08Aug: Spectral sweep through the Ca K line

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08Aug: Spectral sweep through the Ca K line

Post by p_zetner »

Hi Everyone.

Here is an animated spectral sweep through the Ca K line from data taken during my 08Aug imaging session. The wavelength offsets (in milliAngstroms) are marked in the upper left hand corner of each frame. The frame rate is rather slow this time. I used one second per frame (36 frames) for this animation. The entire sweep covers about 1.7 Angstroms. (You can see why a Lunt CaK filter bandpass of 2.2A will fail to reveal filaments!)

Cheers, Peter.
092854 new spectral series flat-div siz labels siz.gif
092854 new spectral series flat-div siz labels siz.gif (14.08 MiB) Viewed 4671 times


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Re: 08Aug: Spectral sweep through the Ca K line

Post by LTHB »

Hi Peter,

very instructing and useful for judging if a CaK-filter is on band or not, thanks for posting!

Frank


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Re: 08Aug: Spectral sweep through the Ca K line

Post by Montana »

Stunning, however the Lunt is 2.4A so it is even worse than you state, the old PST was 2.2A. I wish there was a filter which was tighter. I think Mark's double stack CaK PST filter is the best I have seen.
This is great for reference material :bow

Alexandra


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Re: 08Aug: Spectral sweep through the Ca K line

Post by marktownley »

That's a stunner Peter! I wish my CaK were that good!


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Re: 08Aug: Spectral sweep through the Ca K line

Post by Merlin66 »

Peter,
Well done mate!
What bandwidth were you using??


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Re: 08Aug: Spectral sweep through the Ca K line

Post by Carbon60 »

Fabulous sequence, Peter. Very enlightening!

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Re: 08Aug: Spectral sweep through the Ca K line

Post by p_zetner »

Thanks for all your comments, Everyone.

Ken: I haven't bench tested this lately but can get a rough estimation from my H alpha results of this August. If I compare the fwhm of a measured spectral line in my instrument with a spectral linewidth of the same line in the BASS2000 spectrum, I get an instrumental bandpass of ~ 0.18 Angstrom (fwhm) near H alpha for my spectrometer in the present configuration (25 micron slit, 500mmf collimator, 2400 l/mm grating, 300mmf camera lens).

Here is a comparison plot of a measured spectrum taken on August 8 and the BASS2000 spectrum.
Ha Comparison BASS2000.png
Ha Comparison BASS2000.png (53.13 KiB) Viewed 4583 times

In this magnified view, I'm comparing the widths of a relatively isolated feature.
Ha Comparison BASS2000 - isolated feature.png
Ha Comparison BASS2000 - isolated feature.png (43.97 KiB) Viewed 4583 times
The BASS2000 spectrum claims a resolution of 0.01 Angstrom and I'm assuming the relation: BASS2000_fwhm^2 + Instrumental_fwhm^2 = Measured_fwhm^2 , which is strict for Gaussian profiles. From the fwhm values quoted on the last plot, the Instrumental_fwhm can be calculated to be ~0.18 Angstroms. This seems to be pretty reasonable when you observe how the images change over the course of the spectral sweep.

It's much more difficult to do this type of comparison in the uv because of the congested nature of the spectrum. Here's an example.
CaK Comparison BASS2000.png
CaK Comparison BASS2000.png (107.62 KiB) Viewed 4582 times

Cheers, Peter.


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Re: 08Aug: Spectral sweep through the Ca K line

Post by Merlin66 »

Peter,
As you appreciate it's always difficult to estimate the resolution from spectra comparison. A spectral reference lamp is the preferred method.
Notwithstanding, I think you're pretty close in your estimates....
I ran your spectroheliograph through the SimSpec_SHG for comparison......
SimSpec SHG_Zetner.xls
(55.5 KiB) Downloaded 102 times
shg_Zetner.JPG
shg_Zetner.JPG (183.53 KiB) Viewed 4542 times


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Re: 08Aug: Spectral sweep through the Ca K line

Post by Valery »

p_zetner wrote: Tue Aug 28, 2018 4:38 pm Hi Everyone.

Here is an animated spectral sweep through the Ca K line from data taken during my 08Aug imaging session. The wavelength offsets (in milliAngstroms) are marked in the upper left hand corner of each frame. The frame rate is rather slow this time. I used one second per frame (36 frames) for this animation. The entire sweep covers about 1.7 Angstroms. (You can see why a Lunt CaK filter bandpass of 2.2A will fail to reveal filaments!)

Cheers, Peter.

092854 new spectral series flat-div siz labels siz.gif
Hi Peter,

Missed this topic, unfortunately. Reading it now. Thanks a lot for such a nice animation - very illustrative!

OK about the band pass which allows to image filaments. What must be the FWHM to see them with confidence?


Valery


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Re: 08Aug: Spectral sweep through the Ca K line

Post by p_zetner »

Ken: Thanks for running the calculation. It seems to be in the ball park, although, I guess, it is a best estimate of resolution as, for example, optical aberrations aren't considered. It's pure laziness that I haven't made a bench test of the resolution recently. I'll do this in the near future.

Valery: To roughly answer your question, I can average together some number of frames around line centre to model the action of a filter of given fwhm bandpass. Ideally, this would be a weighted average to simulate, say, a Gaussian bandpass. I'll work on this. Straightforward averaging of frames is easy in ImageJ and, I suppose, would give a rough answer to your question but I have to figure out how to accomplish the weighted averaging for a more refined answer.

Peter


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Re: 08Aug: Spectral sweep through the Ca K line

Post by christian viladrich »

Hi Peter,
This is great !!
Could you display the individual frames in a single large image, just like you did for Ha ?


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Re: 08Aug: Spectral sweep through the Ca K line

Post by christian viladrich »

It looks like that the filaments are best visible in the range of plus / minus 0.080 A.
This means that a filter should have a FWHM < 0.16 A to show any filaments.


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Re: 08Aug: Spectral sweep through the Ca K line

Post by highfnum »

very cool to see sweep


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Re: 08Aug: Spectral sweep through the Ca K line

Post by p_zetner »

Thanks, John.

As you can imagine, these animations require a lot of work (40 spectroheliograms in this case) but I enjoy the end result.

Cheers.
Peter


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