[CCPPETMR/SIRF] Quaternion (#375)

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Richard Brown

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Apr 29, 2019, 12:34:59 PM4/29/19
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This enables averaging of affine transformation matrices via quaternions

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  https://github.com/CCPPETMR/SIRF/pull/375

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  • Quaternion

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Ashley Gillman

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Apr 29, 2019, 4:58:35 PM4/29/19
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Hey Rich, there might be some issue with the quaternion averaging.

https://arc.aiaa.org/doi/abs/10.2514/1.28949

Looks like you should at least divide by the norm to make sure you have a unit quaternion.

David Atkinson

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Apr 30, 2019, 3:50:02 AM4/30/19
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I haven't looked at what you are doing, but in cases where the mean of a set is not clearly defined, one option is to pick from the set an entry that is closest to your expectation of the most representative. (In other words, pick one of the quaternions.) This approach enables you to choose based on expectations, for example, you could compute deformation fields for each quaternion and then select the quaternion that gives the deformation field closest to the mean deformation field within a region of interest.

Richard Brown

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Apr 30, 2019, 5:38:48 AM4/30/19
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@DANAJK

for example, you could compute deformation fields for each quaternion and then select the quaternion that gives the deformation field closest to the mean deformation field within a region of interest.

It sounds like that involves human input. The current method is simply to align all quaternions with the first:

// Sum up all quaternions. Start with first, loop over rest
    Quaternion result(quaternions[0]);

    for (unsigned i=1; i<quaternions.size(); ++i) {

        // But first, check whether subsequent quaternions need to be inverted.
        // Because q and -q are the same rotation, but cannot be averaged, we have to make sure they are all the same.
        if (quaternions[i].is_quaternion_close(quaternions[0])) {
            result.w += quaternions[i].w;
            result.x += quaternions[i].x;
            result.y += quaternions[i].y;
            result.z += quaternions[i].z;
        }
        else {
            Quaternion flipped = quaternions[i].inverse_sign_quaternion();
            result.w += flipped.w;
            result.x += flipped.x;
            result.y += flipped.y;
            result.z += flipped.z;
        }

@ashgillman

Looks like you should at least divide by the norm to make sure you have a unit quaternion.

Good spot, I put in a normalise method, but forgot to use it!

Richard Brown

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Apr 30, 2019, 7:10:51 AM4/30/19
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@rijobro pushed 1 commit.

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codecov[bot]

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Apr 30, 2019, 7:24:50 AM4/30/19
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Codecov Report

Merging #375 into master will not change coverage.
The diff coverage is n/a.

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@@           Coverage Diff           @@

##           master     #375   +/-   ##

=======================================

  Coverage   50.05%   50.05%           

=======================================

  Files           2        2           

  Lines        1720     1720           

=======================================

  Hits          861      861           

  Misses        859      859

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Richard Brown

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Apr 30, 2019, 10:44:22 AM4/30/19
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@rijobro pushed 6 commits.

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Ashley Gillman

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Apr 30, 2019, 6:39:29 PM4/30/19
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I think David is proposing that you use something more like a median than a mean. If you are trying to choose the best of a few possible transforms for a time point, that could be a good choice to eliminate outliers. If you can measure the distance between two quaternions, you could implement something like a simplified k-medoids algorithm to find a median. But it depends what you are trying to achieve, might not be suitable.

Richard Brown

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May 2, 2019, 6:03:46 AM5/2/19
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Ah I see. Seems overkill for my purpose, so I won't put it in. But you're right, it would be useful to make everything more robust in the future if more people start using it.

Richard Brown

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May 2, 2019, 6:11:34 AM5/2/19
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Closed #375.

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