Mixing
Finding the Quaternion that rotates a coordinate system to match another.
$begingroup$
Let's say I want to figure out the orientation of my cell phone. Assume that the phone has two internal sensors that report orientation (a quaternion), but both are a bit unreliable, so I'd like to use them together.
HOWEVER, the coordinate frames the sensors report a quaternion are not the same.
1) Sensor 1 reports q1, a quat in the following frame:
+x = up, +y = right, +z = forward
2) Sensor 2 reports q2, a quat in the following frame:
+x = down, +y = forward, +z = left
Note that both frames are right handed, and there exists a quaternion that rotates frame 1 to frame 2 (I think).
I'd like to apply a rotation to whatever orientation I'm getting from sensor 1, so the data roughly matches readings from sensor 2.
qx * q1 ~= q2
I'd like to figure out what qx is.
ALTERNATE SOLUTION:
I was able to convert q1 into q2 frame using:
q1_in_q2_frame = [-q1i, q1k, q1j, q1]
But, I'd like to achieve this result by figuring out what qx is, because the coordinate frames are not necessarily like what I said before, they could be more arbitrary. Imagine sensor 2 always resets its frame to whatever orientation the cell phone was in when you turn it on/off, so a new qx has to be calculated.
WHAT I'VE TRIED:
I thought I could just multiply the following equation by q1' from left:
qx * q1 ~= q2
qx * q1 * q1' = q2 * q1'
qx = q2 * q1'
While the above solution gives me a correct mapping for the current values of q1 and q2, qx becomes no longer valid when I rotate the cell phone.
UPDATE:
I think I made an error in my quaternion multiplications.
Instead of qx * q1 ~= q2, I should really have:
qx * q1 * qx' ~= q2
But given q1 and q2, how do I figure out qx now?
3d rotations quaternions
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
$endgroup$
add a comment |
$begingroup$
Let's say I want to figure out the orientation of my cell phone. Assume that the phone has two internal sensors that report orientation (a quaternion), but both are a bit unreliable, so I'd like to use them together.
HOWEVER, the coordinate frames the sensors report a quaternion are not the same.
1) Sensor 1 reports q1, a quat in the following frame:
+x = up, +y = right, +z = forward
2) Sensor 2 reports q2, a quat in the following frame:
+x = down, +y = forward, +z = left
Note that both frames are right handed, and there exists a quaternion that rotates frame 1 to frame 2 (I think).
I'd like to apply a rotation to whatever orientation I'm getting from sensor 1, so the data roughly matches readings from sensor 2.
qx * q1 ~= q2
I'd like to figure out what qx is.
ALTERNATE SOLUTION:
I was able to convert q1 into q2 frame using:
q1_in_q2_frame = [-q1i, q1k, q1j, q1]
But, I'd like to achieve this result by figuring out what qx is, because the coordinate frames are not necessarily like what I said before, they could be more arbitrary. Imagine sensor 2 always resets its frame to whatever orientation the cell phone was in when you turn it on/off, so a new qx has to be calculated.
WHAT I'VE TRIED:
I thought I could just multiply the following equation by q1' from left:
qx * q1 ~= q2
qx * q1 * q1' = q2 * q1'
qx = q2 * q1'
While the above solution gives me a correct mapping for the current values of q1 and q2, qx becomes no longer valid when I rotate the cell phone.
UPDATE:
I think I made an error in my quaternion multiplications.
Instead of qx * q1 ~= q2, I should really have:
qx * q1 * qx' ~= q2
But given q1 and q2, how do I figure out qx now?
3d rotations quaternions
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
$endgroup$
$begingroup$
I think you need to think more carefully about what you know, and what you don't know. To compare two quaternions in two different reference frames, it is the case, as you've pointed out, that you need to multiply one of them by the necessary quaternion to rotate one frame into the other. I would think that on a phone, those two reference frames would be fixed, right?
$endgroup$
– Adrian Keister
Jan 25 at 17:36
add a comment |
$begingroup$
Let's say I want to figure out the orientation of my cell phone. Assume that the phone has two internal sensors that report orientation (a quaternion), but both are a bit unreliable, so I'd like to use them together.
HOWEVER, the coordinate frames the sensors report a quaternion are not the same.
1) Sensor 1 reports q1, a quat in the following frame:
+x = up, +y = right, +z = forward
2) Sensor 2 reports q2, a quat in the following frame:
+x = down, +y = forward, +z = left
Note that both frames are right handed, and there exists a quaternion that rotates frame 1 to frame 2 (I think).
I'd like to apply a rotation to whatever orientation I'm getting from sensor 1, so the data roughly matches readings from sensor 2.
qx * q1 ~= q2
I'd like to figure out what qx is.
ALTERNATE SOLUTION:
I was able to convert q1 into q2 frame using:
q1_in_q2_frame = [-q1i, q1k, q1j, q1]
But, I'd like to achieve this result by figuring out what qx is, because the coordinate frames are not necessarily like what I said before, they could be more arbitrary. Imagine sensor 2 always resets its frame to whatever orientation the cell phone was in when you turn it on/off, so a new qx has to be calculated.
WHAT I'VE TRIED:
I thought I could just multiply the following equation by q1' from left:
qx * q1 ~= q2
qx * q1 * q1' = q2 * q1'
qx = q2 * q1'
While the above solution gives me a correct mapping for the current values of q1 and q2, qx becomes no longer valid when I rotate the cell phone.
UPDATE:
I think I made an error in my quaternion multiplications.
Instead of qx * q1 ~= q2, I should really have:
qx * q1 * qx' ~= q2
But given q1 and q2, how do I figure out qx now?
3d rotations quaternions
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
$endgroup$
Let's say I want to figure out the orientation of my cell phone. Assume that the phone has two internal sensors that report orientation (a quaternion), but both are a bit unreliable, so I'd like to use them together.
HOWEVER, the coordinate frames the sensors report a quaternion are not the same.
1) Sensor 1 reports q1, a quat in the following frame:
+x = up, +y = right, +z = forward
2) Sensor 2 reports q2, a quat in the following frame:
+x = down, +y = forward, +z = left
Note that both frames are right handed, and there exists a quaternion that rotates frame 1 to frame 2 (I think).
I'd like to apply a rotation to whatever orientation I'm getting from sensor 1, so the data roughly matches readings from sensor 2.
qx * q1 ~= q2
I'd like to figure out what qx is.
ALTERNATE SOLUTION:
I was able to convert q1 into q2 frame using:
q1_in_q2_frame = [-q1i, q1k, q1j, q1]
But, I'd like to achieve this result by figuring out what qx is, because the coordinate frames are not necessarily like what I said before, they could be more arbitrary. Imagine sensor 2 always resets its frame to whatever orientation the cell phone was in when you turn it on/off, so a new qx has to be calculated.
WHAT I'VE TRIED:
I thought I could just multiply the following equation by q1' from left:
qx * q1 ~= q2
qx * q1 * q1' = q2 * q1'
qx = q2 * q1'
While the above solution gives me a correct mapping for the current values of q1 and q2, qx becomes no longer valid when I rotate the cell phone.
UPDATE:
I think I made an error in my quaternion multiplications.
Instead of qx * q1 ~= q2, I should really have:
qx * q1 * qx' ~= q2
But given q1 and q2, how do I figure out qx now?
3d rotations quaternions
3d rotations quaternions
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
edited Jan 25 at 20:07
Övünç Tüzel
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
asked Jan 25 at 17:28
Övünç TüzelÖvünç Tüzel
163
163
$begingroup$
I think you need to think more carefully about what you know, and what you don't know. To compare two quaternions in two different reference frames, it is the case, as you've pointed out, that you need to multiply one of them by the necessary quaternion to rotate one frame into the other. I would think that on a phone, those two reference frames would be fixed, right?
$endgroup$
– Adrian Keister
Jan 25 at 17:36
add a comment |
$begingroup$
I think you need to think more carefully about what you know, and what you don't know. To compare two quaternions in two different reference frames, it is the case, as you've pointed out, that you need to multiply one of them by the necessary quaternion to rotate one frame into the other. I would think that on a phone, those two reference frames would be fixed, right?
$endgroup$
– Adrian Keister
Jan 25 at 17:36
$begingroup$
I think you need to think more carefully about what you know, and what you don't know. To compare two quaternions in two different reference frames, it is the case, as you've pointed out, that you need to multiply one of them by the necessary quaternion to rotate one frame into the other. I would think that on a phone, those two reference frames would be fixed, right?
$endgroup$
– Adrian Keister
Jan 25 at 17:36
$begingroup$
I think you need to think more carefully about what you know, and what you don't know. To compare two quaternions in two different reference frames, it is the case, as you've pointed out, that you need to multiply one of them by the necessary quaternion to rotate one frame into the other. I would think that on a phone, those two reference frames would be fixed, right?
$endgroup$
– Adrian Keister
Jan 25 at 17:36
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is a simple algebraic solution in terms of quaternions (to your revised question.)
If $v_1$ and $v_2$ are unit quaternions with real part zero, then $q=frac{v_1(v_1+v_2)}{|v_1+v_2|}$ is a quaternion such that $bar{q}v_1q=v_2$. The caveat is, of course, that $v_1$ and $v_2$ do not point in opposite directions, so that the division is defined. If they point in opposite directions, well, you can take any 180 degree rotation in a plane containing the vectors to achieve your goal.
Why does this work? If you go digging, you'll find the formula for $q$ to rotate an element of $mathbb R^3$ to another (both represented as pure imaginary quaternions with zero real part) then the way to do it is to compute $cos(theta/2)-sin(theta/2)v_3$ where $v_3$ is the (right hand) unit normal to the plane spanned by $v_1$ and $v_2$, and $theta$ is the angle measured between $v_1$ and $v_2$.
Now it turns out that $v_1v_2=-cos(theta)+sin(theta)(v_1times v_2)$ and since its negative also represents the same rotation, we can see we're not far off: our angle is just double what it needs to be. What to do then?
That's where $frac{v_1+v_2}{|v_1+v_2|}$ comes in: it's a new vector which produces the same unit normal as before, but now the angle has been halved.
Now, you say, "but I don't want $q_1$ and $q_2$ to have zero real part, I want them to be arbitrary." But that's OK, because when you conjugate with a unit quaternion, it leaves the real part alone. So all you need to do is to compute $q$ as I described for the pure quaternions parts of $q_1$ and $q_2$, and that will work to satisfy $bar{q}q_1q=q_2$.
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
$endgroup$
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
add a comment |
$begingroup$
Are you familar with Direct Cosine Matrices? You are able to form the DCM by directly taking the dot product of the axes. You can then convert the DCM to a quaternion
Take a look at www.starlino.com/dcm_tutorial.html as an example.
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
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add a comment |
$begingroup$
Not sure if I understand your question. But for your last update you are looking for quaternion $x$ such that the following equation holds for some quaternions $p$ and $q$.
$x q x^* = p$
S.t. $x x^* = 1$
As there is no algebraic solution to this equation the next thing to do is casting the problem to the language of linear algebra.
First notice that the equation $x q x^* = p$ is equivalent to
$x q - p x = 0$
S.t. $x x^* = 1$
The product $x q$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $Q$ representing the quaternion $q$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
The product $p x$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $P$ representing the quaternion $p$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
So we have now:
$Q X - P X = 0$
S.t. $X^T X = 1$
Which is a pretty standard matrix equation. Factorizing the $X$ we get:
$(Q - P) X = 0$
S.t. $X^T X = 1$
We can see that the solution is in the null space of the matrix $A = Q - P$. Using the Simgular Value Decomposition (SVD) of the matrix $A$ we get that $A = U D V^T$ where $U$, $D$ and $V$ are $4 times 4$ square matrices. $D$ is a diagonal matrix containing the singular values.
If all values in the diagonal of $D$ are nonzero, then $A$ is full rank and no solution exists (the null space of $A$ is $0$).
If only one value in the diagonal of $D$ is zero then there is only one solution for $X$ which is the column of $V$ corresponding to the zero diagonal value.
If more than one value is zero then there are multiple solutions for $X$. You should choose one.
The constraint $X^T X = 1$ is fulfilled by SVD since the matrix $V$ is orthonormal.
Finally I will define the $Q$ and $P$ matrices given quaternion $q = q_0 + q_x i + q_y j + q_z k$ and quaternion $p = p_0 + p_x i + p_y j + p_z k$:
$x q = Q X = begin{bmatrix}
q_0&-q_x&-q_y&-q_z\
q_x&q_0&q_z&-q_y\
q_y&-q_z&q_0&q_x\
q_z&q_y&-q_x&q_0
end{bmatrix} X$
$p x = P X = begin{bmatrix}
p_0&-p_x&-p_y&-p_z\
p_x&p_0&-p_z&p_y\
p_y&p_z&p_0&-p_x\
p_z&-p_y&p_x&p_0
end{bmatrix} X$
Note that $Q$ differs from $P$ in that the lower $3 times 3$ submatrix is transposed. This is because the noncommutative nature of multiplication of quaternions.
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
There is a simple algebraic solution in terms of quaternions (to your revised question.)
If $v_1$ and $v_2$ are unit quaternions with real part zero, then $q=frac{v_1(v_1+v_2)}{|v_1+v_2|}$ is a quaternion such that $bar{q}v_1q=v_2$. The caveat is, of course, that $v_1$ and $v_2$ do not point in opposite directions, so that the division is defined. If they point in opposite directions, well, you can take any 180 degree rotation in a plane containing the vectors to achieve your goal.
Why does this work? If you go digging, you'll find the formula for $q$ to rotate an element of $mathbb R^3$ to another (both represented as pure imaginary quaternions with zero real part) then the way to do it is to compute $cos(theta/2)-sin(theta/2)v_3$ where $v_3$ is the (right hand) unit normal to the plane spanned by $v_1$ and $v_2$, and $theta$ is the angle measured between $v_1$ and $v_2$.
Now it turns out that $v_1v_2=-cos(theta)+sin(theta)(v_1times v_2)$ and since its negative also represents the same rotation, we can see we're not far off: our angle is just double what it needs to be. What to do then?
That's where $frac{v_1+v_2}{|v_1+v_2|}$ comes in: it's a new vector which produces the same unit normal as before, but now the angle has been halved.
Now, you say, "but I don't want $q_1$ and $q_2$ to have zero real part, I want them to be arbitrary." But that's OK, because when you conjugate with a unit quaternion, it leaves the real part alone. So all you need to do is to compute $q$ as I described for the pure quaternions parts of $q_1$ and $q_2$, and that will work to satisfy $bar{q}q_1q=q_2$.
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
$endgroup$
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
add a comment |
$begingroup$
There is a simple algebraic solution in terms of quaternions (to your revised question.)
If $v_1$ and $v_2$ are unit quaternions with real part zero, then $q=frac{v_1(v_1+v_2)}{|v_1+v_2|}$ is a quaternion such that $bar{q}v_1q=v_2$. The caveat is, of course, that $v_1$ and $v_2$ do not point in opposite directions, so that the division is defined. If they point in opposite directions, well, you can take any 180 degree rotation in a plane containing the vectors to achieve your goal.
Why does this work? If you go digging, you'll find the formula for $q$ to rotate an element of $mathbb R^3$ to another (both represented as pure imaginary quaternions with zero real part) then the way to do it is to compute $cos(theta/2)-sin(theta/2)v_3$ where $v_3$ is the (right hand) unit normal to the plane spanned by $v_1$ and $v_2$, and $theta$ is the angle measured between $v_1$ and $v_2$.
Now it turns out that $v_1v_2=-cos(theta)+sin(theta)(v_1times v_2)$ and since its negative also represents the same rotation, we can see we're not far off: our angle is just double what it needs to be. What to do then?
That's where $frac{v_1+v_2}{|v_1+v_2|}$ comes in: it's a new vector which produces the same unit normal as before, but now the angle has been halved.
Now, you say, "but I don't want $q_1$ and $q_2$ to have zero real part, I want them to be arbitrary." But that's OK, because when you conjugate with a unit quaternion, it leaves the real part alone. So all you need to do is to compute $q$ as I described for the pure quaternions parts of $q_1$ and $q_2$, and that will work to satisfy $bar{q}q_1q=q_2$.
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
$endgroup$
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
add a comment |
$begingroup$
There is a simple algebraic solution in terms of quaternions (to your revised question.)
If $v_1$ and $v_2$ are unit quaternions with real part zero, then $q=frac{v_1(v_1+v_2)}{|v_1+v_2|}$ is a quaternion such that $bar{q}v_1q=v_2$. The caveat is, of course, that $v_1$ and $v_2$ do not point in opposite directions, so that the division is defined. If they point in opposite directions, well, you can take any 180 degree rotation in a plane containing the vectors to achieve your goal.
Why does this work? If you go digging, you'll find the formula for $q$ to rotate an element of $mathbb R^3$ to another (both represented as pure imaginary quaternions with zero real part) then the way to do it is to compute $cos(theta/2)-sin(theta/2)v_3$ where $v_3$ is the (right hand) unit normal to the plane spanned by $v_1$ and $v_2$, and $theta$ is the angle measured between $v_1$ and $v_2$.
Now it turns out that $v_1v_2=-cos(theta)+sin(theta)(v_1times v_2)$ and since its negative also represents the same rotation, we can see we're not far off: our angle is just double what it needs to be. What to do then?
That's where $frac{v_1+v_2}{|v_1+v_2|}$ comes in: it's a new vector which produces the same unit normal as before, but now the angle has been halved.
Now, you say, "but I don't want $q_1$ and $q_2$ to have zero real part, I want them to be arbitrary." But that's OK, because when you conjugate with a unit quaternion, it leaves the real part alone. So all you need to do is to compute $q$ as I described for the pure quaternions parts of $q_1$ and $q_2$, and that will work to satisfy $bar{q}q_1q=q_2$.
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
$endgroup$
There is a simple algebraic solution in terms of quaternions (to your revised question.)
If $v_1$ and $v_2$ are unit quaternions with real part zero, then $q=frac{v_1(v_1+v_2)}{|v_1+v_2|}$ is a quaternion such that $bar{q}v_1q=v_2$. The caveat is, of course, that $v_1$ and $v_2$ do not point in opposite directions, so that the division is defined. If they point in opposite directions, well, you can take any 180 degree rotation in a plane containing the vectors to achieve your goal.
Why does this work? If you go digging, you'll find the formula for $q$ to rotate an element of $mathbb R^3$ to another (both represented as pure imaginary quaternions with zero real part) then the way to do it is to compute $cos(theta/2)-sin(theta/2)v_3$ where $v_3$ is the (right hand) unit normal to the plane spanned by $v_1$ and $v_2$, and $theta$ is the angle measured between $v_1$ and $v_2$.
Now it turns out that $v_1v_2=-cos(theta)+sin(theta)(v_1times v_2)$ and since its negative also represents the same rotation, we can see we're not far off: our angle is just double what it needs to be. What to do then?
That's where $frac{v_1+v_2}{|v_1+v_2|}$ comes in: it's a new vector which produces the same unit normal as before, but now the angle has been halved.
Now, you say, "but I don't want $q_1$ and $q_2$ to have zero real part, I want them to be arbitrary." But that's OK, because when you conjugate with a unit quaternion, it leaves the real part alone. So all you need to do is to compute $q$ as I described for the pure quaternions parts of $q_1$ and $q_2$, and that will work to satisfy $bar{q}q_1q=q_2$.
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
answered Jan 26 at 13:49
rschwiebrschwieb
107k12103251
107k12103251
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
add a comment |
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
Makes sense. I knew about that result for pure quaternions but didn't realize its validity for regular quaternions. Thank you for the clear explanation!
$endgroup$
– Mauricio Cele Lopez Belon
Jan 26 at 20:06
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
@rshwieb This is a really promising answer, thank you very much! My question is, if q' q1 q operation doesn't change the real part of q1, how does the equation hold if the real part of q1 and q2 is different? It looks like q can accomplish q' q1 q = q2 only if q1.w and q2.w are same. Am I missing something?
$endgroup$
– Övünç Tüzel
Jan 28 at 19:53
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
I have tested the solution with random quaternions, and it only works if q1 and q2 has the same real part. Otherwise q' q1 q takes the real part of q1 as you've mentioned.
$endgroup$
– Övünç Tüzel
Jan 28 at 19:56
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel If the quaternions do not have the same real part, then of course it is impossible to find such a $q$. So, it "only" works in every case where there is a solution, yes.
$endgroup$
– rschwieb
Jan 28 at 20:02
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
$begingroup$
@ÖvünçTüzel THere exists a $q$ such that $q^{-1}xq=y$ iff $x$ and $y$ have the same real parts and the lengths of their pure quaternion parts are equal.
$endgroup$
– rschwieb
Jan 28 at 20:03
add a comment |
$begingroup$
Are you familar with Direct Cosine Matrices? You are able to form the DCM by directly taking the dot product of the axes. You can then convert the DCM to a quaternion
Take a look at www.starlino.com/dcm_tutorial.html as an example.
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
$endgroup$
add a comment |
$begingroup$
Are you familar with Direct Cosine Matrices? You are able to form the DCM by directly taking the dot product of the axes. You can then convert the DCM to a quaternion
Take a look at www.starlino.com/dcm_tutorial.html as an example.
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
$endgroup$
add a comment |
$begingroup$
Are you familar with Direct Cosine Matrices? You are able to form the DCM by directly taking the dot product of the axes. You can then convert the DCM to a quaternion
Take a look at www.starlino.com/dcm_tutorial.html as an example.
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
$endgroup$
Are you familar with Direct Cosine Matrices? You are able to form the DCM by directly taking the dot product of the axes. You can then convert the DCM to a quaternion
Take a look at www.starlino.com/dcm_tutorial.html as an example.
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
answered Jan 25 at 19:00
SRVfender01SRVfender01
112
112
add a comment |
add a comment |
$begingroup$
Not sure if I understand your question. But for your last update you are looking for quaternion $x$ such that the following equation holds for some quaternions $p$ and $q$.
$x q x^* = p$
S.t. $x x^* = 1$
As there is no algebraic solution to this equation the next thing to do is casting the problem to the language of linear algebra.
First notice that the equation $x q x^* = p$ is equivalent to
$x q - p x = 0$
S.t. $x x^* = 1$
The product $x q$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $Q$ representing the quaternion $q$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
The product $p x$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $P$ representing the quaternion $p$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
So we have now:
$Q X - P X = 0$
S.t. $X^T X = 1$
Which is a pretty standard matrix equation. Factorizing the $X$ we get:
$(Q - P) X = 0$
S.t. $X^T X = 1$
We can see that the solution is in the null space of the matrix $A = Q - P$. Using the Simgular Value Decomposition (SVD) of the matrix $A$ we get that $A = U D V^T$ where $U$, $D$ and $V$ are $4 times 4$ square matrices. $D$ is a diagonal matrix containing the singular values.
If all values in the diagonal of $D$ are nonzero, then $A$ is full rank and no solution exists (the null space of $A$ is $0$).
If only one value in the diagonal of $D$ is zero then there is only one solution for $X$ which is the column of $V$ corresponding to the zero diagonal value.
If more than one value is zero then there are multiple solutions for $X$. You should choose one.
The constraint $X^T X = 1$ is fulfilled by SVD since the matrix $V$ is orthonormal.
Finally I will define the $Q$ and $P$ matrices given quaternion $q = q_0 + q_x i + q_y j + q_z k$ and quaternion $p = p_0 + p_x i + p_y j + p_z k$:
$x q = Q X = begin{bmatrix}
q_0&-q_x&-q_y&-q_z\
q_x&q_0&q_z&-q_y\
q_y&-q_z&q_0&q_x\
q_z&q_y&-q_x&q_0
end{bmatrix} X$
$p x = P X = begin{bmatrix}
p_0&-p_x&-p_y&-p_z\
p_x&p_0&-p_z&p_y\
p_y&p_z&p_0&-p_x\
p_z&-p_y&p_x&p_0
end{bmatrix} X$
Note that $Q$ differs from $P$ in that the lower $3 times 3$ submatrix is transposed. This is because the noncommutative nature of multiplication of quaternions.
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
$endgroup$
add a comment |
$begingroup$
Not sure if I understand your question. But for your last update you are looking for quaternion $x$ such that the following equation holds for some quaternions $p$ and $q$.
$x q x^* = p$
S.t. $x x^* = 1$
As there is no algebraic solution to this equation the next thing to do is casting the problem to the language of linear algebra.
First notice that the equation $x q x^* = p$ is equivalent to
$x q - p x = 0$
S.t. $x x^* = 1$
The product $x q$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $Q$ representing the quaternion $q$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
The product $p x$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $P$ representing the quaternion $p$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
So we have now:
$Q X - P X = 0$
S.t. $X^T X = 1$
Which is a pretty standard matrix equation. Factorizing the $X$ we get:
$(Q - P) X = 0$
S.t. $X^T X = 1$
We can see that the solution is in the null space of the matrix $A = Q - P$. Using the Simgular Value Decomposition (SVD) of the matrix $A$ we get that $A = U D V^T$ where $U$, $D$ and $V$ are $4 times 4$ square matrices. $D$ is a diagonal matrix containing the singular values.
If all values in the diagonal of $D$ are nonzero, then $A$ is full rank and no solution exists (the null space of $A$ is $0$).
If only one value in the diagonal of $D$ is zero then there is only one solution for $X$ which is the column of $V$ corresponding to the zero diagonal value.
If more than one value is zero then there are multiple solutions for $X$. You should choose one.
The constraint $X^T X = 1$ is fulfilled by SVD since the matrix $V$ is orthonormal.
Finally I will define the $Q$ and $P$ matrices given quaternion $q = q_0 + q_x i + q_y j + q_z k$ and quaternion $p = p_0 + p_x i + p_y j + p_z k$:
$x q = Q X = begin{bmatrix}
q_0&-q_x&-q_y&-q_z\
q_x&q_0&q_z&-q_y\
q_y&-q_z&q_0&q_x\
q_z&q_y&-q_x&q_0
end{bmatrix} X$
$p x = P X = begin{bmatrix}
p_0&-p_x&-p_y&-p_z\
p_x&p_0&-p_z&p_y\
p_y&p_z&p_0&-p_x\
p_z&-p_y&p_x&p_0
end{bmatrix} X$
Note that $Q$ differs from $P$ in that the lower $3 times 3$ submatrix is transposed. This is because the noncommutative nature of multiplication of quaternions.
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
$endgroup$
add a comment |
$begingroup$
Not sure if I understand your question. But for your last update you are looking for quaternion $x$ such that the following equation holds for some quaternions $p$ and $q$.
$x q x^* = p$
S.t. $x x^* = 1$
As there is no algebraic solution to this equation the next thing to do is casting the problem to the language of linear algebra.
First notice that the equation $x q x^* = p$ is equivalent to
$x q - p x = 0$
S.t. $x x^* = 1$
The product $x q$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $Q$ representing the quaternion $q$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
The product $p x$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $P$ representing the quaternion $p$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
So we have now:
$Q X - P X = 0$
S.t. $X^T X = 1$
Which is a pretty standard matrix equation. Factorizing the $X$ we get:
$(Q - P) X = 0$
S.t. $X^T X = 1$
We can see that the solution is in the null space of the matrix $A = Q - P$. Using the Simgular Value Decomposition (SVD) of the matrix $A$ we get that $A = U D V^T$ where $U$, $D$ and $V$ are $4 times 4$ square matrices. $D$ is a diagonal matrix containing the singular values.
If all values in the diagonal of $D$ are nonzero, then $A$ is full rank and no solution exists (the null space of $A$ is $0$).
If only one value in the diagonal of $D$ is zero then there is only one solution for $X$ which is the column of $V$ corresponding to the zero diagonal value.
If more than one value is zero then there are multiple solutions for $X$. You should choose one.
The constraint $X^T X = 1$ is fulfilled by SVD since the matrix $V$ is orthonormal.
Finally I will define the $Q$ and $P$ matrices given quaternion $q = q_0 + q_x i + q_y j + q_z k$ and quaternion $p = p_0 + p_x i + p_y j + p_z k$:
$x q = Q X = begin{bmatrix}
q_0&-q_x&-q_y&-q_z\
q_x&q_0&q_z&-q_y\
q_y&-q_z&q_0&q_x\
q_z&q_y&-q_x&q_0
end{bmatrix} X$
$p x = P X = begin{bmatrix}
p_0&-p_x&-p_y&-p_z\
p_x&p_0&-p_z&p_y\
p_y&p_z&p_0&-p_x\
p_z&-p_y&p_x&p_0
end{bmatrix} X$
Note that $Q$ differs from $P$ in that the lower $3 times 3$ submatrix is transposed. This is because the noncommutative nature of multiplication of quaternions.
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
$endgroup$
Not sure if I understand your question. But for your last update you are looking for quaternion $x$ such that the following equation holds for some quaternions $p$ and $q$.
$x q x^* = p$
S.t. $x x^* = 1$
As there is no algebraic solution to this equation the next thing to do is casting the problem to the language of linear algebra.
First notice that the equation $x q x^* = p$ is equivalent to
$x q - p x = 0$
S.t. $x x^* = 1$
The product $x q$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $Q$ representing the quaternion $q$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
The product $p x$ can be expressed in linear algebra as a matrix multiplication of a $4 times 4$ matrix $P$ representing the quaternion $p$ and a $4 times 1$ column vector $X$ representing quafernion $x$.
So we have now:
$Q X - P X = 0$
S.t. $X^T X = 1$
Which is a pretty standard matrix equation. Factorizing the $X$ we get:
$(Q - P) X = 0$
S.t. $X^T X = 1$
We can see that the solution is in the null space of the matrix $A = Q - P$. Using the Simgular Value Decomposition (SVD) of the matrix $A$ we get that $A = U D V^T$ where $U$, $D$ and $V$ are $4 times 4$ square matrices. $D$ is a diagonal matrix containing the singular values.
If all values in the diagonal of $D$ are nonzero, then $A$ is full rank and no solution exists (the null space of $A$ is $0$).
If only one value in the diagonal of $D$ is zero then there is only one solution for $X$ which is the column of $V$ corresponding to the zero diagonal value.
If more than one value is zero then there are multiple solutions for $X$. You should choose one.
The constraint $X^T X = 1$ is fulfilled by SVD since the matrix $V$ is orthonormal.
Finally I will define the $Q$ and $P$ matrices given quaternion $q = q_0 + q_x i + q_y j + q_z k$ and quaternion $p = p_0 + p_x i + p_y j + p_z k$:
$x q = Q X = begin{bmatrix}
q_0&-q_x&-q_y&-q_z\
q_x&q_0&q_z&-q_y\
q_y&-q_z&q_0&q_x\
q_z&q_y&-q_x&q_0
end{bmatrix} X$
$p x = P X = begin{bmatrix}
p_0&-p_x&-p_y&-p_z\
p_x&p_0&-p_z&p_y\
p_y&p_z&p_0&-p_x\
p_z&-p_y&p_x&p_0
end{bmatrix} X$
Note that $Q$ differs from $P$ in that the lower $3 times 3$ submatrix is transposed. This is because the noncommutative nature of multiplication of quaternions.
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
answered Jan 25 at 22:21
Mauricio Cele Lopez BelonMauricio Cele Lopez Belon
839411
839411
add a comment |
add a comment |
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$begingroup$
I think you need to think more carefully about what you know, and what you don't know. To compare two quaternions in two different reference frames, it is the case, as you've pointed out, that you need to multiply one of them by the necessary quaternion to rotate one frame into the other. I would think that on a phone, those two reference frames would be fixed, right?
$endgroup$
– Adrian Keister
Jan 25 at 17:36