Talk:Trilinear interpolation

The equations for i1, i2, j1, j2 in the example are all the same. This seems like an error. Am I missing something? Skorkmaz 19:53, 4 December 2007 (UTC)[reply]

They're not quite the same. For instance, the equation for

i_{1}

uses

\lfloor y\rfloor

(y rounded downwards to the nearest integer) while the equation for

i_{2}

uses

\lceil y\rceil

(y rounded upwards to the nearest integer).

Oh, that's quite a subtle notation. Thanks. Skorkmaz (talk) 11:17, 5 December 2007 (UTC)[reply]

Thanks for the pictures you added. However, I don't understand the second picture. Shouldn't the blue lines be parallel to the edges of the cube? -- Jitse Niesen (talk) 20:56, 4 December 2007 (UTC)[reply]

The blue lines don't need to be paralel because the end points of the blue lines (C00, C10, C01, C11) are found by interpolation of the cubes lower and upper planes. May be I should make them look less parallel, especially the lower one. Skorkmaz (talk) 11:17, 5 December 2007 (UTC)[reply]

Then I don't understand the picture at all. What are the points C00 etc. and the lines supposed to mean? -- Jitse Niesen (talk) 12:52, 5 December 2007 (UTC)[reply]

Added explanation to article. Skorkmaz (talk) 21:51, 5 December 2007 (UTC)[reply]

This article is incomplete and confusing. The notation is odd and inconsistant. The figures do not have axes labeled. Sorry to rain on someone's parade, but this article need a lot of work. neffk (talk) 18:07, 2 November 2010 (UTC)[reply]

Order of accuracy

Hi all,

Under the section "Compared to linear and bilinear interpolation", it it stated that "The order of accuracy is 1 for all these interpolation schemes,..."

I was under the impression that trilinear interpolation, as well as linear and bilinear, is second-order accurate. See for example "An algorithm for tracking fluid particles in numerical simulations of homogeneous turbulence", Yeung, P.K., Pope, S.B., Journal of Computational Physics, 1988. Quote: "This scheme is well known to be second-order accurate, i.e., the magnitude of the interpolation error decreases asymptotically as (∆x)^2 as the grid spacing ∆x tends to zero."

In Ferzigers book "Computational methods for fluid dynamics", linear interpolation is also stated to be second-order accurate.

Perhaps the meaning of the sentence in this wiki-article is that the linear interpolation method is first order (polynomial of first order), but then the order of accuracy should be of second order. Any thoughts on this?

Best regards,

Reine — Preceding unsigned comment added by Zulugestapo (talk • contribs) 10:25, 7 April 2011 (UTC)[reply]

130.233.46.38 (talk) 18:29, 24 August 2016 (UTC)[reply]

I agree with this comment. For each dimension, we can use the Cauchy remainder^[1] and achieve an upper bound proportional to the square of the distance between points multiplied by some constant depending on the smoothness of the function. This is actually given for the 1D linear interpolation in Wikipedia^[2] : R_T< (x1-x0)^2*C which is the definition of a second order accurate function, i.e. R_T = O( (x1-x0)^2 ) in big-oh^[3] notation.

130.233.46.38 (talk) 18:29, 24 August 2016 (UTC)[reply]

To do

Given the above information, the article should be amended to include the quote from Yeung and Pope, properly cited. For the moment, I'll just correct it to read "These interpolation schemes all use polynomials of order 1, giving an accuracy of order 2, …". yoyo (talk) 20:57, 30 September 2018 (UTC)[reply]

References

Changed variable names for clarity

Hi All,

I changed the variables so as to match the image. I got confused for a bit and I figured this modification might helpful to some people. Did I miss anything or is their a convention I should have followed? — Preceding unsigned comment added by Iamcarlos (talk • contribs) 05:25, 20 April 2012 (UTC)[reply]

visualization

The article on bilinear interpolation has a very neat visualization: Suppose there was a cross-hair on the point in question, dividing the rectangle (formed by the four surrounding points) into four sub-rectangles. The interpolation is then the weighted average of corners' values, where each corner is weighted by the area of the sub-rectangle opposite from it, as a fraction of the main rectangle's area.

I haven't verified it, but I suppose the same kind of visualization (involving the volumes of 8 cuboids instread of the areas of 4 rectangles) would hold for trilinear interpolation. If so, it would be a great addition to the article, since it would be very easy to remember the formula in this way (which the article doesn't even provide, for its unwieldiness, presumably). --Tennenrishin (talk) 15:04, 12 June 2013 (UTC)[reply]

Clarify paragraph

>For an arbitrary, unstructured mesh (as used in finite element analysis), other methods of interpolation must be used; if all the mesh elements are tetrahedra (3D simplices), then barycentric coordinates provide a straightforward procedure.

Why do other methods have to be used? What are they? Where is the mesh even coming from? Or is that only specific to computer graphics?

The reason why i'm asking is: I'm looking into interpolation for characteristic curves, and to me it's entirely unclear if this is relevant to me or not.

[1] ttp://mathworld.wolfram.com/CauchyRemainder.html

[2] ttps://en.wikipedia.org/wiki/Linear_interpolation

[3] ttps://en.wikipedia.org/wiki/Big_O_notation

[1]

[2]

[3]