How does Multitracer consider the outer row of voxels?

[aredolfi]

Dear Multitracer Staff,

My name is Alberto Redolfi and I'm a researcher at IRCCS-FBF neuroimaging Hospital in Brescia (Italy).

Aim of this mail is to survey an important technical aspect related to the volume measurement obtained through the Multitracer software.
As you know, in a tracing session, different softwares treat differently the outer row of voxels.
Therefore, I would like to know if Multitracer considers them as inside, under, or outside the trace and in which way the final area/volume estimation is influenced.

Looking forward to hearing from you,
Best regards,

Alberto.

[arhammond]

The area is the area enclosed by connecting the centers of the voxels that constituted the trace when it was drawn. Note that tracing at higher magnification levels increases the degree to which subvoxel resolution (relative to the original input voxels) can be encoded.

Roger Woods

[Replies have been subsequently posted by arhammond for user reference]

[arhammond]

Therefore, for example, consider that the voxel resolution of the image I'm drawing is 1*1* 1 mm, and consider I'm intrested in the Volume of the Hippocampus, in this case the outer row of voxels traced trough Multireacer would then be considered and the contribution of these voxels on the Hippocampus volume would then equal to 0.5*0.5*0.5 mm. Is that right? Moreover, could you assure me that considering a higher magnification this does not influence the final measurements of the volume? I mean if I trace the volume of Hippocampus with a 1x, 2x or 4x the final hippocampus volume remains exactly the same, is that right?

Thanks in advance for this very useful explanations,

Best regards,
Alberto

[Replies have been subsequently posted by arhammond for user reference]

[arhammond]

It is not the case that each voxel would contribute 0.5*0.5*0.5 mm. Even for a simple shape like a square, the path that you take through a voxel (i.e., a corner versus an edge) will alter the contribution of the voxel. Moreover, when you digitize a trace, if you are drawing quickly, the locations that are encoded are not necessarily even in contiguous voxels, so it's possible to pass through a voxel in a way that is neither a 90 degree corner, nor a 180 degree edge (think of connecting the starting an ending positions of a knight's move in chess). In addition, the areas from each slice are used to estimate a volume in a way that is different from the way that the areas are calculated and there are in fact three different options:
(see http://air.bmap.ucla.edu/MultiTracer/calculating.html).
Probably contrary to your intuition, areas and volumes are not counted by any sort of voxel-counting strategy. As stated previously, the areas calculated as those that would be enclosed by connecting the centers of the encoded voxels by straight lines. If you want to actually see the encoded points, you can export the tracings as UCF's and look at them with a text editor:
(http://air.bmap.ucla.edu/MultiTracer/exporting.htm).

The tracings that you see on screen at any particular resolution simply represent the best approximation to the conceptual line connecting these points.

If this still doesn't make sense, I would suggest getting a piece of
graph paper and placing a tiny dot in each of these coordinate
locations:

1, 1
3, 6
5, 4

Then take a ruler and connect the dots in the order specified and also connect the first and last dot. Now lightl shade in every square on the graph paper that the lines that you drew pass through. You will see that different squares contribute different amounts to the enclosed area. However, the "true" triangle that I had in mind with these instructions was actually:

1.4, .6
3.2, 5.6
4.6, 4

but this couldn't be encoded precisely by specifying the centers of the squares. If I go to 5x magnification, then you'd encode these locations:

7, 30
16, 28
23, 20

The area of this triangle is not exactly 25 times as large as the earlier one because it encodes the "true" triangle more accurately.

When tracing at different resolutions, a "perfect" tracer will generally get different results because the higher resolution images allow the "true" location of the edge to be encoded more accurately at higher resolution.

Roger

[Replies have been subsequently posted by arhammond for user reference]