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Who can be first to the answer?


Tim

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Here is a pic of Regulus. I took it on a whim, remembering vaguely that there is some kind of nebula nearby, which I appear to have missed completely.

(you know your brain has frozen when the only target you can think of is a star.....)

The question is this; There's a couple of fuzzies, but just to the left of Regulus there is a chain of stars, and in the middle of the chain is a fuzzy dot. Being that close to Regulus I would have thought it would have a name? Anyone have the tools to find out??

Cheers

TJ

11712_normal.jpeg

(click to enlarge)

NB: Hmm, it's a little tricky to see on the jpg, if it's too difficult i'll post a tiff of the area.

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I can't see much on the picture posted (I think I have a dodgy screen) but as Moondog says there's UGC5470 just to the north of Regulus at Mag 11 ish and to the northeast about the sam distance is IC591 a Mag 14 galaxy.

About a degree to the southeast is Abell 928 but at mag 17 I doubt it'd be showing!

James

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Heheh, thanks Paul, I thought that was a dust bunny. However, thats not the fuzzy I meant :hello2:

Try this image. Its still pretty faint mind.

James, what is roughly the limiting magnitude of most telescopes then? Does it increase with more exposure, or is the limit, the limit?

Thanks

11738_normal.jpeg

(click to enlarge)

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I'm not entirely sure to be honest. Certainly imaging allows you to reach much deeper as you're collecting photons over time whereas your eyes only see what falls into them at that time. I would guess that there's no effective limit to the magnitude you can go down to beyond the ability of the CCD chip/film to register the photons.

There's various formula for limiting magnitudes for telescopes but they are all in reference to the eye rather than a CCD.

Someone more knowledgable than me will probably prove me wrong :hello2:

James

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the photographic limit for a telescope is different than the visual limit.

visually it is limited by the aperture of your pupil. Formula's exist for calculation, but 10-12 is about norm.

photographically is always fainter, as the QE of a sensor is higher than that of your eye.

theoretically, if your camera shows shot noise of its entire dynamic range then continually exposing an object for longer and longer will yield fainter and fainter objects.

In reality the camera is never shot noise (sqrt Signal) limited, stars are smeared in to a approximately gaussian shape, reducing the flux sent to each pixel.

I have had several stabs at deriving an equation, but incorporation of all variables (including stellar temp and sky background) adds great levels of complications.

It can be estimated fairly straightforwardlyish

I might give it a go later

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the photographic limit for a telescope is different than the visual limit.

visually it is limited by the aperture of your pupil. Formula's exist for calculation, but 10-12 is about norm.

I think visually the limiting magnitude is affected by the aperture of the observers pupil but not limited by it. I might be wrong though......

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Hmm, I would if I could, but the modify button is gone?

But you shouldnt be too modest, we have a great deal of respect for your experience and abilities, and expertise with these things. Besides, I was hoping you might be first in there and show these fellas a thing or two :hello2:

I'm sorry if the title caused you discomfort, no offence or anything else meant or implied, except a gentle tip of the hat to your usual helpfulness and speedy answers to questions.

Tim

PS, I dont think anyone has the answer as yet???

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I've been having a look on Sloane Sky Survey and the object is there but it has no designation other than SDSS number

http://cas.sdss.org/astrodr7/en/tools/chart/navi.asp

Click that and input ra 152.5008261028206 and dec 11.91636037777 in the boxes top left and it gives a close up view on the right with more info on the object under 'quick look' and 'explore'. Zoom out to confirm the field matches yours. Looks like a face-on spiral in the close-up. Info under 'explore' shows it has a red-light mag of 14.40 and green light mag of 15.08.

Zooming in on the main window reveals even more fainter galaxies (some edge-on) :hello2:

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PS, I dont think anyone has the answer as yet???

PGC 29573 // mag 15.80 // 45"x35"

(way too easy... told you i was a geek :hello2: )

edit..

Btw, UGC 5470 is in there. It's the large 'blister' sitting atop the halo surrounding Regulus... see it?

11.5'x8.7' // mag 11.20

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Thanks Carol, much appreciated.

Having the details of the magnitude is important because it helps me determine exposure times required, and lets me compare one telescope against another, and also one camera against another. :thumbright:

Again, apologies for the hiccup.

TJ

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I have compared telescope camera combinations before and the results are always the same POINTLESS!

i found that the WO 66mm f/4.7 compared with a 8'' newt f/4.5 was ten times faster but to compare objects, well they get silly.

1 second on the cats eye = 1hour and ten mins on M31 !!!!!!!

you can't compare

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It only took me 23 seconds to get the designation of the galaxy from Sky Tools. Click the link for more info on the newly released ST3, and scroll down to the section entitled 'SkyTools 3 Pro Edition Features'. Check out the 'Exposure Calculator' sub-section... would that help anyone?

Greg Crinklaw (creator of the program) has a Yahoo group where he stays in constant touch with his customers. In fact, the imagers in the group gave him a lot of input when he was working on ST3, and they also did the Beta testing prior to release.

Greg worked for NASA on the Mars Observer project and developed the image processing software which was to be used to view and analyze images sent back by MO. The orbiter failed as it reached Mars, but his programming was used on its replacement, the Mars Global Surveyor.

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I have compared telescope camera combinations before and the results are always the same POINTLESS!

i found that the WO 66mm f/4.7 compared with a 8'' newt f/4.5 was ten times faster but to compare objects, well they get silly.

1 second on the cats eye = 1hour and ten mins on M31 !!!!!!!

you can't compare

Hmmm, that's very interesting. How did you measure? I had planned to get the pixel ADU value for the centre of the object, see how that increases with exposure length, and then work out how long the exposure would have to be to make that object, say ten times brighter.

That way I would have thought I could get the ADU value of an area of faint nebulosity, and then figure how long I'd need to expose it to really bring it out. I have noticed that on the vast majority of images, the context of the DSO or object is not recorded, which seems a shame, especially when you look at shots like Eddie's M42/B33 widefield. What I dont want to be doing though, is pointlessly spending 10 hours exposing something which is going to take 30 hours to fully bring out. Doing this compare is the only way I can think of to do it. But it wont work???

If it was possible to compare the telescopes, then it would help me decide whether to run a mosaic sequence on say, the 8" Newt, or just stick to the WO72. Say, 4x 2 hours for the Newt or 6 hours on the WO. Is this not realistic thinking??

Carol, thanks for the link to Skytools. I have not heard of that software before you mentioned it. Looks very useful. (Especially when trying to distinguish faint galaxies from dust bunnies :hello2: )

Right, i'm scratching my head now, anybody got an easier way to figure out required exposure lengths? It seemed easier somehow with a DSLR!

Thanks

TJ

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Have to say Tim, I think trying to work out exposure times is an almost pointless exercise. Way too many variables to take into account and that's without the varying sky conditions.

For example I'm hoping to do the RGB on my mono M81/82 image at some point and I think I've worked out the total exposure time will be somewhere around 6 hours, which for me is the longest time I've spent on one image and all things being well, I hope it'll be a pretty decent image. Now, have a look at this: http://blog.deepskycolors.com/archive/2009/01/03/m81-M82-and-Integrated-Flux-Nebula.html . Absolutely gorgeous image no? And that Integrated Flux Nebulosity? Never seen that before. If I could be bothered, I could work out how much exposure time i'd need to get the same amount of data but I can guarantee you I wouldn't get the same result.

For me, go as long as you can be bothered to go with your exposure time. You can always go deeper :hello2:.

Tony..

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JEEPERS would you look at that!!! Amazing image Tony.

See, thats the thing, If I knew that spending 20 hours or whatever would give a reasonable chance of getting that extra data, then i'd do it, but I would hate nothign more than spending the time and getting nothing else worthwhile, when the night could have been spent on another target.

I think i'll just have to move somewhere with many more dark clear nights!

TJ

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