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Greekastronomy

Wizard Nebula - NGC7380 (Full Color)

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album_pic.php?pic_id=9421

-> Telescope: LX200GPS 14" - Hyperstar 3

-> CCD: ATIK 16HR

-> Guiding Telescope: SkyWatcher 4"

-> Guiding Camera: Meade DSI 2

-> Guiding Software: PHD Guiding

-> Filter: Astronomik Ha, O3, S2

-> Processing: MaximDl / AstroArt / PhotoShop CS3

-> Exposure: Ha = 5hrs 10m / O3 = 2hrs 20m / S2 = 3hrs

-> Date: 2, 4, 5, 25 August 2009

-> Place: Athens Greece

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Another beautiful bit of magic from you Anthony.

Really spectacular image.

Ron.:)

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why so red?

have you mapped Ha to red? if so, then it is dominating the image.

what does the image look like in wavelength ordered pallette. You should be able to discern clearly blues, greens and reds...

however, the processing to me looks good, only thing is a simple colour balance issue.

tis a nice image though, cant argue with that.

best wishes

paul

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Thank u all. Paul, this is not a Hubble Palette. I have write it on the title (Full Color). Its a true color image with a new technique that from narrowband u can create full real color image. If u see the Wizard from other astrophotographer with the LRGB technique u will see the same results. Before 3 weeks i have post here the Hubble color Wizard. Thanks again!

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there's a lot to like about your image

as to the technique, it is easy to reassemble the emission lines and place them into their proper positions in the color spectrum.

[sII] and Halpha are both red. So you can mix them together with your preferred weightings.

[OIII] spans both green and the blue part of the spectrum, while Hbeta is a strong line found in the blue.

An interesting fact of ionized hydrogen is that the ratio of Halpha to Hbeta is fixed at approximately 3:1 in favor of Halpha, So you can synthesize the Hbeta line by using a scaled version of Halpha

now you have what you need to know in order to assemble things correctly to make a pseudo true color image. But why do i say pseudo-true color? because the stars are inherently broadband and they are not accurately reproduced using this method. But the nebulosity is accurately reproduced.

so the prescription becomes:

Red = [sII] + Ha

Green = [OIII]

Blue = [OIII] + Hbeta and that is approximated by [OIII] + 0.3*Halpha

here's more on the topic including a spectroscopic plot and example images:

http://www.narrowbandimaging.com/synthetic_rgb_page.htm

you can use facts from science/physics to assist you in making better use of your time under the stars....

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btw, here are some tricolor emission line images of NGC7380 I took several years ago using the wavelength ordered palette (aka Hubble palette)

http://www.narrowbandimaging.com/ngc7380_ap180_cm2_s2hao3_page.htm

http://www.narrowbandimaging.com/ngc7380_ap180_cm10_page.htm

http://www.narrowbandimaging.com/ngc7380_e180_6303_s2hao3_page.htm

and a widefield with neighbors

http://www.narrowbandimaging.com/ngc7380wide_p150_6303_baader_s2hao3_page.htm

The last one, the very widefield was taken using a 150mm medium format (Pentax 6x7) camera lens. The image was used along with others for finding new PNe. More on that later....there's a paper pending on that work :)

post-17284-133877397859_thumb.jpg

post-17284-133877397865_thumb.jpg

post-17284-13387739787_thumb.jpg

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Exactly, richard. Well explained and very good detailed analysed. I have one question though, is the blue better represented with [OIII] + 0.3*Halpha or the Hbeta alone?

Thanks.

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Exactly, richard. Well explained and very good detailed analysed. I have one question though, is the blue better represented with [OIII] + 0.3*Halpha or the Hbeta alone?

Thanks.

Well the [OIII] is equally footed in the blue and green part of the visible spectrum so I argue you need [OIII] in both the green and blue channels.

Clearly Hbeta is squarely in the Blue channel. But you don't need to waste your time taking Hbeta data since it looks like Halpha though: making the observation that I did about the physics of ionized hydrogen recombining and the probability ratios of the n=4 to n=2 transition (hbeta) versus the n=3 to n=2 transitions (halpha) being weighted in favor of the n=3 to n=2 transition by about a factor of three.

http://en.wikipedia.org/wiki/H-alpha

So to get the same signal density in Hbeta as Halpha and it will look the same (except for the stars), you would need to spend about 3x the time exposing. As I said a better use of your time is to simply shoot Halpha and [OIII], and combine as I suggested above noting that a scaled version of Halpha will look virtually identical on a monochrome sensor's response as Hbeta.

use the time you saved to take additional Halpha and [OIII] exposure to get a deeper and smoother image (without resorting to all the hocus pocus smoothing I see ruining otherwise good data) or go on to another target and increase your collection of fine images....

that's the best use of the time under the stars in my mind if the goal is a pseudo true color image of an emission nebula made using emission line filters.

You can always fix the stars by replacing them with a short RGB set that only got stars. I am personally revolted by such an abomination of images, due to my goals of remaining essentially scientifically "pure" but I can completely understand why that would be a good thing to do if you simply wanted to make wallhangers for sale....

Commercially viable prints look very different to me than astro images I would make with my particular set of strange goals.

I say strange because most do it the other way so that leaves me in the minority, but I am perfectly comfortable being in that position. Besides, I've been doing this for a long time and don't really care what people think anymore because I know what I am doing is right and if the rest can't follow along, that's their issue, not mine.

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I wasnt sure what full colour meant.....

in which case, the colour looks pretty good to me!

but as with what richardc said, I would stick O[iII] in the blue channel as their is a line in the 495nm region which is more blue than it is green.....

great looking image I must say.

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well there are actually two lines for [OIII], it is a doublet: one at 500.7 and the other at 496 roughly

The 500.7 line is actually the stronger of the pair....

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indeed there is.

the doublet sort of stradles the blue/green divide....like the mason/dixon line? (my american history aint that hot) :-)

paul

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