Global position system based on position of moon?

May 30, 2007 3:11 AM

If I looked up at the sky some time at night, pointed to the moon and gave you the exact orientation of my arm (the two angles in the north/south and east/west directions) and told you the exact time of night and date it was, would you be able to tell my exact location in the world? If so, how?
posted by DirtyCreature to science & nature (39 comments total)

As long as I knew a few details of how you were doing it (you were pointing directly at the middle of the MMoon for example, and I knew how tall you were or that I could ignore your height), yes.

The position of the moon can be calculated or looked up. You've given a direction, and there's only one line that intersects the centre of the moon and the surface of the Earth. It does intersect the surface twice, except in edge cases, but that's resolvable as you'll only be able to see the Moon from one of those.

Depending on how accurately you want your answer, you'll possibly also need to know the shape of the Earth and do some slightly tricky mathematics to compensate for that, but it's not too much of an issue. You'll also need to measure your arm's position very accurately.

Essentially this is the reverse problem that a lot of telescopes solve every day. They know exactly where they are and when it is, and have to work out how to point exactly at some object. You know exactly when it is and where you're pointing, and need to know where you are.
posted by edd at 3:24 AM on May 30, 2007

I think we'd also need the angle of your arm from the perpendicular.
posted by bricoleur at 3:46 AM on May 30, 2007

On re-reading your post, I guess you meant to include that...
posted by bricoleur at 3:47 AM on May 30, 2007

Ok this is very very approximate but earlier at 9pm, it was very roughly 7 degrees north and 7 degrees west. Where am I?
posted by DirtyCreature at 3:54 AM on May 30, 2007

(from the vertical that is ...)
posted by DirtyCreature at 3:56 AM on May 30, 2007

You mean 2100 UTC? 9pm is a bit ambiguous for such questions.
posted by edd at 4:01 AM on May 30, 2007

Yeah, 9 p.m. where, exactly? Of course, that does sort of ruin your thought experiment.
posted by Civil_Disobedient at 4:08 AM on May 30, 2007

Wouldnt telling you the UTC give you a huge hint? Looks like I cant tell you the UTC time or youd be able to work it out just from that information.
posted by DirtyCreature at 4:08 AM on May 30, 2007

Tell us the time in GMT then.
posted by handee at 4:09 AM on May 30, 2007

Same goes for GMT wouldnt it? Now that I told you 9pm my time? Are you sure it can't be worked out using my local time?
posted by DirtyCreature at 4:14 AM on May 30, 2007

I think there would be a different place consistent with those angles in each time zone, so no.
posted by handee at 4:23 AM on May 30, 2007

It might be possible with your local time but it'd be a whole load harder. UTC only gives us the hint because you already told us it was 9pm local. We can 'forget' that. Plus I'm not going to work through the maths anyway. Here's how I would do it:

Get the position of the moon in alt-az coordinates. How far above the horizon and how far round from north in other words.

Get the position of the moon in RA and dec from here using your UTC time.

There are conversion formulae on the page I linked before in which you now have a and A (the alt and az), declination δ, and you want to know φ - your latitude. You will also get the hour angle H which will, given the RA α let you calculate your LST and hence your longitude.
posted by edd at 4:26 AM on May 30, 2007

Same goes for GMT wouldnt it? Now that I told you 9pm my time? Are you sure it can't be worked out using my local time?

GMT offers no hint of your location, all it does is remove the timezone from the time information.

If you want to offer only a local time, then it is possible to narrow your position to a line that plots around the earth for 24 hours, at which point the best you could do would be to see if the line intersects a major city and make an educated guess based on that.

This is because you haven't given the time at which you pointed.

If, on the other hand, you pick up the phone, place a call, and say "I'm pointing at the phone RIGHT NOW", then the recipient, no matter where in the world either of you are, has a time, and thus can give a location. Same goes if you provide a time without a location via GMT.
posted by -harlequin- at 4:34 AM on May 30, 2007

"I'm pointing at the phone RIGHT NOW"

That actually reads "pointing at the moon", not phone. Your computer display is faulty :)
posted by -harlequin- at 4:35 AM on May 30, 2007

GMT would have been a good time to give, but it looks like you didn't do that. How about UTC?
posted by Civil_Disobedient at 5:27 AM on May 30, 2007

By the way - it hasn't come up in this thread, but it's worth mentioning that it's an extremely popular misconception that just as the sun transits the sky during the day, the moon transits the sky during the night (and, thus, if someone says "I'm pointing at the moon right NOW", they must be in a location where it's nighttime).
posted by dmd at 5:43 AM on May 30, 2007

Look at the wikipedia page on celestial navigation. The short answer is that if you only measure off of one celestial body, you will know that you are somewhere on the LOP (line of position) of that body. You will need another measurement from another body to confirm your postion.
posted by Uncle Jimmy at 6:16 AM on May 30, 2007

Uncle Jimmy: That problem of only knowing you are somewhere on a LOP only happens if you don't know which way is north and only have the altitude of the celestial object I believe. The questioner does however know this.
posted by edd at 6:26 AM on May 30, 2007

It would be almost impossible to get the right angle with your arm, and even a second of inaccuracy (in the angles) could throw the location off by tens of miles.
posted by wackybrit at 6:34 AM on May 30, 2007

Dirty Creature,

You won't tell us what time zone "9PM" is?!

The moon doesn't move much, relative to a day. Imagine the moon at one point and the sun at another point and the earth near the moon. The earth spins, and at one direction of it (relative to where the sun is), every place on earth spins around to 9PM in turn.

The moon still hasn't moved much. To the extent that the moon did move over the course of every timezone spinning around to 9PM, then one could be able to tell where you are, but you'd have to pretty darned exact about where the moon is, and even then we'd give several possible answers because you're not on the exact 9PM line but somewhere in that few degrees of stripe that makes up your timezone.

So, if you're making a puzzle, don't say "9PM local time, which is UTC plus N", but instead say "K hours UTC". That makes it possible to solve without giving too much away.
posted by cmiller at 6:43 AM on May 30, 2007

Yeah, you've blown it by giving us local time. The problem with "local time" is that it's accurate with respect to celestial time only at one longitude within your time zone. Try again, and only tell us the UTC, or even better the UT1.
posted by nicwolff at 8:46 AM on May 30, 2007

This is how mariners used to navigate in the 18th and 19th centuries. But the moon is a terrible choice for this. First, it moves, which makes the calculation a lot more complex. Second, it presents a disk which is half a degree wide, which makes it nearly impossible to take an accurate fix -- what part of the moon do you use?

So they used stars instead.
posted by Steven C. Den Beste at 9:46 AM on May 30, 2007

Ok for those interested in the puzzle, I will do it again some time in the night. Just to make it more interesting so that you can't derive any information about the 9pm time frame, I will determine a time using a random number generator between the hours the moon is visible and if that time is in the middle of the night, I will set my alarm to wake up at that time to do the experiment. I will also tell you some time after the event so you can't derive any information from the time frame. All you need to tell me is what country I am in. City would be a plus but I doubt my measurement would be accurate enough for that.
posted by DirtyCreature at 12:32 PM on May 30, 2007

Use a sextant, to get accurate measurements of the angle.
posted by -harlequin- at 2:56 PM on May 30, 2007

Australia
posted by smcniven at 3:15 PM on May 30, 2007

smcniven: "Australia"

No it's not and I bet you only said that because I had posts about Australian cricket. I've sinced moved countries. You need to tell me why you think its the country Im in too.
posted by DirtyCreature at 4:31 PM on May 30, 2007

I don't think I understand what you mean by 7 degrees north and 7 degrees west from the vertical. If you want to give ALT/AZ or RA/DEC measurements, then maybe we can figure it out.

What is the purpose of the brain teaser? I don't get it.

I mean, the moon is full tonight, right? A full moon rises roughly at sunset and sets at sunrise anywhere in the world. Given the information that you told us (location of moon in sky and your local time) we could easily give you the phase of the moon.
posted by achmorrison at 8:13 PM on May 30, 2007

Has to be alt-az. RA and dec are on the celestial sphere - a star has the same RA and dec observed anywhere in the world - so it won't tell us anything that the UTC doesn't.
posted by edd at 2:28 AM on May 31, 2007

Sorry - unrelenting cloud cover - hardly got a chance to see the moon. But I would guess by observing what I did see of its path, that at around 11:00 UTC, May 31, it would almost be directly overhead - with a few degrees tilt (10 degrees maybe?) from the vertical towards the north (in other words, point vertically up to the sky and then rotate your arm 10 degrees to the north - thats where it should be approximately). Sorry I can't get much more accurate than that - I dont have equipment and am not familiar with the terminology and its been very cloudy. However, circumstances might make it pretty clear to you anyway.

Soooooo ..... which country am I in?
posted by DirtyCreature at 3:22 AM on May 31, 2007

It's almost directly overhead, and only off that in the northern direction, so that means its hour angle is 0 - directly overhead. This means your LST is about 1 hour, or about 1am, as the moon has a RA of about 1 hour at the moment (if I got that calculator I posted working - 16 degrees, and 15 is 1 hour).

Solving for your latitude I get about -35 degrees.

I think that puts you in South Africa, probably west coast.
posted by edd at 4:06 AM on May 31, 2007

edd: "Solving for your latitude I get about -35 degrees.

I think that puts you in South Africa, probably west coast."

Latitude pretty close. Longitude way off. But that might well be me and my crappy estimation.
posted by DirtyCreature at 4:15 AM on May 31, 2007

Well, it's unfortunately not the best placed object for locating you, since it's nearly directly overhead. That makes it very hard for you to measure the azimuthal direction - how round you have to turn round from north before you are 'facing' the object. It'd generally be better to pick an object far from overhead so you've got a much better sense of the orientation, but obviously you don't have much ability to relocate the Moon :-)
posted by edd at 4:32 AM on May 31, 2007

OP, why not pick a specific, definite star (North Star, etc) as the point of reference? For the purposes of your experiment it should be an equivalent test, save all the probable inaccuracies that are obviously resultant from using the moon as the point of reference.
posted by sprocket87 at 5:20 AM on May 31, 2007

edd: I know RA and Dec are fixed for stars, but the Moon moves in its orbit, so you could use RA and Dec for the Moon in this case. It may be harder, but it still could be done.

The original questions though: If I looked up at the sky some time at night, pointed to the moon and gave you the exact orientation of my arm (the two angles in the north/south and east/west directions) and told you the exact time of night and date it was, would you be able to tell my exact location in the world? If so, how?

The answer would be, yes, if you gave the exact position, down to the arcsecond in ALT/AZ of the Moon, then it could be done. Your examples have been way too vague so far to get even close.

As for how to do this, I would approach the problem in the following manner: There should be a set of three equations, for the three variables: altitude, azimuth and time. You make it harder, because you will only give us the local time. But, that just means that after I get an arbitrary "position" from solving the equations (which off hand I don't know how to set up, much less solve. They will involve spherical trig, I'm guessing...) I have to check 24 time zones to see which one has the Moon at your specified position at your local time. This part is easy, because any star chart program will have the moon position at whatever time I want.

The title of your post was the question: Global position system based on position of moon?

The short answer is no. The long answer is, yes if the measurement is accurate and given enough time.

I'm still curious as to what the point of the question is. You can tell a lot more info by looking at the Moon by including the phase of it. What are you trying to accomplish by having people guess where you are? It was obvious to me that you were in the Southern Hemisphere just because you said the Moon was a few degrees in the North, which you don't generally see in the Northern Hemisphere. But, other than that you are not specifying closely enough the position of the Moon in the Sky, I think.
posted by achmorrison at 9:28 AM on May 31, 2007

Erm. I did use the RA and dec in my calculation earlier. The point is that RA and dec are the same for any object at a given time, regardless of where you observe it from. It's the difference between that 'global' coordinate system and the local alt-az system (at a given time) that lets you determine the position of the observer.

So if DirtyCreature had only told us the RA and dec we'd at best only have a vague handle on the time he made his observation.
posted by edd at 3:44 PM on May 31, 2007

Okay, I can agree that's fair to say.

I was thinking about this question a little bit more today (and way more than I should anyway) and I realized that the position of the Moon, regardless of the complexity of the calculations, is a poor choice for a GPS reference, because anyone on the planet will not see it for roughly half the day on any given day and even less than that on the days when it is crescent or new.
posted by achmorrison at 3:58 PM on May 31, 2007

There is no point to this question - other than it feels kind of nifty that you could in theory locate where you are simply by knowing where the moon is at a given time and have a paper table of charts. (ie with no sophisticated equipment)

What I don't get is why you can't tell me where I am more easily? Surely there is only one place in the world where the moon appears directly overhead at any point in time?

For those still interested, I will give it another try tonight. Seriously the clouds were horizon-to-horizon last night. What I do know is the moon started way off around the east-south-east and almost rose to overhead some time in the night. I wont be able to give you extremely accurate measurements. However, I will try to find the time when the moon is exactly due east in the sky, and I will measure as closely as possible the angle a line drawn from the ground to the moon makes with the ground (or vertical if you prefer) using simple geometric measure devices.
posted by DirtyCreature at 6:33 PM on May 31, 2007

What I don't get is why you can't tell me where I am more easily? Surely there is only one place in the world where the moon appears directly overhead at any point in time?

Because you didn't tell us what that point in time was. You said 9:00pm, meaning your local 9:00pm. That leaves 24 time zones to check.

But, take your directly overhead example. Here's how I look at it. Yesterday, the Moon was full. Everyone in the world saw a full Moon. It rose roughly at sunset and set roughly at sunrise. Therefore, it was at it's highest point in the sky at midnight, no matter where in the world you are. If you said that at midnight (your local time) it was right above you, then I'm not sure I could guess where you are because I don't trust your measurement of the position in the sky. Small errors in measurement will lead to large inaccuracies of guessing your location.

The Moon's orbital inclination is roughly five degrees, give or take, I think. The Earth's axial tilt is 23 degrees. For the Moon to have been truly at your zenith, your latitude must be 28 degrees. Except, I'm willing to bet that it was not directly overhead. It's unlikely that can accurately estimate the position of objects in the sky to within several degrees without instruments.

It's an interesting question, no doubt. Thanks for asking.
posted by achmorrison at 11:52 PM on May 31, 2007

Sorry all. Wall to wall cloud again. Looks like the gods dont want you to know where I live.
posted by DirtyCreature at 12:14 PM on June 1, 2007