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A question regarding using FWHM as a measurement of seeing

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#1 hallzhuu

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Posted 13 December 2018 - 09:17 PM

Hello everyone, 

 

I learned that we often use FWHM value as a numerical representation of seeing. My question is that since on the same photograph, stars of different magnitude have different brightnesses and thus different FWHM values, how can we determin seeing? Do we obtain a mean value from all stars? 

 

Also, do exposure affect FWHM values? 

 

Thanks very much for answering my questions! 


Edited by hallzhuu, 13 December 2018 - 09:18 PM.


#2 TOMDEY

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Posted 13 December 2018 - 09:32 PM

If the stars used in the characterization have images that are nowhere saturated, prior to calibration, and then are properly calibrated... the FWHM should be invariant.  Tom


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#3 Stonius

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Posted 13 December 2018 - 09:56 PM

Imagine looking down on a cone from above as it sits in a shallow pool of opaque fluid (coffee? Milk? whatever). You can't see the bottom of the cone, only the top.

 

You can see the slope of the sides of the cone though, can't you?

 

Now imagine more cones at varying depths in the liquid. Some poke out just a little, some a lot, but all still have the same slope to the sides of the cone.

 

How much of the cone is above the liquid is how bright the star is.

How steep the slope of the sides of the cones are is the seeing.

 

That's how you can have different brightnesses, while they all share the same seeing characteristics.

 

Hope that helps :-)

 

Markus


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#4 Stonius

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Posted 13 December 2018 - 10:01 PM

Also, do exposure affect FWHM values? 

Yes. Longer exposures allow atmospheric seeing effects, wind, tracking errors, etc to 'build up' over time.

 

Shorter exposures will generally have better resolution, depending on a bunch of variables.

 

Cheers

 

Markus



#5 gregj888

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Posted 13 December 2018 - 10:11 PM

Professional seeing monitors calculate Ro and To usually using a DIMM setup.

 

AFAIK FWHM is more of a relative measurement.  The scope used and camera can also make a difference.  I downloaded the SBIG Software and have been playing with it, a QHY5Liim and a camera 135mm lens.  You can also use something like Metaguide.  This gives a simple and convenient way to discuss seeing over time.  I would take my measurements day over day, but wouldn't want to compare it to someone else's as a point of argument.  

 

This may help:

     http://www.eaae-astr...sure_seeing.pdf


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#6 ks__observer

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Posted 14 December 2018 - 05:18 AM

Yes. Longer exposures allow atmospheric seeing effects, wind, tracking errors, etc to 'build up' over time.

 

Shorter exposures will generally have better resolution, depending on a bunch of variables.

If you stack an hour of subframes the total errors will be same if you have 5 or 50 subframes.

You can, as a friend explained, cull the worst out if you have shorter franes.



#7 driven

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Posted 10 May 2019 - 11:49 PM

This is a question that has bothered me as well.  TOMDEY echoes what I have heard others say, that the FWHM should be the same for all unsaturated and properly calibrated stars in an image.  Stonius provided an analogy of cones rising from an opaque fluid, and that model matches the way I think about the problem.

 

The FWHM is the full width of the circle that contains all the pixels of a star image that are at least half as bright as the brightest pixel.  In the cone model, it would be the diameter of the cone measured at the point half way between the tip and the base.  If we assume that the seeing is the same for all stars in an image then the slope of the sides of the cones would also be the same.  Brighter stars equate to taller cones.  The base of the cone made by a brighter star would be wider than the base of a dimmer star cone, and likewise the midpoint of a bright star cone would be wider than the midpoint of a dimmer star cone.

 

The cone model does not exactly match reality since the decay of brightness from the center is not linear, it is a sinc function if I remember correctly.  Regardless, any shape that is scaled equally in width and height would share the same characteristics as the cone model.  Cones are easy to visualize and should convey the concepts adequately.

 

So, a FWHM that is invariant with brightness is incompatible with the cone model as I imagine it.  Can anyone explain which one is correct, and why?



#8 CygnusBob

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Posted 11 May 2019 - 02:42 PM

Well, actually the FWHM values can vary at different positions in the FOV in short exposures.  As you trace rays backward up through the atmosphere they will pass through different turbulent regions.  So the total wavefront error will vary.  However, if most the turbulence is occurring close to the ground, then the wavefront error will be more or less the same across the FOV.



#9 gregj888

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Posted 12 May 2019 - 10:38 AM

I think FWHM is calculated from the derived slope not the actual pixels (maybe with some averaging)  though I've never seen much difference.

 

Beyond that, I don't think I understand the conundrum?

 

1) The brighter stars in my experience are bigger at the base.

2) It's a model, individual instances may vary...  but both should yield similar results.  As long as you stick with one you should be good.



#10 driven

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Posted 12 May 2019 - 11:05 PM

My conundrum is trying to figure out how an FWHM measurement of stars of varying brightness can be the same, assuming the seeing was the same across the image.  Brighter stars have a larger footprint in the image, and as far as I understand, half of a bright star is larger than half of a dim one.  How then can the FWHM be the same?



#11 DmitriNet

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Posted 13 May 2019 - 11:42 AM

My conundrum is trying to figure out how an FWHM measurement of stars of varying brightness can be the same, assuming the seeing was the same across the image.  Brighter stars have a larger footprint in the image, and as far as I understand, half of a bright star is larger than half of a dim one.  How then can the FWHM be the same?

FWHM is a volatile measure.  Under even ideal conditions it should be only approximately the same because of shot noise.

Also, one has to look at how exactly it is calculated under less than perfect conditions.   Due to shot noise and seeing per se  the maximum value among all pixels can change substantialy (~25%) for the same star in a matter of seconds.

It is just not a good measure.  FWHF is more stable.


Edited by DmitriNet, 13 May 2019 - 01:02 PM.


#12 ks__observer

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Posted 13 May 2019 - 12:51 PM

I think the question was geared more toward why bright stars and dim stars have the same FWHM even though brighter stars appear to have a much larger visual footprint than dimmer stars.

Star light hits the aperture as a point source.

When it hits the aperture it creates a circular diffraction pattern.

The angle the light spreads out is fixed for a given aperture and wavelength.

This is called the Airy Pattern.

This then gets distorted from seeing.

As discussed above, the pattern forms a cone shape -- FWHM, with smaller outlying wave crests -- rings.

Brighter stars take up a larger visual footprint because the amplitude in the outer rings is much greater -- and probably star saturation.

For dim stars the pattern is there in the pixels even if you can't visually see the pattern.



#13 gregj888

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Posted 14 May 2019 - 12:27 AM

Driven,

 

Check the profile rather than the footprint.  For the FWHM to be the same, the footprint and height can't grow linearly.  

 

Sorry, not on my imaging computer at the moment.



#14 CygnusBob

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Posted 14 May 2019 - 11:49 AM

driven

 

If one star is twice as bright as another, the base appears to be twice as wide.  However, the half height position is twice the number of counts above background and thus the width is the same. 



#15 gregj888

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Posted 17 May 2019 - 10:36 PM

Here are two triangles, one twice as big as the other.  Note the 50% point of the larger one is wider than the smaller one.

 

This just shows that it's not a strict triangle and you need to check the profile.  

 

 

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  • triangle.png


#16 ks__observer

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Posted 18 May 2019 - 12:35 AM

The triangle base stays the ssme but only the height varies with star magnitude.



#17 gregj888

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Posted 18 May 2019 - 01:01 AM

The triangle base stays the ssme but only the height varies with star magnitude.

That will work of course, but I notice brighter stars seem bigger in my images.  This isn't something I've worried about or looked at, so I don't know the answer.  I have looked at the speckle cloud on a larger scope and in that cases the Aire disk isn't a concern.  I need to measure at a couple of stars, that will tell the tale pretty quickly.



#18 ks__observer

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Posted 18 May 2019 - 01:58 PM

This is a little technical but is a good overview:

https://www.telescop..._resolution.htm



#19 driven

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Posted 20 May 2019 - 11:30 PM

This is a little technical but is a good overview:

https://www.telescop..._resolution.htm

 

Great discussion of telescope resolution!

 

So now armed with better information let me have another go.  The size of the Airy disc depends only upon the geometry of the telescope and is independent of the brightness of a star.  So, using the cone model it is not true that the cones have the same shape.  The only way for the width of a cone at half its height to be the same as the width at half height of a shorter cone is for the bases of those cones to also be the same size.  The cone base would be the Airy disc, the width at half height would be the FWHM, and the height would be the brightness.  This would explain why brighter stars would have the same FWHM as dimmer stars.

 

But then the question is now the reverse - if the FWHM of stars of varying brightness is the same (assuming uniform seeing across the frame and no geometric distortion), how is it that brighter stars appear larger?

 

ks__observer explains why in post #12, and that makes sense to me.  But then I go look at Hubble images, free from atmospheric image degradation, and I don't see diffraction rings around brighter stars.  Some have what look more like halos, but no rings.  Where did the rings go?

 

My apologies to hallzhuu if this is straying too far from the OP



#20 ks__observer

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Posted 21 May 2019 - 04:03 AM

This topic is in the "observing" forum, but it might be better in the astrophotography forum.

I do a fair amount of imaging.

Brighter stars tend to saturate the pixels.

Each photon when it hits the silicon ccd is converted to an electron that gets stored in a capacitor until the exposure is completed.  At the end of the exposure, the camera measures the voltage to determine the brightness.

For bright stars, often the capacitor is overwhelmed and cannot collect more photons -- saturation.

My friend who does astrophotography had basically the same question as you.

He took one of his images.

He looked at non-saturated bright stars and very dim stars -- dim stars you can barely see in the photo.

He graphed the pixel value across the stars.

He got the same Airy disk profile in width for all his stars -- height varied.

He even graphed out slight bumps for the rings even on the dim stars.

We don't visually see the stars in astrophotography pics as the same size, but the computer pixel value shows the stars to be all the same width -- the computer sees the same size.


Edited by ks__observer, 21 May 2019 - 04:06 AM.


#21 555aaa

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Posted 21 May 2019 - 01:07 PM

The "cones" aren't equal. They are steeper for brighter stars. The reason that the bright stars look "fatter" than the dim stars is just a consequence of the way you are displaying the brightness. If you have say 10th mag stars and 15th mag stars in your image, the 10th mag stars are 100x brighter. That is a very common situation. Suppose when they get to your display they are an 8 bit brightness value. If it was linear, the faint star would have an RGB value of (2,2,2) and the bright star would have a value of (255,255,255) if it is just clipping, and you will never even see the faint star. So to see the faint stars you totally blow out the bright ones or use a very nonlinear stretch and the result of that is that the bright stars look bloated. They really aren't and if you had a star profile plotting tool you'd see this. They really are the same width in terms of pixels, as long as they are not saturated. Take one of your images and look at it in a true linear representation and you'll see that all you can see are some teeny-tiny dots which are the very brightest stars. If it's a good image, they really are small! 


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#22 driven

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Posted 21 May 2019 - 10:43 PM

Aha!

 

That was the fact I was overlooking this whole time!  I didn't take into account that the screen stretch would bloat the brighter stars.  If I adjust the screen stretch to match the brightness of any individual star, bright or dim, they all look remarkably similar (but not at the same time).

 

So, to get back to the OP, the relative seeing can be determined from the FWHM of any non-saturated star in the image.  They should all be about the same, regardless of star brightness or exposure time.  Worse seeing creates larger FWHM values, as would worse focus, so seeing can only be compared from different images that are all in focus. 

 

Saturated stars (those that exceed the maximum pixel value: around 14000 for a Canon DSLR, 255 for an 8 bit CCD, or 65535 for a 16 bit CCD) will have a larger apparent FWHM because the peak brightness of the star is chopped off, making the half brightness level be farther down the "star cone" and thus wider than it would be if it were not saturated.




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