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1 16th November 13:05
taxidriver
External User
 
Posts: 1
Default Least squares bspline curve fitting


Hi everybody,

i have an question about bspline curve fitting with least squares method,
where
- min |A*x = B|, where x are the unknown control-points, and
- A is the basis matrix with dimensions N x M , where N is the number of
discretized points of my ****nal curve, M is the desired number of
control-points,
- B are the matrix of discretized points ( 1, N, 1, 3)

BasisFunction-Matrix A will be generated with the householder-method,
- Knots are fixed, generated from the ****nal curve, from 0...1, with the
first and last times of Order (you know what i mean),

My problem is, i have a solver where i can put in the parameters, but when i
scale my Matrix A with more Columns, e.g. with more control-points as my
****nal curve, the solver will throw an unhandled exception,

when i debug and take a look to the generated Matrix A, then the last
row-vector, has at the last Order-positions, an undefined value "-1.#IND" ,
like this,

Matrix 7 X 6 // means, 7 discretized points, desired control-points 6
Order 2

1 0 0 0
0 0
0.633787 0.345427 0.020786 0 0
0
0.254618 0.62267 0.122712 0 0
0
0 0.225531 0.72055 0.0539196 0
0
0 0 0.188748 0.736911
0.0743408 0
0 0 0 0.196459
0.664286 0.139255
0 0 0
0 -1.#IND -1.#IND


Can anyone help me, is it possible that the knot-vector is the reason for
that ? or is it ok that the basisfunction has this value , i mean, because
of this, the householder method does not compute the desired control-points
X as long as X is larger than the number of controlpoints of the ****nal
curve.


thanks in advance!
Taxidriver
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2 16th November 13:06
spellucci
External User
 
Posts: 1
Default Least squares bspline curve fitting


In article <44145943$1@news.bea.com>,
"Taxidriver" <ulardoo@web.de> writes:

looks as if the knot vector defining the B-spline basis is erroneous, since
the B-spline for the last knot couldn't be evaluated correctly.
hth
peter
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