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George
2006-Feb-22, 05:02 PM
Googling has not help me find the equation that converts phase angle into illumination. I am trying to obtain the reduction in flux due to either of these parameters.

Is there a resonable formula for spherical bodies? Does it vary much with each planet?

:doh:

grant hutchison
2006-Feb-22, 07:31 PM
You're looking for something called the phase function, which might be a useful phrase to Google. It describes how reflected flux varies with the phase angle. Unfortunately, different surfaces have different phase functions, so that clouds are different from ice, which is different from dust.
Lambertian and Lommel-Seeliger approximations are used: again, useful Google-fodder.

Grant Hutchison

Jeff Root
2006-Feb-22, 07:42 PM
Is the "limb darkening" seen in the Sun's photosphere included in
the phase function, or is it considered a separate phenomenon?

-- Jeff, in Minneapolis

Edit to add: Hey Grant, I have 334 posts, you have 668.
Exactly twice as many!

George
2006-Feb-22, 08:39 PM
Thanks Grant. I am searching now and have already gone "fadder in the fodder" with your advice.

Space.com (http://www.space.com/searchforlife/seti_fade_011101.html)has a smidgen.
"The phase function is how much light is reflected off the planet's surface with illumination angle. With our Moon the phase function is quite "steep." This means that the Moon gets much brighter during its full phase than with its quarter phase (the full Moon is 9 times brighter than the quarter Moon). This is because its surface is rough."
There still should be some reasonably simple equation which works for spherical bodies. I realize surface conditions will make a sizeable difference, yet I would guess the gas planets would not have this problem.

I'll keep a lookin'.

George
2006-Feb-22, 08:48 PM
Is the "limb darkening" seen in the Sun's photosphere included in the phase function, or is it considered a separate phenomenon?
Wouldn't it be an issue for closer objects only?

I also suspect the apparent diameter of the sun, or other large light source, also affect the phase function, right?

For instance, for Mercury, I learned, thanks to courteous help from JPL, that Mars at 99.5% illumination, still has a phase angle of 7.7 degrees.

grant hutchison
2006-Feb-22, 09:12 PM
Is the "limb darkening" seen in the Sun's photosphere included in the phase function, or is it considered a separate phenomenon?I'm not sure how easily the mathematics of these functions would transfer from illuminated bodies to luminous objects.


... Mars at 99.5% illumination, still has a phase angle of 7.7 degrees.Isn't that just the behaviour of any illuminated sphere? cos(7.7o) = 0.99. So one half of the disc is 100% illuminated and the other half is 99% illuminated. Result: 99.5% illumination.
But if you want to know the total brightness of that 99.5% disc, you need the phase function to tell you how the reflected light varies in brightness across the illuminated area.

Grant Hutchison

George
2006-Feb-22, 10:27 PM
I'm not sure how easily the mathematics of these functions would transfer from illuminated bodies to luminous objects.

Isn't that just the behaviour of any illuminated sphere? cos(7.7o) = 0.99. So one half of the disc is 100% illuminated and the other half is 99% illuminated. Result: 99.5% illumination.
Wow. You nailed it on the head! I checked the ephemeris, Horizons (http://ssd.jpl.nasa.gov/horizons.html), for several phase angles and obtained the same result. See, I sensed it shouldn't be too hard.


But if you want to know the total brightness of that 99.5% disc, you need the phase function to tell you how the reflected light varies in brightness across the illuminated area. Yep, enter the empirical, so back to the Google fodder. :)

I wonder if it is only tricky near 0 deg. Also, would the gas planets have much heiligenschein (http://www.sundog.clara.co.uk/droplets/heilig.htm)?

grant hutchison
2006-Feb-22, 10:37 PM
Also, would the gas planets have much heiligenschein (http://www.sundog.clara.co.uk/droplets/heilig.htm)?Might do. Think of the bright glow that sometimes surrounds the shadow of your aeroplane when it's cast on a cloud deck below.

Grant Hutchison

George
2006-Feb-22, 10:44 PM
Might do. Think of the bright glow that sometimes surrounds the shadow of your aeroplane when it's cast on a cloud deck below.
I don't suppose that is due to water properties, either. So, smoke would produce the "holy light", too. Hmmm...that might explain the expression..."holy smoke". [my bad]

grant hutchison
2006-Feb-22, 11:16 PM
I don't suppose that is due to water properties, either. So, smoke would produce the "holy light", too.Well, the gegenschein (http://antwrp.gsfc.nasa.gov/apod/ap990625.html) is "dry" heiligenschein on interplanetary dust. Water droplets, however, produce "wet" heiligenschein, which is accentuated by the reflective properties of the droplets.

Grant Hutchison

George
2006-Feb-22, 11:30 PM
Well, the gegenschein (http://antwrp.gsfc.nasa.gov/apod/ap990625.html) is "dry" heiligenschein on interplanetary dust. Water droplets, however, produce "wet" heiligenschein, which is accentuated by the reflective properties of the droplets.
Interesting. This is encouraging. This might make the brightness equation for phase angle that much less variable between objects.

I wonder if it is just a cos function as you figured, except for the first few degrees?

My hope is to tweak my little magnitude equation to match JPL's ephemeris. I am very close now with most, but I vary too much with Mercury, especially.

grant hutchison
2006-Feb-22, 11:50 PM
I wonder if it is just a cos function as you figured, except for the first few degrees?Certainly, years ago when I wrote a model of the solar system on my old Sinclair Spectrum, I just used (1+cos(α))/2 as my correction for phase angle when calculating planetary magnitudes, and it worked okay-ish for all but Saturn (where I had to add a factor for the ring tilt).

Grant Hutchison

Jeff Root
2006-Feb-23, 12:02 AM
Think of the bright glow that sometimes surrounds the shadow of
your aeroplane when it's cast on a cloud deck below.
Correct me if I'm remembering/surmising wrong, but that would
be caused by two related geometric effects, the first of which
you just described quantitatively: The fact that cloud droplets
nearly opposite the Sun are almost fully illuminated from your
vantage point, while droplets off to the sides are less fully
illuminated; and the fact that shadows cast by droplets nearly
opposite the Sun are mostly hidden by those droplets, while
shadows cast by droplets off to the sides have a good chance
of being in your line of sight.

-- Jeff, in Minneapolis

grant hutchison
2006-Feb-23, 12:12 AM
The fact that cloud droplets
nearly opposite the Sun are almost fully illuminated from your
vantage point, while droplets off to the sides are less fully
illuminatedBut droplets off to the sides are also fully illuminated: it's just that their backscattered light is missing your eyes, whereas the droplets exactly opposite the sun are scattering it back towards you. That's "wet" heiligenschein. The same effect is exploited by the glass beads on road signs, which throw the light of your headlights back towards you.
"Dry" heiligenschein is caused by the shadowing effect you mentioned, and presumably also has a role to play in most cases of wet heiligenschein, too.

Grant Hutchison

George
2006-Feb-23, 04:06 AM
Certainly, years ago when I wrote a model of the solar system on my old Sinclair Spectrum, I just used (1+cos(α))/2 as my correction for phase angle when calculating planetary magnitudes, and it worked okay-ish for all but Saturn (where I had to add a factor for the ring tilt).
The Sinclair Spectrum. :). I haven't heard that mentioned in 20 years. I remember seeing it, but never using one. [I saved my old Apple IIe, Odyssey, TI49a, and slide rule.]

I am sure your formulae works great, and does answer the thread title needs. My method is more cut and paste from the Horizons JPL ephemeris, which gives illumination percentage, too. So the problem is still how much brightness is reduced with illumination?

The brightness is, apparently, not linear with illumination. From what little I've found, it seems logarithmic.

Jeff Root
2006-Feb-23, 06:34 AM
But droplets off to the sides are also fully illuminated: it's just that
their backscattered light is missing your eyes, whereas the droplets
exactly opposite the sun are scattering it back towards you.
That's "wet" heiligenschein.
Oh, so that's what you meant by that.

It doesn't seem to be as strong an effect as the shadowing effect,
though-- Clouds in all directions look bright white where the Sun
is shining on them, and dark where it isn't. They're just a little
brighter opposite the Sun. I'm arguing (for no good reason) that
the droplets en masse appear to act as though they are white
to about the same degree as they act like they are clear.



The same effect is exploited by the glass beads on road signs, which
throw the light of your headlights back towards you.
Invented and made by 3M. And used in making my favorite movie
(2001: A Space Odyssey). Sheesh!

-- Jeff, in Minneapolis

grant hutchison
2006-Feb-23, 12:27 PM
It doesn't seem to be as strong an effect as the shadowing effect,
though-- Clouds in all directions look bright white where the Sun
is shining on them, and dark where it isn't. They're just a little
brighter opposite the Sun.I'm not sure that lets you diagnose which is the predominant effect. Dry heiligenschein is pretty weak with spherical objects like water droplets: it's much better with spikey dust, as on the moon or in interplanetary space. Wet heiligenschein is rather weak with airborne droplets: it works better with a higher refractive index and a reflective surface for the drop to sit on.

Grant Hutchison

George
2006-Feb-23, 03:40 PM
The following is a plot of the change in apparent magnitude for Jupiter as it approaches 100% illumination (phase angle per Grant's cos functon, which matches JPL's Ill. to phase angle values).

The "?schein" is not very apparent. This could be because it is not incorporated in the ephemeris calculations, or gas planets don't have much "schein", or I messed up.

http://img111.imageshack.us/img111/6287/jupiterschein3pn.th.jpg (http://img111.imageshack.us/my.php?image=jupiterschein3pn.jpg)

Mercury is much trickier, of course.

George
2006-Feb-25, 12:52 AM
But if you want to know the total brightness of that 99.5% disc, you need the phase function to tell you how the reflected light varies in brightness across the illuminated area.
If you're still around, JPL gave me the factor for Mars at .016*I. "I" is the phase angle in degrees, not Illumination. This correted the inverse square law equation I use nicely with a delta of < .01 magnitude.