MCNP6.2 - Combination of transformations

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SUMMARY

The discussion focuses on combining multiple transformations in MCNP6.2, specifically regarding the order and reference frames of rotations. The user clarifies that when applying transformations TR1 and TRCL, the reference frame for TRCL is the new origin established by TR1. The final transformation TR3 must be defined to ensure orthogonality of planes to the channel's axis, with the correct order being TR1 followed by the rotational part of TRCL. The correct transformation card for TR3 is provided, ensuring accurate axial segmentation of the tilted channel.

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  • Familiarity with axial segmentation techniques in MCNP
  • Basic linear algebra concepts related to transformations
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This discussion is beneficial for MCNP users, computational physicists, and anyone involved in modeling complex geometries with transformations in MCNP6.2.

19matthew89
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TL;DR
It's unclear which reference frame is used for consecutive transformations and so how to combine TR
Hi everyone.

I am struggling understanding how to combine more than one transformations, especially rotations. This stems mainly form the fact that it's unclear to me what reference frame is used to define the transformations angle if two consecutive transformations are applied. If I have a rotation TR1 and then another rotation TR2, will the reference for rotation TR2 be the original reference frame or the tilted one defined by TR1.
I give the example of interest.
I have a channel which is tilted and translated with respect to center (translation doesn't matter, but let's say of (0 0 1)).
I want to have axial segmentation of this tilted channel to have tallies axially.

The model (too complicated to be changed so I'll stick to how it defines the channel) defines the surface of the tilted and translated channel in two steps

First a transformation TR1 of all the surfaces (let's assume phi=30°, theta=10° and no psi, i.e. psi=0°) is defined and applied in the definition of the surfaces.
E.g. TR1 reads
Code:
TR1= 0                  0              0   
         0.85286867     -0.5         0.15038294
         0.49240396     0.8660254    0.08682363
        -0.17364727      0.0       0.98480791

and for instance a plane 11 (originally parallel to XY plane) is defined via
Code:
11 PZ 1 10.0
.

Then another successive transformation for the cell defining the channel is applied with TRCL, and this rotation is about the z axis (which one? the original one?) and of -60° so it reads:
Code:
TRCL= 0.0         0.0         1.0
       0.5       0.8660254     0.0
      -0.8660254    0.5        0.0
        0.0         0.0        1.0
Now my question. In order to have a series of planes orthogonal to the axis of the channel, do I need to define another transformation, let' say TR3, given by the rotational part of TRCL_rot and TR1 (i.e TRCL_rot*TR1), and then apply that to the planes, such as
Code:
11 PZ 3 10.0
.
?

Or the last one is not necessary because the rotation of TRCL will occur around the new z axis defined by TR1 and so the orthogonality of the planes is already guaranteed by the first transformation TR1?

Thanks in advance

P.S. I have a related question about the orders of the transformation. Is the rotational part applied first and then the result being translated, or first translated and then rotated? I ask because in the latter case, if the orginal reference frame is considered, rotation phi of 180° after a translation is equivalent to bringing the item on the other side of the origin.
 
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Ok,
I solved it and found out that if a transformation is applied the reference for the following transformation will be the new origin, given by the first transformation.

About the order I solved it and it's actually the opposite of what I was expecting.
For the example given above, indeed a new transformation TR3 has to be defined with: the right translation vector, but the order of the rotation is TR1*TRCL_rot, with TRCL_rot the rotational part of the TRCL card defined.

So the right transformation card reads:

Code:
TR3        0         0         1 
       0.85944703  0.48860593  0.15038294
       -0.50379801  0.85944704  0.08682363
       -0.08682363 -0.15038295  0.98480791
 
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