Tait-Bryan angles orientation representation with rotation order $xyz$, the rotation matrix is
begin{equation}label{eq:euler-model-matrix}
%hspace*{-6cm}
begin{split}
R &= R_Z(psi) R_Y(theta) R_X(phi) \
&=
left[
begin{array}{ccc}
cos{psi} & -sin{psi} & 0 \
sin{psi} & cos{psi} & 0 \
0 & 0 & 1 \
end{array} right]
left[
begin{array}{ccc}
cos{theta} & 0 & sin{theta} \
0 & 1 & 0 \
-sin{theta} & 0 & cos{theta} \
end{array} right]
left[
begin{array}{ccc}
1 & 0 & 0 \
0 & cos{phi} & -sin{phi} \
0 & sin{phi} & cos{phi} \
end{array} right]
\
&=
left[
begin{array}{ccc}
cos{theta}cos{psi} & cos{psi} sin{theta} sin{phi} - cos{phi} sin{psi} & cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} \
%
cos{theta}sin{psi} & cos{phi} cos{psi} + sin{theta} sin{phi} sin{psi} & cos{phi}sin{theta} sin{psi} - cos{psi} sin{phi} \
%
-sin{theta} & cos{theta}sin{phi} & cos{theta}cos{phi}
end{array}
right]
end{split}
end{equation}
With the $zyx$ rotation order, the same matrix is
$$
R =
left[
begin{array}{cccc}
cos{theta'}cos{phi'} & -cos{theta'}sin{phi'} & sin{theta'} \
cos{psi'}sin{phi'}+ cos{phi'} sin{theta'}sin{psi'} & cos{phi'}cos{psi'} - sin{theta'}sin{phi'}sin{psi'} & -cos{theta'}sin{psi'} \
sin{phi'}sin{psi'}- cos{phi'}cos{psi'}sin{theta'} & cos{psi'}sin{theta'}sin{phi'} + cos{phi'}sin{psi'} & cos{theta'}cos{psi'} \
end{array}
right]
$$
What you're asking is how to convert this to the $zyx$ rotation order. So let's assume that you have calculated the rotation matrix $R$ with the $xyz$ rotation order (the first method). We need to now interpret the angles as if you had created the matrix with the second method. It's relatively easy to extract the angles from the second matrix:
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-R_{01}}{R_{00}}} \
theta' =& arcsin{R_{02}} \
psi'=& arctan{frac{-R_{12}}{R_{22}}} \
end{array}
right.
$$
Note that the entries here are from the second $R$-matrix. I've added the prime symbol to these angles to denote that they are the angles of the $zyx$-rotation order. The non-prime angle symbols are from the $xyz$-rotation order. Now we can plug in the values from the first matrix to obtain
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-cos{psi} sin{theta} sin{phi} + cos{phi} sin{psi}}{cos{theta}cos{psi} }} \
theta' =& arcsin{left( cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} right)} \
psi'=& arctan{frac{-cos{phi}sin{theta} sin{psi} + cos{psi} sin{phi} }{cos{theta}cos{phi}}} \
end{array}
right.
$$
You might want to double-check these results yourself, as there are quite many terms.
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
$endgroup$
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
add a comment |
$begingroup$
With Tait-Bryan angles orientation representation with rotation order $xyz$, the rotation matrix is
begin{equation}label{eq:euler-model-matrix}
%hspace*{-6cm}
begin{split}
R &= R_Z(psi) R_Y(theta) R_X(phi) \
&=
left[
begin{array}{ccc}
cos{psi} & -sin{psi} & 0 \
sin{psi} & cos{psi} & 0 \
0 & 0 & 1 \
end{array} right]
left[
begin{array}{ccc}
cos{theta} & 0 & sin{theta} \
0 & 1 & 0 \
-sin{theta} & 0 & cos{theta} \
end{array} right]
left[
begin{array}{ccc}
1 & 0 & 0 \
0 & cos{phi} & -sin{phi} \
0 & sin{phi} & cos{phi} \
end{array} right]
\
&=
left[
begin{array}{ccc}
cos{theta}cos{psi} & cos{psi} sin{theta} sin{phi} - cos{phi} sin{psi} & cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} \
%
cos{theta}sin{psi} & cos{phi} cos{psi} + sin{theta} sin{phi} sin{psi} & cos{phi}sin{theta} sin{psi} - cos{psi} sin{phi} \
%
-sin{theta} & cos{theta}sin{phi} & cos{theta}cos{phi}
end{array}
right]
end{split}
end{equation}
With the $zyx$ rotation order, the same matrix is
$$
R =
left[
begin{array}{cccc}
cos{theta'}cos{phi'} & -cos{theta'}sin{phi'} & sin{theta'} \
cos{psi'}sin{phi'}+ cos{phi'} sin{theta'}sin{psi'} & cos{phi'}cos{psi'} - sin{theta'}sin{phi'}sin{psi'} & -cos{theta'}sin{psi'} \
sin{phi'}sin{psi'}- cos{phi'}cos{psi'}sin{theta'} & cos{psi'}sin{theta'}sin{phi'} + cos{phi'}sin{psi'} & cos{theta'}cos{psi'} \
end{array}
right]
$$
What you're asking is how to convert this to the $zyx$ rotation order. So let's assume that you have calculated the rotation matrix $R$ with the $xyz$ rotation order (the first method). We need to now interpret the angles as if you had created the matrix with the second method. It's relatively easy to extract the angles from the second matrix:
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-R_{01}}{R_{00}}} \
theta' =& arcsin{R_{02}} \
psi'=& arctan{frac{-R_{12}}{R_{22}}} \
end{array}
right.
$$
Note that the entries here are from the second $R$-matrix. I've added the prime symbol to these angles to denote that they are the angles of the $zyx$-rotation order. The non-prime angle symbols are from the $xyz$-rotation order. Now we can plug in the values from the first matrix to obtain
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-cos{psi} sin{theta} sin{phi} + cos{phi} sin{psi}}{cos{theta}cos{psi} }} \
theta' =& arcsin{left( cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} right)} \
psi'=& arctan{frac{-cos{phi}sin{theta} sin{psi} + cos{psi} sin{phi} }{cos{theta}cos{phi}}} \
end{array}
right.
$$
You might want to double-check these results yourself, as there are quite many terms.
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
$endgroup$
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
add a comment |
$begingroup$
With Tait-Bryan angles orientation representation with rotation order $xyz$, the rotation matrix is
begin{equation}label{eq:euler-model-matrix}
%hspace*{-6cm}
begin{split}
R &= R_Z(psi) R_Y(theta) R_X(phi) \
&=
left[
begin{array}{ccc}
cos{psi} & -sin{psi} & 0 \
sin{psi} & cos{psi} & 0 \
0 & 0 & 1 \
end{array} right]
left[
begin{array}{ccc}
cos{theta} & 0 & sin{theta} \
0 & 1 & 0 \
-sin{theta} & 0 & cos{theta} \
end{array} right]
left[
begin{array}{ccc}
1 & 0 & 0 \
0 & cos{phi} & -sin{phi} \
0 & sin{phi} & cos{phi} \
end{array} right]
\
&=
left[
begin{array}{ccc}
cos{theta}cos{psi} & cos{psi} sin{theta} sin{phi} - cos{phi} sin{psi} & cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} \
%
cos{theta}sin{psi} & cos{phi} cos{psi} + sin{theta} sin{phi} sin{psi} & cos{phi}sin{theta} sin{psi} - cos{psi} sin{phi} \
%
-sin{theta} & cos{theta}sin{phi} & cos{theta}cos{phi}
end{array}
right]
end{split}
end{equation}
With the $zyx$ rotation order, the same matrix is
$$
R =
left[
begin{array}{cccc}
cos{theta'}cos{phi'} & -cos{theta'}sin{phi'} & sin{theta'} \
cos{psi'}sin{phi'}+ cos{phi'} sin{theta'}sin{psi'} & cos{phi'}cos{psi'} - sin{theta'}sin{phi'}sin{psi'} & -cos{theta'}sin{psi'} \
sin{phi'}sin{psi'}- cos{phi'}cos{psi'}sin{theta'} & cos{psi'}sin{theta'}sin{phi'} + cos{phi'}sin{psi'} & cos{theta'}cos{psi'} \
end{array}
right]
$$
What you're asking is how to convert this to the $zyx$ rotation order. So let's assume that you have calculated the rotation matrix $R$ with the $xyz$ rotation order (the first method). We need to now interpret the angles as if you had created the matrix with the second method. It's relatively easy to extract the angles from the second matrix:
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-R_{01}}{R_{00}}} \
theta' =& arcsin{R_{02}} \
psi'=& arctan{frac{-R_{12}}{R_{22}}} \
end{array}
right.
$$
Note that the entries here are from the second $R$-matrix. I've added the prime symbol to these angles to denote that they are the angles of the $zyx$-rotation order. The non-prime angle symbols are from the $xyz$-rotation order. Now we can plug in the values from the first matrix to obtain
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-cos{psi} sin{theta} sin{phi} + cos{phi} sin{psi}}{cos{theta}cos{psi} }} \
theta' =& arcsin{left( cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} right)} \
psi'=& arctan{frac{-cos{phi}sin{theta} sin{psi} + cos{psi} sin{phi} }{cos{theta}cos{phi}}} \
end{array}
right.
$$
You might want to double-check these results yourself, as there are quite many terms.
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
$endgroup$
With Tait-Bryan angles orientation representation with rotation order $xyz$, the rotation matrix is
begin{equation}label{eq:euler-model-matrix}
%hspace*{-6cm}
begin{split}
R &= R_Z(psi) R_Y(theta) R_X(phi) \
&=
left[
begin{array}{ccc}
cos{psi} & -sin{psi} & 0 \
sin{psi} & cos{psi} & 0 \
0 & 0 & 1 \
end{array} right]
left[
begin{array}{ccc}
cos{theta} & 0 & sin{theta} \
0 & 1 & 0 \
-sin{theta} & 0 & cos{theta} \
end{array} right]
left[
begin{array}{ccc}
1 & 0 & 0 \
0 & cos{phi} & -sin{phi} \
0 & sin{phi} & cos{phi} \
end{array} right]
\
&=
left[
begin{array}{ccc}
cos{theta}cos{psi} & cos{psi} sin{theta} sin{phi} - cos{phi} sin{psi} & cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} \
%
cos{theta}sin{psi} & cos{phi} cos{psi} + sin{theta} sin{phi} sin{psi} & cos{phi}sin{theta} sin{psi} - cos{psi} sin{phi} \
%
-sin{theta} & cos{theta}sin{phi} & cos{theta}cos{phi}
end{array}
right]
end{split}
end{equation}
With the $zyx$ rotation order, the same matrix is
$$
R =
left[
begin{array}{cccc}
cos{theta'}cos{phi'} & -cos{theta'}sin{phi'} & sin{theta'} \
cos{psi'}sin{phi'}+ cos{phi'} sin{theta'}sin{psi'} & cos{phi'}cos{psi'} - sin{theta'}sin{phi'}sin{psi'} & -cos{theta'}sin{psi'} \
sin{phi'}sin{psi'}- cos{phi'}cos{psi'}sin{theta'} & cos{psi'}sin{theta'}sin{phi'} + cos{phi'}sin{psi'} & cos{theta'}cos{psi'} \
end{array}
right]
$$
What you're asking is how to convert this to the $zyx$ rotation order. So let's assume that you have calculated the rotation matrix $R$ with the $xyz$ rotation order (the first method). We need to now interpret the angles as if you had created the matrix with the second method. It's relatively easy to extract the angles from the second matrix:
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-R_{01}}{R_{00}}} \
theta' =& arcsin{R_{02}} \
psi'=& arctan{frac{-R_{12}}{R_{22}}} \
end{array}
right.
$$
Note that the entries here are from the second $R$-matrix. I've added the prime symbol to these angles to denote that they are the angles of the $zyx$-rotation order. The non-prime angle symbols are from the $xyz$-rotation order. Now we can plug in the values from the first matrix to obtain
$$
left{
begin{array}{ll}
phi' =& arctan{frac{-cos{psi} sin{theta} sin{phi} + cos{phi} sin{psi}}{cos{theta}cos{psi} }} \
theta' =& arcsin{left( cos{phi} cos{psi} sin{theta} + sin{phi} sin{psi} right)} \
psi'=& arctan{frac{-cos{phi}sin{theta} sin{psi} + cos{psi} sin{phi} }{cos{theta}cos{phi}}} \
end{array}
right.
$$
You might want to double-check these results yourself, as there are quite many terms.
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
answered Jan 21 at 8:37
Matti P.Matti P.
2,1811414
2,1811414
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
add a comment |
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
I actually now realised that these results are probably not for the intrinsic case, but rather for the extrinsic case. Well, the idea is the same. But in order to to get the transformation matrices for the intrinsic case, you just have to flip the rotation order (from $xyz$ to $zyx$) and the corresponding angles.
$endgroup$
– Matti P.
Jan 21 at 8:41
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
$begingroup$
Thanks for your detailed explanation. you are my life saver!
$endgroup$
– Zen
Jan 21 at 9:08
add a comment |
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