## Quantitative comparisons of macrophotography with and without Lieberkühn reflectors

In order to quantitatively examine the effect of the the 3D printed Lieberkühn reflectors I described previously, I came up with two image quality metrics relevant to their use, both measured on the “dark side” of the image subject. The metrics we will look at today are average intensity and, as a measure of contrast, the standard deviation of pixel values on a line trace.

I’ll be using the Megachile photo from the previous post for these analyses.

The first line trace begins at right eye and extends back behind the wing:

If we plot these values together, we see that the photo taken with the Lieberkühn (values in black) is brighter and brings out a lot more detail, while the photo taken without is relatively flat and dark. We see similar results for second and third traces, taken across the tegula and along the wing.

If we compare the average values:

octave3.2:25> mean(RE523(:,2)) %with Lieberkühn, right eye trace
ans = 102.00
octave3.2:26> mean(RE524(:,2))%w/o Lieberkühn, right eye trace
ans = 54.093
octave3.2:28> mean(AT523(:,2))%with Lieberkühn, trace across tegula
ans = 81.553
octave3.2:27> mean(AT524(:,2))%w/o Lieberkühn, trace across tegula
ans = 51.288
octave3.2:29> mean(W523(:,2))%with Lieberkühn, across the wing
ans = 103.85
octave3.2:30> mean(W524(:,2))%w/o Lieberkühn, across the wing
ans = 53.045

We see that taken together, the averages of the plots from the photo taken with the Lieberkühn are about 80% brighter than those without.

mean(Lieberkühn)/mean(no Lieberkühn) = 1.8142

octave3.2:56> std(RE523(:,2)))%with Lieberkühn, right eye trace
ans = 20.316
octave3.2:55> std(RE524(:,2))%w/o Lieberkühn, right eye trace
ans = 7.3926

octave3.2:54> std(AT523(:,2))%with Lieberkühn, trace across tegula
ans = 17.737
octave3.2:53> std(AT524(:,2))%w/o Lieberkühn, trace across tegula
ans = 13.227

octave3.2:52> std(W523(:,2))%with Lieberkühn, across the wing
ans = 12.746
octave3.2:51> std(W524(:,2))%w/o Lieberkühn, across the wing
ans = 8.2902

Using standard deviation as a metric for image detail, we get an increase of about 75% in standard deviation over the dark photo by using the reflector.

octave3.2:30> (20.316+17.737+12.746)/(7.3926+13.227+8.2902)
ans = 1.7572

The averages, standard deviation etc. may seem a bit redundant at this point; you don’t need to plot a pixel-value profile to see that the image with the reflector is much brighter and more detailed than the photo taken without.