Comment on Orion Nebula without a star tracker video

Orion Nebula WITHOUT a Star Tracker or Telescope, Start to Finish, DSLR Astrophotography

I am trying to estimate the time and cost for myself. Today is 19 Jun 2021 for me. Amazon link to Canon EOS 5D Mark III body only is $899.99 + shipping and tax – never used Canon, lots of screens and menus

Manfrotto 055 Aluminum 3 Section Tripod is $269.95
Rokinon 85M-C 85 mm F1.4 lens for Canon is $249 but says not zoomable – hours to figure out. Video shows zooming
Bahtinov Mask Focusing Mask for Telescopes is $22.99 but there are many diameters
Deep Sky Stacker is free but it has to be installed and learned. “AHD and Bayer Drizzle” and “bilinear interpolation” could use some explanations.
Adobe Photo Shop – not sure what he is using – probably the 12 month subscription $179.91
I live in a large city and never see any stars at all. Infinite time delay.
So if I order all these, figure how to use them. A few days maybe. But still no sky to take pictures of. That business of focusing to get the star dots as small as possible. A bit iffy. I can’t see small screens that well. Setting the threshold for statistical selection of “best” stars. Problematic.
Moving and refocusing.
Knowing where to find Orion’s Nebula (M42, NGC 1976) and that means finding Orion’s belt. Then Orion’s sword. says “It is seen as the middle “star” in the “sword” of Orion, which are the three stars located south of Orion’s Belt.” “visible to naked eye” “magnitude +4”
What was all that ISO business? When I searched for “stars”+”magnitude+4″+”ISO” I found and it sort of helped.
ISO 800 for 2 minutes in a 35 mm camera with F 1.4 lens? Try finding an explanation of ISO, let alone tradeoffs and picking “good” images. I kept searching. Finally I searched for “magnitude” “f 1.4” “starlight” and found “Stellar apparent magnitude vs exposure times” that was rather horrible. I searched “magnitude” “starlight” “exposure” “iso” and found Sky and Telescope (respectable) at
two minutes (120 seconds) for magnitude 4
10 minutes (600 seconds) for magnitude 16 says (I am a pretty good mathematician but Wikipedia should learn how to put in put in calculator, how rude). 16 minus 4 is 12 apparent magnitude steps. They don’t number the equations so I can’t point you to what I am looking at.
Went to and found F2/F1 ~ 2.512^(16-4) = 2.512^12 = (100)^(12/5) = 63095.73
I think that 2 minutes up to 10 minutes gives me five times as many photons. This factor of about 60,000 difference in the watts/meter^2 flux of energy is hiding the fact that even though you can see the stars in the photo, they are not as bright and as many pixels.
The total pixels times their respective intensities has to add up in the respective photos. There should be about 63000 times more total intensity (pixels times their intensities) in an area of the sky that overall is 12 apparent magnitudes larger.
So just sum the intensities in one image, do the same for another of the same size. Take the ratio of Intensities and call it “relative brightness”
The ratio of the intensities (Intensity1/Intensity2) take the log10 of that and divide by (5/2)= 2.5 to get the apparent magnitude differences. Then use the tables at where they call the ratio of Intensities “brightness” to see what times are required for other stars and things.
I really appreciate the time you took to make this video. It gave me great insight into what faces someone who wants to take pictures of the sky (assuming they can see it). This is a great video for walking through everything for someone who has a Canon camera and wants to try taking pictures of the sky. It is literally “astro”photography.
Richard Collins, Director, The Internet Foundation
Richard K Collins

About: Richard K Collins

Director, The Internet Foundation Studying formation and optimized collaboration of global communities. Applying the Internet to solve global problems and build sustainable communities. Internet policies, standards and best practices.

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