On Tue, 30 Jun 2009 13:51:15 -0600
Charles R Harris <[email protected]> wrote:
> On Tue, Jun 30, 2009 at 12:26 PM, Nils Wagner
> <[email protected]>wrote:
>
>> On Tue, 30 Jun 2009 11:10:39 -0600
>> Charles R Harris <[email protected]> wrote:
>> > On Tue, Jun 30, 2009 at 10:56 AM, Charles R Harris <
>> > [email protected]> wrote:
>> >
>> >>
>> >>
>> >> On Tue, Jun 30, 2009 at 10:40 AM, Nils Wagner <
>> >> [email protected]> wrote:
>> >>
>> >>> On Tue, 30 Jun 2009 10:27:05 -0600
>> >>> Charles R Harris <[email protected]> wrote:
>> >>> > On Tue, Jun 30, 2009 at 5:11 AM, Nils Wagner
>> >>> > <[email protected]>wrote:
>> >>> >
>> >>> >> On Tue, 30 Jun 2009 11:22:34 +0200
>> >>> >> "Nils Wagner" <[email protected]>
>>wrote:
>> >>> >>
>> >>> >>> Hi all,
>> >>> >>>
>> >>> >>> How can I build the following product with numpy
>> >>> >>>
>> >>> >>> q_i = \varepsilon_{ijk} q_{kj}
>> >>> >>>
>> >>> >>> where \varepsilon_{ijk} denotes the permutation
>> >>>symbol.
>> >>> >>>
>> >>> >>> Nils
>> >>> >>>
>> >>> >> Sorry for replying to myself.
>> >>> >> The permutation symbol is also known as the
>> >>>Levi-Civita
>> >>> >>symbol.
>> >>> >> I found an explicit expression at
>> >>> >> http://en.wikipedia.org/wiki/Levi-Civita_symbol
>> >>> >>
>> >>> >> How do I build the product of the Levi-Civita
>>symbol
>> >>> >>\varepsilon_{ijk} and
>> >>> >> the two dimensional array
>> >>> >> q_{kj}, i,j,k = 1,2,3 ?
>> >>> >>
>> >>> >
>> >>> > Write it out explicitly. It essentially
>> >>>antisymmetrizes
>> >>> >q and the three off
>> >>> > diagonal elements can then be treated as a vector.
>> >>> >Depending on how q is
>> >>> > formed and the resulting vector is used there may
>>be
>> >>> >other things you can do
>> >>> > when you use it in a more general expression. If
>>this
>> >>>is
>> >>> >part of a general
>> >>> > calculation there might be other ways of
>>expressing
>> >>>it.
>> >>> >
>> >>> > Chuck
>> >>>
>> >>> Hi Chuck,
>> >>>
>> >>> Thank you for your response.
>> >>> The problem at hand is described in a paper by
>>Angeles
>> >>> namely equation (17c) in
>> >>> "Automatic computation of the screw parameters of
>> >>> rigid-body motions.
>> >>> Part I: Finitely-separated positions"
>> >>> Journal of Dynamic systems, Measurement and Control,
>> >>>Vol.
>> >>> 108 (1986) pp. 32-38
>> >>>
>> >>
>> >> You can solve this problem using quaternions also, in
>> >>which case it reduces
>> >> to an eigenvalue problem. You will note that such
>>things
>> >>as PCA are used in
>> >> the papers that reference the cited work so you can't
>> >>really get around that
>> >> bit of inefficiency.
>> >>
>> >
>> > Here's a reference to the quaternion approach:
>> >
>>http://people.csail.mit.edu/bkph/papers/Absolute_Orientation.pdf.
>> >You can
>> > get the translation part from the motion of the
>> >centroid.
>> >
>> > If you are into abstractions you will note that the
>> >problem reduces to
>> > minimising a quadratic form in the quaternion
>> >components. The rest is just
>> > algebra ;)
>> >
>> > Chuck
>>
>> It turns out that the product is simply an invariant of
>>a
>> 3 \times 3 matrix.
>>
>> from numpy import array, zeros, identity
>> from numpy.linalg import norm
>>
>>
>> def vect(A):
>> """ linear invariant of a 3 x 3 matrix """
>> tmp = zeros(3,float)
>> tmp[0] = 0.5*(A[2,1]-A[1,2])
>> tmp[1] = 0.5*(A[0,2]-A[2,0])
>> tmp[2] = 0.5*(A[1,0]-A[0,1])
>>
>> return tmp
>
>
> Out of curiosity, where did the .5 come from? It is not
>normally part of the
> Levi-Civita symbol.
>
> Chuck
Hi Chuck,
It's my fault. The components of the invariant q are given
by
q_i = 0.5 \varepsilon_{ijk} q_{kj}
Nils
_______________________________________________
Numpy-discussion mailing list
[email protected]
http://mail.scipy.org/mailman/listinfo/numpy-discussion
