granthutchison wrote:chornedsnorkack wrote:To the contrary - this demonstrates the error in your calculations.
Seriously? You resuscitate a decade-old thread for this?
The alternative would have been post a completely new thread on Alpha Centauri, without reference to the inspiring thread.
granthutchison wrote:My "calculation" involved nothing more complex than dividing by 500, and you haven't demonstrated an error. In fact, you have the logic completely reversed - you should look for the flaws in your own argument if it comes into conflict with such a simple bit of arithmetic.
I admit that your calculation is simple. This makes it easier to search for the error in it.
granthutchison wrote:Here are the flaws:
1) You quote a value for illuminance, but the visibility of stars is a matter of luminance.
2) This is important because the light path in your example is completely different from the scenario under discussion - and the specific distribution of sky luminance strongly affects the resulting illuminance.
3) The difference between zenith luminance at sunset and daytime zenith luminance is not as large as the difference between sunset illuminance and daytime illuminance (see 2 above).
While the distribution is indeed different, it does give some hard constraints.
My calculation for comparison:
The illuminance with Sun at zenith, Sun and blue sky combined is 129 000 lux
Of which the diffuse illumination over the whole blue sky shines 20 000 lux into shades - the rest is direct rays from the disc.
The illuminance with Sun just under horizon is 400 lux.
This is 50 times less than the diffuse illumination of blue sky with Sun at zenith.
An unknown but allegedly small fraction of this 400 lux illuminance of horizontal ground with Sun just under horizon is the glow near horizon just above Sun.
It may be a small fraction, but it cannot be zero, much less negative.
We therefore get a hard upper bound on the luminance of zenith sky with Sun under horizon. Not more than 1/50 the daytime value, and can be less.
granthutchison wrote:4) The zenith sky at sunset would indeed be grey (specifically, it would have the same spectral distribution as the sun) if it were not for the enhanced absorptive effect of ozone because of the long horizontal light path at sunset. The blue zenith at sunset is the colour of ozone, not the result of Rayleigh scattering.
It would be white.
But the point is, the luminance of sky with Sun just under horizon is sufficient for colour perception. It is possible to see blue sky overhead, pink and yellow at various places near horizon.
This is not possible in moonlit sky. At full Moon, Rayleigh scattering in sky causes scattered light in sky outshining the ordinary skyglow, and hampering observation of dim stars and diffuse nebulae.
The light of Moon is nearly as white as sunlight - very slightly yellower. Rayleight scattering could be expected to scatter mainly blue light into moonlit sky with full moon at zenith.
It does. The spectrum of moonlit sky is provably blue.
Spectrum is - but the sky is not. The colour of moonlit sky is gray because it is too dim to see colour.
The sky illuminated by Alpha Centauri high in sky, however, at the estimate of 500 times dimmer than Sun, is at a minimum 10 times less bright than overhead sky with Sun just under horizon. And 800 times brighter than moonlit sky.
One argument of mine is that the luminance of sky with Alpha Centauri high in sky should suffice for colour perception - by comparison with twilit sky at similar luminances.
Now, back to your calculation.
As I argued, twilit sky is at least 50 times dimmer than daytime sky.
If stars magnitude -2,7 were visible in daytime sky then stars of at least +1,5 should be visible in twilit sky - as soon as the whole disc of Sun is under horizon. But are they?
Since the computation is clear and obvious, the only possible place for error is the value. Magnitude -2,7.
Maybe one issue is the definition of "visible". Large phase Moon is readily seen in daytime sky - noticed on casual scan of sky as a "cloud" that does not move and has its shape.
Venus, magnitude -4? Not so readily. Not unless you specifically look for it knowing where to look.
If at sunset, stars are visible to magnitude +1,5, then Jupiter is currently high in sky, far from Sun. And so is Aldebaran. Are they both visible at the immediate moment of sunset?
