I’ve used micatoolbox and QCPA for data collection and analysis and am reaching the manuscript writing stage (hooray). In my outputs, my understanding is that the “Area” column is the total number of pixels that each cluster covers (but please correct me if I’m wrong), and this is calibrated by creating the scale bar in the mspec image. I’d like to convert that value to millimeters or some other SI unit of measure to have a more realistic discussion of my color patch sizes. Can this be done? Or is the Area value actually millimeters-squared and not pixels after all?
The QCPA measures area in two different ways. To translate PT (Average patch size, estimated based on the transition matrix) to mm or cm, please see the supplementary material of the QCPA paper. Else, the output of the particle analysis is measured in pixels (a precise measure), which can similarly be translated to mm/cm using your size standard in the resized image.
Yes, but you might want to confirm that the ratio is reflective of the final resolution of the image used when generating the pattern statistics (rather than the initial resolution prior to size reduction due to acuity modelling). Once you’re confident with the ratio, multiply CAA:PT.vrt and CAA:PT.hrz with the ratio to translate to a 2D distance for each vertical and horizontal patch size. Alternatively, using the particle analysis outcome (which reports the area in square pixels I think) you can similarly transform reported cluster areas into mm/cm once you have calculated your ratio.
Thank you for getting back to me! If you don’t mind me asking you to confirm that I’m understanding you correctly, here’s an example I worked:
The scale bar I drew in my mspec is 50 mm, and when I save it, micatoolbox pastes text over the scale bar line segment saying “Scale Bar: 1496.1083:50”. I assume that is my “pixel/distance ratio” mentioned in the QCPA paper’s supplemental file. After running the QCPA, the 1 patch of interest from my ROI returned an Area of 2522. So, using the above scale bar ratio, that gives me a patch size of 84.29 square mm. Obviously without knowing my subject, you can’t confirm that that sounds like a logical value, but is the method correct?