There’s a basic guideline for shooting the moon... and there’s a more complex formulat that works out what happens when the moon (or any celestial object) is near the horizon.
First the basic guideline: The Looney 11 Rule
That’s the real name... I didn’t make that up. Looney being a play on Lunar.
Anyway, the moon is lit by the Sun. And since the rule for daytime exposures (shooting any subject in full mid-day sun) is called the “Sunny 16” rule (if you use f/16, then you’ll get a correct exposure if you set the shutter speed to the inverse of your ISO setting. E.g. at ISO 100 use 1/100th sec. At ISO 200 use 1/200th sec, etc.
You’d think you could use that rule, but it turns out the moon isn’t highly reflective. Only about 1/8th of the light that hits it (about 12%) reflects back. It’s roughly the reflectivity of an old black rubber tired (turned gray) or a worn asphalt road (not fresh). Also, you’re shooting thorugh the Earth’s atmosphere. A typical subject we shoot could be anywhere from a few feet to a few miles of air between the camera at the subject. For the moon, it’s the height of the entire atmosphere. Dust particles, moisture, and other scattering effects slightly dim the moon and all in, you need to add just about 1 full stop of light (going from f/16 to f/11 is one full stop of aperture).
So the Looney 11 rule says that if you set the aperture to f/11, then you’ll get the correct exposure if you set the shutter speed to the inverse of the ISO (e.g. at ISO 100, use 1/100th sec.). Most people tend to over-expose the moon by trusting their camera. The “problem” is the camera sees all the dark sky around the moon and assumes it needs to increase the exposure to find the details in all that blackness. So really you want to exposure *just* for the disk of the moon (e.g. if you used spot metering on the surface of the moon only, you’d get a better exposure than you would if you let the camera use it’s default ‘evaluative metering’ which tries to sample most of the frame.
If I could offer one single piece of advice... it would be to (a) know the Looney 11 rule and (b) use “spot metering” to meter the disk of the moon.
But be careful with (b) because that’s really only going to work well if your lens has a long focal length (short focal length lenses will meter more than the size of the sun itself and average in a lot of the background black sky resulting in an over-exposure.)
But you mentioned you thought possible the moon was a bit more red due to smoke or other effects that alter the transparency of the air. This effect is called “extinction”.
The concept of “extinction” is simple... but the rules to determine it are very very complicated. You can do a search for “atmospheric extinction” to find articles that will go into detail.
But the simple idea is this:
If you are at sea level then the amount of air directly above your head (drawing a line straight up to the top of the atmosphere) is 1 air mass.
If you were to draw a line at a 45° angle to the top of the atmosphere then you would be looking through at 1.4x air masses — you are no longer looking through the shortest possible distance to exit the atmosphere).
The atmosphere is mostly transparent... so this doesn’t really effects the exposure in any meaningful way (unless the air is really loaded with things making it less transparent ... dust, smoke, water vapor, etc.)
It turns out, if you had no trees or buildings on the land and had a perfectly flat view to the horizon... and looked at a “star” that was just at the horizon line... you’re looking through about 40 air masses (and that *does* add up in terms of impacting your exposure in a very noticeable way).
The complicated bit is that there no set rule in terms of how it impacts exposure because it really depends on the transparency of the air... and that varies.
If an object (such as the moon) is more than about 30° above the horizon, the effect is small. But the closer it gets to the horizon, ithe faster it ramps up. When an object gets below 10° above horizon, it starts to ramp up quickly... and below 5° it gets quite extreme (this is why, as the sun kisses the horizon at sunset, there’s a very noticeable difference in brightness from the bottom of the sun to the top of the sun and usually the sun is heavily distorted.)
Tim Campbell
5D III, 5D IV, 60Da
