Hexpentaquaternions: a two-hand quaternion algebra
- From: "Peter M Jack" <math@xxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 30 Jan 2006 13:30:05 +0000 (UTC)
A new method for solving quaternion equations.
Suppose we're given a quaternion equation
A[1]qB[1] + A[2]qB[2] + ... + A[n]qB[n] = C
and required to solve for q,
where A[k],B[k],C, q, are all quaternions from
Hamilton's right-hand system ij = +k. So,
q = q0.1 + q1.i + q2.j + q3.k
C = c0.1 + c1.i + c2.j + c3.k
A[k] = a[k,0].1 + a[k,1].i + a[k,2].j + a[k,3].k
B[k] = b[k,0].1 + b[k,1].i + b[k,2].j + b[k,3].k
and the right-hand basis elements obey the rules
i^2 = j^2 = k^2 = -1,
ij = -ji = +k
jk = -kj = +i
ki = -ik = +j
To solve the equation, all we need to do is
convert the B parameters into "left-hand" quaternions B'
so that we can move them over to the other side of the
q parameter, and write
A[1]B'[1]q + A[2]B'[2]q + ... + A[n]B'[n]q = C
then we can aggregate the factors
(A[1]B'[1] + A[2]B'[2] + ... + A[n]B'[n]) . q = C
and invert the equation
q = (A[1]B'[1] + A[2]B'[2] + ... + A[n]B'[n])^-1 . C
that's it.
Details can be found in my new paper,
"Hexpentaquaternions: a two-hand quaternion algebra."
reference # hypcx-20060129a
available online here,
http://www.hypercomplex.com/research/emgrav/abs20060129a.html
The .pdf, .dvi, .djvu versions of the paper have active url links to
references around the web, but in the .ps paper these links are inactive.
enjoy
pmj
.
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