analemma?

PS: notice that the person responsible for that has posted to that thread--and had another image at APOD just two days ago!

=D> =D> =D> , hhEb09'1 - or may we call you "1 mile"?

It sure made solving the question easier that we have an old thread that started to get going again a short while ago with just this subject and with a lot of participation on your part - or rather, your earlier incarnation... :wink:

Incidintally, the web site that gave me the idea was, of course, aristarchus'. I didn't know he also belonged to this pack.

So, what's next?

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I kinda like Grey's idea better, but hey it's not up to me.
I think he pretty much showed up when we first started discussing his websiteHere's one I was hoping you guys could answer for me, I don't know the answer: How much does the brightness of the Milky Way vary, across its image? I'm not even sure how to quantize that, so make something up--if you just use same-size areas at a certain level, I'd imagine a lot of the variance would be attributed to individual bright stars, but I don't know really.

That's nice, so we can still stick with ATP. Saves memory.

Ugh.... :-k

I venture this suggestion - it is extremely crude, but maybe a good first approximation:

Axel Mellinger has produced an astounding all-sky panorama of the Milky way. He used individual 35 mm slide frames which he stitched together producing the panorama. There is a technical article describing his method.

In this article you find a table of exposure times. Now assuming that the sky was a good mag 6 dark at every opportunity (sounds reasonable: Mellinger went to great lengths to obtain his shots from dark places); that the film wasn't plagued by Schwarzschild effect too much (sounds reasonable for the film used); and assuming that the limit of detection of a star on the film was a reasonably uniform mag 9-10 (Mellinger cites this value as "typical"), we can perhaps conclude that the brightness of the Galactic band differs by maybe not more than two to five times - Exposures, at least, differed by a factor of only two(22 to 45 min). Add to that a x2.5 for one magnitude difference in detection limit, and you get a variance of x5 - a bit more than two /f stops.

I suppose that this would give you an estimate of the "overall" brightness, as you are dealing with 28 mm images on 35 mm films - 51 segments in all; not some huge panorama into which thousands of individual images at, say, 200 mm, have gone. These would be influenced by individual bright stars, but probably not the wide angle ones.

What do you think?

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Non sunt multiplicanda entia praeter necessitatem.

I was thinking of Pace (it's Latin you know), but yeah that works
Please?

Well, see above - I chose to edit my last post with my idea instead of writing a new one - I am playing WAG so much I did not want to give too much reason to be accused of artificially upping my post counts... 8-[

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Non sunt multiplicanda entia praeter necessitatem.

So where are we here? Is there a question on the table or what?

Yes, there is, and an attempt at an answer. The third post up from yours contains both. It is not a quiz question though, but a genuine one. hhEb09'1 will decide which attempt serves him best.

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Non sunt multiplicanda entia praeter necessitatem.

I'll ask another. The analemma, as discussed above, intersects itself. What two dates? (and what is the astronomical significance, if any)

Was my answer that bad?

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Non sunt multiplicanda entia praeter necessitatem.

13 April and 30 August?

According to the diagram on this page, these two dates and the solstices are the four dates where the equation of time is zero.

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- There must be a new moon out, she said. He's always bad then.

One might add that these dates are not the equinoxes because of the combined effect of the Obliquity of the Ecliptic and the Law of Areas.

http://tinyurl.com/dbuvd

Which page is that Eroica? That url seems to point to the higher level frame.

The zero point can be any date (like choosing the origin) but I think if you choose one of the solstices to be zero, then the other can not be. Still, even the analemma there seems to show the crossing to be off the zero axis.
But the analemma itself (or, equivalently, the equation of time) is the result of that combined effect. If the orbit were a perfect circle, I believe that the analemma would cross at the equinoxes. So, in that sense, you can blame it on the Law of Areas, not the combined effect.

PS:I'm still looking at that. I'm trying to figure out, just by looking at that photo montage, how one area can only be a magnitude or two brighter than others. Some look awash in light, in that photo.

True, as shown on the plots at the URL I give.

ops: Summation Effect
You're right. I misread the curve. The other two points don't quite coincide with the solstices. The first one seems to be about a week before the Summer Solstice, and the later one is a few days after the Winter Solstice.

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- There must be a new moon out, she said. He's always bad then.

So who's turn is it?

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I don't know, but I'll ask one. This follows on from a point that was raised earlier (possibly in this thread, I don't recall):

What are the respective sizes of the shadows which the Galilean satellites cast on Jupiter during transits?

(Diameters of the umbras [Edit: make that shadows] to the nearest Km. Assume that the centres of Sun, moon, umbra and Jupiter are all in a line. Use average distances as given in the Planetary Fact Sheets.)

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- There must be a new moon out, she said. He's always bad then.

Io - 1822 km
Europa - 1568 km
Ganymede - 2634 km
Callisto - 2582 km

Edit typo (not the numbers)

2 nd Edit: these numbers represent the radiiof the respective shadows.

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Since the Sun is so much larger than the moons, shouldn't the umbras be smaller than the moons? For instance, during an annular solar eclipse here on earth, the umbra is non-existent.

](*,) Yes, of course! Strike that.

Any other takers? Arneb's figures are very different from my own (but that only means that at least one of us is way off! )

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- There must be a new moon out, she said. He's always bad then.