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# What is the formula for spurious disc size? Topic Page: 1 2

The trouble with Dawes's desciribing the resolution of a star as measured by radii from center to first minima is that it still leaves the actual spurious disc size a little vague to me.

Minus the minima - is there a formula for arriving at the exact size of the spurious disc per resolution/aperture of scope? Radii or diameter is fine.

Pete

"I got here the same way the coin did" - Anton Chigurh "Noth doth not noth" PWP “We are the universe contemplating itself” - Carl Sagan 70mm TV Ranger 150mm SCT, Celestron 8" F/9 PARKS reflector, Clements Mirror, Spectrum Coatings Plossls, Orthos, TV Barlows DBK21au http://www.cloudynights.com/photopost/showgallery.php?ppuser=64701&cat=500
Pete, my understanding is that the radius from the center of the pattern to the first minimum is controlled by a formula related to the aperture of the telescope. This Rayleigh Criterion is given, in arc seconds, by the formula:

Radius to first minimum = 13.8/D D is in cm.

Inside this radius exists the 'spurious disc' to which you refer. I'm not aware of any formulae which give its radius because there are two parameters which control its size:

The brightness of the star. brighter = larger radius ( assuming a given telescope size, say, 8-inches, the very brightest stars like Sirius e.g. will have a spurious disc radius approaching the Rayleigh radius. The dark space between the spurious disc and the first ring will be very small. A sixth magnitude star in the same scope will have a much smaller spurious disc radius relative to the Rayleigh radius so the dark space between the spurious disc and the first ring will be wider. But the radius to the center of the dark ring will be exactly the same for both stars because this is controlled by the aperture of the telescope only.))

The size of the central obstruction: Larger obs, = smaller spurious disc radius (because the obstruction takes some of the light from the spurious disc and puts it into the rings. An 8-inch refractor and an 8 -inch SCT will have exactly the same Rayleigh radius but the amount of light in the central spurious disc and the apparent width of the first dark ring will appear slightly different in the two scopes. The obstructed scope will show a slightly smaller spurious disc,a slightly wider first dark space and a slightly brighter first ring when compared to the refractor. <assume both scopes are looking at the same star> )

This is a complex area of discussion.

Dave

David Cotterell Toronto, Ontario "If an observer actually sees an object, there is no point in referring to a formula to find out whether he ought to see it; and if he fails to detect it, no formula will ensure his success." - W.H. Steavenson 8" f/15.5 TEC Maksutov - 16" f/5 Teeter/Zambuto Dob -  TEC 140 -  AT 65EDQ APO Refractor - Astro-Physics Mach 1 GTO Mount -  iOptron ZEQ25 mount - Discmount DM6 -  Canon 60Da

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Quote:

The trouble with Dawes's desciribing the resolution of a star as measured by radii from center to first minima is that it still leaves the actual spurious disc size a little vague to me.

Minus the minima - is there a formula for arriving at the exact size of the spurious disc per resolution/aperture of scope? Radii or diameter is fine.

Pete

Well, Dawes Limit is an observational limit for double star resolution, so it doesn't quite give you the size of the visible 'spurious' disk that you see in the telescope. There is a certain visible "threshold"of light intensity that you hit for the apparent edge of that disk, and that angular distance depends a bit on how bright the star is. On average, the spurious disk radius (in arc seconds) may be very roughly about half the Rayleigh Limit of 5.45/D, where D is the aperture in inches (138.4/D for D in millimeters). For really bright stars, the spurious disk will be somewhat larger than this figure, and for really faint stars, it may be somewhat smaller. The only fairly precise figure for any disk size is the one given by the Rayleigh limit, and that only gives the radius of the central maximum of the diffraction pattern from the first zero of the pattern (the middle of the first dark ring out from the central disk) to the center of the diffraction pattern. Sometimes, this radius is called the radius of the diffraction disk, rather than the radius of the actual visible "spurious" disk. Clear skies to you.

David W. Knisely . . . . . . "If you aren't having fun in this hobby, you aren't doing it right." Hyde Memorial Observatory http://www.hydeobservatory.info Prairie Astronomy Club http://www.prairieastronomyclub.org
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To appreciate the problem of assigning a size to the 'disk', study the brightness profile of the diffraction pattern. It resembles a Gaussian flanked by 'ringing' wavelets. And for obstructed apertures the pattern's relative intensity profile evolves with percentage of obstruction.

Since the central peak in the pattern (and indeed the entire pattern itself) has a continuous slope in the intensity plot, any 'edge' must be defined by the width of the peak at some chosen relative intensity. This is why the spurious disk is seen as larger for brighter stars; one sees 'farther down the slope' and nearer to the first minumum.

Given these factors, it can be seen why the non-arbitrary distance from central peak to first minimum is so useful.

Home-made and modified binoculars My Gallery (mostly DIY stuff) **UPDATED Jun 22, 2013** Simple minds discuss people. Good minds discuss events. Great minds discuss ideas. - Hyman Rickover
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the diffraction pattern generated by a star image is fuzzy at the boundaries. the basic calculation is:

radius of diffraction element R = k*lambda/D.

k = a constant that is defined for the specific feature you want to calculate. (a table listing the values of k for all the principal features at their peak values ... rings at their brightest, intervals at their darkest ... is provided ---> here .)

lambda = the specific wavelength of light you are using to generate the diffraction pattern (different wavelengths refract at different angles, producing different dimensions in the diffraction artifact.) normal value is a visual "green" at peak photopic sensitivity, which is ~550 nm. for scoptopic vision, you'd use ~500-510 nm. in millimeters = 0.00055 or 0.00051.

D is the diameter in millimeters of the objective or entrance pupil, whichever is smaller. (a central obstruction actually makes the diffraction dimensions smaller; this is usually ignored.)

in most treatments the airy disk radius is simply k = 1. the value k = 1.22 commonly used is actually the darkness minimum of the first diffraction interval. to get the airy disk boundary, it might be reasonable to take the midpoint between the two (1.11). depends on why the number matters and what you intend to use it for.

"computing your resolution limit" is simply the game of choosing a specific wavelength of light and value of k. so k = 1.22 for rayleigh, 1.00 for dawes and 0.7 for sparrow.

nearly all telescopes above 6" or so are limited not by their optics but by the atmosphere in 19 nights out of 20, at best. like everything else in optics outside a modulation transfer function, resolution limits are conceptual and not empirical: what you see is what you get.

i believe bright images are larger not due to imaging fainter diffraction rings at larger radii, but because the high luminance magnifies the inherent dispersion, scatter and flare in the optical system.

bruce

Meade 305 ƒ/10.4 LX200 • Royce 250 ƒ/19 Dall Kirkham • AstroTech 250 ƒ/8 RC • AstroTech 250 ƒ/4 "RFT" Newtonian • Meade 80 ƒ/6 ED APO • TEC 140 ƒ/7 APO • Orion 180 ƒ/15 Mak Cass ... Denkmeier II binoviewer - Nikon 8x60 binocular

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I did step right into that one didn't I. Forgot that the visible disc does definately vary with brightness.

My upshot of the question was to coe up with a formula describing the smallest disc I can resolve AS a disc of true surface area as opposed to merely a spurious disc. I did resolve Titan, Europa and Io.... Can I resolve Vesta or Ceres as a discrete disc as well? I wanted a hard number for a roughly 6th mag spurious disc. My angular res, atleast per Dawes is .576... does that mean in theory I can resolve a disc of .577? I know its Dawes which isnt so much as laws of light but "laws" of observation.

Ive always wanted to know the smallest RESOLVABLE disc size... for example sake - 6V.

Pete

Pete

"I got here the same way the coin did" - Anton Chigurh "Noth doth not noth" PWP “We are the universe contemplating itself” - Carl Sagan 70mm TV Ranger 150mm SCT, Celestron 8" F/9 PARKS reflector, Clements Mirror, Spectrum Coatings Plossls, Orthos, TV Barlows DBK21au http://www.cloudynights.com/photopost/showgallery.php?ppuser=64701&cat=500
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drollere wrote:

Quote:

in most treatments the airy disk radius is simply k = 1. the value k = 1.22 commonly used is actually the darkness minimum of the first diffraction interval. to get the airy disk boundary, it might be reasonable to take the midpoint between the two (1.11). depends on why the number matters and what you intend to use it for.

"computing your resolution limit" is simply the game of choosing a specific wavelength of light and value of k. so k = 1.22 for rayleigh, 1.00 for dawes and 0.7 for sparrow.

Unfortunately, Dawes Limit does *not* involve wavelength and never really has. It has always been a purely visual empirical observational limit based on the double star observations by W.R. Dawes and his own idea of what constitutes resolution. His original limit was stated as r = 4.56/D, where r is the separation in arc seconds and D is the aperture in inches. You can read the original paper on-line via the archives, but no wavelength had ever been involved. Oh, if you want to play around with the numbers, you can come up with a formula that for 550nm gives about the same figures as Dawes Limit does, but what you end up with is not really Dawes limit.

W.R. Dawes Paper on Dawes Limit (see p. 234-235)

As for the Rayleigh limit, that is r = 1.22*(Lambda)/D, where r is the angle in radians, Lambda is the wavelength of the light, and D is the aperture in compatible units. The 1.22 factor out in front comes from doing the diffraction analysis for a circular aperture. However, as stated before, there is no accurate formula for the actual angular diameter of the visible spurious disk of a star as seen through the telescope. Clear skies to you.

David W. Knisely . . . . . . "If you aren't having fun in this hobby, you aren't doing it right." Hyde Memorial Observatory http://www.hydeobservatory.info Prairie Astronomy Club http://www.prairieastronomyclub.org
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Its too bad Dave. Seems like its almost right there. I could almost see a spread sheet for various magnitude stars for given apertures showing the apparent reduction or swelling of spurious disc size. Im onboard with what you are saying, just a shame those values cant be arrived at.

Thanks tho!

Pete

"I got here the same way the coin did" - Anton Chigurh "Noth doth not noth" PWP “We are the universe contemplating itself” - Carl Sagan 70mm TV Ranger 150mm SCT, Celestron 8" F/9 PARKS reflector, Clements Mirror, Spectrum Coatings Plossls, Orthos, TV Barlows DBK21au http://www.cloudynights.com/photopost/showgallery.php?ppuser=64701&cat=500
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George Airy (of Airy pattern fame) discussed the variance of the diameter of the spurious disk in relation to source intensity (90%~50%) way back in 1834 within his original paper defining the Airy pattern (see article beginning page 238 :

The rapid decrease of light in the successive rings will sufficiently explain the visibility of two or three rings with a very bright star and the non-visibility of rings with a faint star. The difference of the diameters of the central spots (or spurious disks) of different stars (which has presented a difficulty to writers on Optics) is also fully explained. Thus the radius of the spurious disk of a faint star, where light of less than half the intensity of the central light makes no impression on the eye, is determined by making n= 1,616, or s=1.17/a:
whereas the radius of the spurious disk of a bright star, where light of 1/10th the intensity of the central light is sensible, is determined by making n = 2,73, or s = 1.97/a.

The general agreement of these results with observation is very satisfactory. It is not easy to obtain measures of the rings; since when a is made small enough to render them very distinct as to form and separation, the intensity of their light (which varies as a^) is so feeble that they will not bear sufficient illumination for the use of a micrometer. Fraunhofer however obtained measures agreeing pretty well (as to proportion of diameters, &c.) with the results above.
(s=seconds arc, a= aperture, radius in inches.)

As an example take a 6" aperture, radius of the spurious disk 90% illuminated (with visible rings) is 1.97/a (radius inches) or a diameter of 1.313"arc. On the other end of the scale, when only the FWHM--defined by Airy as 1.17/a-- is visible (dim star lacking visible rings) the diameter of the spurious disk would be 0.78"arc. The ratio is 1.68:1.

To illustrate:

Airy didn't discuss magnitudes in relation to the above (maybe someone has since, I don't know) but he did utilize a wavelength in his calculations (0.000022inch or 558nm). This represents the midpoint of the spectrum visible to the human eye and a figure around 555nm has been in use for over 150 years as the frequency specification in all white light (full spectrum) visual resolution formulae. If you are going to use any hard figure to divide by aperture in order to come up with an angular resolution limit, controlling wavelength must be specified either explicitly (as in the Airy/Rayleigh limit cases) or implicitly (as in the case of Dawes limit.) Defining a resolution limit without frequency is like defining speed without time. 60miles per. Per what? Second? Minute? Hour? All the parameters must be specified in order for the numerical figure of either to be meaningful.

Mardi 4" achromat, ETX-70, 8"cat. Whitepeak Lunar Observatory Website

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Quote:

Unfortunately, Dawes Limit does *not* involve wavelength and never really has. It has always been a purely visual empirical observational limit based on the double star observations by W.R. Dawes and his own idea of what constitutes resolution. His original limit was stated as r = 4.56/D, where r is the separation in arc seconds and D is the aperture in inches. You can read the original paper on-line via the archives, but no wavelength had ever been involved. Oh, if you want to play around with the numbers, you can come up with a formula that for 550nm gives about the same figures as Dawes Limit does, but what you end up with is not really Dawes limit.

i'm not really following the nuance here. the diffraction pattern creates the visual issue that dawes attempted empirically to resolve. if the diffraction pattern can be specified in some other way, that doesn't mean it's no longer "dawes test", it only means that the test has been put into a different currency.

one can quibble further and say that dawes' test only applies to two stars of equal 6th magnitude as viewed through a 6 inch telescope, and anything else is not "dawes' test". the only way you can step out of that restriction is to generalize from the effects of light generally. which is what the basic formula i gave does.

dawes took a particular notion of a visual pattern that he wanted to define ... rayleigh used a different notion, sparrow used a different notion. pick your k.

Quote:

The 1.22 factor out in front comes from doing the diffraction analysis for a circular aperture.

that's correct. but on the one hand, you can also specify values of k for any aperture shape (square, triangular, elliptical, apodizing mask), and at any cutoff boundary between "light" and "dark". and, on the other hand -- what proportion of all telescopes do not have circular entrance pupils?

as i said: these are all arbitrary moves. pick your conception of what a "spurious disk" really is, pick your k, pick your wavelength, pick your aperture. have fun.

the web page i linked to has all the details. clear skies.

bruce

Meade 305 ƒ/10.4 LX200 • Royce 250 ƒ/19 Dall Kirkham • AstroTech 250 ƒ/8 RC • AstroTech 250 ƒ/4 "RFT" Newtonian • Meade 80 ƒ/6 ED APO • TEC 140 ƒ/7 APO • Orion 180 ƒ/15 Mak Cass ... Denkmeier II binoviewer - Nikon 8x60 binocular

Astronomical Files from Black Oak Observatory

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Frequency (wavelength) is necessary to consider here simply because the diffraction pattern's dimensions scale with wavelength. The smaller the wavelength, the smaller the pattern. The effect in 'white' light is a blended, overlapping series of colored patterns scaling with wavelength which further blurs the overall aspect, even with perfect optics under ideal conditions.

Home-made and modified binoculars My Gallery (mostly DIY stuff) **UPDATED Jun 22, 2013** Simple minds discuss people. Good minds discuss events. Great minds discuss ideas. - Hyman Rickover
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drollere wrote:

Quote:

i'm not really following the nuance here. the diffraction pattern creates the visual issue that dawes attempted empirically to resolve. if the diffraction pattern can be specified in some other way, that doesn't mean it's no longer "dawes test", it only means that the test has been put into a different currency.

There are a few problems with Dawes limit. First, Dawes has his own opinion based on his visual observations of what exactly qualifies as "resolved". Second, the separation involved here doesn't exactly involve any physics in terms of the locations within a diffraction pattern, but is only a separation limit arrived at by pure observations of various double stars with known angular separations. Third, there is no wavelength dependence, as Dawes used the full bandwidth of the his own eye. The Rayleigh and Sparrow limits both are based on specific locations and profiles in the diffraction pattern at given wavelengths, which is why they are generally used. This does not invalidate Dawes Limit, but the limit is not one based purely on the physics of light, as the Rayleigh limit is. Clear skies to you.

David W. Knisely . . . . . . "If you aren't having fun in this hobby, you aren't doing it right." Hyde Memorial Observatory http://www.hydeobservatory.info Prairie Astronomy Club http://www.prairieastronomyclub.org
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[quote

As an example take a 6" aperture, radius of the spurious disk 90% illuminated (with visible rings) is 1.97/a (radius inches) or a diameter of 1.313"arc. On the other end of the scale, when only the FWHM--defined by Airy as 1.17/a-- is visible (dim star lacking visible rings) the diameter of the spurious disk would be 0.78"arc. The ratio is 1.68:1.

To illustrate:

.

The diagrams you posted are absolutely 'spot' on. The boundary line between the green and pink zones moves up and down with the magnitude of the star. The 'sides' of the light curve of the spurious disc are slanted so the pink/green boundary intersects a wider zone as it moves downward. If we could get a formula for this widening as a function of the brightness of the star (for a given aperture) we could generate a graph which would give a visual representation of the diameter of the spurious disc.

But we would eventually need a LOT of graphs because in this hobby we have scopes from 50mm to 1000mm or more aperture, some without central obstruction and some with obstructions varying from as little as 10% right upt o 40% or more.

Quite a homework assignment for someone!

The diffraction patterns shown are excellent as well as they show a 6" scope's view of a star like Rigel on the left and some anonymous 6th mag star on the right. Look at the difference in the size of the spurious discs!

Great post, Mardi!!

Dave

David Cotterell Toronto, Ontario "If an observer actually sees an object, there is no point in referring to a formula to find out whether he ought to see it; and if he fails to detect it, no formula will ensure his success." - W.H. Steavenson 8" f/15.5 TEC Maksutov - 16" f/5 Teeter/Zambuto Dob -  TEC 140 -  AT 65EDQ APO Refractor - Astro-Physics Mach 1 GTO Mount -  iOptron ZEQ25 mount - Discmount DM6 -  Canon 60Da

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Hi Pete

Quote:

My upshot of the question was to coe up with a formula describing the smallest disc I can resolve AS a disc of true surface area as opposed to merely a spurious disc. I did resolve Titan, Europa and Io.... Can I resolve Vesta or Ceres as a discrete disc as well? I wanted a hard number for a roughly 6th mag spurious disc. My angular res, atleast per Dawes is .576... does that mean in theory I can resolve a disc of .577?

This is highly relevant to my interests! Recently, after an observation of the Jovian moons through my 80mm f/15 Vixen achromat, I've been wondering about the same thing. During said observation I could clearly tell that Ganymede was bigger than the others and that the others had different albedo. They also had slightly different diameters, but I couldn't tell if this was because I could truly tell that they were physically different in size or if I was fooled by their differences in brightness.

Now, things get extremely interesting if we consider the fact that even Ganymede is much smaller than the diameter of the airy disk of a 80mm. The *radius* of the first minima in arc seconds is determined by the following formula(1)

s = 2.76 / a

Where s is the radius of the first minima in arc seconds and a is the radius of the objective is inches. The diameter of the first minima is thus TWICE the value of s for any given aperture. In the case of my 80mm, it follows that the diameter of the first minima is a whopping 3.56"...

The airy disk is naturally going to be smaller, since it never meets the first minima at all. A bright star, where light almost out to the first minima can be seen has a spurious (airy) disk, whose radius can be approximately determined by the following formula:

s = 1.97 / a

For my 80mm, that equals a diameter of 2.54".

A faint star, where only light brigther than half the intensity is seen by the eye has a spurious disk whose radius can be approximately determined(1) as

s = 1.17 / a

Or in the case of my 80mm, with a diameter of 1.5". This is below the diameter of Ganymede and since the surface brightness of any star is much higher than that of any of the Jovian moons, I must conclude that it is indeed possible to resolve Ganymede with a 80mm. But how about the other moons? Is it possible that I saw them as true disks and not spurious disks? I am not sure. I think the comparatibly low surface brightness of their surfaces can work to our advantage here and that it might indeed be possible to see them as disks as it should be possible to magnify them enough to make the spurious disk small enough that the real surface behind should be visible, if you can follow my argument.

Consider this: If Ganymede had had a very high albedo and been much brighter than now, its spurious disk would have been much larger and it would have shown an airy disk BIGGER than 1.7", precluding resolution in an 80mm.

Or would it? Is my theory wrong?

As the surface of any resolvable body seen in a telescope is made up of a large number of faint spurious disks, is the limit of resolvability linked to surface brightness? The dimmer the surface, the better the resolvability? Is this why the Jovian moons never show up as huge disks in even small telescopes? A 60mm shows them as equal-sized orbs at low power, but they don't grow wildly with increased power, as we might have believed at first. Is it possible to resolve Ganymede in even a 60mm, due to the limited surface brightness? Just increase the power enough for it to grow faint enough for the true sizes to become apparent?

Ladies and Gentlemen, I eagerly await your response. Experimentation under real-world conditions would be highly valuable.

Sources:

1)http://en.wikipedia.org/wiki/Airy_disk

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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As Ganymede at opposition is around 1.8"arc diameter, differentiating between a 1.5"arc spurious disk and the actual disk of Ganymede would seem to be to be a pretty tall order in visual acuity. Here's a visual example:

6:5 ratio, (1.8":1.5") Ganymede at opposition on left, spurious disk of 80mm (at FWHM) on right.

Now consider that one must make this size difference "call" from memory, with no "reference" spurious disk for comparison, unlike the illustration above (unless an about equal magnitude star happens to be in the high power field with Ganymede...

A filar micrometer would probably help!

re; extended detail, visual resolution is not so much about the angular size of the collection of spurious disks comprising an extended image as it is about the contrast between those points. Lower surface brightness means lower contrast which means lower resolution.

Mardi 4" achromat, ETX-70, 8"cat. Whitepeak Lunar Observatory Website

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Hi photonovore

Thanks for your reply. It remains a fact that I saw a distinct size difference between Ganymede and the rest of the moons. The other three were the same size or so near, I couldn't tell for sure, due to seeing. This must mean that their visible diameters have been below 1.5". Callisto is around 1.6" if I recall, but Io and Europa are near 1".

All I can say for now is that I hope for some clear, steady nights soon so that I can experiment some more.

I am familiar with the fact that resolution of detail on an extended surface is all about contrast, but what if the problem was *resolution of the surface itself*? Here a fainter surface would mean a smaller spurious disk and thus smaller diameter.

Hmm. I must make some drawings to better illustrate what I'm talking about.

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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Perhaps by "resolution of the surface itself" you mean resolution of the actual planetary disk as distinct from the spurious disk (as formed by the planetary disk)? If so I do follow what you are trying to say and yes, it does make sense to me--re; a very small (near point source) extended object (like a jovian moon), if one sees a disk sensibly larger than the predicted spurious disk for that object (considering the state of FWHM seeing of course) then the disc would likely be real. The problem i see in is being able to say with surety which sort of disk you are actually seeing, since the difference is so minute...(and usually lacking in helpful high-power-field comparison objects).

Mardi 4" achromat, ETX-70, 8"cat. Whitepeak Lunar Observatory Website

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pete ... to the issue of whether you can see ganymede or not ... not to worry.

the diffraction pattern is wave generated by light from *all parts* of the image. it affects high frequency detail at the edges, the same as it affects the "point" radiance of a star.

the apparent diameter of ganymede would be 1/2 of the "spurious disk diameter" you choose (using your scholarly adjudication of the correct value of k) on each side of the object, plus the actual diameter of the object. therefore, if ganymede is only 1" wide, and the spurious disk is 1" wide, you'll see 2" in contrast to the 1" of a star.

in addition ... the diffraction rings generated by the "point" wavefronts across the disk of the satellite overlap and cancel each other, so a second sign that you're seeing a disk instead of a star is that the disk won't have defined diffraction rings around it, only a diffuse glare.

another way to think about it is that glare diffracts into shadow, so if you can see the shadows of the moons on jupiter as widths rather than points, then you can probably also see the disk of the moon.

whether you actually visualize the disk in the image as a disk in your eye has more to do with the magnification you're using to bring the disk into your eye's visual resolution limit. as a double star guy you know all about that.

bruce

Meade 305 ƒ/10.4 LX200 • Royce 250 ƒ/19 Dall Kirkham • AstroTech 250 ƒ/8 RC • AstroTech 250 ƒ/4 "RFT" Newtonian • Meade 80 ƒ/6 ED APO • TEC 140 ƒ/7 APO • Orion 180 ƒ/15 Mak Cass ... Denkmeier II binoviewer - Nikon 8x60 binocular

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Hi Mardi

Quote:

yes, it does make sense to me--re; a very small (near point source) extended object (like a jovian moon), if one sees a disk sensibly larger than the predicted spurious disk for that object (considering the state of FWHM seeing of course) then the disc would likely be real.

Yes, yes! That's what I'm talking about!

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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Hi drollere

I'm not Pete, but I think you're replying to my post about observing Ganymede in the 80mm. If not, please forgive me.

Quote:

the apparent diameter of ganymede would be 1/2 of the "spurious disk diameter" you choose (using your scholarly adjudication of the correct value of k) on each side of the object, plus the actual diameter of the object. therefore, if ganymede is only 1" wide, and the spurious disk is 1" wide, you'll see 2" in contrast to the 1" of a star.

This can't be true. I speculated long and hard about this one and if it was true, then Ganymede - and the rest of the satellites - would appear much larger. Let's say we choose 1" as the diameter of the spurious disk. Ganymede is around 1.7" to 1.8" in itself, then the visible diameter would be 2.7" to 2.8" in diameter. Jupiter is near 45" now and if we choose 2.7", then the diameter of Ganymede would appear to be 1/17th the diameter of Jupiter. This sounds huge. The real size relationship is more like 1:26.

Another indication this is not so, is by watching one of the satellites disappear behind Jupiter. You will see the disk starting to get eclipsed as soon as it hits the edge of Jupiter. If there was some "diffraction size increase", then you would see the disk of the satellite extend beyond the edge of Jupiter and a little bit inside the edge, even as it passed behind Jupiter's disk, but we don't.

Quote:

in addition ... the diffraction rings generated by the "point" wavefronts across the disk of the satellite overlap and cancel each other, so a second sign that you're seeing a disk instead of a star is that the disk won't have defined diffraction rings around it, only a diffuse glare.

Yes! Now we're talking. The question here is whether this happens even if the disk is perhaps only half the size of the spurious disk of a star of the same magnitude, because it is not a infinitesily small point. The point is that I can't recall ever having seen diffraction rings around the Jovian satellites. This might be because scopes too small to resolve them as disks don't show diffraction rings around point sources of that magnitude, but it might also be because of interference as you mention in the passage I quoted. I don't know, I merely speculate.

All this has got me really excited and I know what I will turn my scopes to the first next time it gets clear.

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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Quote:

the apparent diameter of ganymede would be 1/2 of the "spurious disk diameter" you choose (using your scholarly adjudication of the correct value of k) on each side of the object, plus the actual diameter of the object. therefore, if ganymede is only 1" wide, and the spurious disk is 1" wide, you'll see 2" in contrast to the 1" of a star.

I don't think I am understanding you. You seem to be saying that there is such a thing as "different sized point sources"--as a disk of 1"arc would be unresolvable in an 80mm scope--it would be a point source IOW. But since it is otherwise measurable as 1" arc (by another, larger telescope) you propose that it's diffraction pattern is based upon that diameter--in the smaller scope--where it is seen as a somehow "larger" point source (having a correspondingly larger diffraction pattern and spurious disk) than a simple star?

So you propose that in an 80mm scope you would see Ganymede (as a point source) as 2" in diameter, where in a larger scope you could see it --as a disk--at it's true diameter--of 1" arc?

If so that doesn't make any sense to me...

Mardi 4" achromat, ETX-70, 8"cat. Whitepeak Lunar Observatory Website

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Quote:

Quote:

the apparent diameter of ganymede would be 1/2 of the "spurious disk diameter" you choose (using your scholarly adjudication of the correct value of k) on each side of the object, plus the actual diameter of the object. therefore, if ganymede is only 1" wide, and the spurious disk is 1" wide, you'll see 2" in contrast to the 1" of a star.

I don't think I am understanding you. You seem to be saying that there is such a thing as "different sized point sources"--as a disk of 1"arc would be unresolvable in an 80mm scope--it would be a point source IOW. But since it is otherwise measurable as 1" arc (by another, larger telescope) you propose that it's diffraction pattern is based upon that diameter--in the smaller scope--where it is seen as a somehow "larger" point source (having a correspondingly larger diffraction pattern and spurious disk) than a simple star?

So you propose that in an 80mm scope you would see Ganymede (as a point source) as 2" in diameter, where in a larger scope you could see it --as a disk--at it's true diameter--of 1" arc?

If so that doesn't make any sense to me...

I do not think that you can apply a formula for a point source to anything other than a point source. There is no star that we can see anything more than a point source with a telescope of a meter or less in diameter. Just last night I was looking at the Jovan moons with my 4" TV102 refractor and could see the apparent disk of each moon. Ganymede was definately the largest with Io being the smallest at 179x. I could even see the apparent differences in size at 129x. Ganymede I would guess to be a little less than 2" in diameter with Io being about 1" in diameter. If I were seeing a disk from each point along the edge, the disk would have been much larger than they appeared to be, maybe 3" in diameter ( one half the spurious disk + 1.8"moon disk + one half the spurious disk = about 3" diameter for Ganymede). There must be a formula for objects other than a point source. How else could we see detail on the moon and planets that is smaller than 2" in width?

Buddy

C-11 SCT, XT10i Dob, C-6 SCT, ETX125PE Mak-Cass, TV102, & AT66

"We the People are the rightful master of both congress and the courts - not to overthrow the constitution, but to overthrow the men who pervert the constitution." Abraham Lincoln"

The heavens declare the glory of God; the skies proclaim the work of his hands." Psalms 19:1

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So much has been said. Mardi, Dave, Droll, Jensen and all others...

For the record the case of Ganymede through my 70mm is too tough for me to call. I just barely [with my scope standing on the tips of its resolution toes] was able to make Uranus as different from a star. Its funny, when I did the side by side with my 8" reflector and my 70mm on this object, it was a kind of simple but sensitive matter to see the disc with the 70mm. Years later with no big brother 8 to show the way I had to fish out the disc of the planet on my own [with 70mm]. What a different task. If I saw green itd been so much easier, alas, it was on ly grey blue to me - like 10gajillion other stars. So with variable seeing it took succesive nights when the skhy finally calmed enough for me to hang on to the actual disc sighting of the planets surface area barely. My scope resolves 1.65" the planet was 3.4". My gawd was it ever a slight detail to make.

Having done it I felt like I just hit a home run. I felt pumped. I felt like the heavens presented me a challenge and I defeated it with this little scope. The a-b comparo with the large reflector afforded no such sense of accomplishment as the planet was already ID'd. Coolest thing Ive done with the little scope in a while.

But back to Ganymede...

Having seen the difficulty of making a true discerning ob of Uranus's disc at 100x through the 70mm, it tempers what I would expect to see of a 1.7" disc. In fairness the seeing needs to cooperate and I cant say Ive had that enough to contemplate a 1.7" challenge. Whats peculiar is that I see the differfent sizes of the moons - is this merely the minimal magnitude differences producing differing spurious disc sizes? Im inclined to believe so.

I think the only real way I can make the 1.7" disc of ganymede with any sense of trueness is to have a like-magnitude star in the same 100x field.

Can you see it with an 80mm? Im such a pro-you can do it- kind of observer I want to say yes. Based on my Uranus experience though - and Mardis illustration, you have your work cut out for you.

I DID resolve the .95 second dis of Titan conclusively as an extended object through my 8" at 364x. The seeing required untold failed attempts that offered otherwise mediocre seeing on Saturn. This was a very severe test of amospheric steadiness [for me and that scope]. Was Titan for my 8 more difficult than Uranus for my 2.7" refractor - seeing aside?

Lets look....

1.65"res versus 3.4" Uranus - actual disc 2X larger than Dawes limit.

.576"res versus .95" Titan - actual disc 1.64X larger than
8 inch apt. dawes limit.

Projecting Ganymede through my 70mm...

1.65"res versus 1.7" Ganymede disc... 1.03X larger than Dawes for the 70mm aperture.

OOUUCCHH.

For me its a total loss. If Titan was any lesson - and certainly Uranus*, this just isnt possible for me. Still I see the varying sizes of the moons. I think Im merely seeing the spurious disc show.

Can the magic be had with 80mm over my 70mm? Id like to think so, but only you can make the call.

Pete
*My 70mm Neptune was too too unambiguous to call. Frankly Id opt for Neptune before Id tackle Ganymede, alas the dimness cuts out the light you need at just the magnification required to see the disc. Life is cruel!!!

"I got here the same way the coin did" - Anton Chigurh "Noth doth not noth" PWP “We are the universe contemplating itself” - Carl Sagan 70mm TV Ranger 150mm SCT, Celestron 8" F/9 PARKS reflector, Clements Mirror, Spectrum Coatings Plossls, Orthos, TV Barlows DBK21au http://www.cloudynights.com/photopost/showgallery.php?ppuser=64701&cat=500
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to the issue of whether you can see ganymede or not ... not to worry.

the diffraction pattern is wave generated by light from *all parts* of the image. it affects high frequency detail at the edges, the same as it affects the "point" radiance of a star.

the apparent diameter of ganymede would be 1/2 of the "spurious disk diameter" you choose (using your scholarly adjudication of the correct value of k) on each side of the object, plus the actual diameter of the object. therefore, if ganymede is only 1" wide, and the spurious disk is 1" wide, you'll see 2" in contrast to the 1" of a star.

At one time I made exactly the same argument. But I now know that is incorrect. In fact this same discussion resides in the archives. Also this discussion resides in the archives of the planetary forum, see Seeing Jupiter's Moon as Disks.

Extended objects can be thought of as being made up of an infinite number of points. This might lead one to the conclusion as it is stated above. It seems somewhat reasonable at this step one could say then the points along the edges of the extended object have a diffraction pattern the same as if they were a stellar point of equal magnitude as the object and therefore the object is as wide as the spurious disk from points in the central area plus one half spurious disk on each side of the extended object.

However, a bit deeper thought shows us this cannot hold to be true.

What is missing, and of the utmost importance, is that the integrated magnitude of the whole extended object comes from an infinite number of points, all those points in the central area plus all the points along the edges of the extended object, but whose individual magnitudes decline inversely with the number of points and is therefore infinity lower than the integrated magnitude of the whole extended object.

For sake of discussion (and by no means intended to be an accurate representation of Ganymede, but intended only to represent the problem) let us say that Ganymede is made up of 1000 points of light, each point being of magnitude 12, but integrated as a whole all points show an extended object at mag5. [ One argument might be; well lets assume only a very few points all at mag 6, integrated to equal mag 5 disk. That fails, as we could then cut out all the light of the entire spurious disk from the center mag 6 point and you would find yourself with a very thin ring incapable of supporting mag 5 points around the edge. ]

Now if we simply take the actual points along the edge, or a ring of virtual points (keeping in mind the ring represents the diameter of the extended object) and black out the central area of light, would the diameter of the object we see be twice the size of the ring itself? Of course the answer is No.

Once we take this step, we find that the points are so faint as to have a diminishingly small spurious disk (refer to the slope reference to magnitude creating varying spurious disk sizes as shown in Mardi's graphs above) and in fact they do not add one half the width of the (resolution limit) spurious disk to the to the overall object dimension. I don't doubt that they may add some tiny fraction of an arcsecond to the overall width of the extended object, perhaps measured in several hundreths of an arcsecond. But whatever it may be, due to the integrated magnitude being broken into small parts and being much lower, it is small.

edz

Teach a kid something today. The feeling you'll get is one of life's greatest rewards. member#21
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Hi EdZ

Quote:

Insert lots of clever stuff Ed just said in his post above

YES! I knew I had it! I haven't posted it here, but was about to, but you got there before me. I came to the exact same conclusion as you, after reading a document by Chris Lord

http://www.brayebrookobservatory.org/BrayObsWebSite/BOOKS/contrast%26definition.\pdf

I posted my findings in the forum of the 60mm club, as we coincidentally had a discussion going about the resolvabiility of the Jovian satellites. I should have posted it here as well, but for some reason, just didn't get around to do that.

Here's my first post in that thread, which included the above link:

Quote:

It seems resolution is more a function of contrast AND aperture, rather than
aperture alone, as most sources will have us to believe.

If I understand the math and theory correctly, we can resolve the Jovian moons
in small telescopes because they are finite-sized spots, with finite surface
brightness, with incredibly high contrast to the background. Stars are
infinitely small (relatively speaking) and have infinite surface brightness so
they obey the diffraction law in a somewhat different way.

The contrast with the background is really high for the edge of a planet or
moon, so the resolution we can achieve must be pretty high. In other words, we
are trying to resolve a tiny white dot *with a finite surface brightness* on a
pitch-black background. If we assume the contrast to be 0.99, the resolution in
arcseconds for a 60mm is as high as 0.67"! Since we are dealing with
finite-sized areas, we are also dealing with the brightness-area relationship,
in other words, the more we magnify, the dimmer the surface of said area gets
and the contrast is lowered. Still, the magnification can be pretty extreme,
before the Jovian satellites are magnified out of visibility in the eyepiece, so
even with very high powers, the contrast can probably safely be assumed to be
over 0.98, in which case the resolution of a 60mm is still below 1" and all the
moons thus resolvable. If I set the contrast to 0.99 it should be possible for
me to resolve TITAN in my 80mm! WOW! I've never even considered that! I must try
it some day.

Another important lesson from the document is that the MTF value of the
telescope plays a very crucial role in how well it performs on extended sources.
The higher the value, the better it performs and can resolve smaller detail. The
MTF value is also affected by the atmosphere and other factors, so the more
perfect the whole system, the better. Again, our small 60mm's can perform close
to theory, simply because they have an easier time matching theory, by being
easier to make close to perfection. It is also much more likely to be possible
to find a night of perfect seeing with a 60mm rather than with a larger scope,
etc.

And in a reply to another member:

Quote:

The problem of using stars in this analogy is that stars have much, much higher
surface brightness than planets. In this case, this leads to an erroneous
result. Per Lord's paper, when we are dealing with contrast levels, resolving
ability and finite surfaces, we are dealing with a much more complex set of
phenomena than when simply dealing with multiple stars. This is also why the law
of resolving power was accurately described in theory long before the same was
done for planetary details.

We are not dealing with diffraction of point sources, but of the ability of the
telescope (and in the end, the eye of the observer) to detect a disk of modest
(relatively) brightness, contrasting with the background. There is a very large
difference here.

If you took a small point of any of the Jovian satellites, the diameter of a
average star, as seen if we could resolve it, say 0.0001", then that point would
be utterly invisible, because its surface brightness is much lower than that of
a star. It therefore shows almost infinitely faint diffraction rings. Add up
enough of these tiny areas side by side and they become detectable in the
telescope. Their combined brightness is now perhaps on par with the single star,
which is only 0.0001" in diameter, but the overall surface brightness is not
larger than that tiny point we started with. Therefore the satellite still shows
no diffraction rings! It simply can't!

If it is to show diffraction rings, it must be smaller than the smallest the
instrument in question can resolve, given its MTF curve and have a surface
brightness high enough to make this possible.

Responses from several 60mm club members are that they can indeed see the Jovian moons as tiny disks, of different sizes, when observing them at high power in 60mm and larger scopes.

Exciting stuff.

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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My understanding of this is that the resolution of a disk is limited by the PSF of the aperture. IOW, no real disk is resolvable as such when its apparent diameter is less than the size of the spurious disk for that magnitude (as a point source) and aperture. So, it would not be possible for a 60mm objective to show any of the jovian moons as real disks. What is possible (using a 60mm) is observing differing sizes of *spurious* disks, their sizes varying according to magnitude differences.

Mardi 4" achromat, ETX-70, 8"cat. Whitepeak Lunar Observatory Website

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My understanding of this is that the resolution of a disk is limited by the PSF of the aperture. IOW, no real disk is resolvable as such when its apparent diameter is less than the size of the spurious disk for that magnitude (as a point source) and aperture. So, it would not be possible for a 60mm objective to show any of the jovian moons as real disks. What is possible (using a 60mm) is observing differing sizes of *spurious* disks, their sizes varying according to magnitude differences.

I agree.
Anything smaller than the resolution limit (use Rayliegh 138/Dmm or 5.45/Din) of the scope cannot be resolved as larger than the spurious disk.
edz

Teach a kid something today. The feeling you'll get is one of life's greatest rewards. member#21
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Do you mean that when one of the moons starts to transit Jupiter that that little round disk that I see against the surface of the planet is not the disk of that moon but only the spurious disk of a point source of light? Watch one and tell me thats what it is. If all I ever saw were the moons against the dark sky, I could maybe agree that the size differance was due to the magnitude differance in the moons, but when they enter or leave the shadow of Jupiter and you see what looks like a phase of the moon, or start or end a transit and you see that little disk against the surface of Jupiter, I find it hard to believe it's only a spurious disk.

Buddy

C-11 SCT, XT10i Dob, C-6 SCT, ETX125PE Mak-Cass, TV102, & AT66

"We the People are the rightful master of both congress and the courts - not to overthrow the constitution, but to overthrow the men who pervert the constitution." Abraham Lincoln"

The heavens declare the glory of God; the skies proclaim the work of his hands." Psalms 19:1

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Do you mean that when one of the moons starts to transit Jupiter that that little round disk that I see against the surface of the planet is not the disk but only the spurious disk of a point source of light? Watch one and tell me thats what it is. If all I ever saw were the moons against the dark sky, I could maybe agree that the size differance was due to the magnitude differance in the moons, but when they enter or leave the shadow of Jupiter and you see what looks like a phase of the moon, or start or end a transit and you see that little disk against the surface of Jupiter, I find it hard to believe it's only a spurious disk.

You have not provided enough information in your question.
Are you talking about the moon disk itself?
What size scope are you refering to?

edz

Teach a kid something today. The feeling you'll get is one of life's greatest rewards. member#21
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I'm with Buddy here. January 29th, 2002, I used my Vixen 80/1200 achromat at 171x to watch Io glide behind Jupiter's disk. I could see the tiny moon slowly disappear. At one time I estimated that I could still se Io when 4/5th of it was covered behind Jupiter. The visible part was a tiny, tiny crescent, a small bulge on Jupiter's limb. The seeing was very good. There is no way the spurious disk theory can explain this observation. Io is much smaller than the Airy disk of a 80mm, so how come I could see it as a crescent?

Just now I relived that moment, thanks to Stellarium. It was a very eerie feeling. I blew up the magnification really high, so that the true size of Io was shown in respect to Jupiter (if the power is not high enough, Stellarium will only show large, bright disks for the moons), set the time to run natural and stood back several feet from the screen. I couldn't get further away than what was equivalent to around 300x, but I could again, with just one eye open, see Io when it was about 4/5ths covered. This is of course just a very rough simulation, since there is no way the computer screen can mimick the actual brightness and contrast ratios between Jupiter, Io and the background sky, nor can it recreate the seeing conditions or the effects of diffraction. Still, the image looked very natural and quite like what I remember seeing through my late friend Per Darnell's 178mm apochromat at 300x-400x.

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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Do you realize an 80mm scope is showing you approx a 2 arcsecond sprious disk. Even larger if the spurious disk is 60% of the Airy disk, which it very well could be.

edz

Teach a kid something today. The feeling you'll get is one of life's greatest rewards. member#21
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Do you realize an 80mm scope is showing you approx a 2 arcsecond sprious disk. Even larger if the spurious disk is 60% of the Airy disk, which it very well could be.

Hi EdZ

I am fully aware of this, which is why these observations of the Jovian satellites with small telescopes intrigue me so much. There is something very different going on than when we're making airy disks of stars. I believe Chris Lord's paper describes it well, if one thinks about it. It is all about MTF, contrast and surface brightness.

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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As Ganymede at opposition is around 1.8"arc diameter, differentiating between a 1.5"arc spurious disk and the actual disk of Ganymede would seem to be to be a pretty tall order in visual acuity. Here's a visual example:

6:5 ratio, (1.8":1.5") Ganymede at opposition on left, spurious disk of 80mm (at FWHM) on right.

Now consider that one must make this size difference "call" from memory, with no "reference" spurious disk for comparison, unlike the illustration above (unless an about equal magnitude star happens to be in the high power field with Ganymede...

Can't speak for anyone else, but I clearly see the difference here.
This jives with my experience viewing Jupiter and it's moons one night recently in my 60mm f/20. I was able to clearly see size differences, and could tell which was Io by it's different color.
As I could hardly believe what I was seeing, I hung the 100mm f/13 on the mount, and verified what I was seeing.

Doug Truck Stop Astronomer Hunter, Fisher, Outdoorsman Carton Scope Club Six Inch Or Larger Refractors
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Quote:

Quote:

Do you mean that when one of the moons starts to transit Jupiter that that little round disk that I see against the surface of the planet is not the disk of that moon but only the spurious disk of a point source of light? Watch one and tell me thats what it is. If all I ever saw were the moons against the dark sky, I could maybe agree that the size differance was due to the magnitude differance in the moons, but when they enter or leave the shadow of Jupiter and you see what looks like a phase of the moon, or start or end a transit and you see that little disk against the surface of Jupiter, I find it hard to believe it's only a spurious disk.

You have not provided enough information in your question.
Are you talking about the moon disk itself?
What size scope are you refering to?

edz

Yes, I am talking about the moons disk seen against Jupiter in the background. I had not mentioned the shadow, but it too is a nice round disk that's about the same size as the moons disk when they are both transiting at the same time. I also made reference to a moon exiting or entering an eclipse from Jupiters shadow and being able to see what appears to be phases as the shadow darkens the moon. I am using my 4" TV102 refractor (@ 125x with 7mm & 175x with 5mm) and my AT66 ED refractor (@ 114x, much harder to see though).

Buddy

C-11 SCT, XT10i Dob, C-6 SCT, ETX125PE Mak-Cass, TV102, & AT66

"We the People are the rightful master of both congress and the courts - not to overthrow the constitution, but to overthrow the men who pervert the constitution." Abraham Lincoln"

The heavens declare the glory of God; the skies proclaim the work of his hands." Psalms 19:1

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I believe some of you could gain a better understanding if you would take the time to perform the geometric calculations. Several very knowledgable people in this thread have done so in the past. They have the background and have provided proper discussion here.

A 100mm scope is more than enough to resolve some of the moons. Those resolved will vary with the distance in AU to Jupiter.

A 66mm scope will acually show a spurious disk that is "larger than" that seen by the 100mm scope would show of a resolved disk, of even the 100mm image resolved of the largest moon.

The size of the spurious disk would be so large in a small scope (aproxx 2.5" in a 66mm scope) that you would easily think you are looking at a resolved disk. David and I have both engaged in previous conversations (one that comes to mind is Izar with 60-50-40mm scopes) that discuss the size of the spurious disk.

A 2.5" spurious disk partially hidden in eclipse by the planet shadow may easily be visible, given appropriate power, regradless of scope. Actually, it may be easier to see this in a smaller scope due to the fatteneing of the spurious disk.

Shadow transits much of the time do not appear round. In fact at the extreme limbs they are very much elongated and can be significantly larger than the moon.

Teach a kid something today. The feeling you'll get is one of life's greatest rewards. member#21
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Hi EdZ

I have done many calculations of the airy disk sizes in my different telescopes, but something is going on here, that the math of the spurious disk can't really explain.

Io is, with a diameter of around 1", much smaller than the spurious disk of a 6th-magnitude star as seen with an 80mm achromat, right? If so, it should not be resolvable, following the theory of the spurious disk.

But if this is so, then shouldn't Io always appear as a round dot, no matter what shape (crescent, etc) it really had, when being occulted by Jupiter or Jupiter's shadow? If we say that Io is going to be occulted by Jupiter, as I observed it, January 29th, 2002, would it then not be seen as an ever fainter, round spurious disk, on the edge of Jupiter, getting fainter as it was being occulted, but always staying round, since it is much smaller than the spurious disk the 80mm would show, given Io's magnitude?

This is not what I observed. Io was gradually covered by Jupiter and clearly showed itself as a little disk with a part if it missing, covered by Jupiter. The very last part seen was ( - shaped and appeared as a very slight bulge on Jupiter. Very, very thin. How can the theory of the spurious disk explain this observation?

I maintain my belief that we are dealing with something entirely different here. The Jovian Satellites are tiny disks with finite surface brightnesses and we are therefore dealing with MTF curves and contrast ratios here. The resolving abilities of a telescope of finite-sized objects with varying degrees of contrast is very different than when we are dealing simply with spurious disks of stellar objects.

If only it was clear outside, I could go out and try my various telescopes on the actual objects, under real-world conditions. Shame on me that I haven't paid much attention to the Jovian moons in the past.

Clear skies!
Thomas, Denmark

"You're not afraid of the dark, are you?" - Riddick "The best scientists are humble. They seek to understand, not to ensure their legacy, but merely to understand." - Mori

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"...But if this is so, then shouldn't Io always appear as a round dot, no matter what shape (crescent, etc) it really had, when being occulted by Jupiter or Jupiter's shadow? If we say that Io is going to be occulted by Jupiter, as I observed it, January 29th, 2002, would it then not be seen as an ever fainter, round spurious disk, on the edge of Jupiter, getting fainter as it was being occulted, but always staying round, since it is much smaller than the spurious disk the 80mm would show, given Io's magnitude?

This is not what I observed. Io was gradually covered by Jupiter and clearly showed itself as a little disk with a part if it missing, covered by Jupiter. The very last part seen was ( - shaped and appeared as a very slight bulge on Jupiter. Very, very thin. How can the theory of the spurious disk explain this observation?..."

That is similar to what I am talking about. What I saw last fall was not the moons disk being occulted by Jupiter, But an eclipse of the moon by Jupiters shadow. This occured out from the planet in space, where Jupiters shadow graduly covered the moons disk surronded by black sky. I could see what appeared to be phases where the shadow was only covering up part of the moon. You would not see that if it were only a spurious disk would you?

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I believe some of you could gain a better understanding if you would take the time to perform the geometric calculations. Several very knowledgable people in this thread have done so in the past. They have the background and have provided proper discussion here.

A 100mm scope is more than enough to resolve some of the moons. Those resolved will vary with the distance in AU to Jupiter.

A 66mm scope will acually show a spurious disk that is "larger than" that seen by the 100mm scope would show of a resolved disk, of even the 100mm image resolved of the largest moon.

The size of the spurious disk would be so large in a small scope (aproxx 2.5" in a 66mm scope) that you would easily think you are looking at a resolved disk. David and I have both engaged in previous conversations (one that comes to mind is Izar with 60-50-40mm scopes) that discuss the size of the spurious disk.

A 2.5" spurious disk partially hidden in eclipse by the planet shadow may easily be visible, given appropriate power, regradless of scope. Actually, it may be easier to see this in a smaller scope due to the fatteneing of the spurious disk.

Shadow transits much of the time do not appear round. In fact at the extreme limbs they are very much elongated and can be significantly larger than the moon.

But arn't those geometric calculation for a point source of light? I agree when we are talking about Izar, it's a star or point source. But do they apply when the object is not a point source? It appears to me that they may not apply to anything other than a point source. If my 100mm scope can only resolve the true disk of only the larger moons of Jupiter, that would mean that some are seen as true disks and some would be larger spurious disks, Correct? But that is not what i am seeing. Starting with Ganymede, the largest, what I am seeing are disk of varying size for each moon that is something less than the size of Ganymede. If you are correct, those that are spurious disk's would be larger than Ganymede, Correct?

Buddy

C-11 SCT, XT10i Dob, C-6 SCT, ETX125PE Mak-Cass, TV102, & AT66

"We the People are the rightful master of both congress and the courts - not to overthrow the constitution, but to overthrow the men who pervert the constitution." Abraham Lincoln"

The heavens declare the glory of God; the skies proclaim the work of his hands." Psalms 19:1

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If you are correct, those that are spurious disk's would be larger than Ganymede, Correct?

entirely dependant on the size of the scope

edz

Teach a kid something today. The feeling you'll get is one of life's greatest rewards. member#21
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If my 100mm scope can only resolve the true disk of only the larger moons of Jupiter, that would mean that some are seen as true disks and some would be larger spurious disks, Correct?

I thought that I did list the size of my telescope. It's my 4" TV102 (100mm) that you said could only resolve the larger moons.

Even so you did not address the origional question. Are not those geometric calculations for a point source of light? It would appear that they may not apply to anything except for a point source of light. How do those geometric calculations apply to an extended surface?

Buddy

C-11 SCT, XT10i Dob, C-6 SCT, ETX125PE Mak-Cass, TV102, & AT66

"We the People are the rightful master of both congress and the courts - not to overthrow the constitution, but to overthrow the men who pervert the constitution." Abraham Lincoln"

The heavens declare the glory of God; the skies proclaim the work of his hands." Psalms 19:1

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