I am considering having a Nipkow disk manufactured locally, only the manufacturer has a minimum hole size capability of 2mm. I was originally going to do the holes by hand, or have a larger hole created, and then attaching 'slap on' hole patches which I could then position as required.
But I got to thinking... what if I were to use the 2mm holes, how could I space the holes to make it work? And then I started thinking "well, a bit of overlap isn't going to hurt"... and then "actually, a bit of overlap is going to help"... "what is the optimal overlap?"
The attached diagram shows what I think should happen. Given three Nipkow holes, conceptually 'side by side' (over time, they are), 'a', 'b', and 'c', it would appear ideal to me if the light passing through the non-overlapping section (B') is equal to the light passing through the overlapping section (A').
But then I realised that small parts of B' overlap the area of the overlapping section A', even though the section itself isn't in the overlap. So I added the red lines to divide the sections more clearly. Now the task is to ensure that the same amount of light is visible in each red horizontal section, and furthermore to have the width of the horizontal sections the same, or as close as possible. I don't know if the light passing through the sections is equivalent when the sections are the same width -- I highly doubt it -- but there must be some optimal 'width' vs 'light' setting that would make each section appear roughly the 'same'.
So I thought I'd ask anyone if they have any thoughts on how to calculate (a) the light that goes through each section, and (b) the width of the sections (ie: spacing of the holes) and (c) the problems with this idea/approach.
Where I'm leading towards is to actually use 2mm holes, but overlap each hole by (say) 1/3 of the diameter of the hole by shifting it towards the next hole (as shown in the bottom right of the image). This would give me equally spaced 'areas', an unknown light value through each, and (say) a final image width of 42+mm.
Using this technique, the extreme left/right pixel will appear slightly different, but that's not an issue for me.