Hi Leo & Sam,

Thanks for your replies.
The bilinear integration method which Sam wrote was just I intended.
I could open the PDF document on Mac OS X and iOS (iPhone/iPad).

Additionally, I found a similar problems on Temporal Statistics filter.
I prepared two transiend data sets.

> Case 1:
> t=1.0, scalar=1
> t=2.0, scalar=2
> t=4.0, scalar=4
> 
> Case 2:
> t=1.0, scalar=1
> t=1.1, scalar=2
> t=4.0, scalar=4

Both Temporal Average Values are 2.3333.
And if I apply Temporal Interpolation filter and Temporal Statistics filter 
again,
there are different results.
The reason is the Temporal Static filter is just calculating sum average,
and transient time values are not considered.

I'd like to get time-weighted average values by Temporal Statistics filter.


Magician


On 2012/06/11, at 9:36, Yuanxin Liu wrote:

> Hi, Sam, 
>   Would you mind reposting this?  The pdf file seems corrupt.
> thanks,
> Leo
> 
> On Sun, Jun 10, 2012 at 7:38 PM, Samuel Key <samuel...@bresnan.net> wrote:
> Hi Leo,
> 
> Though likely more arithmetically intense than you would like, I have 
> attached a 3-page write-up on how to do an exact integration over the surface 
> of a bi-linear isoparametric coordinate based quadrilateral in three-space. 
> It is easily adapted to a flux integration as well.
> 
> If it helps, please free to incorporate it into ParaView.
> 
> Sam Key.
> 
> 
> 
> 
> On 6/10/2012 12:55 PM, Yuanxin Liu wrote:
>> Hi, 
>>    This is not really a bug, but just what the linear precision integration 
>> does.  As you can see in vtkIntegrateAttributes.cxx:215, the quad cell is 
>> basically cut into two triangles along a diagonal, integration is done for 
>> each triangle and the results are added up.   So, depending where your "1" 
>> falls on a diagonal vertex, you will get different results.
>> 
>> Leo
>> 
>> 
>> On Sat, Jun 9, 2012 at 8:57 AM, Magician <f_magic...@mac.com> wrote:
>> Hi all,
>> 
>> 
>> Now I found a bug of the filter.
>> 
>> My ParaView's version is 3.14.1.
>> I define node names as below:
>> > c--d
>> > |  |
>> > a--b
>> 
>> Case 1: a=1.0, b=0.0, c=0.0, d=0.0
>> Integration Value = 0.33333
>> 
>> Case 2: a=0.0, b=1.0, c=0.0, d=0.0
>> Integration Value = 0.16666
>> 
>> Case 3: a=0.0, b=0.0, c=1.0, d=0.0
>> Integration Value = 0.16666
>> 
>> Case 4: a=0.0, b=0.0, c=0.0, d=1.0
>> Integration Value = 0.33333
>> 
>> So Integration Values are calculated as (2*a+b+c+2*d)/6.
>> I think it's not a good result.
>> 
>> 
>> Magician
>> 
>> 
>> On 2012/06/08, at 12:54, Magician wrote:
>> 
>> > Hi all,
>> >
>> >
>> > I'm now comparing integration results of each postprocessors.
>> > I made a simple Quadratic VTK model as below:
>> >> # vtk DataFile Version 2.0
>> >
>> >> QuadCell
>> >> ASCII
>> >> DATASET UNSTRUCTURED_GRID
>> >> POINTS 4 FLOAT
>> >> 0 0 0
>> >> 1 0 0
>> >> 0 1 0
>> >> 1 1 0
>> >> CELLS 1 5
>> >> 4 0 1 3 2
>> >> CELL_TYPES 1
>> >> 9
>> >> POINT_DATA 4
>> >> SCALARS scalar FLOAT 1
>> >> LOOKUP_TABLE DEFAULT
>> >> 0.0
>> >> 1.0
>> >> 2.0
>> >> 4.0
>> >
>> > When I apply Point Data to Cell Data filter,
>> > scalar values are interpolated as 1.75.
>> > It may be calculated as (0.0+1.0+2.0+4.0)/4.0.
>> >
>> > But when I apply Integrate Variables filter,
>> > scalar value is integrated as 1.833....
>> > The area value of quadratic cell is 1.0, so I thought
>> > integrated values should be equal to cell data.
>> >
>> > How the integrated value is calculated?
>> >
>> >
>> > Magician

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