I’m curious if, the iconic Moon drawings correspond with an actual latitude and longitude. Assuming the local time is midnight, where might this be? If not Midnight, when would this be accurate for the Northern Hemisphere? Here is one illustration at Amazon.com Amazon.com
It’ll never look like that from anywhere. The “horns” of the Moon are always on opposite ends of a diameter.
Yes, the shape of the terminator is wrong. Also there are stars visible through the unlighted part of the Moon, which is obviously impossible.
I understand the impossibility of a ‘Face’ on the Moon as well. I guess it’s just poetic license, as it were.
One thing you can sort of work with is the angle of the crescent in the sky. The lit side of the moon always points towards the Sun. This means that the crescent is pretty much symmetric about the line of the ecliptic in the sky (at least for naked eye visual purposes, the inclination of the Moon’s orbit adds a bit of wiggle room to this, but it is close enough). The ecliptic crosses the sky at different angles defending upon your latitude and the time of year. In equatorial regions you would probably never see the moon at the angle depicted as low in the sky as the iconic pictures present the moon. The horns would always be pointing closer to up or down when the moon was low in the sky. In very high latitudes the opposite is true. The ecliptic is low, and the horns will tend to be much closer to vertical, with seasonal variation giving you nearly 45º of change.
The pic linked to would suggest a passably high latitude. But again, lots of seasonal variation.
Nice artwork, but where’s the cow?
For the same reason, since the left side of the moon is lit and there are stars in the sky, one can conclude that it is early morning before dawn.
. . . or late evening after sunset, and the observer is in the southern hemisphere.
Which is how I always see it 
ETA. Rereading my description, it is a bit confusing. When I refer to the angle of the horns, I don’t mean the angle they roughly point, rather the angle of a line drawn between them.
Also note the lower left of the Moon is partially covering a cloud. Ergo a weirdly broken off fragment of the Moon is inside the atmosphere.
So, Wednesday, 7:38 UTC, Doomsday.
Also the moon’s nose looks like it’s around 400 miles long. I don’t know if the moon face requires oxygen (it seems it would if it has a “human” nose, though the nose could have evolved through the eons to be purely decorative, and function more for mating purposes now than respiration) A 400 mile long functioning nose would certainly be able to intake exponentially more oxygen than a normal size nose, perhaps sufficiently powerful to inhale enough of the trace lunar surface oxygen to make some use of it. The amount of oxygen required to keep the moon face alive may be much less than one might expect, since it is only a face, although somewhere within would need to be lungs of some sort, to power the nose.
If there’s a way to determine longitude just by looking at the earth’s moon, it’s news to me.
The ability to determine longitude from astronomical observations was valuable for hundreds of years, and even then, it was tricky to do at all and harder still to do so with much consistency. If one could determine longitude by observing the earth’s moon (as opposed to those of other planets), it would have been a huge deal prior to the invention of the shipboard chronometer. It would definitely have been described in Dava Sobel’s fab book, Longitude.
The traditional way to determine longitude at sea by observing the moon, without a chronometer, is to measure the angle between the moon and a planet or star and work out the exact time from that. It was no secret: “The method was published in 1763 and used until about 1850 when it was superseded by the marine chronometer. A similar method uses the positions of the Galilean moons of Jupiter.”, just more trouble than it was worth once accurate chronometers didn’t cost a fortune.
Most of you are answering based on the normal phases of the moon in a typical month. But the OP’s shape is a reasonable (read: exaggerated) of how it can look during a lunar eclipse. For more details, see Quercus’ link to XKCD.
It’ll never look like that during a lunar eclipse, either. The shadow of the Earth is bigger than the Moon, not smaller.
Right. That’s what I said.
The lunar distance method didn’t just require looking at the moon, but also measuring the angular distance between the moon and something else and then using a set of lookup tables to determine longitude. It was a hassle and obviously unworkable on cloudy nights. I was mistaken about “hundreds of years,“ but it’s been a while since I read the book I mentioned.