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alpal

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About alpal

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  1. That's a very impressive image and a very impressive lens you have. That's one of the most expensive camera telephotos you can buy. Well done. cheers Allan
  2. I've been think about this for a few days. I am looking for a simple explanation. For instance if we go here and calculate the size of the airy disc for old my 8" f6 Newt. http://www.wilmslowastro.com/software/formulae.htm#Airy we get 1.26 arc seconds. Using this calculator https://astronomy.tools/calculators/telescope_capabilities the Dawes limit is 0.57 arc seconds and the Rayleigh limit is 0.68 arc seconds. Considering that the best seeing I've had is 2.9 arc seconds FWHM then an obstruction of say 20 to 30% would be negligible in effect. That would explain w
  3. Yes - 20 to 25% obstruction is almost negligible. Many of the large astrographs have a 50% obstruction but still seem to produce nice images.
  4. Hubble's secondary mirrors has a very small obstruction which is easy to do at f 24. https://en.wikipedia.org/wiki/Hubble_Space_Telescope https://www.nasa.gov/mission_pages/hubble/story/index.html Primary Mirror Diameter: 94.5 inches (2.4 m) secondary Mirror Diameter: 12 inches (0.3 m) Obstruction percentage 0.3/2.4 = 12.5 %
  5. Thanks Vlaiv, thanks for trying to explain these issues. I'm not sure why telescopes with such large secondary obstructions still seem to work so well.
  6. Why? Well yes - that article was a little bit biased as the author is the Australian and New Zealand Takahashi dealer - Claudio Voarino: https://www.astronomy-electronics-centre.com.au/ cheers Allan
  7. Thanks Vlaiv, there is a good article here: https://www.astronomy-electronics-centre.com.au/article_resolvingpower.htm QUOTE
  8. Because you wrote "Ok, I'll show you why it is not a very good solution" I still haven't found the software or calculator that I remember seeing many times about 10 years ago.
  9. On the contrary - the article gives a telling picture that I post here now:
  10. It looks like a very good solution. I think it is referring to splitting a double star? The maximum ratio of energy between the diffraction rings is 5.25.
  11. But those numbers don't agree with my original calculations. It also says on page 11. http://s000.tinyupload.com/index.php?file_id=91035876064397637171 Table I. Values for obstruction and contrast factor. The left-hand column is obstructions in percentage of secondary to primary mirror in linear diameters. Next column gives the central spot energy percentage, next the energy in the rings and then the contrast factor values taken from the reference text. I can't get a copy of the reference text but I understand the 5.25 number. Hurlburt, H.W., "Improvement
  12. The one which is 80% would have a more blurry view as per page 13 of the link I gave above and give again here: http://s000.tinyupload.com/index.php?file_id=91035876064397637171
  13. I found something on my computer about contrast factor not contrast ratio and I saved the pdf here with file hosting online: http://s000.tinyupload.com/index.php?file_id=91035876064397637171 It says on page 11. From the graphs and tables published in the referenced articles on Newtonian improvements in the Journal of the Association of Lunar and Planetary Observers (J.A.L.P.O.) a general equation can be arrived approximating the"contrast factor" value for your system (See Table I): CF = 5.25 - 5.1x - 34.1x 2 + 51.1x 3 where x is the obstruction ratio or secondary/prima
  14. Thanks Vlaiv, Yes - I'm assuming that "relative peak diffraction intensity" is contrast ratio ? Oh dear - we have a definition problem.
  15. Thanks Vlaiv, the graph is here: and the formula here: reduction in the relative peak diffraction intensity - and the energy content of the central maxima - by a factor (1-ο2)2, as per here: https://www.telescope-optics.net/obstruction.htm so an obstruction of say 20% is (1-ο2)2 = (1-o.22)2 = 0.9216 or 92% contrast ratio. so an obstruction of say 50% is (1-ο2)2 = (1-0.52)2 = 0.5625 or 56% contrast ratio. Can you confirm those calculations please? That is actually all I was after. Whatever Newt. design program I used to use had that
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