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Say, for sake of argument, someone was randomly transported in time and space. Would it be possible for them to determine their location on Earth and the current time using just observations of the stars?

  • They have no charts or equipment (not even anything to write with, so all calculations must be done in their head), but have a working knowledge of constellations and the location of other celestial bodies.

  • Their tolerance for accuracy is pretty wide:

    • The location can be as vague as "the north-west of the North American coast", "central Australia", etc. (other things, such as geographical landmarks and seasonal conditions can refine this, but we're not interested in that);

    • The time can be +/-100 years. (If it's possible to do better with mental calculations, then all well and good!)

If this is possible, could you outline the procedure for doing so?

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  • $\begingroup$ Dear Xophmeister - Great question - I added the experimental physics tag - ingeniuous methods of measurement will be of interest to an experimentalist $\endgroup$ – WetSavannaAnimal Aug 5 '13 at 14:45
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    $\begingroup$ Dear Xophmeister - You may well have done so already, but if you're interested in this kind of problem, read "Longitude" by Dava Sobel. She concentrates on John Harrison and his marine chronometer, and barely mentions Euler and Mayer, but hers is a wonderful summary of the problem and the people involved. $\endgroup$ – WetSavannaAnimal Aug 6 '13 at 0:59
  • $\begingroup$ @WetSavannaAnimalakaRodVance Added to my wish list :) $\endgroup$ – Xophmeister Aug 6 '13 at 9:07
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Century

First, you'd have to watch through a night to see if Polaris wobbles - currently, the radius is about 1° I think, but that changes with precession (and nutation, but that's small enough to ignore).

Once you know that, you can try to find a point in the sky that stays still all the time (like Polaris nearly does in our time). This is celestial pole, the direction the axis of earth points to. It moves in a circle with ~23° radius ( = obliquity of the ecliptic, currenty ~23,4°). The center of the circle is roughly between Polaris and Vega, that makes it easy.

Then, you have to see whether Polaris is approaching or departing it on its way on the "precession circle".
If you can measure the angle $\alpha$ it has travelled on its way on this circle, you can estimate which century you have been transported to: $year = 2000 + \alpha \frac {26000} {360}$
If you take exact values and do all the calculations and especially the measurements very very exact, you might get more than just the century.

If you really want to prepare, print this and pin it somewhere you will see it every day: https://en.wikipedia.org/wiki/File:Precession_N.gif

Beware that if you are transported more than 13,000 years back or forth, you will guess wrong.
In this case, you'll have to take the proper motion of the stars into account - you'll have to know not only the sky but also the motion of the stars (or at least a few) very very good. Which is the first thing that is really hard to do.

Maybe you'd have to invent something with sticks for the measurements.

Year

I guess you need to know the planets positions (especially Jupiter because it's bright and not to fast) of the time you are transported to to derive the year - but you'd have to be a walking ephemeris for this.

Time of the year

The time of the year is actually much easier than the year - most of the time, you can even feel it.
If you still need to calculate it, it is a direct function of the position of the sun in the sky:
You need to know the orientation of the ecliptic in the sky and the equinox of our time - in Pisces, you can see it in the image in that Wikipedia link: the intersection of "0°" and the ecliptic.
You measure the time in hours $h$ that passes from either the spring or the fall equinox (of our time) is at midheaven until the sun is at midheaven (noon).
The current date of the year, measured (roughly) from january to december in values from 0 to 1 is: $ y = h/24-0.2+\frac \alpha {360}$
If you took the fall equinox, you have to add or substract 0.5, whichever fits.
Without the $-0.2$, it would be measured from spring equinox (~ march 23rd) to spring equinox.

Geographical latitude

The angle of the celestial pole mentioned above and the ground (as long as it's horizontal) helps you calculate the geographical latitude $ \phi = 90°-angle$.

Geographical longitude

Sorry, that's the thing I could not do.

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  • $\begingroup$ It would be nice if you added some references for further reading. $\endgroup$ – babou Aug 5 '13 at 20:03
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    $\begingroup$ I added more detailed information, with links $\endgroup$ – Pharaoh Aug 5 '13 at 21:35
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This is likely stretching your requirements a little bit, but I find this related method ingenious and surprising.

You can actually find your longitude as well as the other data from @Pharoh's answer if you can remember very well how the night sky looked at your original position on Earth. The method is the Lunar Distance Method (see Wiki page with this name), and was proposed by Tobias Mayer and Leonard Euler (it had to be someone with a brain the size of a planet) as a solution to the longitude problem (see the Wikipedia page "Longitude Prize"). Here's how it works.

  1. You measure the "lunar distance" - the difference between the altitudes (spherical co-ordinate wontedly written $\theta$) of the Moon and some other well known celestial body.
  2. Crucial to the method is the fact: any observer anywhere on Earth who can see the two compared bodies (e.g. Moon and Regulus in the picture below) at the same time as you will observe the same value for this measurement.
  3. If you can recall what the same angle was at different times of day at your original location on Earth, you can therefore infer what time it is at your original location, to within about half an hour (the Moon's path across the sky is quite swift - it shifts about a degree (its own diameter) each hour. You can find your local time from the altitude of known stars, so you know can work your longitude out.

As I said, it's a bit of a stretch: in practice, the "lunar distance" as a function of time of day at Greenwich was tabulated in nautical almanacs. It is also a function of date, so the almanacs gave figures for hundreds of years. But if you had a memory for such things like Rain Man, you might pull it off.

enter image description here

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  • $\begingroup$ Rather than memorising an almanac, which seems a bit impractical, to say the least, would it not be possible to calculate/estimate the tables manually? (After all, tables don't exist for all time!) Either way, I may be misunderstanding, but wouldn't you need to know the date reasonably accurately -- at least within a few days, rather than a century -- for this method to work at all? $\endgroup$ – Xophmeister Aug 6 '13 at 8:59
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    $\begingroup$ I'm not sure: I think I agree with you that the dates would need to be known within a few days. You're absolutely right but I always said this was a stretch on your requirements: I just wanted to show that you can reckon longitude without a chronometer, which is something that I only learnt recently and is really quite surprising and ingenious. $\endgroup$ – WetSavannaAnimal Aug 6 '13 at 13:20
  • $\begingroup$ All the same, it was definitely worth mentioning -- it's pretty interesting and, as you say, ingenious -- thank you :) $\endgroup$ – Xophmeister Aug 6 '13 at 13:45

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