Starship Asterisk*

by **Chris Peterson** » Sat Jul 15, 2017 6:31 pm

neufer wrote:
Chris Peterson wrote:
neufer wrote:
let's take the APOD's full moon.

You claim that the APOD's full moon is dimmer than magnitude -4.74 ( = -12.74 + 40 x 0.2 mags)
when tiny Venus gets to magnitude -4.9 :!:

I don't buy it.
Why not? Using S&T's calculator, the expected magnitude of the lowest Moon image here ranges from about zero at 510 nm to -5 at 656 nm. That seems absolutely reasonable here. The magnitude we're talking about here isn't the integrated magnitude, it's the point magnitude.
I have no idea what "point magnitude" even means :!:

It means that these calculations look at simple intensity. You need to be careful how you apply them to non-point sources where the magnitude no longer reflects the intensity, but is integrated over a finite surface area.

by **neufer** » Sat Jul 15, 2017 6:24 pm

Chris Peterson wrote:
neufer wrote:
let's take the APOD's full moon.

You claim that the APOD's full moon is dimmer than magnitude -4.74 ( = -12.74 + 40 x 0.2 mags)
when tiny Venus gets to magnitude -4.9

I don't buy it.
Why not? Using S&T's calculator, the expected magnitude of the lowest Moon image here ranges from about zero at 510 nm to -5 at 656 nm. That seems absolutely reasonable here. The magnitude we're talking about here isn't the integrated magnitude, it's the point magnitude.

I have no idea what "point magnitude" even means

A zenith full moon is magnitude -12.74 because it is looms large in our sky (while Venus doesn't).

I appreciate your clarification that you are using S&T's calculator
with aerosol extinction which I have thus far ignored.

However, I don't think that aerosol extinction is all that important in the APOD:

Note that the middle moon in the APOD extends from roughly 0.25º to 0.75º above the horizon.

By my calculations:

1) 0.25º above the horizon corresponds to ~34 atmospheres of path
2) 0.75º above the horizon corresponds to ~26 atmospheres of path.

Hence the lower end of the middle moon encounters ~8 extra atmospheres of extinction.

Ergo: at 0.2 mags/atmosphere the the lower end of the moon should
be 1.6 magnitudes (~4+ times) darker than the top end of the moon).

This seems unreasonable to me just eyeballing it.

by **Chris Peterson** » Sat Jul 15, 2017 2:04 am

neufer wrote:
Chris Peterson wrote:
neufer wrote: Your just stated RED horizon extinction number of 8 magnitudes = 40 x 0.2 magitudes
would make rising Sirius appear as a red star dimmer than +6.4 magnitude.
Okay, that's reasonable for Sirius on the horizon, under clear coastal conditions. But you seem to think that the Egyptians observed the rising of Sirius on the horizon. That's not quite what a heliacal rising is. They noted the earliest day they could see Sirius before dawn. So let's consider something much more likely- typical desert transparency and a typical heliacal rising altitude of 3°. Now we're looking at an extinction of 2 mag at 620 nm. Or 2.5 mag at the extraordinarily low altitude of 2°. Which is consistent with seeing Sirius very low against an early dawn sky- AKA "heliacal rising". So I don't see the problem.
Okay, then let's take the APOD's full moon.

You claim that the APOD's full moon is dimmer than magnitude -4.74 ( = -12.74 + 40 x 0.2 mags)
when tiny Venus gets to magnitude -4.9 :!:

I don't buy it.

Why not? Using S&T's calculator, the expected magnitude of the lowest Moon image here ranges from about zero at 510 nm to -5 at 656 nm. That seems absolutely reasonable here. The magnitude we're talking about here isn't the integrated magnitude, it's the point magnitude.

The red absorption rate maybe somewhat more than the 0.05 magnitudes suggested by Sky & Telescope
but you have given absolutely no references for your sky high value of 0.20 magnitudes.

I'm using S&T's calculator, linked on the very page you provided! For the current summer conditions at Boston, the zenithal extinction is 0.42 mag at 510 nm, 0.26 mag at 656 nm. If we consider the yearly average conditions, it's 0.32 mag at 510 nm, 0.18 mag at 656 nm.

The same calculator, applied to my winter observing conditions, gives a zenithal extinction of 0.19 mag at 510, 0.10 mag at 656, which is consistent with the values I use when I correct my photometry measurements. So I'm pretty confident that the calculator is accurate (and since the underlying calculations are visible, I'm even more confident).

by **neufer** » Fri Jul 14, 2017 11:13 pm

Chris Peterson wrote:
neufer wrote:
Chris Peterson wrote: At sea level, under better than average transparency, the extinction for deep red (Ha) is around 0.15 mag.
Where our eyes are more sensitive (about 600 nm), it's around 0.2 mag.
Your just stated RED horizon extinction number of 8 magnitudes = 40 x 0.2 magitudes
would make rising Sirius appear as a red star dimmer than +6.4 magnitude.
Okay, that's reasonable for Sirius on the horizon, under clear coastal conditions. But you seem to think that the Egyptians observed the rising of Sirius on the horizon. That's not quite what a heliacal rising is. They noted the earliest day they could see Sirius before dawn. So let's consider something much more likely- typical desert transparency and a typical heliacal rising altitude of 3°. Now we're looking at an extinction of 2 mag at 620 nm. Or 2.5 mag at the extraordinarily low altitude of 2°. Which is consistent with seeing Sirius very low against an early dawn sky- AKA "heliacal rising". So I don't see the problem.

Okay, then let's take the APOD's full moon.

You claim that the APOD's full moon is dimmer than magnitude -4.74 ( = -12.74 + 40 x 0.2 mags)
when tiny Venus gets to magnitude -4.9

I don't buy it.

The red absorption rate maybe somewhat more than the 0.05 magnitudes suggested by Sky & Telescope
but you have given absolutely no references for your sky high value of 0.20 magnitudes.

by **Chris Peterson** » Fri Jul 14, 2017 7:35 pm

MarkBour wrote:(I'm guessing we probably don't equally look through 40 times as much oxygen, nitrogen, hydrogen, water vapor, ozone, etc. considered individually.)

Actually, we pretty much do, given that most of the causes of extinction are dominant in the first few kilometers of the atmosphere, so things stay pretty linear except possibly within a degree or two of the horizon.

by **Chris Peterson** » Fri Jul 14, 2017 7:22 pm

neufer wrote:
Chris Peterson wrote: At sea level, under better than average transparency, the extinction for deep red (Ha) is around 0.15 mag.
Where our eyes are more sensitive (about 600 nm), it's around 0.2 mag.
Your just stated RED horizon extinction number of 8 magnitudes = 40 x 0.2 magitudes
would make rising Sirius appear as a red star dimmer than +6.4 magnitude.

Okay, that's reasonable for Sirius on the horizon, under clear coastal conditions. But you seem to think that the Egyptians observed the rising of Sirius on the horizon. That's not quite what a heliacal rising is. They noted the earliest day they could see Sirius before dawn. So let's consider something much more likely- typical desert transparency and a typical heliacal rising altitude of 3°. Now we're looking at an extinction of 2 mag at 620 nm. Or 2.5 mag at the extraordinarily low altitude of 2°. Which is consistent with seeing Sirius very low against an early dawn sky- AKA "heliacal rising". So I don't see the problem.

by **MarkBour** » Fri Jul 14, 2017 6:49 pm

Thanks for the pointers and info. I'll go look at "extinction" this weekend. With a lot of luck, I may even gain enough understanding to follow the debate I unwittingly kicked off. Perhaps not, though. If you delve into the components and the whys of atmospheric extinction, I'm sure it gets very complex very fast. My initial post ventured a terribly crude estimate, which I now see is flawed in a couple of ways, for I was not imagining that toward the horizon one looks through 40 times as much air ("40 air masses", was the term used) as one does at the zenith. That's remarkable. And then, since the composition of the atmosphere changes with altitude, when one gets into the details, I'm doubtful it is a simple multiplication anyway, though that may be a decent approximation, for all I know. (I'm guessing we probably don't equally look through 40 times as much oxygen, nitrogen, hydrogen, water vapor, ozone, etc. considered individually.)

by **neufer** » Fri Jul 14, 2017 6:10 pm

Chris Peterson wrote:
At sea level, under better than average transparency, the extinction for deep red (Ha) is around 0.15 mag.
Where our eyes are more sensitive (about 600 nm), it's around 0.2 mag.

Your just stated RED horizon extinction number of 8 magnitudes = 40 x 0.2 magitudes
would make rising Sirius appear as a red star dimmer than +6.4 magnitude.

by **Chris Peterson** » Fri Jul 14, 2017 3:06 pm

neufer wrote:
Chris Peterson wrote:
neufer wrote:
You seem to be implying that much of the extinction is due to something other than Rayleigh scattering
and that the red light extinction rate is much higher than ~0.05 magnitudes per atmosphere
(or about 2 magnitudes on the horizon).
No. You seem to be inferring something that I'm not implying.
Chris Peterson wrote:
neufer wrote:
Please just tell us what the red light extinction rate is then.
At sea level, under better than average transparency, the extinction for deep red (Ha) is around 0.15 mag. Where our eyes are more sensitive (about 600 nm), it's around 0.2 mag.
I repeat:
neufer wrote:
You seem to be implying that much of the extinction is due to something other than Rayleigh scattering
and that the red light extinction rate is much higher than ~0.05 magnitudes per atmosphere
(or about 2 magnitudes on the horizon).
Your just stated RED horizon extinction number of 8 magnitudes = 40 x 0.2 magitudes
would put rising Sirius at less than +6.4 magnitude.

No, I didn't. As previously noted, that would only be the case if the formula for extinction yielded constant values independent of wavelength. At short wavelengths, the rising Sirius is dimmer than +6.4 mag. Its total intensity can be reduced by 11 magnitudes without it becoming invisible.

by **neufer** » Fri Jul 14, 2017 2:49 pm

Chris Peterson wrote:
neufer wrote:
You seem to be implying that much of the extinction is due to something other than Rayleigh scattering
and that the red light extinction rate is much higher than ~0.05 magnitudes per atmosphere
(or about 2 magnitudes on the horizon).
No. You seem to be inferring something that I'm not implying.

Chris Peterson wrote:
neufer wrote:
Please just tell us what the red light extinction rate is then.
At sea level, under better than average transparency, the extinction for deep red (Ha) is around 0.15 mag. Where our eyes are more sensitive (about 600 nm), it's around 0.2 mag.

I repeat:

neufer wrote:
You seem to be implying that much of the extinction is due to something other than Rayleigh scattering
and that the red light extinction rate is much higher than ~0.05 magnitudes per atmosphere
(or about 2 magnitudes on the horizon).

Your just stated RED horizon extinction number of 8 magnitudes = 40 x 0.2 magitudes
would put rising Sirius at less than +6.4 magnitude.

by **Chris Peterson** » Fri Jul 14, 2017 2:26 pm

neufer wrote:
Chris Peterson wrote:
neufer wrote: Sky and Telescope assumes that Rayleigh scattering constitutes about 90% of the (clear sky) extinction:
Did you try using their spreadsheet to calculate the actual numbers?
No. So what?

So you're not considering the complete physics behind extinction.

You seem to be implying that much of the extinction is due to something other than Rayleigh scattering
and that the red light extinction rate is much higher than ~0.05 magnitudes per atmosphere
(or about 2 magnitudes on the horizon).

No. You seem to be inferring something that I'm not implying.

Please just tell us what the red light extinction rate is then.

At sea level, under better than average transparency, the extinction for deep red (Ha) is around 0.15 mag. Where our eyes are more sensitive (about 600 nm), it's around 0.2 mag.

by **neufer** » Fri Jul 14, 2017 2:16 pm

Chris Peterson wrote:
neufer wrote:
Chris Peterson wrote:
You are misinterpreting things if you think that Rayleigh scattering is the only component of atmospheric extinction.
Sky and Telescope assumes that Rayleigh scattering constitutes about 90% of the (clear sky) extinction:
Did you try using their spreadsheet to calculate the actual numbers?

No. So what?

(I was born & raised a slide rule man and I don't need no stinkin' spreadsheet.)

Chris Peterson wrote:
neufer wrote:
Chris Peterson wrote:Your point here escapes me.
If red extinction on horizon were significantly greater than the Sky & Telescope Rayleigh scattering
amounts it would have been very difficult for the Egyptians to observe the rising of Sirius.

Your own stated horizon extinction number of 11 magnitudes
would put rising Sirius at +9.4 magnitude
No it wouldn't, because that's just the average extinction across the spectrum. As I noted, the wavelength dependence results in significant color changes as you go through more air mass. You can reduce the total intensity of Sirius by 11 magnitudes and still have a visible source at longer wavelengths.

You seem to be implying that much of the extinction is due to something other than Rayleigh scattering
and that the red light extinction rate is much higher than ~0.05 magnitudes per atmosphere
(or about 2 magnitudes on the horizon).

Please just tell us what the red light extinction rate is then.

by **Chris Peterson** » Fri Jul 14, 2017 1:58 pm

neufer wrote:
Chris Peterson wrote:You are misinterpreting things if you think that Rayleigh scattering is the only component of atmospheric extinction.
Sky and Telescope assumes that Rayleigh scattering constitutes about 90% of the (clear sky) extinction:

Did you try using their spreadsheet to calculate the actual numbers?

neufer wrote:
Chris Peterson wrote:Your point here escapes me.
If red extinction on horizon were significantly greater than the Sky & Telescope Rayleigh scattering
amounts it would have been very difficult for the Egyptians to observe the rising of Sirius.

Your own stated horizon extinction number of 11 magnitudes
would put rising Sirius at +9.4 magnitude

No it wouldn't, because that's just the average extinction across the spectrum. As I noted, the wavelength dependence results in significant color changes as you go through more air mass. You can reduce the total intensity of Sirius by 11 magnitudes and still have a visible source at longer wavelengths.

by **neufer** » Fri Jul 14, 2017 1:48 pm

Chris Peterson wrote:
neufer wrote:
So you disagree with Sky & Telescope :

[b][color=#0000FF]Rayleigh scattering Sky & Telescope[/color][/b]
Not particularly.

You are misinterpreting things if you think that Rayleigh scattering is the only component of atmospheric extinction.

Sky and Telescope assumes that Rayleigh scattering constitutes about 90% of the (clear sky) extinction:

http://www.skyandtelescope.com/astronomy-resources/transparency-and-atmospheric-extinction/ wrote:
<<Extinction at 510 nanometers, the bluish green wavelength to which human night vision is most sensitive, is what matters most to deep-sky observers. According to Daniel W. E. Green's article Magnitude Corrections for Atmospheric Extinction, for at observer at sea level viewing a star directly overhead, Rayleigh scattering reduces the star's brightness by 0.145 magnitude at this wavelength. (Other sources yield values closer to 0.140 magnitude.)

Ozone absorption removes another 0.016 magnitude, so the total extinction at sea level
is roughly 0.16 magnitude for a star directly overhead if the air contains no impurities whatsoever.>>

Chris Peterson wrote:
neufer wrote:
Chris Peterson wrote:
The extinction for red light, even deep red like Ha, never gets close to 0.05 magnitudes at sea level.
Violet extinction at zenith = 1.0 - 10^{(-0.28 x 0.4)} = 23%
Red extinction at zenith = 1.0 - 10^{(-0.05 x 0.4)} = 4.5%

Red extinction on horizon = 1.0 - 10^{(-0.05 x 0.4 x 40)} = 84%

Egyptians were able to observe the rising of Sirius after all.
Your point here escapes me.

My (clear sky) magnitude extinction numbers are consistent with Sky & Telescope's graph above.

If red extinction on horizon were significantly greater than the Sky & Telescope Rayleigh scattering
amounts it would have been very difficult for the Egyptians to observe the rising of Sirius.

Your own stated horizon extinction number of 11 magnitudes
would put rising Sirius at +9.4 magnitude:

Chris Peterson wrote:
At the horizon, you're looking through about 40 air masses, giving an attenuation of more than 11 magnitudes,
with a significant variation due to the dependence of Rayleigh scattering on wavelength.

by **Chris Peterson** » Fri Jul 14, 2017 1:16 pm

neufer wrote:
Chris Peterson wrote:
neufer wrote: The first Google hit for "atmospheric extinction" gives an extinction for zenith at sea level of about 0.28 magnitudes for violet light but only about 0.16 magnitudes for "510 nanometers, the bluish green wavelength to which human night vision is most sensitive." Red light appears to be around 0.05 magnitudes. (Hence there is a wide variation over the visible spectrum.)
No. This is a very small variation. The ratio is significant (about a factor of two between red and blue), but it only manifests once you have quite a few air masses. At the zenith you're looking at about a tenth of a magnitude between the two, which is visually insignificant in terms of altering the apparent color of an object. (The extinction for red light, even deep red like Ha, never gets close to 0.05 magnitudes at sea level.)

[b][color=#0000FF]Rayleigh scattering Sky & Telescope[/color][/b]
So you disagree with Sky & Telescope :?: :

Not particularly. You are misinterpreting things if you think that Rayleigh scattering is the only component of atmospheric extinction.

Violet extinction at zenith = 1.0 - 10^{(-0.28 x 0.4)} = 23%
Red extinction at zenith = 1.0 - 10^{(-0.05 x 0.4)} = 4.5%

Red extinction on horizon = 1.0 - 10^{(-0.05 x 0.4 x 40)} = 84%

Egyptians were able to observe the rising of Sirius after all.

Your point here escapes me.

by **neufer** » Fri Jul 14, 2017 11:03 am

Chris Peterson wrote:
neufer wrote:
Chris Peterson wrote:
What you're looking at is called atmospheric extinction. Google it for lots more information. It is very small at the zenith- at sea level, the extinction for visible light is only about 0.28 magnitudes, with only a tiny variation over the visible spectrum.
The first Google hit for "atmospheric extinction" gives an extinction for zenith at sea level of about 0.28 magnitudes for violet light but only about 0.16 magnitudes for "510 nanometers, the bluish green wavelength to which human night vision is most sensitive." Red light appears to be around 0.05 magnitudes. (Hence there is a wide variation over the visible spectrum.)
No. This is a very small variation. The ratio is significant (about a factor of two between red and blue), but it only manifests once you have quite a few air masses. At the zenith you're looking at about a tenth of a magnitude between the two, which is visually insignificant in terms of altering the apparent color of an object. (The extinction for red light, even deep red like Ha, never gets close to 0.05 magnitudes at sea level.)

: [b][color=#0000FF]Rayleigh scattering Sky & Telescope[/color][/b]

So you disagree with Sky & Telescope

:

Violet extinction at zenith = 1.0 - 10^{(-0.28 x 0.4)} = 23%
Red extinction at zenith = 1.0 - 10^{(-0.05 x 0.4)} = 4.5%

Red extinction on horizon = 1.0 - 10^{(-0.05 x 0.4 x 40)} = 84%

Egyptians were able to observe the rising of Sirius after all.

(The zenith extinction for 510 nanometer visible light could well be about 0.28 magnitudes in modern polluted skies, however.)

by **Chris Peterson** » Fri Jul 14, 2017 5:15 am

neufer wrote:
Chris Peterson wrote:
MarkBour wrote:I was looking for a diagram that showed the amount of reduction across the visible-light spectrum as one looks at it through increasing amounts of the Earth's atmosphere.
What you're looking at is called atmospheric extinction. Google it for lots more information. It is very small at the zenith- at sea level, the extinction for visible light is only about 0.28 magnitudes, with only a tiny variation over the visible spectrum.
The first Google hit for "atmospheric extinction" gives an extinction for zenith at sea level of about 0.28 magnitudes for violet light but only about 0.16 magnitudes for "510 nanometers, the bluish green wavelength to which human night vision is most sensitive." Red light appears to be around 0.05 magnitudes. (Hence there is a wide variation over the visible spectrum.)

No. This is a very small variation. The ratio is significant (about a factor of two between red and blue), but it only manifests once you have quite a few air masses. At the zenith you're looking at about a tenth of a magnitude between the two, which is visually insignificant in terms of altering the apparent color of an object. (The extinction for red light, even deep red like Ha, never gets close to 0.05 magnitudes at sea level.)

by **neufer** » Fri Jul 14, 2017 2:02 am

Chris Peterson wrote:
MarkBour wrote:
I was looking for a diagram that showed the amount of reduction across the visible-light spectrum as one looks at it through increasing amounts of the Earth's atmosphere.
What you're looking at is called atmospheric extinction. Google it for lots more information. It is very small at the zenith- at sea level, the extinction for visible light is only about 0.28 magnitudes, with only a tiny variation over the visible spectrum.

The first Google hit for "atmospheric extinction" gives an extinction for zenith at sea level of about 0.28 magnitudes for violet light but only about 0.16 magnitudes for "510 nanometers, the bluish green wavelength to which human night vision is most sensitive." Red light appears to be around 0.05 magnitudes. (Hence there is a wide variation over the visible spectrum.)

by **Chris Peterson** » Thu Jul 13, 2017 11:32 pm

MarkBour wrote:I was looking for a diagram that showed the amount of reduction across the visible-light spectrum as one looks at it through increasing amounts of the Earth's atmosphere.

What you're looking at is called atmospheric extinction. Google it for lots more information. It is very small at the zenith- at sea level, the extinction for visible light is only about 0.28 magnitudes, with only a tiny variation over the visible spectrum. At the zenith, stars look almost the same from the Earth's surface as they do from space. As the altitude decreases, the air mass increases and so does the attenuation. At 45°, there's still only about 0.4 magnitudes of loss, still too small to visually detect color shifts (but a concern for photometry). At the horizon, you're looking through about 40 air masses, giving an attenuation of more than 11 magnitudes, with a significant variation due to the dependence of Rayleigh scattering on wavelength.

by **Boomer12k** » Thu Jul 13, 2017 11:10 pm

Very nice.... AND.... "The Energizer Bunny Rabbit in The Moon"

:---[===] *

by **MarkBour** » Thu Jul 13, 2017 7:42 pm

(Prompted by noting the very familiar color change in the moon images.)

I was looking for a diagram that showed the amount of reduction across the visible-light spectrum as one looks at it through increasing amounts of the Earth's atmosphere. I found this page from California Polytechnic:
http://www.calpoly.edu/~rfield/Thermalstructure.htm
Which has a nice diagram at:
http://www.calpoly.edu/~rfield/Thermals ... age002.gif (the left image below),

The upper curve shows the spectrum at the point of reaching our atmosphere, and the lower curve the spectrum at the ground, when the Sun is directly overhead. The lower curve peaks near yellow, as we know, but the percentage difference from the upper curve to the lower curve is greater for the blue-violet end of the spectrum than it is for the red end. At the blue-violet edge, for example, it looks like about a 33% drop to the lower curve, whereas at the red end, about an 8% drop to the lower curve. I looked for a plot of the spectrum at sunrise or sunset, but could not find one. If one could predict it by simply doubling or tripling the percentage dropouts "across the board" to represent the greater amount of atmosphere in the way, then I have very roughly sketched in a couple of lower curves in the second diagram to represent this. (The two faint curves drawn in white on the image at the right.) So, this would produce a reddened image; I wonder how close my curves come to actual observations.

An interesting feature of some of the graphs I found was that some also showed the effect of going down under about 10 meters of water. The University of Colorado, http://lasp.colorado.edu/home/sorce/ins ... m/science/ had a figure at: http://lasp.colorado.edu/home/sorce/fil ... /fig01.gif

: Capture2.gif (42.02 KiB) Viewed 5399 times

This shows much more than the visible spectrum, so it is harder to tell what is going on, but it appears that water very quickly filters out reds far more than blues, so the light turns bluer as one goes down, but then, not very far below the surface, it is just going to get really dark throughout. This is obviously what people observe.

Anyway, I struggled with this some, and finally got back to what every school child knows.

As I said, though, It would be nice to have some better charts for this. Initially, I was wondering how you can tell atmospheric reddening from other kinds of reddening by looking more closely at the whole spectrum, but I had trouble finding even this much data on atmospheric absorption from a quick internet search.

by **heehaw** » Thu Jul 13, 2017 7:35 pm

And no mooning jokes, please!

by **ta152h0** » Thu Jul 13, 2017 7:47 am

not easy to do for an amateur with a handheld on a tripod and a 1000mm telephoto. I tried with soe success..

by **APOD Robot** » Thu Jul 13, 2017 4:08 am

Full Moon and Boston Light

Explanation: This well-planned telephoto timelapse captures July's Full Moon rise across outer Boston Harbor, Massachusetts, planet Earth. In the foreground, the historic terrestrial beacon is known as Boston Light. July's Full Moon is known to some as a Thunder Moon, likely a reference to the sounds of the northern summer month's typically stormy weather. But the eastern sky was clear for this video sequence. Near the horizon, the long sight-line through atmospheric layers filters and refracts the moonlight, causing the rising Moon's reddened color, ragged edges and distorted shape.

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