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Embryonic heart

MartinB

By MartinB
in Imaging - Deep Sky

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MartinB

MartinB

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Got this data on Saturday. Badly needs some flats but hope to get these during the day at Kelling. Will take my white T shirt. Was taken with an ED80 and WO 0.8 flattener and a QHY8 with 13nm Ha filter. Thought the spacing was ok but the field didn't turn out very flat so more work to do there.

15x800 secs subs - 3hours 20mins

The neb was close to the moon which has probably hit the contrast a bit. Hope to add some unfiltered colour data at Kelling with a bit of luck. The inside of the ventricle is darker than the sky background which I think is an artefact from the gradient removal. Should be able to sort that out when I've got the flats.

8703_normal.jpeg

(click to enlarge)

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Ant

Ant

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the stars seem a bit elongated . or is it my old eyes lol ,

Yeah I would agree that if you look closely the stars are elongated slightly towards the corners.

But if you have to look for problems then IMO there aren't any.

Fantastic shot Martin, very 3D looking!

Ant

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MartinB

MartinB

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Thanks guys. Certainly are elongated stars in the corners, the field flatness wasn't good at all so obviously need to tweak the spacing between chip and reducer. I'll never completely get rid of the elongated stars in the corners with this set up unfortunately. That's the problem with these big chips, much more of a faff than the good old SXV H9, which would have also mopped up the Ha better.

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centroid

centroid

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Wow!!, and image from Chesterfield!!

Its a long time since we saw one of those :smiley: .

Still you heven't forgotten how to do it Martin, as given the conditions, that's a very reasonable result indeed.

As to the flatness out at the extremities of the image, don't rely on the calculated spacing being 100% correct, as if you used mine, or Ron Wodaski's calculator, bear in mind that these assumes that the 'flattener' of Focal Reducer's stated focal length is correct.

The mathematical formula is 100%v accurate, if the data fed into it is 100% correct.

I find that with my particular Celestron f/6.3, I have to reduce the spacing a little from the calculated value, or the I see star distortion, right out at the edges. Probably another Celestron 6.3, would be different again, due to manufacturing tolerances.

Dave

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MartinB

MartinB

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The mathematical formula is 100%v accurate, if the data fed into it is 100% correct.

I find that with my particular Celestron f/6.3, I have to reduce the spacing a little from the calculated value, or the I see star distortion, right out at the edges. Probably another Celestron 6.3, would be different again, due to manufacturing tolerances.

Dave

The ED80 has been demonstrated to by pretty good at 50-55mm for an APS sized chip like the QHY8 just curling up at the edges and corners but flat across the rest of the field. My image started to curve not far out from the centre. I will have to try adjusting and see how it goes. Bern is sending me a variable spacer and using CCD inspector I hope to be able to get close.

Was just great to be out again after a long long time.

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MikeD

MikeD

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Great job Martin. :smiley:

The inside of the ventricle is darker than the sky background which I think is an artifact from the gradient removal.

I think it looks about the same as in my shot, the same areas are the darkest part of the image, could it be dark nebula?

Mike.

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centroid

centroid

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Peter,

With Focal Reducers (FR) , its not just a matter of attaching them to the back of the scope, and then a camera on the back end of the FR.

In order to achieve the stated f/ratio for a particular FR, it is necessary to know the manufacturers stated focal length of the reducer.

You then enter this, along with the native focal ratio of the scope, and the distance between the CDD and the FR, into an equation, and the answer will be the resulting f/ratio.

If the distance is not correct, then the stated focal ratio for that FR, will not be achieved.

For example, with my f/10n SCT and the Celestron f/6.3 FR in line, the theoretical spacing is 105mm. However, in practice I use 100mm.

You'll find a calculator Rod Wodaski's website which is a 'freebie'.

I wrote a collection of little calculator applications in Visual Basic, which reside on my obsy PC, and that's what I use.

HTH.

Dave

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MartinB

MartinB

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Billy, the formula I know is to calculate the plate scale or sampling rate in arc seconds per pixel.

plate scale (arc seconds per pixel) = 2062.65xp/f where p is pixel size in microns and f is focal length in mm. Of course, this causes a problem if you want to know the focal ratio you are achieving with a reducer because you are left with 2 unknowns - the plate scale and the effective focal length with the reducer.

As Dave says you can use Ron's FOV calculator to compare your FOV with the predicted and play around with the focal length setting until it matches your FOV. You will then easily be able to work out your focal ratio.

A neater way is to plate solve your image. This relies on software which uses a large star data base to match your image with the data base. I believe you can do this in The Sky 6 pro edition but I use Maxim. This then tells you your plate scale. I haven't checked it with the Heart subs yet but will have a look when I get back from Kelling.

I think it looks about the same as in my shot, the same areas are the darkest part of the image, could it be dark nebula?

Mike.

May be Mike, if the whole field has some background nebulosity, that would explain it.

Its a good image despite the 'lack' of flatness - as you know, I am experimenting with almost the same setup and am hoping that CCD inspector will assist me too. I'd be very interested to hear how you get on, Martin.

I'll hopefully have some more info after this weekend Steve

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