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George
2009-Nov-16, 12:06 AM
I'd like to check my estimate on a luminosity question with those better at this than I am.

If I turned off the Sun, how many O-class stars (assume a 40,000 Solar luminosity for these) placed 3 light years away would be required to keep the sky appearing blue?

I went with the assumption that the sky is weakly blue at sunset, so I am using an air mass of 40. This may be pushing it, but not by much, I think. [I'm ignoring the Chappuis effect also, but it isn't critical for this calculation.]

The Solar constant varies with our our orbit, but let's use 1350 w/m^2 with 1000 w/m^2 being the amount that reaches our surface from the zenith (i.e. 1 Atm). This is a reduction of about 25% per Atm. [This is a bolometric value, admittedly, so it may be less attenuation for the visible portion of the spectrum, perhaps. Anyone know?] Let's use the 25% for now.

[I am also ignoring the fact that the O-class stars are far more prodigious in the blue end of the spectrum than the Sun is, but it won't be hard to adjust this value later.]

crunch... crunch...

I get 9 O-class stars required. Am I close?

bebe7
2009-Nov-16, 12:34 AM
I'd like to check my estimate on a luminosity question with those better at this than I am.

If I turned off the Sun, how many O-class stars (assume a 40,000 Solar luminosity for these) placed 3 light years away would be required to keep the sky appearing blue?

I went with the assumption that the sky is weakly blue at sunset, so I am using an air mass of 40. This may be pushing it, but not by much, I think. [I'm ignoring the Chappuis effect also, but it isn't critical for this calculation.]

The Solar constant varies with our our orbit, but let's use 1350 w/m^2 with 1000 w/m^2 being the amount that reaches our surface from the zenith (i.e. 1 Atm). This is a reduction of about 25% per Atm. [This is a bolometric value, admittedly, so it may be less attenuation for the visible portion of the spectrum, perhaps. Anyone know?] Let's use the 25% for now.

[I am also ignoring the fact that the O-class stars are far more prodigious in the blue end of the spectrum than the Sun is, but it won't be hard to adjust this value later.]

crunch... crunch...

I get 9 O-class stars required. Am I close?

Interesting question...reminds me off a question that went...if the Sun all of a sudden was totally not there, how long would it take for the earth to fall out of orbit?

George
2009-Nov-16, 01:12 AM
Interesting question...reminds me off a question that went...if the Sun all of a sudden was totally not there, how long would it take for the earth to fall out of orbit? I can't remove the Sun, only its yellow color. :)

grant hutchison
2009-Nov-16, 01:41 AM
Apparent visual magnitude of the Sun: -26.8.
Apparent visual magnitude of the full Moon: -12.6.
Apparent visual magnitude of an O5V at 3 light years: -10.9.
That means you need ~2,300,000 O stars to match the illuminance of the Sun, and about five to match the full Moon.

How blue do you want the sky to be?

Grant Hutchison

grant hutchison
2009-Nov-16, 02:24 AM
Here are a couple of ways you might come at it.

1) A patch of daytime blue sky an arc-minute across (matching the size of a diffraction-limited star to the naked eye) has a brightness equivalent to a visual magnitude of -3.5. We know we lose colour perception of stars somewhere between magnitude 2 and magnitude 3. So if we make that patch ~6 magnitudes dimmer, then that should take it to the threshold of blueness. That would imply a scattered source 6 magnitudes dimmer than the Sun, or about magnitude -21.

2) The illuminance from a clear sunset sky (which you're taking as faintly blue) is about 800 lx. A sky with a full moon in it comes in at 0.2 lx (including the moon itself). So we need something very roughly 4000 times brighter than the full moon: about nine magnitudes. That takes us up to magnitude -22.

So somewhere between visual magnitude -21 and -22 seems reasonable. Call it 10.5 magnitudes brighter than an O5V at three lightyears. So we need about 16,000 stars.

Grant Hutchison

George
2009-Nov-16, 04:37 PM
Apparent visual magnitude of the Sun: -26.8.
Apparent visual magnitude of the full Moon: -12.6.
Apparent visual magnitude of an O5V at 3 light years: -10.9.
A -10.9 star at 3 lightyears would have a visual luminosity gain over the Sun of about 15,000. This seems a little low. I see, however, that my 40,000 luminosity figure, taken from a star table, is a bolometric one. I think 25,000 would be ok to consider since I am aiming for the brightest stars.


That means you need ~2,300,000 O stars to match the illuminance of the Sun, and about five to match the full Moon. Yes, my numbers match (using about 15,000 as the luminosity of the O star).

Using a visual luminosity of 25,000 Suns, I get 1,440,000 stars required to match the Sun's brilliance upon our sky. 3 to match the Moon.


How blue do you want the sky to be? I'd like to start with the minimum luminosity for photopic vision. Perhaps even 1 cd/m^2, which is borderline with mesopic vision, I think.


2) The illuminance from a clear sunset sky (which you're taking as faintly blue) is about 800 lx. A sky with a full moon in it comes in at 0.2 lx (including the moon itself). So we need something very roughly 4000 times brighter than the full moon: about nine magnitudes. That takes us up to magnitude -22. I think 100 lux is more than enough to see color fairly easily, so perhaps 500x brighter than the Moon is sufficient, or -19.35 mag.

If so, this would require, I think, about 1500 O-class stars with luminosities of 25,000 Suns.

In my approach above, however, I tried to take the twilight sky intensity when the Sun is going over the horizon. I'm using 40 Atmospheres as the no. of air masses the light must pass (optical path?) to be overhead. [Is this too much since overhead is not ground level?]

I also used 1.35 as the number of times in brightness the Sun is above our atmosphere (1350/1000 in w/m^2), which is bolometric, admittedly. At AM40, this reduces the light by 163,400x, which is a 13 mag. reduction.

If this reduced level is close, then only about 8 O-class stars at 25k Sun luminosity are required.

Further, the blue to red photon flux ratio of a 40,000K star vs. the Sun is about 4 to 1 (using a Planck distribution). For a blue sky, this should cut the number of required stars by this amount. Thus, in the chance that my preceeding 8 star value is remotely correct, only 2 O-class stars may be needed for a very dim but blue sky.

grant hutchison
2009-Nov-16, 05:48 PM
I think 100 lux is more than enough to see color fairly easily ..Not really the issue, however, if I understand you correctly. We're not interested in whether we can distinguish colour by the integrated light of the whole sky, but whether we can discern colour in a patch of that sky.


Thus, in the chance that my preceeding 8 star value is remotely correct, only 2 O-class stars may be needed for a very dim but blue sky.So that's something in the vicinity of a full moon. Does the sky look blue to you when there's a full moon?

Grant Hutchison

grant hutchison
2009-Nov-16, 06:20 PM
I'd like to start with the minimum luminosity for photopic vision. Perhaps even 1 cd/m^2, which is borderline with mesopic vision, I think.One can go lower than that, with colour discrimination tests. However, I doubt if you're talking about being able to tell the difference between a greenish black and a bluish black. You want something that looks unequivocally blue, yes?
If that's your chosen value, it's a thousandth of the luminance of the blue daylight sky, which would mean we need to knock 7.5 magnitudes off the solar apparent magnitude: -19. 1,700 stars of magnitude -10.9, and then allow your jigger factor for spectra.

Grant Hutchison

hhEb09'1
2009-Nov-16, 06:22 PM
of a full moon. Does the sky look blue to you when there's a full moon?
Yeah, you'd expect there'd be a peak at blue no matter what the number, but it wouldn't be detectable with our traditional analog sensors.

George
2009-Nov-16, 06:42 PM
So that's something in the vicinity of a full moon. Does the sky look blue to you when there's a full moon Using AM40 and the liberal, admittedly, ~ 75% attenuation per atmosphere, we would need a -13.7 mag. Moon, or almost 3 Moons to produce a dim but blue sky.

Does the sky look blue when I see three full Moons? Yes, but the pink elephants might be a contributing color contrast factor. ;) [I'm joking because I don't really engage in the pink elephant making activities.]

I would guess that we wouldn't need 900 full Moons (using the 21 mag. estimate you gave earlier), but I don't know why your approach would not be correct.

George
2009-Nov-16, 06:47 PM
Yeah, you'd expect there'd be a peak at blue no matter what the number, but it wouldn't be detectable with our traditional analog sensors. Yep, and any camera with some exposure time will reveal this. Here (http://www.opticsforkids.org/teachersparents/articles/pdfs/what%20color%20is%20the%20night%20sky.pdf) is one example.

George
2009-Nov-16, 06:57 PM
One can go lower than that, with colour discrimination tests. However, I doubt if you're talking about being able to tell the difference between a greenish black and a bluish black. You want something that looks unequivocally blue, yes? Yes, genuinely blue, but near the lowest level of brightness.


If that's your chosen value, it's a thousandth of the luminance of the blue daylight sky, Why 1/1000?

grant hutchison
2009-Nov-16, 07:01 PM
Why 1/1000?You wanted 1 cd.m-2. The blue daylight sky comes in at 1000 cd.m-2. Turn the atmospheric illumination down 1000-fold, and you have the necessary reduced luminance of the blue sky.

Grant Hutchison

grant hutchison
2009-Nov-16, 07:20 PM
Using AM40 and the liberal, admittedly, ~ 75% attenuation per atmosphere, we would need a -13.7 mag. Moon, or almost 3 Moons to produce a dim but blue sky.In which case a gaze shielded from the moon, combined with a bit of dark adaptation, should show up a patch of midnight blue sky. It was an honest enquiry. :)

Grant Hutchison

George
2009-Nov-16, 10:46 PM
You wanted 1 cd.m-2. The blue daylight sky comes in at 1000 cd.m-2. Turn the atmospheric illumination down 1000-fold, and you have the necessary reduced luminance of the blue sky. Easy enough. :) [I missed the 1000 cd/m^2 as the normal blue sky value.]

The difference between this approach (1 cd/m^2), which seems the more appropriate one, and the AM40 attempt is ~ 5.5 mag.; not insignificant. The latter approach makes little sense because it puts us into the scoptic vision range. Perhaps the mesopic range could be a possibility.

I should be able to dig-up more info on sky illuminance this evening, but I think the 1 cd/m^2 will likely hold.

Based on this value for illuminance for a "midnight" blue sky:
For stars with a 25,000x Solar luminance, 40,000K temp. (4x blue photon flux increase):

At 3 lyrs... ~ 360 stars required
At 2 lyrs... ~ 160 stars
At 1 lyr.... ~ 40 stars

Thanks for the help, Grant!

George
2009-Nov-16, 10:48 PM
In which case a gaze shielded from the moon, combined with a bit of dark adaptation, should show up a patch of midnight blue sky. It was an honest enquiry. :) Midnight blue? :) This time of year, I hear where you're at it gets late early! ;) [I think this is a Yogi malaprop.]

Hornblower
2009-Nov-17, 12:27 AM
George if I understood you correctly, you calculated an estimate for the luminance of the overhead sky at sunset that was down by a factor of 163,400 from full daylight. I just now measured it with my exposure meter at the published sunset time, and it was down only a factor of 16. If I misunderstood you, please clarify what your reasoning was.

The blue tint definitely was weakened but I think it was due to atmospheric extinction of the blue end of the solar spectrum, not because of the reduced brightness. As I write this, I am measuring the brightness of pastel blue and beige surfaces illuminated only by my computer monitor across the room. It is down a factor of 10,000 from the daytime sky, and I can still see the blue distinctly. When I mask the screen down to a sliver about 1/40 of the full screen, I am getting a good approximation of full moonlight. Now the color is indistinct. I think I still see some blue, but it is not an unbiased test. I would say that my lower limit for reliable color perception is somewhere between these two levels, perhaps about 1/50,000 of the daytime sky brightness. That is in reasonably good agreement with the consensus in this discussion so far.

In my experience a clear sky in full moonlight looks more of a cool gray than a real blue.

neilzero
2009-Nov-17, 02:42 AM
Hi Grant: The blue luminosity/magnitude of an O5V star would be significantly less than the visual luminosity -10.9. The ultra violet, and Xray luminosity would be dangerous to humans, before we had enough O5V stars to produce a definite blue tint. Even the Equator of Earth might cool enough to snow oxygen and nitrogen. A few minutes after sun set the the sky is near full brightness at one mile altitude = at what altitude does the blue scattering mostly occur? The side effects of F5V stars would be less drastic, but we would need millions of them, unless they were closer than one light year.
What does the V mean in O5V? Neil

George
2009-Nov-17, 05:08 AM
George if I understood you correctly, you calculated an estimate for the luminance of the overhead sky at sunset that was down by a factor of 163,400 from full daylight. That was an extreme method where I tried using a reduction based on an AM40 attenuation at 0.75 per atmosphere.

The better method was using an illuminance of 1 cd/m^2, which is about the bottom of the normal photopic (color), not mesopic, range for the eye. We might cut this a little by taking into consideration the Purkinje effect, where the eye's rods perceive dim light as bluer than it really is. Perhaps we could argue that this effect would allow us to go to the lower end of the mesopic range, which is a little less than 0.01 cd/m^2.

We used 1000 cd/m^2 for normal blue sky, which meant we would cut luminosity by 1000 for the blue sky that was acceptable (i.e. 1 cd.m^2). However, now that I'm back at the house and have some references, a clear daytime sky is about 10,000 cd/m^2; which seems to match the value you just obtained. [1,000 cd/m^2 is correct, but for an overcast sky, apparently.]

Taking the product of the two values ( eye sensitivity and reduced sky brightness), we have a 1 million luminosity reduction allowable.


I would say that my lower limit for reliable color perception is somewhere between these two levels, perhaps about 1/50,000 of the daytime sky brightness. That is in reasonably good agreement with the consensus in this discussion so far. That is within the broad range.

Perhaps 100,000x, in lieu of 1 million, might be a fair maximum value.

So.... at 3 lyrs (189,720 AU), we need a star with a luminosity of about 36 billion (3.6e10) times that of the Sun to match the Sun. But we, hopefully, can still see a dark blue sky if it is reduced by 100,000x. This reduces our luminosity requirement to 360,000.

Since we are using a bright star of 25,000 Suns (visual), this puts us at only needing 14 of these stars.

A final reduction comes from the fact that there are about 4x as many blue photons in this light than found in sunlight due to our distant star's hot temp. of 40,000K. Yet this is a Planck comparison and the Sun does kick out slightly more blues than a true blackbody, so lets say our O-class star has only a 3x advantage.

The net result (in this round :) ) is the need for about 5 very bright O-class stars to produce a dark blue sky. [Perhaps even less if we can justify using an eye sensitivity near the bottom end of the mesopic range. If so, then only a single 12,000 luminosity star is needed.]

Is this about right?

grant hutchison
2009-Nov-17, 10:54 AM
We used 1000 cd/m^2 for normal blue sky, which meant we would cut luminosity by 1000 for the blue sky that was acceptable (i.e. 1 cd.m^2). However, now that I'm back at the house and have some references, a clear daytime sky is about 10,000 cd/m^2; which seems to match the value you just obtained. [1,000 cd/m^2 is correct, but for an overcast sky, apparently.]1000 cd.m-2 is correct for the blue part of blue sky, too. 10,000 cd.m-2 is what you get closer to the horizon, where the blue starts to be dominated by white. So it's a standard reference figure for photographers who have a lot of sky in the background, but I'm not sure it does the job you want.

Grant Hutchison

grant hutchison
2009-Nov-17, 11:12 AM
Hi Grant:It's George's scenario, so I'm not sure why these comments are addressed to me. But here goes:
The blue luminosity/magnitude of an O5V star would be significantly less than the visual luminosity -10.9.I guess that depends on what the zero-point for blue magnitude is. (Magnitudes are ratios, rather than absolute values.)
The ultra violet, and Xray luminosity would be dangerous to humans, before we had enough O5V stars to produce a definite blue tint.I think that's right.
Even the Equator of Earth might cool enough to snow oxygen and nitrogen.George should perhaps have kept the sun turned on and deployed his O stars during the night.
A few minutes after sun set the the sky is near full brightness at one mile altitude = at what altitude does the blue scattering mostly occur?Where the air is densest: ground level.
What does the V mean in O5V?It's the luminosity class. Class V is a main-sequence star.

Grant Hutchison

George
2009-Nov-17, 12:42 PM
1000 cd.m-2 is correct for the blue part of blue sky, too. 10,000 cd.m-2 is what you get closer to the horizon, where the blue starts to be dominated by white. So it's a standard reference figure for photographers who have a lot of sky in the background, but I'm not sure it does the job you want.
Agreed.

I think I was too conservative on the "blue" photon flux ratio and suspect 3.5x is more likely, yet still slightly conservative. [I suppose and actual spectrum would be nice to have for a 40,000 K star, but I suspect it will be very close to a Planck result for the blue range of its spectrum.]

So here's a possible range of blues using a max. blue sky of 10k cd.m-2; 25,000x luminosity O-class star; 40,000K temp.; 3.5x for blue photon ratio advantage; 3 lyrs. distance:

Using Grant's Blue sky level at 1000 cd.m-2.... ~ 41,000 stars needed. This would be slightly less bright than a normal bright blue sky.

[Using Hornblower's 50k reduction]
Blue sky at 0.2 cd.m-2....... ~ 8 stars needed. This would be a dark but somewhat blue sky.

Perhaps even one star (sky luminance of 0.02 cd.m-2) might cause blue to be seen in this dark sky.

George
2009-Nov-17, 12:45 PM
George should perhaps have kept the sun turned on and deployed his O stars during the night. It was just going to be a temporary shut-down for practice just in case some bird drops a bread crumb onto the Sun or other potential shut-down scenario.

Of course, I would expect funding due to the solution it brings to GW. ;)