Largest Full Moon an Ice-Bound photo opp ?

[Chip]

"The moon's average distance from us is about 238,855 miles (384,400 km). Tonight it will be just 221,560 miles (356,567 km) away. It will be 14 percent bigger in our sky and 30 percent brighter than some other full moons during the year, according to NASA."

It's just damp here, getting cold tonight and we'll see the same moon. I'm hoping some of you who are ice-bound do a little night hiking and get some great pics ! Good luck !

 

[DougPaul]

There is a prior thread in the photography forum: http://www.vftt.org/forums/showthread.php?t=26722

(It may only be visible to members of the forum.)

Info at http://news.nationalgeographic.com/news/2008/12/081211-biggest-brightest-moon.html.

Doug

 

[Chugach001]

Yep, my son and I skinned to the top of a local ski resort last night for a moonlight ski down and were just out in the yard. Wonderfully bright.

The math never quite works - it will be 7% closer so how could it possibly be 14% larger and 30% brighter? I remember staying up late at the Cirque of the Towers at around 10,000' in Wyoming for some super moon event and it was regular-good but not worth losing sleep over. Tonight is regular awesome plus 7% if you ask me.

J

 

[DougPaul]

The math never quite works - it will be 7% closer so how could it possibly be 14% larger and 30% brighter?

EDIT: The following analysis is incorrect. See my later posts in this thread. The photo, however, is still good.

7% closer => ~7% bigger diameter
7% bigger diameter => ~14% more area

14% more area => 14% more total reflected light
7% closer => ~14% more of the reflected light reaches a spot on the ground (inverse square law)
14% increase * 14% increase => ~30% increase

Doug

 

[Chugach001]

Hmm, well that makes sense.
Thanks!

 

[forestgnome]

a classic DougPaul response...

Excellent image of Moon!!!

happy trails

 

[moonrock]

7% closer => ~7% bigger diameter
7% bigger diameter => ~14% more area

14% more area => 14% more total reflected light

7% closer => ~14% more of the reflected light reaches a spot on the ground (inverse square law)
14% increase * 14% increase => ~30% increase

Doug

Hmmm (scratch-scratch):

That's only a 14(.5) % greater APPARENT area, as seen from Earth.
But the "total reflected light" from the moon does not change, because the moon is NOT (7 %) closer to the SUN.

The latter part (inverse square law) makes sense, if you treat the moon as a point source of light:
when the source is 7 % closer, 0.93 squared = 0.865 area of earth surface would receive the same moonlight (at least, from directly overhead) that 1.0 area would "normally" (on average) intercept.

[If EARTHlight (sunlight reflected from earth to moon and back to earth) were visible during a new moon (vs crescent moon, when we normally see it), it would (approximately) obey both of the above arguments, because the moon would be closer to both light source and observer.]

So the moon is not brighter; it's just closer.
I THINK that's right;
anybody who's taken physics (since 1978 !) care to weigh in ?

MR

 

[jniehof]

So the moon is not brighter; it's just closer.
I THINK that's right;
anybody who's taken physics (since 1978 !) care to weigh in ?

I think you're right. In radiative transfer terms, the specific intensity (ergs/sec/sq. cm/sr) of the moon remains the same. The only thing that's changing from our end is that the moon's filling a larger solid angle (the sr bit), which is inverse square.

DougPaul's explanation made sense to me at first, though, so it seems a likely source of the error.

(I took radiative transfer in 2005, IIRC. I think that satisfies your requirement.)

 

[Tom Rankin]

I noticed that the moon can vary from 29.3 to 34.1 minutes of arc. This is a difference of 15%. So squaring 1.15 ==> ~ 1.3. This might be the answer to the dilemma.

 

[nartreb]

"The moon's average distance from us is about 238,855 miles (384,400 km). Tonight it will be just 221,560 miles (356,567 km) away."

So it's about 7.8% closer than average, or about 15.6% closer than at apogee.

"It will be 14 percent bigger in our sky and 30 percent brighter than some other full moons during the year, according to NASA."

Note that they didn't say "than average". Probably there's only one full moon at apogee, so to get two or three ("some other") you need to compare against not-quite-apogee.

 

[DougPaul]

Moonrock is correct: my analysis is wrong--I was driven by attempting to get NASA's 30% brighter number, but I used incorrect data (7% closer rather than 12.3% closer) and ended up modeling the same effect twice.

The Moon was 12.3% closer than the max distance (356566km/406601km = .877 = -12.3%), 14% larger than minimum (406601km/356566km = 1.14 = +14% bigger*). So applying inverse square (1/(.877^2) = 1.30 = +30% brighter).

* This assumes that the tangent function is linear, which is approximately true for small angles.

Distances from http://news.nationalgeographic.com/news/2008/12/081211-biggest-brightest-moon.html

Finally found where I got the 7% from--Chugach001's "7% closer" not realizing that it was 7% closer than average rather than 7% closer than maximum.

We could also figure in the eccentricity of the earth's orbit (which has ~3% difference between max and min)...

Doug

 

[DougPaul]

"It will be 14 percent bigger in our sky and 30 percent brighter than some other full moons during the year, according to NASA."

Note that they didn't say "than average". Probably there's only one full moon at apogee, so to get two or three ("some other") you need to compare against not-quite-apogee.

I also found the statement "14 percent bigger" to be ambiguous: it wasn't clear to me whether they talking about the linear dimension or the area.

(It was linear dimension.)

Doug

 

[nartreb]

"The moon's average distance from us is about 238,855 miles (384,400 km). Tonight it will be just 221,560 miles (356,567 km) away."

I also have to point out the deceptive air of precision in this sentence. The distance is given down to the kilometer. In fact it must vary by more than 30km during the course of the night, as the moon will only be at its exact nearest point for an instant. (Change of distance of roughly 56,000 km over the course of a year => average change 152km per 24hrs. [I'm falsely assuming that the moon's perigee cycle corresponds to a solar year, but this is good enough for a ballpark ] Call a night 12 hrs (ballpark) and take into account that the moon "retraces its steps" in terms of distance on this particular night. ) Also, is this distance from center to center, or surface to surface? The phrase "distance from us" seems to me to imply the latter, in which case kilometer precision is ludicrous, since actual distance varies with latitude (and also with time == longitude).

Edit: from the NatGeo article: "The moon's farthest apogee for the year will occur a couple weeks later on December 26, when the natural satellite will be 252,650 miles (406,601 kilometers) from Earth. "

So my method of estimating the change in distance over one night was way too conservative. The moon will actually change distance by 50,000km in just fourteen days, so the change in distance during the night should be closer to 1,800km.

 

[Grumpy]

I never doubted that the NASA announcement -- 30% brighter and 14% bigger -- related to appearance. That was good enough for me to recognize that this event would represent an extra good opportunity to get a decent photo of a full moon, if the weatherman cooperated by delivering adequately clear skies.

I also think the inverse square law applies to reflected light sources. Any photographers here ever "bounce" a flash off a studio umbrella to illuminate a subject? What happens to your required f/stop when you double the umbrella-to-subject distance?

G.

 

[DougPaul]

I also think the inverse square law applies to reflected light sources. Any photographers here ever "bounce" a flash off a studio umbrella to illuminate a subject? What happens to your required f/stop when you double the umbrella-to-subject distance?

Strictly speaking, inverse square applies only to point sources (and is close enough if a source is much smaller than the distance).

One way of analyzing it would be to break the umbrella up into lots of little pieces, apply the inverse square law to each piece and add to get the total light. The sum would only be close to inverse square if the umbrella were far away compared to its size.

Doug

 

[moonrock]

Whatever the numbers say, it was pretty impressive, as was that photo.

BTW, the bright portions are (really-beaten-up) anorthosite, similar bedrock composition to the central Adirondack High Peaks.
But when you reduce the sun intensity to that seen on earth, lunar rock is reportedly dark gray - like asphalt.

Interesting statistic: 29.3 to 34.1 minutes of arc (My thumb is accurate to the nearest 2 degrees).
Amazing how horizon content can make the moon "seem" bigger, even though it's optically the same.

 

[Grumpy]

Strictly speaking, inverse square applies only to point sources (and is close enough if a source is much smaller than the distance).

One way of analyzing it would be to break the umbrella up into lots of little pieces, apply the inverse square law to each piece and add to get the total light. The sum would only be close to inverse square if the umbrella were far away compared to its size.

Doug

The other way would be to fall back on your experience. Doing that, if you have used this kind of lighting system you will realize the inverse square law works in recalculating f/stop (or flash power setting) when you change umbrella (light source) to subject distance.

Actually, you probably used a guide number system, which derives directly from the inverse square law and may be a little easier to apply in some cases. After determining the best base exposure for your umbrella-flash combo you can work directly from there.

Guide Number (GN) = (Light-Subect Distance in ft) X (f/stop used to achieve optimum exposure).

Say your best test exposure comes in at 1/60 sec at f/16 at ISO 200 with flash at half power and umbrella 4 ft from the subject. That produces a guide number (GN) of 64. Move the umbrella and flash out to 8 ft, and the exposure drops back to 1/60 sec at f/8 using the guide number. Now check that against the inverse square law.

You also will notice that the exposure difference (falloff) across the subject (on a plane from closest to farthest from the light) between highlights and shadows is less pronounced as you move the flash and umbrella farther back. (Assuming you've pretty well killed the ambient light in your studio.) That's also the inverse square law at work.

As a working photographer, I may know the underlying science, but tend to look at things like the inverse square law from their usefulness to me in practice. Guide numbers derived from the inverse square law are useful; for Q&D I also could remember that halving or doubling my umbrella-flash distance from the subject will produce a two f/stop (4X) difference in exposure.

G.

 

[jniehof]

But when you reduce the sun intensity to that seen on earth, lunar rock is reportedly dark gray - like asphalt.

The intensity of the Sun on the Moon is (to within atmospheric losses...which are relatively small) the same as on the Earth. But reflecting even a small fraction of sunlight is very bright. There's an interesting explanation of what makes lunar reflectivity complicated.

The Voyager image of Earth and Moon together probably gives the best reference for how the Moon "really" looks, since there's a point of reference in the frame. (Even then, the Moon had to be brightened.)

EDIT: Bah, NSSDC's page of Earth-Moon images is much better, including an original-brightness version of that.