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Asbytec
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Re: Plato's challenge new [Re: Mare Nectaris]
      #6372315 - 02/12/14 01:13 PM

Quote:

But remember - there is one reference for even the most eagle-eyed observers: Chuck Norris does not *see* craterlets on Plato; they *reveal* themselves to him.






Chuck Norris trains a lot. Seek and ye shall find.


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mikewirths
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Re: Plato's challenge new [Re: NeilMac]
      #6372421 - 02/12/14 02:05 PM

thanks for the "likes" guys! I think I can do better this year (if the seeing cooperates) I've made some improvements, like a 1/30th wave secondary,better camera, and a Zeiss Abbe barlow

cheers

Mike


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Asbytec
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Re: Plato's challenge new [Re: mikewirths]
      #6373207 - 02/12/14 09:29 PM

Mike, according to measurements using quickmap, you are resolving craters down to 0.3 miles across. I have not done the math, but surely that's less than 0.5" arc - the average best the atmosphere can put up, they say. In terms of aperture, you are resolving down to 0.3 * 18, or 5.5/D(inches). That is totally consistent with observing 'g' in a 6" at 0.93 miles * 5.9" ~ 5.5/D as well. In fact, 'g' appears the same as your image. It's just a small with a dimmer floor as the smallest craters you imaged in crater form. In crater form, no less, and in excellent seeing, of course.

What's interesting is you are seeing shadows at about half that angular diameter. Pit shadows might be be crater form, but they are lacking the brighter rim. So, to see the rim, you need a larger crater consisting of a pit and a rim which means the angular diameter is larger than just a dark pit.

Now, that's interesting because I think I noted above having barely glimpsed dark pits on craters smaller than 'g' at ~0.9 miles. So, extended object resolution is a very complex thing depending on how you define it: crater form with rim, pit shadow only, or just a speck. And if so, then without doing the math, yet, 1.5 mile is about Raleigh limit. Down to about 0.75 miles would be 0.5R (the same empirical limit used for splitting very tight equal doubles.) I am not sure how 0.5R applies or correlates between extended object and point source resolution, but the thought they may have a similar empirical limit is interesting.

I think as long as the shadow and rim is 5.5D, you /could/ see it in crater form if seeing permitted. The aperture is capable of that, conditions might not be. At higher sun angles, the pit shadow shrinks and disappears. We loose the crater. We might resolve it in terms of angular resolution, but it's not visible to us because the contrast is gone. But, we should be able to see pits down to about 3/D because 5.5/D includes room for the brighter rim and the pit shadow. But just barely, fleetingly, and only in perfect seeing (cooled and collimated, of course.)


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azure1961p
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Re: Plato's challenge new [Re: Asbytec]
      #6373255 - 02/12/14 09:58 PM

Well it makes sense - though I haven't done it - knowingly. I've seen those pepper specks wavering but never made a concerted effort for measure. Though I can't join in on these observations due to a number of seasonal reasons - I'm saving it all for better sky's and snow free ground!

It makes sense - the same sense though of a different dynamic in resolving doubles as elongated down to .5/Raleigh.

Pete


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Asbytec
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Re: Plato's challenge new [Re: azure1961p]
      #6373325 - 02/12/14 10:29 PM

I think so, too, Pete, need to think on it some more. I believe 9/D is a great rule of thumb for what we can see on an average night. It's consistent with the big four being difficult at times. However, I think we can do better down to 5.5/D as an aperture - diffraction - limit rather than a seeing limit.

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Asbytec
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Re: Plato's challenge new [Re: Asbytec]
      #6373650 - 02/13/14 07:05 AM

Crunching some numbers for a 150mm aperture viewing the moon at 250,000 miles (400,000km) on 9 Feb 1125UT, Seeing 8/10 Pickering, Transparency 4/5, at 320x with 6mm TMB II.

Raleigh is 0.92" arc, 1.1 Miles or 1.8 Km
CO is 0.83" arc, 1.0 Miles or 1.6 Km

Dawes is 0.77" arc, 0.9 Miles or 1.5 Km
CO is 0.70" arc, 0.8 Miles or 1.4 Km

Lambda/D is 0.76" arc, 0.9 Miles or 1.5 Km
CO is 0.68" arc, 0.8 Miles or 1.3 Km

Raleigh/2 is 0.46" arc, 0.6 Miles or 0.9 Km
CO is 0.42" arc, 0.5 Miles or 0.8 Km
------------------------------------------------------------
Crater diameter below were measured from quickmap.

Crater 'g' is 0.67" arc, 0.8 Miles or 1.3 Km - Seen in crater form
Shadow is 0.54" arc, 0.7 Miles or 1.0 Km - Seen

Crater 'h' is 0.57" arc, 0.7 Miles or 1.1 Km - Larger not seen in crater form
Shadow is 0.43" arc, 0.5 Miles or 0.8 Km - Larger possibly seen once or twice as fleeting dark spot
------------------------------------------------------------
Seeing crater 'g' seems to correlate nicely with the CO modified Lambda/D limit just below Dawes. That appears to be impossibly small, yet it was surely seen regardless of what I think about it. Observing 'g' might actually make sense as this is where the FWHM of two 6th magnitude stars just begin to touch. It is the limit of resolution in that sense. I'm not sure of the correlation as the moon is an extended object. Still, it does not change the fact 'g' was observed and is reasonably consistent with the smallest craters Mike resolved in his aperture.

Possibly observing a dark spot at 'h' (presumibly the larger of the two) seems to /suggest/ resolution approaching 0.5R as modified. This is near an empirical limit for viewing very tight, bright, and equal doubles. Right now that is just interesting and not conclusive. But, it's interesting if 'h' shadow was about 80% of it's pit diameter on the date of the observation. (That figure was assumed for 'g', as well.) Despite 'h' being a double crater, I saw neither nor did I see them as one elongated form. Crater 'g', however, was seen clearly at least 3 times during this observation.

In any case, the observation of 'g' and lack of observation of 'h' seems to suggest the 0.1 mile difference in their diameter (as measured on quickmap and is an approximation) might define an extended object resolution limit near this angular diameter. And it's not so much about being eagle eyed as it is with the aperture and conditions. If it wasn't on the focal plane, it would not have been seen.

By the way, 'e' was held in crater form for long periods of time, 'g' was fleeting in crater form (as 'f' might be if confirmed), 'h' was not much more than a blurry something that /may have/ offered a dark spot from time to time. Crater 'i' was another potential dark spot about the same diameter as 'h'. Below makes some intuitive, progressive sense to me:

Crater 'e' at 1.0 mile across: easily held for long periods.
Crater 'g' at 0.8 mile across: fleeting, but crater form.
Crater 'h' at 0.7 mile across: the larger nor the pair not seen except as /possibly/ a rare and very fleeting dark spot only possibly glimpsed. I know that sounds weird as maybe the pair could be seen as one. Wish I could say they were, but they were not.

Edited by Asbytec (02/13/14 07:51 AM)


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azure1961p
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Re: Plato's challenge new [Re: Asbytec]
      #6373868 - 02/13/14 10:37 AM

Hi Norme,

Great report...

That E and G were held in crater form, ie; shadow, rim, brighter sunlit side of crater, would clearly suggest the 9/D rule is in need of some revision. I think it still holds merit but not with any particularly stringent threshold rule.
I know Ive seen crater "spots" well past but without certain measure how far.

I could see how H might appear as a transient blemish - that's the ragged edge Norme. The peril here though is in the double crater slurring together to form a pseudo crater form - somewhat like two threshold faint stars of fine seperation appearing as one faint star - brightness combined.

Harold Hill whose got the excellent portfolio of lunar drawings book in an interview I believe stated with his ten inch reflector he was gliding stars down to a third of a mile or so in Plato. It was a sky and tel article interview - wish I had the certain number - it was clearly under half a mile though.

The 80% pit shadow would seem to be a player here - got all crater involved Id think. I don't really know - the 80% shadow fill would seem optimum but at some point the low sun angle darkens the surrounding terrain so much its a matter of diminishing returns. The angle of the terrain could be a big player here too. Its something to ponder as there's a lot involved potentially to create that exceptional view.

Pete


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nirvanix
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Re: Plato's challenge new [Re: azure1961p]
      #6373900 - 02/13/14 10:52 AM

Quote:



Harold Hill whose got the excellent portfolio of lunar drawings book in an interview I believe stated with his ten inch reflector he was gliding stars down to a third of a mile or so in Plato. It was a sky and tel article interview - wish I had the certain number - it was clearly under half a mile though.

Pete




Thanks Pete,
What exactly did he mean by 'gliding stars'? I take it he spotted something like a speckle of light that he knew to be a crater? That gives me encouragement to go after the smaller craters in Plato when seeing allows.


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Asbytec
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Re: Plato's challenge new [Re: nirvanix]
      #6373925 - 02/13/14 11:16 AM

Yea, Pete, the low angle contrast is something I am still mulling over. That was very evident when Nirv and I observed Plato on the 8th. On that night, 'e' was faintly seen as a tiny fleeting crescent of light. All of them were. They were much cleaner the following night. On the 8th, just as Plato was leaving the terminator, the contrast between it's floor and the crater pits was just too low (for me) to see. However, the faintly lit rims could be seen.

I am not sure 9/D has to be revised. We just need to differentiate what it is telling us. It's a great rule of thumb for resolution in average seeing - an atmospheric limit. It is not a diffraction limit based on Raleigh or anything else.

That limit seems to be much smaller. If you perform the math above on Mike's image, you will get consistent results with a much smaller diffraction limit on the order of 5.5/D. That's what his resolution of craters 0.3 miles in diameter proves - it was on the focal plane. It was resolved. As for highly processed images, there may be a difference in processing bright low contrast objects like Jupiter and bright high contrast objects like the moon. But, Mike's images prove such small resolution is possible. I hope to have proved it can be seen.

Edited by Asbytec (02/13/14 11:48 AM)


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azure1961p
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Re: Plato's challenge new [Re: nirvanix]
      #6374233 - 02/13/14 02:04 PM

Quote:

Quote:



Harold Hill whose got the excellent portfolio of lunar drawings book in an interview I believe stated with his ten inch reflector he was gliding stars down to a third of a mile or so in Plato. It was a sky and tel article interview - wish I had the certain number - it was clearly under half a mile though.

Pete




Thanks Pete,
What exactly did he mean by 'gliding stars'? I take it he spotted something like a speckle of light that he knew to be a crater? That gives me encouragement to go after the smaller craters in Plato when seeing allows.




Glimpsing craters!!!! - gliding stars" was my creative auto-correct at work - lol - sorry for the confusion.
Pete


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David Knisely
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Re: Plato's challenge new [Re: Asbytec]
      #6374265 - 02/13/14 02:34 PM

After a little revision (and using the LRO PDS Archive Interface for better projection), here are the diameters of the craterlets in Plato that I labeled in my original article:

LUNAR ORBITER RECONNAISSANCE CAMERA CRATER DIAMETERS
FOR CRATER PITS INSIDE LUNAR CRATER PLATO
(diameters +/- 0.03 km)

Craterlet A: 2.48 km (1.54 miles), Craterlet B: 2.00 km (1.24 miles)
Craterlet C: 2.20 km (1.37 miles), Craterlet D: 1.91 km (1.19 miles)
Craterlet W (west wall): 3.13 km (1.94 miles),

Craterlet e: 1.73 km (1.07 miles), Craterlet f: 1.46 km (0.91 miles)
Craterlet g: 1.40 km (0.87 miles),
Craterlet h (triple crater feature 2.24 km x 1.19 km (1.39 miles x 0.74 miles))
components "h-1": 1.19 km, "h-2": 1.08 km, "h-3": 0.79 km

Craterlet i: 1.27 km (0.79 miles), Craterlet j: 1.09 km (0.68 miles)
Craterlet k: 0.95 km (0.59 miles), Craterlet l: 0.94 km (0.58 miles)
Craterlet m: 0.91 km (0.57 miles), Craterlet n: 0.87 km (0.54 miles)
Craterlet o: 1.10 km (0.68 miles)
Craterlet p (triple crater feature approx. 1.8 km x 1.5 km (1.1 mi. x 0.9 mi.))
(components: p1: 1.27 km, p2: 1.04 km, p3: 0.57 km)

Craterlet q (doublet): 0.78 km (0.48 miles)
Craterlet r: 1.19 km (0.70 miles)

Craterlet h is a triple (2.4 km x 1.19 km roughly), so it is difficult to resolve as anything but an irregular albedo feature. Again, you cannot use Dawes Limit for extended detail on the moon. All you can do is see what you can see. Clear skies to you.


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Sarkikos
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Re: Plato's challenge new [Re: David Knisely]
      #6374538 - 02/13/14 05:35 PM

Great info, David. Accurate to +/- 0.03 km! Thanks.

Mike


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Asbytec
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Re: Plato's challenge new [Re: Sarkikos]
      #6374603 - 02/13/14 06:16 PM Attachment (11 downloads)

David, that 'g' is closer to 0.9 miles makes sense, it's actually more consistent with Mike's image than being 0.8 miles. I got 1.3km (0.8 Miles) from rim to rim using quickmap. Thanks for the work to get a better measurement. That changes the results above slightly from obstructed lambda/D to obstructed Dawes criterion. (And it's actually consistent with my tightest double star with dark space split to date: STT507 is 7 and 8 magnitude pair reported at ~0.7" arc or maybe slightly wider near 0.72" arc.)

Thinking more about this, it may have been that Plato's floor actually improves the chance of resolution. If we can think of extended objects as an infinite number of spurious discs, then Plato's darker floor would consist of smaller discs than those comprising brighter lunar surface.

One would think at such small scales, however, that higher contrast would be needed in line with their spacial frequency. Indeed 'g's floor was faintly grey. Apparently we can get away with some unknown level of high contrast instead of 100% contrast required at frequencies near and above Dawes or Raleigh. I'm not sure what contrast level is required, but it does not seem to be 100%. Maybe just being a "high contrast" lunar object is good enough.


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azure1961p
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Re: Plato's challenge new [Re: Asbytec]
      #6374858 - 02/13/14 08:47 PM

Norme the dark floor would not be to an advantage however if the brighter partitions of the craft puts are that much darker as well.

Mike, Id think the error is greater than .03 simply and it has nothing to do with absolute accuracy of the image resolution but that ambiguous transition between plain, crater rim and crater floor. Fresher craters make easier work of this but its still has this null zone that can be challenging in knowing where to consistently draw the line. When you get into the details it gets hard to get better accuracy a lot of times than a quarter mile. Defined rilles can make this An easier task. My finds anyway. Id guess accuracy is generally between .10 and .20 depending in target.

Pete

Edited by azure1961p (02/13/14 08:48 PM)


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Asbytec
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Re: Plato's challenge new [Re: azure1961p]
      #6374897 - 02/13/14 09:07 PM

Yea, Pete, our findings can only be as accurate as our measurements of various features and their characteristics, especially as seen from a distance. I am not sure how accurate these tools are. For a 6", there is some variation of about 0.1 mile difference between Raleigh (1.0), Dawes (0.9) and Lambda/D (0.8). Raleigh could be 1.1 to 0.9, Dawes 1.0 to 0.8, etc. So, our best guess is to measure as best we can and work with as few significant digits as possible.

Still, so far, 'g' can be seen and 'h1' cannot. So whatever size they are seems to hint they define boundaries to the diffraction limit - until someone reports 'h1' as resolved. Then we convert them to some resolution criteria so we can apply them to different apertures. Then seeing decreasing with aperture at increased resolution begins to complicate things further.

Yea, it is all about having enough image contrast transferred to the focal plane, a certain object contrast, and the feature has to be large enough to be seen. For a diffraction limit, per experience and Mike's images, that seems to be near 5.5/D on a very good night. As for contrast, well, might need some way to measure it. However, empirically, one day out of the terminator, there was not enough contrast - for me. One more day's increase in sun angle provided enough contrast - for me, on that night, in the conditions present. So, whatever those levels of contrast were seems to define what is needed. I am sure it was not 100% during either observation.

Edited by Asbytec (02/13/14 09:29 PM)


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mikewirths
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Re: Plato's challenge new [Re: Asbytec]
      #6374939 - 02/13/14 09:24 PM

Hi Guys,

Interesting discussion! Of course any image is a static representation of many tiny moments of best seeing, so the visual experience is much more dynamic. But I wonder if these tiny "at the limit" features that are more spots than resolved craterlets are also bound by the quality of a given nights transparency (and not just how good the seeing is). A mountaintop observing session with equal good seeing compared to say a really good night in the Fla keys should be superior in terms of ferreting out the smallest low contrast spots.

just my 2 cents

Mike


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Asbytec
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Re: Plato's challenge [Re: mikewirths]
      #6374973 - 02/13/14 09:36 PM

Mike, great point...and as you can tell, I find the discussion fascinating, too.

Yea, transparency seems to be a factor, of course depending on how good or bad it is. You cannot see anything with overcast skies - exaggerating to make a point.

Particulates, as I understand it, cause humidity to become more opaque and scatter(?) light. This scattered light would seem to dim contrast. And if significant, likely could affect detection of smaller, grey crater floors presumably already hampered by diffraction contrast loss.

By the way, I'd think memory of that observation is much like your static image being that it consists of three still frames of 'g', as well.


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David Knisely
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Re: Plato's challenge [Re: Asbytec]
      #6375381 - 02/14/14 02:51 AM

Asbytec wrote:

Quote:

David, that 'g' is closer to 0.9 miles makes sense, it's actually more consistent with Mike's image than being 0.8 miles. I got 1.3km (0.8 Miles) from rim to rim using quickmap.




The "line" tool in Quickmap appears to underestimate linear size by as much as 3.7 percent. I use the PDS Archive Interface and then use the scale on the lower-left of the images to get diameters. Most of the problem with the diameter estimates is getting a precise location for the opposing rims, as with the highest resolution images, determining the crest of each rim can be a little imprecise (hence, the +/- 0.03 km figure, which might be a hair liberal, as the error could be as much as 0.05 km or even slightly more). Clear skies to you.


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Asbytec
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Re: Plato's challenge [Re: David Knisely]
      #6375385 - 02/14/14 03:03 AM

Agreed, David. It's not so hard to know resolution and estimate how large that would be at a distance. It's a little more difficult to measure something and determine resolution. Anyway, we do the best we can. I find this whole observing challenge and the discussion fascinating.

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azure1961p
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Re: Plato's challenge [Re: David Knisely]
      #6375593 - 02/14/14 08:48 AM

Quote:

Asbytec wrote:

Quote:

David, that 'g' is closer to 0.9 miles makes sense, it's actually more consistent with Mike's image than being 0.8 miles. I got 1.3km (0.8 Miles) from rim to rim using quickmap.




The "line" tool in Quickmap appears to underestimate linear size by as much as 3.7 percent. I use the PDS Archive Interface and then use the scale on the lower-left of the images to get diameters. Most of the problem with the diameter estimates is getting a precise location for the opposing rims, as with the highest resolution images, determining the crest of each rim can be a little imprecise (hence, the +/- 0.03 km figure, which might be a hair liberal, as the error could be as much as 0.05 km or even slightly more). Clear skies to you.





Exactly - its a little dicey but I did enjoy doing that on some areas . Id like to do it here but alas a lousy laptop.

Pete


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