quaternion renormalization
2
$begingroup$
I am trying to create 4 symbolic equations that correspond to the 4 elements of a normalized quaternion. The equation is for quaternion multiplication
delta q = 0.5*q*[0, omega]^T
and then I normalize the output. My solution is:
import sympy as sy
q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n = sy.symbols("q_r q_i q_j q_k
w_ib_l w_ib_m w_ib_n")
q1 = [q_r, q_i, q_j, q_k]
q2 = [0, w_ib_l, w_ib_m, w_ib_n]
def quat_equ(q_l,q_w):
q_r_d = 0.5 * (q_l[0] * q_w[0] - q_l[1] * q_w[1] + q_l[2] * q_w[2] +
q_l[3] * q_w[3])
q_i_d = 0.5 * (q_l[0] * q_w[1] + q_l[1] * q_w[0] + q_l[2] * q_w[3] -
q_l[3] * q_w[2])
q_j_d = 0.5 * (q_l[0] * q_w[2] + q_l[2] * q_w[0] + q_l[3] * q_w[1] -
q_l[1] * q_w[3])
q_k_d = 0.5 * (q_l[0] * q_w[3] + q_l[3] * q_w[0] + q_l[1] * q_w[2] -
q_l[2] * q_w[1])
for i in range(1):
q_r_d *= 1 / sy.sqrt(q_r_d**2 + q_i_d**2 + q_j_d**2 + q_k_d**2)
q_i_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_j_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_k_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
return [q_r_d, q_i_d, q_j_d, q_k_d]
quat0 = quat_equ(q1, q2)[0]
quat1 = quat_equ(q1, q2)[1]
quat2 = quat_equ(q1, q2)[2]
quat3 = quat_equ(q1, q2)[3]
qq = [1., 0., 0., 0., 1., 1., 1.]
var = [q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n]
q0 = sy.Subs(quat0, var, qq).doit()
q1 = sy.Subs(quat1, var, qq).doit()
q2 = sy.Subs(quat2, var, qq).doit()
q3 = sy.Subs(quat3, var, qq).doit()
When I try to re-normalize the quaternion more than 2 times this method becomes very slow. The symbolic equation also becomes massive because the re-normalization adds on basically 4 extra symbolic equations, which slows down later processes a lot. I couldn't think of a better method to create a normalized quaternion that 1) initializes faster and 2) is a smaller symbolic equation. I don't use symbolics in python much so most of Sympy is new to me.
python symbolic-math sympy
asked 6 hours ago
BotoBoto
111
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Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
2
$begingroup$
I am trying to create 4 symbolic equations that correspond to the 4 elements of a normalized quaternion. The equation is for quaternion multiplication
delta q = 0.5*q*[0, omega]^T
and then I normalize the output. My solution is:
import sympy as sy
q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n = sy.symbols("q_r q_i q_j q_k
w_ib_l w_ib_m w_ib_n")
q1 = [q_r, q_i, q_j, q_k]
q2 = [0, w_ib_l, w_ib_m, w_ib_n]
def quat_equ(q_l,q_w):
q_r_d = 0.5 * (q_l[0] * q_w[0] - q_l[1] * q_w[1] + q_l[2] * q_w[2] +
q_l[3] * q_w[3])
q_i_d = 0.5 * (q_l[0] * q_w[1] + q_l[1] * q_w[0] + q_l[2] * q_w[3] -
q_l[3] * q_w[2])
q_j_d = 0.5 * (q_l[0] * q_w[2] + q_l[2] * q_w[0] + q_l[3] * q_w[1] -
q_l[1] * q_w[3])
q_k_d = 0.5 * (q_l[0] * q_w[3] + q_l[3] * q_w[0] + q_l[1] * q_w[2] -
q_l[2] * q_w[1])
for i in range(1):
q_r_d *= 1 / sy.sqrt(q_r_d**2 + q_i_d**2 + q_j_d**2 + q_k_d**2)
q_i_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_j_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_k_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
return [q_r_d, q_i_d, q_j_d, q_k_d]
quat0 = quat_equ(q1, q2)[0]
quat1 = quat_equ(q1, q2)[1]
quat2 = quat_equ(q1, q2)[2]
quat3 = quat_equ(q1, q2)[3]
qq = [1., 0., 0., 0., 1., 1., 1.]
var = [q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n]
q0 = sy.Subs(quat0, var, qq).doit()
q1 = sy.Subs(quat1, var, qq).doit()
q2 = sy.Subs(quat2, var, qq).doit()
q3 = sy.Subs(quat3, var, qq).doit()
When I try to re-normalize the quaternion more than 2 times this method becomes very slow. The symbolic equation also becomes massive because the re-normalization adds on basically 4 extra symbolic equations, which slows down later processes a lot. I couldn't think of a better method to create a normalized quaternion that 1) initializes faster and 2) is a smaller symbolic equation. I don't use symbolics in python much so most of Sympy is new to me.
python symbolic-math sympy
asked 6 hours ago
BotoBoto
111
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Why isfor i in range(1)written as a loop? It's indentation is wrong, which makes this broken code. (A good way to post code is to paste it into the question editor, highlight it, and press Ctrl-K to mark it as a code block.)
$endgroup$
– 200_success
5 hours ago
$begingroup$
Note that MathJax is available for writing equations in your question.
$endgroup$
– 200_success
5 hours ago
$begingroup$
In addition to what 200_success alreay said about broken code, the symbol definition is also no valid Python code. Your code could also greatly benefit from some comments to give reviewers a few hints which parts you suspect or know to be slow.
$endgroup$
– Alex
3 hours ago
add a comment |
2
2
2
$begingroup$
I am trying to create 4 symbolic equations that correspond to the 4 elements of a normalized quaternion. The equation is for quaternion multiplication
delta q = 0.5*q*[0, omega]^T
and then I normalize the output. My solution is:
import sympy as sy
q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n = sy.symbols("q_r q_i q_j q_k
w_ib_l w_ib_m w_ib_n")
q1 = [q_r, q_i, q_j, q_k]
q2 = [0, w_ib_l, w_ib_m, w_ib_n]
def quat_equ(q_l,q_w):
q_r_d = 0.5 * (q_l[0] * q_w[0] - q_l[1] * q_w[1] + q_l[2] * q_w[2] +
q_l[3] * q_w[3])
q_i_d = 0.5 * (q_l[0] * q_w[1] + q_l[1] * q_w[0] + q_l[2] * q_w[3] -
q_l[3] * q_w[2])
q_j_d = 0.5 * (q_l[0] * q_w[2] + q_l[2] * q_w[0] + q_l[3] * q_w[1] -
q_l[1] * q_w[3])
q_k_d = 0.5 * (q_l[0] * q_w[3] + q_l[3] * q_w[0] + q_l[1] * q_w[2] -
q_l[2] * q_w[1])
for i in range(1):
q_r_d *= 1 / sy.sqrt(q_r_d**2 + q_i_d**2 + q_j_d**2 + q_k_d**2)
q_i_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_j_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_k_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
return [q_r_d, q_i_d, q_j_d, q_k_d]
quat0 = quat_equ(q1, q2)[0]
quat1 = quat_equ(q1, q2)[1]
quat2 = quat_equ(q1, q2)[2]
quat3 = quat_equ(q1, q2)[3]
qq = [1., 0., 0., 0., 1., 1., 1.]
var = [q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n]
q0 = sy.Subs(quat0, var, qq).doit()
q1 = sy.Subs(quat1, var, qq).doit()
q2 = sy.Subs(quat2, var, qq).doit()
q3 = sy.Subs(quat3, var, qq).doit()
When I try to re-normalize the quaternion more than 2 times this method becomes very slow. The symbolic equation also becomes massive because the re-normalization adds on basically 4 extra symbolic equations, which slows down later processes a lot. I couldn't think of a better method to create a normalized quaternion that 1) initializes faster and 2) is a smaller symbolic equation. I don't use symbolics in python much so most of Sympy is new to me.
python symbolic-math sympy
asked 6 hours ago
BotoBoto
111
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to create 4 symbolic equations that correspond to the 4 elements of a normalized quaternion. The equation is for quaternion multiplication
delta q = 0.5*q*[0, omega]^T
and then I normalize the output. My solution is:
import sympy as sy
q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n = sy.symbols("q_r q_i q_j q_k
w_ib_l w_ib_m w_ib_n")
q1 = [q_r, q_i, q_j, q_k]
q2 = [0, w_ib_l, w_ib_m, w_ib_n]
def quat_equ(q_l,q_w):
q_r_d = 0.5 * (q_l[0] * q_w[0] - q_l[1] * q_w[1] + q_l[2] * q_w[2] +
q_l[3] * q_w[3])
q_i_d = 0.5 * (q_l[0] * q_w[1] + q_l[1] * q_w[0] + q_l[2] * q_w[3] -
q_l[3] * q_w[2])
q_j_d = 0.5 * (q_l[0] * q_w[2] + q_l[2] * q_w[0] + q_l[3] * q_w[1] -
q_l[1] * q_w[3])
q_k_d = 0.5 * (q_l[0] * q_w[3] + q_l[3] * q_w[0] + q_l[1] * q_w[2] -
q_l[2] * q_w[1])
for i in range(1):
q_r_d *= 1 / sy.sqrt(q_r_d**2 + q_i_d**2 + q_j_d**2 + q_k_d**2)
q_i_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_j_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
q_k_d *= 1 / sy.sqrt(q_r_d ** 2 + q_i_d ** 2 + q_j_d ** 2 + q_k_d ** 2)
return [q_r_d, q_i_d, q_j_d, q_k_d]
quat0 = quat_equ(q1, q2)[0]
quat1 = quat_equ(q1, q2)[1]
quat2 = quat_equ(q1, q2)[2]
quat3 = quat_equ(q1, q2)[3]
qq = [1., 0., 0., 0., 1., 1., 1.]
var = [q_r, q_i, q_j, q_k, w_ib_l, w_ib_m, w_ib_n]
q0 = sy.Subs(quat0, var, qq).doit()
q1 = sy.Subs(quat1, var, qq).doit()
q2 = sy.Subs(quat2, var, qq).doit()
q3 = sy.Subs(quat3, var, qq).doit()
When I try to re-normalize the quaternion more than 2 times this method becomes very slow. The symbolic equation also becomes massive because the re-normalization adds on basically 4 extra symbolic equations, which slows down later processes a lot. I couldn't think of a better method to create a normalized quaternion that 1) initializes faster and 2) is a smaller symbolic equation. I don't use symbolics in python much so most of Sympy is new to me.
python symbolic-math sympy
python symbolic-math sympy
asked 6 hours ago
BotoBoto
111
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
BotoBoto
111
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
BotoBoto
111
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
BotoBoto
111
asked 6 hours ago
BotoBoto
111
111
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Boto is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Why isfor i in range(1)written as a loop? It's indentation is wrong, which makes this broken code. (A good way to post code is to paste it into the question editor, highlight it, and press Ctrl-K to mark it as a code block.)
$endgroup$
– 200_success
5 hours ago
$begingroup$
Note that MathJax is available for writing equations in your question.
$endgroup$
– 200_success
5 hours ago
$begingroup$
In addition to what 200_success alreay said about broken code, the symbol definition is also no valid Python code. Your code could also greatly benefit from some comments to give reviewers a few hints which parts you suspect or know to be slow.
$endgroup$
– Alex
3 hours ago
add a comment |
1
$begingroup$
Why isfor i in range(1)written as a loop? It's indentation is wrong, which makes this broken code. (A good way to post code is to paste it into the question editor, highlight it, and press Ctrl-K to mark it as a code block.)
$endgroup$
– 200_success
5 hours ago
$begingroup$
Note that MathJax is available for writing equations in your question.
$endgroup$
– 200_success
5 hours ago
$begingroup$
In addition to what 200_success alreay said about broken code, the symbol definition is also no valid Python code. Your code could also greatly benefit from some comments to give reviewers a few hints which parts you suspect or know to be slow.
$endgroup$
– Alex
3 hours ago
1
1
$begingroup$
Why is
for i in range(1) written as a loop? It's indentation is wrong, which makes this broken code. (A good way to post code is to paste it into the question editor, highlight it, and press Ctrl-K to mark it as a code block.)$endgroup$
– 200_success
5 hours ago
$begingroup$
Why is
for i in range(1) written as a loop? It's indentation is wrong, which makes this broken code. (A good way to post code is to paste it into the question editor, highlight it, and press Ctrl-K to mark it as a code block.)$endgroup$
– 200_success
5 hours ago
$begingroup$
Note that MathJax is available for writing equations in your question.
$endgroup$
– 200_success
5 hours ago
$begingroup$
Note that MathJax is available for writing equations in your question.
$endgroup$
– 200_success
5 hours ago
$begingroup$
In addition to what 200_success alreay said about broken code, the symbol definition is also no valid Python code. Your code could also greatly benefit from some comments to give reviewers a few hints which parts you suspect or know to be slow.
$endgroup$
– Alex
3 hours ago
$begingroup$
In addition to what 200_success alreay said about broken code, the symbol definition is also no valid Python code. Your code could also greatly benefit from some comments to give reviewers a few hints which parts you suspect or know to be slow.
$endgroup$
– Alex
3 hours ago
add a comment |
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1
$begingroup$
Why is
for i in range(1)written as a loop? It's indentation is wrong, which makes this broken code. (A good way to post code is to paste it into the question editor, highlight it, and press Ctrl-K to mark it as a code block.)$endgroup$
– 200_success
5 hours ago
$begingroup$
Note that MathJax is available for writing equations in your question.
$endgroup$
– 200_success
5 hours ago
$begingroup$
In addition to what 200_success alreay said about broken code, the symbol definition is also no valid Python code. Your code could also greatly benefit from some comments to give reviewers a few hints which parts you suspect or know to be slow.
$endgroup$
– Alex
3 hours ago