vlaiv

Two things happen here. First you set the time fixed - and this requires you to examine time zone and longitude. If it is noon in England - it will be sunrise in New York. Sun will be seen in different position in the sky at the same time. If you want to compare "apples to apples" in this case - you need to ask - where the Sun will be as it is transiting meridian in local time frame - meaning in both noon at England and in New York. Second thing is moving in latitude - or south / north position. Once you set specific transit time - then you can say at what altitude will transit happen, or when the planet is highest in the sky - how high it will reach depending on where you are located north or south. In your particular case if Mars is at 40 degrees altitude at 19h UTC in UK - it will be in the same place at 16:48 UTC on Cyprus (I took that you are at 0 lat in UK, while Cyprus is 33 degrees east, so it leads by 2.2 hours). Then if you are say at 53 degrees North in UK and Cyprus is at 35 degrees - at 16:48 UTC on Cyprus (which is GMT+3, so take that into account with local time) - Mars will be 40 + (53 - 35) = 58 degrees altitude. For any other time - calculation complicates and it's best to use planetarium software to figure out where planet will be. https://theskylive.com/planetarium?objects=sun-moon-mars-mercury-venus-jupiter-saturn-uranus-neptune-pluto&localdata=34.7768|32.4245|Paphos%2C+Cyprus|Asia%2FNicosia|0&obj=mars&h=16&m=48&date=2023-06-09#ra|8.942793329865122|dec|18.728398611111096|fov|80 Here we go - at 19:48 local time Mars is said to be at 45 degrees (so a bit lower than we calculated - but it's setting fast, at 19:05 was at 53 degrees altitude).

Regardless of its poor spot diagram? That translates into ~1.64" at 790mm of focal length and that is RMS. Corresponding FWHM is x2.355 or ~3.86". Further that means sampling at 2.4"/px, and this is limited by optics alone. By the way, those 6.3um are not at the edges of the FOV - it is pretty much like that across the field:

What are parameters to judge the win? Will integration time come into play? There is x4 difference in time between 10" and 5" which translates into x2 difference in SNR. If we compare apples to apples - 10" might be marginally better in terms of resolution but it will certainly be quite a bit better in terms of SNR in a given time frame.

Most likely it is, but you can measure it if you have calipers. It needs to be 42mm outer diameter on thread and 0.75mm pitch (distance between thread peaks). If it is - you'll need T2 - 1.25" adapter like this one: https://www.firstlightoptics.com/adapters/baader-125-t2-eyepiece-holder.html

Well, in that case it might be worth optimizing secondary, but do consider camera upgrade in the future.

What are you going to image? Planets or DSO? You won't have any issues with contrast as contrast can be adjusted for images. It is only important for visual - where you can "tweak" the image. Resolution will be minimally impacted for planetary (most people do planetary images with SCTs which have up to 40% central obstruction) and virtually not at all for DSO because resolution is governed by other factors - seeing, mount performance and aperture size. If you take two 6" scopes - one with large secondary and one with optimized secondary - you might be able to see the difference at the eyepiece when viewing planets in good seeing - but otherwise, scopes will perform the same and you won't be able to distinguish them.

I think that 150mm is very short distance between FP and secondary. From optical axis to the surface of the OTA, there is at least 80mm of distance if not more. Actually - it is 180mm according to TS website. It has 175mm of inner diameter of OTA tube and 2.5mm wall thickness. So there is total of 90mm "eaten up" by OTA alone. That places focal plane at just 50mm above surface of the OTA, and I don't think that focuser is very low profile. Focuser that comes with that OTA has 58mm to 2" connection: and on top of that you need to account 2" to sensor distance for your camera. All in all - I'd put FP to secondary distance to be at least 170mm - 175mm

I'm not very confident that FDM printed plastic will be air tight as well. Maybe look into putting some sealing paint / resin over it?

That really depends on hardware used and calibration performed. Ideally - you want hardware with very stable bias that does not use its own internal dark subtraction (like modern DSLR cameras do) and you want to perform dark scaling. With "well behaving" hardware above should work and give you results that are close to temperature matched darks. Best way to check if your hardware is going to be well behaved is to test it. To establish bias stability do the following: - take set of bias subs, then turn off power to the sensor, power it back on and do another set of bias subs. Measure mean and standard deviation of all subs. They should look the same between subs in each batch and between batches. - take set of bias subs and set of darks with increasing exposure length. Stack each to their respective stack and measure mean value of each. Plot in spreadsheet - they should form as close to linear function as possible (there will be some distortion because of temperature change as sensor heats up and cools down) if two above behave well - it should be possible to do dark scaling as part of calibration process. That is algorithm that will automatically "adjust" for different temperature (will select suitable multiplicative constant based on pixel statistics).

Hi and welcome to SGL. Nice capture indeed. @Kon here regularly captures ISS with 8" dob without tracking if I'm not mistaken. Those images are simply stunning given equipment and technique. Worth checkout out those captures.

Not really sure about that. Depends in which direction outside of the galaxy we are talking about, but I doubt it even for the shortest path. Sun has apparent magnitude of 4.83 at 10 parsecs. For the sake of argument - let's say that NELM from outside of the galaxy is ~9, so the difference between the two is ~5 magnitudes. This allows for only x100 difference in intensity. Now, intensity drops with square of distance, so this is really only sqrt(100) = x10 of distance. In another words, Sun will be seen as mag9 star from 10 x 10 parsecs away. That is 100pc or 326 light years. Now some info on Milky Way - thickness of the thin disk is only 220–450 pc (718–1,470 ly). It looks like Sun is situated smack in the middle of it (here is scheme from Wiki): So there is at least 110pc to the edge of thin disk - and we could argue that outside of the galaxy is at least outside of thick disk, let alone galactic halo. Given the above - I doubt that Sun is visible from outside of our galaxy.

There is something seriously wrong with that scope This is moon thru ST102 - single shot with DSLR at prime focus: Yes, there is some violet - blue halo (not very visible against black sky) and there is some yellowing on the inside of the limb - but nowhere near as in your shot. Btw ST102 is shot focal length fast achromat - 4" F/5. It should be really colorful scope - but in reality it is not as bad as your image shows.

Just remember that observing is a skill that you improve with practice. We actually learn how to see those faint objects and it is important to spend time under stars. On numerous occasions I've read and later witnessed that myself - in the beginning what we call bright DSOs - don't seem bright at all - but there comes a time, after observing faint and very faint stuff that one gets surprised by how bright those targets really are in comparison.

Yes - it is very important for conditions to be right. Good transparency and good position in the sky - which is usually overhead near the zenith - unless there is particular patch of the sky that is darker (like when you are on the edge of the city and there is one direction with lower LP). I just dug up one of my old reports. Bortle 7 / red zone, 4" fast achomatic refractor: M8, M20, M17, M16, M18, M24, M2, M28, M22, NGC6638 and NGC6642 (Very faint, peripheral vision), M25, M26 and M11 This was summertime, I was on the edge of white / red zone and observing in the direction of the least LP (due south - which helped).

I'm not sure that you'll be able to see the sun and visible stars from the outside of our galaxy.

I think that there are two versions or possibly more. I was looking at these two, and they seem to be of a different size: https://zhumell.com/products/zhumell-z130-portable-altazimuth-reflector-telescope and this one https://zhumell.com/products/zhumell-z114-portable-altazimuth-reflector-telescope Note that top one has two "handle slots" and looks bigger compared to bottom one. It holds 130mm scope so I guess it is in Heritage 130P class. Smaller one is probably in Heritage 100 class Not sure where that Omegon one falls between these two. They have specs on their website - it says that base is 195mm and height is 250mm. Maybe this will help: http://www.waloszek.de/astro_omegon_basis_e.php (it looks like it is about the same size as Heritage 100 - so smaller).

But it is obvious that it outperforms the TOA. If you look at the images - C11 image is sharper and more contrasty. It goes a bit deeper and I think it was easier to process by looking at both results. All of that in ~1/3 of time? By most discussed metric by amateur astronomer images - which is "speed of the scope" and how to get a good image (which is equal to high SNR first and foremost) - C11 wins hands down. If that metric is not as important to you - well, that is another matter. As you've put it - it is much easier for you to image with TOA and based on that - only you can say if difference is worth it to you or not.

I knew it was possible from Bortle scale definition: That is for Bortle 1 sky. It is supposed to be averted vision target in Bortle 3 skies and even Bortle 4 (when high in the sky). https://en.wikipedia.org/wiki/Bortle_scale but never saw it myself without a pair of binoculars or a telescope. I managed once to get a glimpse from Bortle 7 with pair of 35mm bins - but I lost it and could not find it again that night.

Just in case you want to purchase a scope just to get the mount - here is alternative scope which you might prefer over small newtonian: https://www.telescope.com/Orion-StarMax-90mm-TableTop-Maksutov-Cassegrain-Telescope/p/102016.uts

You don't need to purchase scope with it just to have a mount (although, we all love good excuse to purchase another scope): https://www.astroshop.eu/alt-azimuth-without-goto/omegon-mini-ii-dobsonian-mount/p,53648

This brings on an interesting topic - how many galaxies can be actually seen with a naked eye? M31, M33, LMC, SMC are those that I can think of, but apparently - there is more: https://en.wikipedia.org/wiki/List_of_galaxies#Naked-eye_galaxies

Think about what this means. This means that for any given pixel you choose in the background on the image - you can expect it to be in certain range of values with certain probability. I'll give you idea with Gaussian distribution because Poisson distribution for large numbers behaves pretty much like Gaussian - and Gaussian is more familiar: There is some mean value - central value. When we say that background is 900 - that is what we mean - average value of background pixels is 900. Then there is some noise. If we have SNR of roughly 30 - this means that noise is 30e. Noise is sigma in above distribution graph. Standard deviation. About 68% of all pixels will have value in +/- one sigma range - which means that roughly 68% pixels will have values between 870 and 930. About 95% of all pixels will have values in two sigma range - which means that roughly 95% of all pixels of the background will fall between 840 and 960 in their value. If we take these pixel values they will look like this: 891, 914, 902, 907, 899, .... They won't be exactly 900 - but will scatter according to above graph. When we take some number of such samples on average (non bias selection) and we perform their average - that average value will fall closer to true mean value - closer to 900 than individual samples. Noise can be thought of as scatter around the mean value. When we perform averaging of samples to get new values - we improve SNR. What does that mean? Well - average value remains the same (averaging won't change actual average value) - but scatter will be reduced. If you have 10000 above random values and you divide them into groups of say 100 numbers in each group and you average every group - each group will then become single value - and you will have 100 values. Those 100 values will be more closely grouped around whole average. This is what SNR improvement is - you reduced scatter of averaged values. If you average whole 10000 - you will get single value so we can't really talk about scatter - but we can talk about uncertainty - or range around true mean where this value falls. If you have 10000 numbers with average of 900 and standard deviation of 30 and you average those numbers - you will get result that is in 900 +/- 0.3 with 68% certainty, or in 900 +/- 0.6 with 95% certainty. Makes sense now?

You start with exposure that you have in order to determine exposure that you need to swamp the read noise. You don't need much light at all since you'll be averaging larger number of pixels in the background. Say that you have significantly lower average background signal - say that you have only something like 50e. That is SNR of 7 (sqrt(50) = ~7e of noise 50/~7 = ~7 SNR). Again if you average 10000 samples (100x100 pixels) - you get boost of x100 in SNR or SNR of 700. In this case, your average signal will be 50 +/- 0.07e - again, plenty of accuracy for what you need. Even if you have something very low - lower than read noise, say your signal is 1e on average. SNR will be 1 because sqrt(1) = 1 and 1/1 = 1 If you average 10000 such pixels - you will get SNR of 100 or your final value will be 1 +/- 0.01 Even if we account for read noise - make it 2e again and see what happens: Signal is 1e, shot noise is 1e, read noise is 2e. Total noise is sqrt(1*1+ 2*2) = sqrt(5) = ~2.236, so SNR is 1/2.236 = ~0.447. After averaging 10000 pixels SNR will be 44.7, so noise in final result will be 1/44.7 = ~0.022. We still have only 2% of error in our measurement of average background level. You don't need much light to have good estimation of average signal if you use that many samples - 10000 pixels. It's like stacking 10000 images. We manage to get good images by stacking only hundreds of subs and even less.

Noise, by definition is random deviation from some mean value. If you take some number of noisy samples and you average them - you will get value that is close to that average/mean value. More samples you take - closer to expected value you will be with your average. This is how stacking works - we take bunch of images that are noisy and average them and get much cleaner image - one that is much closer to true / average value of signal. When you measure sky background value in the image - you average bunch of pixels that have same mean value. Yes, each will be slightly different in true value due to noise - but since they all have same mean value - which is level of sky background - you will get value that is very close to what you want to measure. It's like stacking thousands of subs to get correct image but we "stack" thousands of pixels instead (thousands of single values). Here is an example. Say that your mean background value is 900 electrons. Just from Poisson noise - we will have noise level of 30. Add some read noise to it - let's say that read noise is 2e, so total noise is sqrt(30*30+2*2) = 30.066 (see here how small read noise compared to large background noise barely affects total noise levels). Thus we have SNR of 900/ 30.066 = ~29.93. This really means that we can expect to see 900 +/- 30.066 as background value about 67% of the time (values between ~870 and 930). But if we take square that is only 100x100 pixels - that is 10000 pixels, and average those - we get SNR improvement that is equal to x100 (square root of number of averaged samples). This means that our SNR is no longer ~29.93 but rather ~2993 (increased by factor of x100). so we have 900 / some_uncertainty = 2993 from there some_uncertainty = 900 / 2993 = ~0.3 Number that we get by taking average is actually 900 +/- 0.3e Yes, there will still be some residual error of measurement - but it is so small that we can just take our value to be correct value. We can easily say that 900.3e is actually 900e - it won't change our results by any significant amount. Larger patch of the sky you take - better estimate of background signal value you get (more precise). Only problem is that with larger patches of sky we also take some stars. This is why it is better to use median of this then true mean - as median is far less sensitive to outliers (which would be anomalous pixel values due to star light falling on those pixels).

Depends on how you measure. If we want to calculate exposure length to swamp the read noise - we do it by measuring average background value - not noise. We measure sky signal level. Sky signal level will not depend on read noise. It might depend on bias level, and that is why we need to calibrate it - to remove that bias level / offset. Once we measure sky background level - then we calculate sky background noise. If we were to measure standard deviation of image - we would find that it is a bit higher then calculated value - precisely for the reason you are mentioning - there are other noise sources that add up to total noise levels in the image (both thermal and read noise). However - those don't affect sky background value and sky background value is directly related to the amount of noise coming from it - because it follows Poisson distribution and for it - noise is equal to square root of signal value.