Spline problem [Matlab]

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From: J Mathgeek on 2 Jun 2010 09:40

We've got a spline problem:

A cable is represented by a number of fixed points.
In the middle between each of these points we know the
direction of the cable (a "compass direction measurement").

The cable is being represented by a cubic spline through the fixed points.
However, we also need to utilize the direction information.
I wonder if it is possible to use a derivative constraint or something?
Or, if the SLM - Shape Language Modeling could do this:
http://www.mathworks.com/matlabcentral/fileexchange/24443-slm-shape-language-modeling

Any ideas or suggestions?

Regards,
Jan

From: Bruno Luong on 2 Jun 2010 17:23

From: John D'Errico on 2 Jun 2010 17:53

"J Mathgeek" <jmathgeek(a)gmail.com> wrote in message <hu5n06$dbi$1(a)fred.mathworks.com>...
> We've got a spline problem:
>
> A cable is represented by a number of fixed points.
> In the middle between each of these points we know the
> direction of the cable (a "compass direction measurement").
>
> The cable is being represented by a cubic spline through the fixed points.
> However, we also need to utilize the direction information.
> I wonder if it is possible to use a derivative constraint or something?
> Or, if the SLM - Shape Language Modeling could do this:
> http://www.mathworks.com/matlabcentral/fileexchange/24443-slm-shape-language-modeling
>

SLM will do it easily enough, as long as you have the
optimization toolbox.

Your compass direction is just a slope, or at least you
can turn it into one.

Make sure you have enough knots, otherwise it will not
be an exact fitting spline. A knot at each listed point,
plus an additional knot at the intermediate points will
give you an exact solution.

For example, here is such a spline, fit to an exponential
function, where the intermediate values are given as
slopes.

X = 0:.2:1;
Y = exp(X/2);
Xint = 0.1:0.2:0.9;
Yprime = .5*exp(Xint/2);

slm = slmengine(X,Y,'knots',0:.1:1,'xyp',[Xint',Yprime']);
plotslm(slm)

I guess you will need to take my word for it that the
curve fits very nicely, and goes through the points
as provided. Or you can try the above example for
yourself.

HTH,
John

From: Bruno Luong on 2 Jun 2010 18:07

Same example as John, but using the other package: ;-)

X = 0:.2:1;
Y = exp(X/2);

Xint = 0.1:0.2:0.9;
Yprime = .5*exp(Xint/2);

slope = struct('p',1,'x',Xint,'v',Yprime);
pp = BSFK(X,Y,[],[],[],struct('Animation',1,'pntcon',slope))

% Bruno

From: Jan Vidar on 7 Jun 2010 07:22

What if the slopes are at known lengths along the spline from the fixed points,
(e.g., slope no. 1 is 2.5 meters along the spline from fixed point no. 1 and 3.9 meters along the spline from fixed point no. 2), would that be possible to do?

Jan

"Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <hu6km8$kqg$1(a)fred.mathworks.com>...
> Same example as John, but using the other package: ;-)
>
> X = 0:.2:1;
> Y = exp(X/2);
>
> Xint = 0.1:0.2:0.9;
> Yprime = .5*exp(Xint/2);
>
> slope = struct('p',1,'x',Xint,'v',Yprime);
> pp = BSFK(X,Y,[],[],[],struct('Animation',1,'pntcon',slope))
>
> % Bruno

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