On Mar 20, 9:09 am, eyuc...@gmail.com wrote:
> Hi all,
>
> I was having problems on making 2D interpolations on irregular data
> points. I read the past posts on this news group, I read the articles
> in David's website, and I read Ken's sample chapter on this subject.
>
> However, there are still some aspects that I would like to ask about.
>
> 1. Is it possible to construct an irregular data interpolation scheme
> that reduced to bilinear interpolation at the limit of regular grid?
>
> Let me elaborate on this a little bit: The interpolation schemes for
> regular grids all take the neighboring 4-pts to do the job, whereas
> that for irregular grids takes only 3. Sometimes, we do have cases
> where the grid is just slightly distorted and a reasonable 4-pt
> interpolation *seems* possible.(namely, it is possible to define the
> "neighboring four points," and one starts to wonder if a 4-pt scheme
> is superior than the 3-pt scheme.
>
> If 3-pt scheme is superior, then we shouldn't be using 4-pt scheme
> even in the case of regular grids. If 4-pt scheme is better, there
> should be a way to do 4-pt irregular interpolation, at least when some
> conditions are met, right?
>
> 2. There is a very good property about 1D linear interpolation, which
> can be shown as follows:
>
> IDL> x=dindgen(5)
> IDL> y=[1.2,3.2,4.5,6.1,6.2] ; just some arbitrary irregular spaced
> data.
> IDL> plot, x, y
> IDL> print, interpol(y,x,2.1)
> 4.6599998
> IDL> print, interpol(x,y,4.6599998)
> 2.0999999
>
> that is, INTERPOL(x,y,INTERPOL(y,x,?))=? (? is just any number or
> vector.)
> I believe this really lies in the fact that the inverse of linear
> transformation is still a linear transformation. Therefore, a linear
> mapping from x to y is the inverse of the linear mapping from y to x.
> (you can add a /spline keyword and INTERPOL(x,y,INTERPOL(y,x,?))=?
> won't be true anymore)
>
> Thus, I can claim that linear interpolation in 1-D is a good scheme
> (even better than spline) in the sense that I can answer the question
> "what x should I use in order to get y?" in a relative simple way when
> I use IDL. (if you use spline, then you'll probably need to solve for
> a cubic equation to get the job done)
>
> In 2D case, what is the corresponding "good scheme" ? My intuition
> tells me that the answer is the TRIGRID method, but I would like to
> discuss with you guys before I feel too certain about it. (Or, shall
> we suggest ITT built a "backtrace" keyword in all their interpolation
> routines?)
>
> 3. What is the real difference between INTERPOLATE and BILINEAR? It
> seems to me that INTERPOLATE can do everything BILINEAR does, with
> more accuracy.
>
> 4. What is the real difference between GRIDDATA and TRIANGULATE
> +TRIGRID? Does the capability of GRIDDATA covers TRIGRID?
>
> 5. In GRIDDATA, there is an option "/linear," However, I wonder what
> does linear means when we only have 3 neighbouring points?
I was thinking about the same problem and came to this conclusion:
Imagine a distorted quadrilateral like
C------D
| P /
| /
A\ /
\B
where a,b,c,d also serve as actual function values at the nodes with
coordinates A_x,A_y and so on.
The bilinearily interpolated value p at position P is given by
p = a + (b-a)x + (c-a)y + (a-b-c+d)xy (see http://en.wikipedia.org/wiki/Bilinear_interpolation)
with [x,y,xy]^T = M * [P_x-A_x,P_y-A_y,(P_x-A_x)(P_y-A_y)]^T (this is
my "magic")
and M =
[1 0 1] [ B_x-A_x C_x-A_x D_x-
A_x ]^-1
[0 1 1] * [ B_y-A_y C_y-A_y D_y-A_y ]
[0 0 1] [(B_x-A_x)(B_y-A_y) (C_x-A_x)(C_y-A_y) (D_x-A_x)(D_y-A_y)]
M maps the distorted qudrilateral to a nice unit square which is then
connected to the alternative bilinear interpolation function found at
http://en.wikipedia.org/wiki/Bilinear_interpolation.
Hope it helps!
Cheers
Philip
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