Hi Cory,
you are correct. This tiny change solves the problem! I hope that our
discussion will also allow others to review their output code with
respect to this change.
Thank you very much for clarification and your help (and your great
software as well).
Best regards
Philipp Engels
On
Philipp,
Great, I'm glad to hear it is working again as you expect.
- Cory
On Mon, Oct 13, 2014 at 3:17 AM, Philipp E pspeac...@gmail.com wrote:
Hi Cory,
you are correct. This tiny change solves the problem! I hope that our
discussion will also allow others to review their output code with
Classification: UNCLASSIFIED
Caveats: NONE
Thanks JB. I will dig into the compilation process and see if there is anything
I missed.
Thanks
-simon
-Original Message-
From: Biddiscombe, John A. [mailto:biddi...@cscs.ch]
Sent: Friday, October 10, 2014 6:10 PM
To: Su, Simon M CTR USARMY
Classification: UNCLASSIFIED
Caveats: NONE
Hi Cory,
Thanks for the info. I will update my build script.
Thanks
-simon
-Original Message-
From: Cory Quammen [mailto:cory.quam...@kitware.com]
Sent: Saturday, October 11, 2014 6:04 PM
To: Su, Simon M CTR USARMY ARL (US)
Cc:
You can create a wavelet source and then display it as a volume.
On Fri, Oct 10, 2014 at 3:45 PM, B.W.H. van Beest b...@xs4all.nl wrote:
Can perhaps somebody point me how to address the following.
When I create a new source (say: a box) I get an a box that
can be displayed via its
Bertwim,
I'm not sure that the volume renderer can handle VTK_HEXAHEDRON
elements. Try the Tetrahedralize filter on your source and see if
the Volume representation works.
Thanks,
Cory
On Mon, Oct 13, 2014 at 10:19 AM, B.W.H. van Beest b...@xs4all.nl wrote:
Hi,
I'm struggling to get a proper
No, it doesn't work. The filter creates tets, which are visible in the
wireframe representation,
but when I select Volume representation the image disappears.
I can't imagine it is so difficult to render a volume in Paraview. I
must be doing something wrong
Regards,
Bertwim
On 10/13/2014
do you have any scalars? those are not optional for volume rendering.
Additionally you may need to adjust the transfer function,
On 10/13/2014 10:22 AM, B.W.H. van Beest wrote:
No, it doesn't work. The filter creates tets, which are visible in the
wireframe representation,
but when I select
Aashish,
Thanks for the suggestion.
However, doing this gives me all kinds of opengl errors (invalid
instruction).
Besides that, the wavelet gives me a cube indeed, but I need to
understand when
I can have a volume rendering. I created new Source's (e.g. tri-prism, a
wedge,
a box with
So when you created a wavelet source and change the representation to
volume you got the errors? Can you post those errors?
Thanks,
On Mon, Oct 13, 2014 at 1:33 PM, B.W.H. van Beest b...@xs4all.nl wrote:
Aashish,
Thanks for the suggestion.
However, doing this gives me all kinds of opengl
Yes, I do set a scalar for each cell. I left them out in the example, as
also with a scalar I couldn't get a volume display.
I'm not sure what you mean by adjusting the transfer function.
Regards,
Bertwim
On 10/13/2014 07:28 PM, Burlen Loring wrote:
do you have any scalars? those are not
Hi Aashish,
This is what I do:
1. start paraview
2. Sources - Wavelet - Apply.
A cube is displayed (Outline representation).
3. Select 'Volume representation
cube disappears. Then:
=
Generic Warning: In
What graphics card / graphics driver version / OS this system has?
- Aashish
On Mon, Oct 13, 2014 at 1:41 PM, B.W.H. van Beest b...@xs4all.nl wrote:
Hi Aashish,
This is what I do:
1. start paraview
2. Sources - Wavelet - Apply.
A cube is displayed (Outline representation).
3.
Aashish,
I do see errors in the wavelet example, but not in the other cases. If
ok with you , I'll send the full code (it is not much, ~100K) as a
zipped tar file by email. There is no other data.
Regards,
Bertwim
On 10/13/2014 07:43 PM, Aashish Chaudhary wrote:
Bertwim,
If you are seeing
Cory, Bertwin, et al.;
The exact volume of an arbitrarily oriented, arbitrary 8-vertex
hexahedron can be computed with a very elementary construct.
First, place an average-coordinate point at the center of the
hexahedron; this point will be a common vertex point for a set of
tetrahedrons
Samuel,
Thanks for taking the efforts to dive into this.
In fact, it is a good reminder to me to try a single tetraeder first. My
primary problem seems to be that
I cannot get volume rendering to work.
Kind Regards,
Bertwim
On 10/13/2014 07:17 PM, Samuel Key wrote:
Cory, Bertwin, et al.;
The
OS: OpenSuSe 13.1
# lspci | grep -i vga
00:02.0 VGA compatible controller: Intel Corporation 4th Gen Core
Processor Integrated Graphics Controller (rev 06)
01:00.0 VGA compatible controller: Advanced Micro Devices, Inc.
[AMD/ATI] Mars [Radeon HD 8670A/8670M/8750M]
# glxinfo
name of display:
Hi Joe,
I've seen this before and there is a bug report in mantis here:
http://www.paraview.org/Bug/view.php?id=13382
There has also been a patch posted to the vtk-developers list a while back
although I haven't tested it:
http://markmail.org/message/zg3damrg27p5sgue
Regards,
Paul
On 10
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