For a 6″ f/12 Newtonian, a sphere suffices…
I mentioned that I was searching for my 6″ f/12 that I made years ago. Still have not found it, but I was wondering: how good is a 6″ f/12 sphere? I recall hours of polishing to try to get to a nice, smooth null, but don’t remember if I ever quantitatively figured out how good such a mirror would actually perform. So, I used tex, the same program that I used yesterday to analyze the real test data for a 6″ f/6.4, but instead entered the data for a 6″ f/12. To simulate a perfect sphere, I set all the zones to be zero: all measuring the same. Here’s the output:
TEXEREAU MIRROR TEST SHEET Comments: six inch Optical diameter: 6 Readings per zone: 1 Radius of curvature: 144 f/D: 12.00 Diffraction disc: 0.000316224 1 ZONE 1 2 3 4 5 2 h(x) 1.3416 1.8974 2.3238 2.6833 3.0000 3 h(m) 0.6708 1.6195 2.1106 2.5035 2.8416 4 hm**2/R 0.0031 0.0182 0.0309 0.0435 0.0561 5 hm/4f 0.0023 0.0056 0.0073 0.0087 0.0099 6 D1 0.0000 0.0000 0.0000 0.0000 0.0000 7 D2 0.0000 0.0000 0.0000 0.0000 0.0000 D3 0.0000 0.0000 0.0000 0.0000 0.0000 8 D123 0.0000 0.0000 0.0000 0.0000 0.0000 9 D123 + 0.0423 0.0423 0.0423 0.0423 0.0423 0.0423 10 Lamda c 0.0392 0.0241 0.0114 -0.0012 -0.0137 11 Lamda f * 1e5 9.13 13.56 8.35 -1.04 -13.56 12 Lamda f / rho 0.289 0.429 0.264 -0.033 -0.429 13 u * 1E6 -1.27 -1.88 -1.16 0.14 1.88 14 Wavefront -1.18 -1.71 -1.69 -1.12 0.00 Reference parabola: y = -0.288278 * x**2 + 0 Maximum wavefront error = 1 / 12.6 wave at zone 2
Not bad at all. The wavefront error is around 1/13 wave, and the transverse aberrations compared to the Airy disc sizes are all less than one (read from line 12 of the output above). A good null for a 6″ f/12 is indeed a very good telescope: even Texereau would be happy.
What about the classic 6″ f/8? We can do the same experiment here.
TEXEREAU MIRROR TEST SHEET Comments: Optical diameter: 6 Readings per zone: 1 Radius of curvature: 96 f/D: 8.00 Diffraction disc: 0.000210816 1 ZONE 1 2 3 4 5 2 h(x) 1.3416 1.8974 2.3238 2.6833 3.0000 3 h(m) 0.6708 1.6195 2.1106 2.5035 2.8416 4 hm**2/R 0.0047 0.0273 0.0464 0.0653 0.0841 5 hm/4f 0.0035 0.0084 0.0110 0.0130 0.0148 6 D1 0.0000 0.0000 0.0000 0.0000 0.0000 7 D2 0.0000 0.0000 0.0000 0.0000 0.0000 D3 0.0000 0.0000 0.0000 0.0000 0.0000 8 D123 0.0000 0.0000 0.0000 0.0000 0.0000 9 D123 + 0.0635 0.0635 0.0635 0.0635 0.0635 0.0635 10 Lamda c 0.0588 0.0362 0.0171 -0.0018 -0.0206 11 Lamda f * 1e5 20.55 30.51 18.79 -2.34 -30.51 12 Lamda f / rho 0.975 1.447 0.891 -0.111 -1.447 13 u * 1E6 -4.28 -6.36 -3.91 0.49 6.36 14 Wavefront -3.99 -5.77 -5.69 -3.76 0.00 Reference parabola: y = -0.972966 * x**2 + 0 Maximum wavefront error = 1 / 3.7 wave at zone 2
As you can see, this would not meet Texereau’s exacting standards. Even at f/8, we need to exert some work to turn it into an excellent performer.
Addendum: I took the source code for Lindner and Phillips’ program, and cleaned it up a bit, and added it to my source repository. You can get the code here. I like that it duplicates the calculations that are done in Texereau’s book, even though it’s not the most sophisticated program in the world.
I suspect the world would be better if that percentage were even greater.
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Congrats, glad to hear all is well.