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I need to perform a special interpolation, using that kind of basis : $$\varphi_{i,j}(x) = a_i + b_ix + c_i(\cosh(\tau\ x) - 1) + d_i(\sinh(\tau\ x) - \tau\ x)$$ where the $a_i$, $b_i$, $c_i$ and $d_i$ are determined with the following constraints : $$ \begin{array}{|c|c|c|c|} \hline &\varphi_{i,j}(x_i)&\varphi_{i,j}(x_{i+1})&\varphi_{i,j}^\prime(x_i)&\varphi_{i,j}^\prime(x_{i+1})\\ \hline j=0&1&0&0&0\\ \hline j=1&0&1&0&0\\ \hline j=2&0&0&1&0\\ \hline j=3&0&0&0&1\\ \hline \end{array} $$ I need to determinate the minimum tension factor $\tau$, on each interval. I've read this paper, but I don't really understood. Does anyone ever used that kind of interpolation ?

Ps. I apologize for my English.

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The idea with tension splines is that you keep adjusting those $\tau$ values up until you're satisfied with the appearance of your spline... –  J. M. Jul 9 '12 at 12:45
    
In the paper I read, there is a description of a method for select automatically the tension factor. –  Alexandre Hoffmann Jul 9 '12 at 13:08

1 Answer 1

We used these "splines with tension" in a CAD system for many years. As J.M. said, the idea is that the user adjusts the tension values until he/she is happy with the shape of the curve. I read the paper quickly, and didn't see how "optimal" tension values are selected. But, clearly, you first have to decide what you're going to optimise. There is no such thing as a universally "best" tension value.

In answer to your question, the minimum tension value is normally taken to be zero, in which case these splines are just regular cubic splines.

We eventually ditched the idea. Splines with exponential terms don't fit nicely into the CAD world, and there are other ways to get "tension" kinds of shape control, using splines that are simply piecewise polynomials.

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