Changing KMESH/KINTS from azimuthal to polar direction in MCNP6

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SUMMARY

The discussion centers on modeling a cylindrical source in MCNP6 using the FMESH tally in cylindrical coordinates. The user seeks to define the polar angle (theta) in a manner consistent with spherical coordinates, where theta = 0 aligns with the positive z-axis and theta = 180 with the negative z-axis. The MCNP6 version 1 user manual, specifically Section 3.3.5.25 and Section 3.3.5.8, provides guidance on using GEOM=CYL and applying dose conversion factors. However, the FMESH does not support a polar coordinate system as requested, leading to suggestions for alternative approaches.

PREREQUISITES
  • Understanding of MCNP6 version 1 and its functionalities
  • Familiarity with cylindrical coordinates and their application in radiation transport modeling
  • Knowledge of dose conversion factors and health physics standards
  • Experience with FMESH and TMESH tally types in MCNP6
NEXT STEPS
  • Review MCNP6 version 1 user manual, particularly Sections 3.3.5.25 and 3.3.5.8
  • Explore the use of TMESH for alternative modeling of dose distribution
  • Investigate health physics standards for obtaining dose conversion factors
  • Learn about the limitations of FMESH in spherical coordinate systems
USEFUL FOR

This discussion is beneficial for radiation physicists, MCNP6 users, and researchers modeling cylindrical sources in radiation transport simulations.

khary23
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I am modeling a cylindrical source in MCNP6 and would like to use the FMESH tally in cylindrical coordinates. I am looking for the dose to water from the source as a function of radial distance as well as polar angle running from 0 to 180 degrees in the YZ plane not around Z. Is there a way to do this?
 
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Sure. I'm looking at the MCNP6 version 1 user manual and it's Section 3.3.5.25, page 3-67. You use GEOM=CYL then the I direction is radial, J is the z direction, and K is the theta.

To get the dose you need the conversion factors. These allow you to convert from a flux at a particular energy to a dose. You will find a table that gives the energy range split up into a bunch of bins, and dose conversion factor for each bin. You can get those from the appropriate health physics standard. You enter them in the same way you would with a type F4 tally. That's detailed in Section 3.3.5.8, page 3-211.
 
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Hello,
What I want to do is leave I(radial) and J(z) as they are in cylindrical coordinates , but change K so that is is not rotating about the z axis. What I want to do is have zero degrees pointing in the +z direction and 180 degrees in the -z direction and 90* at y =0. I drew a crude picture of what I want to do.
20190814_133656.jpg
 
Do you mean you want to have the axis of the cylinder along a line you specify? That's also in the FMESH description. Whatever the axis of the cylinder is, theta is defined to be the angle around the circumference of the cylinder.
 
No. I do not want theta to be defined as the angle around the circumference of the cylinder. I want theta to be defined as it is in spherical coordinates as the polar angle. Where theta =0 is along the z positive axis and theta = 180 is in the negative z direction.
If it is not possible I can use TMESH just thought it might more elegant to have the mesh and source geometry be the same.
Thanks for your help!
 
Oh. Except your first post said you were using a cylindrical source. And you drew a cylinder. If your source is cylindrical then why don't you want theta defined as the angle round the circumference?

If your source is really spherical, sadly FMESH does not have a spherical polar coordinate option.
 
Yes the source I am modeling is cylindrical. The formalism I am following dictates that (r,theta) specifies a point relative to the source longitudinal axis. Here is a picture of the geometry that the protocol calls for.

1565818597327.png
 
I suppose you could turn the fmesh 90 degrees to the source cylinder. In your drawing you could have the z-axis of the fmesh coming out of the screen. That would give you a circular mesh in the plane of the source. It won't give you the other angle though.
 

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