LDN 1228 - more data made a big difference

[gorann]

This image is the result of having three times more data than in my previous post. Taken over three nights with the RASA8 and ASI2600MC (4 min exposures at gain 100). It made such a difference that I post it separately. So now 6 hours in stead of two hours and even at f/2, faint objects like these "dark" nebulae need integration time. Much less frustrating to process and no NR had to be applied this time. Data collected over three nights 24 - 27 August.

Stacked in PI and processed in PS. No crop so the RASA 8 is quite good at filling up the ASP-C sized chip of the ASI2600. No flats or darks or any calibration done and not needed as far as I can tell.

With regard to capturing photons, 6 hours at f/2 should equal 70 hours at f/7, if I got it right. Someone may correct me like @ollypenrice or @vlaiv......

For comparison I put the two hour version below the new 6 hour one.

LDN 1228 is a dark nebulosity site in Cepheus. Quite some amazing shapes formed by the dark dust - food for fantasy.

I am not totally unhappy for the clouds that have rolled in - have not got much sleep the last nights, not the least because the old Mac Powerbook decided now and then to stop downloading subs, so I had to keep a watch.

 

 

 

 

 

 

Edited August 27, 2020 by gorann

[andrew s]

Very impressive and has a beauty in subtle shades. Regards Andrew 

[gorann]

5 minutes ago, andrew s said:

Very impressive and has a beauty in subtle shades. Regards Andrew 

Thanks Andrew, much appreciated! Clouds have moved in and I look forward to getting a full nights sleep😵

[andrew s]

Just now, gorann said:

Thanks Andrew, much appreciated! Clouds have moved in and I look forward to getting a full nights sleep😵

Good it's not fair if you hover up all the photons with the RASA. They are a great addition to the astrograph options. 

Regards Andrew 

[smr]

Yep, data is everything really. 

In the 2017 Astronomy Photographer of the Year book I was given by my Dad for Christmas, there's one theme amongst all the images in the book, whether winner, runner up or commended... they all have a staggering amount of integration time, on average around 50 hours. Whether a Nebula was imaged with a Canon 600D, 6D or a £6k Mono Camera.

Nice image Goran, lovely dust showing through. Amazing part of the sky.

[gorann]

8 minutes ago, smr said:

Yep, data is everything really. 

In the 2017 Astronomy Photographer of the Year book I was given by my Dad for Christmas, there's one theme amongst all the images in the book, whether winner, runner up or commended... they all have a staggering amount of integration time, on average around 50 hours. Whether a Nebula was imaged with a Canon 600D, 6D or a £6k Mono Camera.

Nice image Goran, lovely dust showing through. Amazing part of the sky.

Thanks, yes there are no shortcuts, although maybe f/2 helps a bit.

[vlaiv]

2 hours ago, gorann said:

With regard to capturing photons, 6 hours at f/2 should equal 70 hours at f/7, if I got it right. Someone may correct me like @ollypenrice or @vlaiv......

In very specific circumstance 6 hours at F/2 will be equal to 73.5 hours at F/7 - if you take your scope and stop it down to F/7 by using aperture mask of ~57mm. Alternatively you can take ~57mm aperture refractor with 400mm FL, attach same camera and yes in ~70 hours (a bit less - no central obstruction and better light transmission) - you'll get same result.

Other scopes at F/7 will probably need less time than that, but FOV might not be the same. Alternatively if one uses different camera - F/7 scope can potentially capture same image in even less time.

[gorann]

Thanks Vlaiv, that was about how I saw it. Maybe f/7 was not a realistic comparison. But there are many 400mm FL refractors at f/5 (or thereabouts) and they would need 37.5 hours.

Edited August 27, 2020 by gorann

[gorann]

4 hours ago, andrew s said:

Good it's not fair if you hover up all the photons with the RASA. They are a great addition to the astrograph options. 

Regards Andrew 

I have no experience with the other low f-value "Astrographs" around but they all are more expensive than the RASA 8, some rediculously expensive, and often with less aperture and most come with star spikes and a need to fiddle with collimation. Seems like the RASA 8 just keeps its factory collimation (even mine that was second hand), and maybe the non-Newtonian construction is less sensitive to collimation. My only issue with it so far is star shapes in the corners which suggest some tilt but they are usually easy to fix in processing, and the RASA 8 is only claimed by Celestron to work well up to a 22 mm image circle and I am pushing it to 28 mm (like in this image) with my APS-C chip, and still vignetting is minor.

The RASA 11 at twice the price provides a larger image circle of 43mm but it weighs 20 kg instead of 8 kg and has a longer FL (620 mm compared to 400 mm) so much less field of view with the same camera. So, with a RASA 11 you need a full frame APS to get the same field of view that I have with the APS-C. I wonder what the impact on image quality would be, if any. Maybe a tiny increase in resolution on a very clear night, but then with a much more expensive camera and mount than with the RASA 8.

Edited August 27, 2020 by gorann

[wimvb]

Great image, Göran.

There’s a weak Ha signal in the 2 h version, that I miss in the 6 h version. A passing Ha cloud, or did you hoover up the last seconds of the sunset as well. 😁

[gorann]

12 minutes ago, wimvb said:

Great image, Göran.

There’s a weak Ha signal in the 2 h version, that I miss in the 6 h version. A passing Ha cloud, or did you hoover up the last seconds of the sunset as well. 😁

Thanks! I could have ovesaturated the red a bit in the 2 h version. The Ha is there but not as prominent after 6 h, now apparently mixed with other light picked up from those structures.

Edited August 28, 2020 by gorann

[vlaiv]

I wonder, why do these dark structures end up being red in images?

Are they red in real life, and if so, why?

On the other hand, could that be imaging artifact?

This is my reasoning - this is gas and dust that is reflective in nature - so very low levels of light scatter of it and that light is predominantly from surrounding stars. We can see them as low intensity reflection nebulae.

Most reflection nebulae are either gray or bluish. This is due to Rayleigh scattering as blue part of spectrum is scattered much stronger. We can see that in this Iris image from APOD:

Original page: http://sai.msu.su/apod/ap051229.html

Why do dark structures appear red then?

Two possible solutions:

1. atmosphere influence. Here we have another "round" of Rayleigh scattering going on, but this time light is scattered away from our targets. This is reason why everything we record is shifted towards reds - lower "temperature" colors (reason why sun looks yellow rather than white and orange/red at sunset / dawn).

2. Light pollution - which is mostly red on cameras - probably because of its spectra and how it interacts with atmosphere. Might change with wide spread use of LEDs.

3. Just an artifact of processing. Trying too hard to show dark structures as signal instead of just being dark / light blocking patches. Too much saturation that goes into reds?

 

Above images are interesting to case study this phenomena - it contains both gray parts and darker parts turning red for some reason. Why do you think this is?

[gorann]

1 hour ago, vlaiv said:

I wonder, why do these dark structures end up being red in images?

Are they red in real life, and if so, why?

On the other hand, could that be imaging artifact?

This is my reasoning - this is gas and dust that is reflective in nature - so very low levels of light scatter of it and that light is predominantly from surrounding stars. We can see them as low intensity reflection nebulae.

Most reflection nebulae are either gray or bluish. This is due to Rayleigh scattering as blue part of spectrum is scattered much stronger. We can see that in this Iris image from APOD:

Original page: http://sai.msu.su/apod/ap051229.html

Why do dark structures appear red then?

Two possible solutions:

1. atmosphere influence. Here we have another "round" of Rayleigh scattering going on, but this time light is scattered away from our targets. This is reason why everything we record is shifted towards reds - lower "temperature" colors (reason why sun looks yellow rather than white and orange/red at sunset / dawn).

2. Light pollution - which is mostly red on cameras - probably because of its spectra and how it interacts with atmosphere. Might change with wide spread use of LEDs.

3. Just an artifact of processing. Trying too hard to show dark structures as signal instead of just being dark / light blocking patches. Too much saturation that goes into reds?

 

Above images are interesting to case study this phenomena - it contains both gray parts and darker parts turning red for some reason. Why do you think this is?

Interesting Vlaiv. I just googled images for LDN1228 and a bunch came up and they are all red / brownish like mine, which was a releif🙂.

Here are examples of LDN1228 from Astrobin:

https://www.astrobin.com/363920/

https://www.astrobin.com/kwj3o9/

https://www.astrobin.com/okyb9k/B/

It is a very faint signal compared to the Iris with no really strong star shining on the dust. However, the nebulosity in the image is not uniformly red. The fainter nebulosity is more yellowish or even bluish on the left side of the image. If it was an artefact of processing faint signals then the faintest parts should be more red but it seems to be the opposite. Also, I have the impression that there was more red signal in the dark nebulosity after 6 hours than 2 hours. Could it not just be that there is some Ha emmision mixed in at various degrees? I could point out that LDN1228 was quite high up in the sky here so less atosphere than for many other objects.

Images of other dark nebulosities are not allways red, some are quite grayish - black. For example, Barnard 174 is another dark nebula in Cepheus and it does not show this reddish colour:

https://www.astrobin.com/71661/C/

https://www.astrobin.com/203604/

Here is an example of Barnard 72 that would have been imaged closer to the horizon and it is geryish black:

https://www.astrobin.com/bijs57/B/

 

 

Edited August 28, 2020 by gorann

[vlaiv]

5 minutes ago, gorann said:

However, the nebulosity in the image is not uniformly red. The fainter nebulosity is more yellowish or even bluish on the left side of the image. If it was an artefact of processing faint signals then the faintest parts should be more red but it seems to be the opposite.

Exactly. Look at these two samples:

Now, let's for a moment do analysis of what we see here. This is my reasoning. In gray nebulosity in above sample we don't see much noise. This leads me to conclusion that this part is high in SNR (not overly, but enough to be rendered smooth). It is gray as well - which means that both R, G and B signal is of equal strength - so SNR is similar if not equal (maybe red is a bit stronger as this patch is more beige than gray).

Now in second image - I can clearly see grain in red. This means that this patch is lower SNR than above patch.

Lower SNR means greater uncertainty in signal precision and also signal being lower. Do we have enough SNR there to be certain it is redish?

In any case - you should not worry of your rendition not being good - I just wondered what would be explanation for these red tones.

 

[gorann]

1 minute ago, vlaiv said:

Exactly. Look at these two samples:

Now, let's for a moment do analysis of what we see here. This is my reasoning. In gray nebulosity in above sample we don't see much noise. This leads me to conclusion that this part is high in SNR (not overly, but enough to be rendered smooth). It is gray as well - which means that both R, G and B signal is of equal strength - so SNR is similar if not equal (maybe red is a bit stronger as this patch is more beige than gray).

Now in second image - I can clearly see grain in red. This means that this patch is lower SNR than above patch.

Lower SNR means greater uncertainty in signal precision and also signal being lower. Do we have enough SNR there to be certain it is redish?

In any case - you should not worry of your rendition not being good - I just wondered what would be explanation for these red tones.

 

Yes, in all astroimages bright areas show less noise than darker ones, but not all dark areas are reddish. So why do the darker areas in LDN1228 allways turn out reddish but not in Barnard 72 or 174?

[andrew s]

It may also depend on the spectra of the illuminating star or stars. Reflection nebula are often illuminated by very hot massive stars with most of their output in the UV. 

Cooler stars have their output mainly in th IR. With other stars filling in. So what you get will depend on the illumination, forward or backward scattering and the nature of any dust in the nebula. For very distant object interstellar reddening may also play a role.

Regards Andrew 

Edited August 28, 2020 by andrew s

[vlaiv]

8 minutes ago, gorann said:

Yes, in all astroimages bright areas show less noise than darker ones, but not all dark areas are reddish. So why do the darker areas in LDN1228 allways turn out reddish but not in Barnard 72 or 174?

Ok, so this is literally the first image in google search for LDN1228:

And this is second:

and this is yours:

In both above images we have grey area that I marked out being darker. Whole nebulosity has this beige or brownish tint to it - and I suspect that is color shift due to atmospheric scattering.

Yours has red in dark areas - I also marked it out.

[gorann]

There are examples with more or less red. It just depends on the level of saturation chosen. If I just turn down the master saturation in PS it looks like this. Not sure if it is more "correct" rendention of the nebula.

[vlaiv]

1 minute ago, gorann said:

Not sure if it is more "correct" rendention of the nebula.

yes, that is a good question.

Did you do color calibration on stars?

[gorann]

22 minutes ago, andrew s said:

It may also depend on the spectra of the illuminating star or stars. Reflection nebula are often illuminated by very hot massive stars with most of their output in the UV. 

Cooler stars have their output mainly in th IR. With other stars filling in. So what you get will depend on the illumination, forward or backward scattering and the nature of any dust in the nebula. For very distant object interstellar reddening may also play a role.

Regards Andrew 

Thanks Andrew for allowing for the possibility of reddish dark nebulae😉

[gorann]

2 minutes ago, vlaiv said:

yes, that is a good question.

Did you do color calibration on stars?

No, no color calibration done.

[gorann]

 

16 minutes ago, vlaiv said:

yes, that is a good question.

Did you do color calibration on stars?

 

Found this IOTD by Maurice Toet. Looks like he also picked up brownsh-red in LDN1228. But maybe I had my red saturation turned up a bit too much. I thought it looked nice.

https://www.astrobin.com/full/z4sajo/0/

[gorann]

So upon suggestions I have turned down the red saturation a bit, so here is a new version. I agree that it may look better. Maybe I am just a sucker for red😉. I finally also gave the darker parts a bit of NR.

[vlaiv]

10 minutes ago, gorann said:

 

 

Found this IOTD by Maurice Toet. Looks like he also picked up brownsh-red in LDN1228. But maybe I had my red saturation turned up a bit too much. I thought it looked nice.

https://www.astrobin.com/full/z4sajo/0/

I guess this is down to atmospheric scattering.

Here is a little experiment:

First our two color targets as I believe look outside of our atmosphere - above one is dust in your image, and other one is reference - GV2 star 6500K - our sun.

We don't see our sun as being white in regular day light - but rather yellowish - this is due to Rayleigh scattering in our atmosphere (sky takes blue color for itself and leaves sun color to be yellowish). We can do this by adjusting image temperature in Gimp for example:

Nice, we already have brownish color for our dark nebula (although it is gray in nature). Let's do something astrophotographers like to do in their processing - let's boost saturation:

In the end, if I compare the two - part of LDN1228 from the image you posted last and this color that I made by simple process of atmospheric scattering and boosting saturation:

Pretty good match, right?

[gorann]

17 minutes ago, vlaiv said:

I guess this is down to atmospheric scattering.

Here is a little experiment:

First our two color targets as I believe look outside of our atmosphere - above one is dust in your image, and other one is reference - GV2 star 6500K - our sun.

We don't see our sun as being white in regular day light - but rather yellowish - this is due to Rayleigh scattering in our atmosphere (sky takes blue color for itself and leaves sun color to be yellowish). We can do this by adjusting image temperature in Gimp for example:

Nice, we already have brownish color for our dark nebula (although it is gray in nature). Let's do something astrophotographers like to do in their processing - let's boost saturation:

In the end, if I compare the two - part of LDN1228 from the image you posted last and this color that I made by simple process of atmospheric scattering and boosting saturation:

Pretty good match, right?

So then, what is the correct rendention Vlaiv? According to your suggestion LDN1228 will be brown when seen from earth, and that is from where we are looking at it and take pictures of it....