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More Clarification Please - Photons

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#1 Galaxyhunter

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Posted 22 October 2019 - 12:25 PM

In the interest not to hijack a another thread,  "Clarification: Is focal ratio the supreme factor in integration time?" I 'm starting this one.   I seen a post on the other thread & I'm wondering about the physics of a photon.  The post from the other thread (used with permission)

 

I recall, probably from the old film days, a truism:

F/ratio applies to extended objects while aperture applies to point sources.

So greater aperture gets you fainter stars no matter what the f/ratio is.

But faster f/ratio gets you more nebulosity.......

Now mix in focal length, pixel size, pixel sensitivity, exposure length, ISO, number of sub-frames, which stacking algorithm used, darks, lights, flats, bias,  and what-have-you, stir it all up well and the result?

The good old 'it depends'.....

Dave

 

I know I'm pretty dense, but the way I see it, a photon of light is just a photon of light.  Vega is a point source ( so they say),  so aperture rules.  If you put a billion others stars around Vega, now it becomes an extended object. The photons emanating from it is still just photons.

 

1: So how does the laws of physics change the photon of light depending on its origin?

2: How does your imaging chip know where that photon of light originates?

 

I realize that I'm dumber than a box of rocks, so explane it to me so a 5 year old can understand.


Edited by Galaxyhunter, 22 October 2019 - 12:26 PM.


#2 Der_Pit

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Posted 22 October 2019 - 01:27 PM

I know I'm pretty dense, but the way I see it, a photon of light is just a photon of light.  Vega is a point source ( so they say),  so aperture rules.  If you put a billion others stars around Vega, now it becomes an extended object. The photons emanating from it is still just photons.

 

1: So how does the laws of physics change the photon of light depending on its origin?

2: How does your imaging chip know where that photon of light originates?

 

I realize that I'm dumber than a box of rocks, so explane it to me so a 5 year old can understand.

OK, I'll have a try.

 

Starting with your star example, a single point source will be (yes, ideally) will be imaged on a single pixel.  If you double the focal length (change F-ratio, at same aperture) nothing happens.  All light in one pixel, as you say.

 

But now you have a whole bunch of stars.  If it is really a lot, you will have more than one being imaged on your pixel.  Say the stars are 0.5" apart, and you have an image scale of 1.5"/px.  So you receive the light of around 9 stars in your pixel.

 

Now you again double the focal length, and your image scale will be 0.75"/px.  And suddenly you will only have 2-3 stars being imaged onto that same pixel.  The intensity from each star collected by your scope is still unchanged, but you have less of them per pixel, and intensity goes down.

 

So the key point of discrimination is, do you still resolve the single stars, or not.  If you don't resolve them anymore, because too many are imaged onto one pixel, one starts calling that 'surface brightness'; in the extreme case of an emission nebula it really is a continuous brightness, but already for galaxies you cannot really see the difference.  But in both cases you no longer have (single) star brightness, but an average over an area, and of course then the total sum depends on the area that you look at.  

 

Does that help disentangle your brain? 



#3 Stelios

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Posted 22 October 2019 - 01:39 PM

Greater aperture gets you *smaller*, not fainter stars. 

Faster F-ratio gets you the same amount of nebulosity but *faster*--you require shorter exposures for same results (other things being equal).

 

The laws of physics don't change a photon of light depending on its origin. A photon will hit your sensor, and likely ("quantum efficiency") be recorded. Subsequent photon recordings from the same source (be it a star or a nebula, who cares, they're photons) pile up in the appropriate pixels (mapping their origin as per the laws of optics), until a limit is reached ("well depth") for each pixel--usually attained with bright stars or overlong exposures--after which there's potential overflow issues. The electrical operation of the camera shutter and the dark current introduce pseudo-photons (aka "noise") into the various photosites and thus the image. 

 

It's nowhere near as confusing as you think, although it most definitely seems that way at first. All the calibration frames (darks, flats, bias) have a purpose. I recommend this book if you really want an understanding of the subject and to learn much more about astrophotography in general.


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#4 the Elf

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Posted 22 October 2019 - 01:50 PM

Very well explained! +1

Some more info, not quite sure what you want/need to know:

Light is an electromagnetic wave. Somehow. Light is a particle. Somehow. This is called a dualism. There are experiments that show an outcome which can be explained by the theory of waves. Laser interferrence patterns are a good example. There are others that show an outcome which can be explained by the particle theory. Charge generation on sensors is a good example. Both are correct. Somehow. In AP the wave behaviour explains the chromatic aberration in refractors and the diffraction limited resolution. It explains the extinction in the atmosphere as well as refraction. It helps to calculate the layers of a filter. The particle behaviour helps to explain the sensor technology and quantum efficiency.

There is no difference in photons depending on the origin in general. There is a minimal portion of energy that depends on the wavelength. One photon represents the smallest amount of energy that can be transported at a time for a given wavelength. So wavelength and energy somehow are two ways to characterize the same thing. Wavelenght can indicate how the photon was generated, e.g. by Hydrogen gas or by Oxigen gas.

The imaging chip is a bucket that collects charges. The charges are genrated by photons, by higher energy impacts like cosmic rays, gamma rays, etc. The high energy impacts can create several charges at a time, photons create only one. Last but not least heat constantly creates chareges. Impurities in the silicon help the effect of heat generated charged. In the image you can't tell if the signal was a photo or heat. Cosmic ray strikes do not happen so often and can be identified as short bright traces. At my place I have one or two in a hundert subs. If you go inside the exploded Tschernobyl reactor you will have plenty of them on eacht image you take. The charges created by photons can come from the targe lightyears away or from the dust right over your head illuminated by the neighbourhood. You just don't know.

In daylight photography you can use polarization filters to remove light that has been reflected on non metallic surfaces. The waves change after something has happend to them. So in a way you can tell where a photo comes from in some cases.

One last thing about the aperture: If you compare two scopes, say one f/8 and the other one f/5.6, the faster one collects the same amount of photons in half the time. Half the time means only half the dark current. From that point of view the larger aperture helps signal to noise ratio but this is not caused by the photons but the problems of the sensor.


Edited by the Elf, 22 October 2019 - 01:53 PM.


#5 TOMDEY

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Posted 22 October 2019 - 01:57 PM

The paradox goes away... goes away entirely... when you examine, just a bit more closely, what the instrumentalists are choosing to call stars or  nebulae. That choice is the ultimate resolution limit of your telescope, limited by that pesky (J1(x)/x)2, more commonly known as the Airy Disc. If your scope is incapable of resolving the stars (true for virtually all of our amateur telescopes)... then, even under the best use conditions... you don't see images of the stars, but little Airy Discs centered where the stars are. If you are using your scope at lowish mags, where even the Airy Discs themselves are unresolved (to your eye or array) then the stars/nebulae casual rule applies. If you go to higher mags... then both the Airy Discs and nebulae lose Radiance/Luminance at the same pace, as magnification increases. This is what we call ~empty magnification~; so we back off on the mag and wind up right back where the star/nebl differentiation applies. No laws of physics are violated... indeed... those laws mandate the transition and relationships!

 

Conclusion: It's actually more the semantics of instrument use, rather than the physics itself... which causes the faux paradox. It's a pretty subtle legerdemain... part of the sociology of science. Your telescope is the street magician, transforming stars into Airy Discs.  Tom

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#6 RedLionNJ

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Posted 22 October 2019 - 02:29 PM

In the interest not to hijack a another thread,  "Clarification: Is focal ratio the supreme factor in integration time?" I 'm starting this one.   I seen a post on the other thread & I'm wondering about the physics of a photon.  The post from the other thread (used with permission)

 

I recall, probably from the old film days, a truism:

F/ratio applies to extended objects while aperture applies to point sources.

So greater aperture gets you fainter stars no matter what the f/ratio is.

But faster f/ratio gets you more nebulosity.......

Now mix in focal length, pixel size, pixel sensitivity, exposure length, ISO, number of sub-frames, which stacking algorithm used, darks, lights, flats, bias,  and what-have-you, stir it all up well and the result?

The good old 'it depends'.....

Dave

 

I know I'm pretty dense, but the way I see it, a photon of light is just a photon of light.  Vega is a point source ( so they say),  so aperture rules.  If you put a billion others stars around Vega, now it becomes an extended object. The photons emanating from it is still just photons.

 

1: So how does the laws of physics change the photon of light depending on its origin?

2: How does your imaging chip know where that photon of light originates?

 

I realize that I'm dumber than a box of rocks, so explane it to me so a 5 year old can understand.

1. It doesn't

2. It doesn't

 

This is because the line in red is just a summary of the two theoretical extremes. Practical reality in most cases lays somewhere between.

 

Let's say the resolution of your sensor is such that one pixel covers 5 arcsec. Also assume 100% QE.

You're pointed directly at Vega and all the incoming photons from Vega are focused onto that one pixel. Let's say 100 pretty energetic photons per second. 

Then you point to the nearby Ring Nebula and count the photons arriving from it at that same pixel in one second. You're likely to have a ton more (after all, they cover the entire area of the pixel) much less energetic pixels.

With both these cases (identical equipment), f-ratio is the same.

 

Now use a much shorter f-ratio. Same aperture and pixel size.

The situation for Vega hasn't changed. Same number of photons on the pixel.

But for the Ring Nebula, being far more than a point source, we now gather far more photons onto that same pixel.

 

Now, instead of changing the f-ratio, let's go back to a fixed f-ratio and change the aperture:

For Vega, we can capture more photons due to the larger aperture.

For the Ring Nebula, we can also capture more photons due to the larger aperture.

 

So for a fixed f-ratio and pixel size, aperture is the determining factor. 

 

No contradictions and the scope/sensor doesn't need to know where each photon came from.



#7 dmdouglass

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Posted 22 October 2019 - 03:08 PM

Galaxyhunter....

 

I am not sure about Physics....

But changes in aperture result in amount of light collected in a specific length of time (that is why large telescopes sometimes are referred to as "light buckets"...  Thus, larger aperture = "brighter" Vega.

 

EDIT ADD:  AND... as the aperture increases (gets bigger).... letting in more light, some of the more "dim" stars around Vega would start to show up....

 

Changes in focal length result in changes of size of image... or in simplistic terms... how big is Vega and where is it.  Longer focal length = smaller Vega.  Conversely, shorter focal length = larger Vega.  

 

EDIT ADD:  AND... as the focal length REDUCES....  the wider the field of view (FOV).

 

Hope that helps.


Edited by dmdouglass, 22 October 2019 - 03:15 PM.


#8 Galaxyhunter

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Posted 22 October 2019 - 06:35 PM

Thank You Gentlemen,  It will take some time to digest this.



#9 Galaxyhunter

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Posted 22 October 2019 - 06:51 PM

1. It doesn't

2. It doesn't

 

This is because the line in red is just a summary of the two theoretical extremes. Practical reality in most cases lays somewhere between.

 

Let's say the resolution of your sensor is such that one pixel covers 5 arcsec. Also assume 100% QE.

You're pointed directly at Vega and all the incoming photons from Vega are focused onto that one pixel. Let's say 100 pretty energetic photons per second. 

Then you point to the nearby Ring Nebula and count the photons arriving from it at that same pixel in one second. You're likely to have a ton more (after all, they cover the entire area of the pixel) much less energetic pixels.

With both these cases (identical equipment), f-ratio is the same.

 

Now use a much shorter f-ratio. Same aperture and pixel size.

The situation for Vega hasn't changed. Same number of photons on the pixel.

But for the Ring Nebula, being far more than a point source, we now gather far more photons onto that same pixel.

 

Now, instead of changing the f-ratio, let's go back to a fixed f-ratio and change the aperture:

For Vega, we can capture more photons due to the larger aperture.

For the Ring Nebula, we can also capture more photons due to the larger aperture.

 

So for a fixed f-ratio and pixel size, aperture is the determining factor. 

 

No contradictions and the scope/sensor doesn't need to know where each photon came from.

I'm not trying to be argumentative,  But isn't the light coming from the Ring Nebula made up of thousands of millions of individual  of "Point Sources" . Each photon has its own intensity lever &  when combined on the sensor, creates the actual ring. I wouldn't hardly think that the entire Ring Nebula is just a single photon coming through space.



#10 sharkmelley

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Posted 23 October 2019 - 02:03 PM

 

 

F/ratio applies to extended objects while aperture applies to point sources.

 

 

I'll try a different way of explaining things.  I'll try to keep it simple and it may or may not help.

 

Consider 2 scopes where the aperture of one has twice the diameter of the other. The aperture area of one has 4x the area of the other. So assuming the extended object fits on the sensor in both cases then the camera on larger aperture scope will record 4x as many photons than on the smaller scope.  This is true for both an extended object and for a star.

 

However, the physics of light means that for a star there is an additional interesting effect.  The image of the star becomes tighter for the larger aperture scope.  To put it another way, resolution increases.  Technically it is all to do with Point Spread Functions and Airy Discs.  So not only does the larger aperture scope collect 4x as much light from the star, it also compresses that light into a smaller star image.

 

In summary, a larger aperture has a beneficial effect for both extended objects and stars, by collecting more light from each.  But the effect is exaggerated for stars by compressing that light into a tighter star image.

 

Mark


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#11 the Elf

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Posted 24 October 2019 - 06:20 AM

Mark, very well explained! Thank you.



#12 Galaxyhunter

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Posted 24 October 2019 - 07:51 PM

I'll try a different way of explaining things.  I'll try to keep it simple and it may or may not help.

 

Consider 2 scopes where the aperture of one has twice the diameter of the other. The aperture area of one has 4x the area of the other. So assuming the extended object fits on the sensor in both cases then the camera on larger aperture scope will record 4x as many photons than on the smaller scope.  This is true for both an extended object and for a star.

 

However, the physics of light means that for a star there is an additional interesting effect.  The image of the star becomes tighter for the larger aperture scope.  To put it another way, resolution increases.  Technically it is all to do with Point Spread Functions and Airy Discs.  So not only does the larger aperture scope collect 4x as much light from the star, it also compresses that light into a smaller star image.

 

In summary, a larger aperture has a beneficial effect for both extended objects and stars, by collecting more light from each.  But the effect is exaggerated for stars by compressing that light into a tighter star image.

 

Mark

Thank You Mark,

 

Okay, I can see & agree to this.  I know the f-ratio theory on the lower the number, the faster the optics, that is not a problem.  So if we go back to the statement that starting all of this -

F/ratio applies to extended objects while aperture applies to point sources.

 

What I can not see is,  How does the f-ratio theory have a bigger effect on a Nebula over a Star?



#13 WadeH237

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Posted 24 October 2019 - 08:20 PM

What I can not see is,  How does the f-ratio theory have a bigger effect on a Nebula over a Star?

Given two telescopes with the same aperture, using identical cameras, a faster scope will have a shorter focal length and will concentrate the light from an object onto a fewer number of pixels.  If you have an extended object that covers many hundreds of pixels, concentrating the light onto a much smaller number of pixels, puts lots more signal into each pixel.  Now imagine a point source, like a star - and assume that all of the light hits just one pixel on each scope.  In that case, the signal of that pixel will be the same in both scopes.

 

Now in actuality, the light from a star is going to cover more than one pixel, unless you are significantly under sampled.  So the focal ration will have some effect, even on stars.  But hopefully you can see the difference between a very small source, and an extended source.



#14 Galaxyhunter

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Posted 24 October 2019 - 09:08 PM

Given two telescopes with the same aperture, using identical cameras, a faster scope will have a shorter focal length and will concentrate the light from an object onto a fewer number of pixels.  If you have an extended object that covers many hundreds of pixels, concentrating the light onto a much smaller number of pixels, puts lots more signal into each pixel.  Now imagine a point source, like a star - and assume that all of the light hits just one pixel on each scope.  In that case, the signal of that pixel will be the same in both scopes.

 

Now in actuality, the light from a star is going to cover more than one pixel, unless you are significantly under sampled.  So the focal ration will have some effect, even on stars.  But hopefully you can see the difference between a very small source, and an extended source.

I just looked at a couple of my images,  the smallest stars I could find (19-20th mag) is covering roughly 4-5 pixels each way.  So in the real world,  there basically is no Point Source.  Or would that be determined by the OTA & camera combination?



#15 TelescopeGreg

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Posted 24 October 2019 - 11:43 PM

I just looked at a couple of my images,  the smallest stars I could find (19-20th mag) is covering roughly 4-5 pixels each way.  So in the real world,  there basically is no Point Source.  Or would that be determined by the OTA & camera combination?

Not to stir things up, but could the 4-5 pixels covered actually be one pixel-worth of star that wandered around a bit due to seeing during the exposure?  And, that assumes the star was perfectly focused, too, or at least focused down to a single pixel-worth of space.



#16 sharkmelley

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Posted 25 October 2019 - 12:54 AM

Thank You Mark,

Okay, I can see & agree to this. I know the f-ratio theory on the lower the number, the faster the optics, that is not a problem. So if we go back to the statement that starting all of this -
F/ratio applies to extended objects while aperture applies to point sources.

What I can not see is, How does the f-ratio theory have a bigger effect on a Nebula over a Star?

The statement F/ratio applies to extended objects while aperture applies to point sources is a misleading statement as it stands. Sorry, I should have made that clearer.

I guess it's a statement that probably made sense in the context of the discussion where it was originally made.

Mark

Edited by sharkmelley, 25 October 2019 - 01:30 AM.


#17 Galaxyhunter

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Posted 25 October 2019 - 05:56 AM

Not to stir things up, but could the 4-5 pixels covered actually be one pixel-worth of star that wandered around a bit due to seeing during the exposure?  And, that assumes the star was perfectly focused, too, or at least focused down to a single pixel-worth of space.

Go ahead & stir things up, anything is possible as I have made more my share of mistakes   blush.gif .  Is this out of focus?  http://hawkeye-obser...00-2/Eagle.html



#18 WadeH237

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Posted 25 October 2019 - 07:03 AM

I just looked at a couple of my images,  the smallest stars I could find (19-20th mag) is covering roughly 4-5 pixels each way.  So in the real world,  there basically is no Point Source.  Or would that be determined by the OTA & camera combination?

Stars really are a point source, even in the real world.  Or at least their angular size is so small as to be effectively a point source.

 

Their light is distributed by the optics of your system (the Airy disc size), by focus, by seeing, by tracking errors, etc.  Image scale is another factor in distribution of the light.  If you are under sampled, such that the light from a star covers just a single pixel, then variations in these things won't change the appearance of the star in an image (unless they are large enough to spread out the light over many pixels).  If you are sampled such that the light is distributed over many pixels, then the appearance will change based on that distribution, possibly to the point where the star is no longer visible.

 

The other point (no pun intended) is that, since they are point sources, the distribution of light is not even between the pixels.  The pixels that capture the center of the image of the star will have much more signal than the pixels that capture the periphery.  This is less true of extended sources.



#19 TOMDEY

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Posted 25 October 2019 - 08:40 PM

I'll try a different way of explaining things.  I'll try to keep it simple and it may or may not help.

 

Consider 2 scopes where the aperture of one has twice the diameter of the other. The aperture area of one has 4x the area of the other. So assuming the extended object fits on the sensor in both cases then the camera on larger aperture scope will record 4x as many photons than on the smaller scope.  This is true for both an extended object and for a star.

 

However, the physics of light means that for a star there is an additional interesting effect.  The image of the star becomes tighter for the larger aperture scope.  To put it another way, resolution increases.  Technically it is all to do with Point Spread Functions and Airy Discs.  So not only does the larger aperture scope collect 4x as much light from the star, it also compresses that light into a smaller star image.

 

In summary, a larger aperture has a beneficial effect for both extended objects and stars, by collecting more light from each.  But the effect is exaggerated for stars by compressing that light into a tighter star image.

 

Mark

Yeeeep! That's a great point, that many/most people are entirely unaware of, That it's quartic (not quadratic) with aperture. And why my 36-inch scope can (at least in principle) produce an Airy Disc ~image of a star~ that is 1300 times more dazzling than that seen in a 6-inch scope, used at the same magnification. And helps explain why the big scope "feels BIG", because mag 3, 2, 1, 0 stars are wonderfully overwhelming... looking like little welding arcs... almost painful to look at!

 

Here's a film image I took of Almach decades ago. It conveys the flavor or looking at a bright star through a BIG scope. >>>    Tom

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#20 Stephen Kennedy

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Posted 25 October 2019 - 08:41 PM

There are a few stars such as the M1-1b Red Super-giant Betelgeuse that can be imaged as extended objects.  Betelgeuse is so large that if you replaced the Sun with it, its surface extends nearly to the orbit of Jupiter.  This can be done using special techniques with some of the 8 to 10 meter aperture telescopes located in essentially perfect seeing sites such as Mauana Kea in Hawaii, the summit of the Canary Islands in the Atlantic and the Atacama high desert in Chile.  Besides the excellent high altitude conditions, these telescopes are able to perform this feat as a result of their enormous apertures.  The HST, operating above the atmosphere and with its huge by amateur standards 2.4 meter main mirror can also do this.  If they ever get launched and built the 6.5 meter mirror of the James Webb Space Telescope and the 30 meter mirror of the TMT could provide even more detailed extended object images of super giant stars in our area of the Milky Way Galaxy.



#21 imtl

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Posted 25 October 2019 - 10:51 PM

There are a few stars such as the M1-1b Red Super-giant Betelgeuse that can be imaged as extended objects.  Betelgeuse is so large that if you replaced the Sun with it, its surface extends nearly to the orbit of Jupiter.  This can be done using special techniques with some of the 8 to 10 meter aperture telescopes located in essentially perfect seeing sites such as Mauana Kea in Hawaii, the summit of the Canary Islands in the Atlantic and the Atacama high desert in Chile.  Besides the excellent high altitude conditions, these telescopes are able to perform this feat as a result of their enormous apertures.  The HST, operating above the atmosphere and with its huge by amateur standards 2.4 meter main mirror can also do this.  If they ever get launched and built the 6.5 meter mirror of the James Webb Space Telescope and the 30 meter mirror of the TMT could provide even more detailed extended object images of super giant stars in our area of the Milky Way Galaxy.

Imaging of these close super giants like Betelgeuse is done with interferometry by using a minimum of 2 telescope like you mentioned.

Interferometry is the only way to reconstruct the image of these stars with the current big telescopes. Done on few few I may add.

 

Antares

 

eso1726a.jpg

 

Betelgeuse

 

Betel_haubois800.jpg




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