Astro Question arising from BlueMoment.com.

[Mark-1]

A man is standing on an unknown Island somewhere in the Northern Hemisphere. There is 24 hours of clear sky and enough ambient light to make out a horizon all the time.

He has no formal astro nav knowledge but attempts to find his position using nothing but common sense:

- He works out his lat from the angle of the Polar Star making no corrections.

- He works out his long from the difference between local mid day and Greenwich mid day, again making no corrections.

Assuming his measurement for the height of the north star & the time of local mid day are perfectly accurate, what is the worst case error in his position using this somewhat Heath Robinson method?

 

[bbg]

One of your assumptions is inconsistent.

Latitude will have an error factor because the Pole Star rotates very slightly around the pole. I would have to go home and check some references and make some calculations to figure out what the range of error would be, but there would be one.

As for longitude, if his "measurement ... of local mid day" is "perfectly accurate", then there will be no corrections necessary, and there will be no error in this component. But I have no idea how he would make such a measurement using "nothing but common sense".

 

[BlondeBimbo]

Not really sure of the question, if it is a general one, then the pole star could be out by 1/2 to circ 1 ¾ deg at various GHA, with the lat and month of year adding up to 2 mins depending upon one’s position in terms of Lat.

Of course there will be further errors due to the latitude and refraction which may subtract say 5 mins as well.

If you are being a little more specific with the being able to see etc, then the total error is likely to be ~ 50mins in Lat (i.e. assuming you are near the pole) as the refraction will be zero (ish)

In terms of Long, again this depends upon the time of year as local meridian can differ from GMT by up to 15 mins in time (4 deg) due to the equation of time.

So up to 4 deg left to right and 50 mins up down

In terms of nautical miles assuming a Lat of > 70deg 4 degLong = 82 nm (or so) and 50 mins Lat = 50 nm (ish)

Too many guesses without really understanding the question though.

My tupenth

BB

 

[sternsheet]

Isn't this a question of Arc to Time? We can, I think, for the purposes of this question assume that the Pole Star variation is insignificant to the answer. There will be a variation, as bbg correctly states. But I don't think this is the point of the question.

What will vary is the difference between mean time and apparent time. Local mid-day will be given in mean time, which may or may not be the same as local mid-day apparent time (see the difference for meridian passage for that day). Whilst all points in that time zone will have the same mean time, there will be a 15 degree difference in apparent time, or 15 x 4 mins (60mins) between each boundary of the time zone. Arc to time for 60 minutes is 15 degrees, so around 15 degrees error max?

So one might assume the answer being looked for is 60 minutes (plus or minus the meridian passage difference for that particular day).

 

[Simes]

BlondeBimbo is right

Latitude can be guessed at from Polaris. So a degree of error is about right.
Using LAN (Local Apparent Noon) would give your position to within a round 26" (I am working from memory here). To correct this further you would need to apply the "Eqn of Time" from NP 136 to adjust "Mean Time" to the correct time that the Sun is overhead at Greenwich.
So you should be able to position yourself to within 4 degrees of Long and around 1 degree of Lat.

Simes

 

[Bilgediver]

To correct this further you would need to apply the "Eqn of Time" from NP 136 to adjust "Mean Time" to the correct time that the Sun is overhead at Greenwich.

I don t think Man Friday has a copy of NP 136 to lend him

 

[Lady Campanula]

Well, let's see....

We'll need to specify some assumptions. The first of these is that an optical instrument is available for the measuring of the angular height-above-horizon of stars. Lets' call this a sextant. If none were available, it would be appropriate to make one.....

The second assumption is that we have a good timepiece, and sufficient time available to carry out the following tasks.

We'll also want to construct a 'gnomon' - a true-vertical post ( plumblines ), and have a couple of other, shorter ones to hand ('gnomes'. We use these, aligned with the direction of Polaris ( like the foresight and rearsight of a rifle ), to define the direction of True North and, by extension, the direction of True South - which is on our own Local Meridian.

Due to Polaris being positioned NOT QUITE above the Pole ( Extended ), but revolving apparently around it in a small circle, of radius ~41', as the Earth rotates, that direction defined by the 'gnomon' pole and a 'gnom' will change cyclically by 1°22'. One should be able to determine the 'mean' direction....

Alternatively, if one can identify the bright star Kochab ( extend arm and 4 fingers, positioning one finger on Polaris. The end of the little finger - 16.5° - should overlay Kochab ), then wait until Kochab is 'level-abeam' Polaris, then the cyclic ( or 'Q' ) error amounts to nil, and Polaris is Due North. Plant a stick, as before, defining North and South. More of this later.....

With respect to Polaris, Latitude = Height Observed above the horizon, subject to several corrections. These are 'Sextant Index Error', corrections for 'Dip of the Horizon' and 'Refraction' ( often given as a combined 'Total Correction' - more later ).

Measure the Height of Polaris when Kochab is 'level-abeam' or at '3/9 o'clock', and there is negligible 'Q' error. Correct for 'S. Index Error' and 'Total Correction', and you have your Latitude. Do this several times, as all good trad navs do, and average your calculation.

The sextant Index Error can be determined in an earlier quiet moment. What about 'Q', which varies as 'height of eye' above the sea and the 'observed altitude'?

Seemples! Draw a line representing 9° to 90° Observed Altitude, and sub-divide it in quarters. Use 'Total' corrections as follows +6' ( for first quarter 9°-20° ), then +3.5', 3', +2.5' respectively....

And the two-three gnom'poles defining South? Calculate Longitude by Equal Altitudes as described in an earlier post, and use the poles-defined direction of Due South to refine your judgement of when the Sun crosses your Local Meridian, which is Local Noon, and so exactly when to time-by-timepiece the sun's maximum altitude.

Or ask a passing Polynesian.....

( With acknowledgments to 'The Lo-Tech Navigator' by Tony Crowley )



Edit: Revisiting this, I 'reckon' that an averaged set of measurements, as above, should produce an 'Ellipse of Probability' about 4.3nm by 7.6nm. There are, of course, arguments that purport to show that the probability of being within the triangle bounded by three position lines is significantly lower than being outside, so it's probabalistically better to work with a Two-Position Line Fix ( average of several sights, each ) than go for three or more - but that's a 'geeky nav' kind of question well suited to a rainy Sunday afternoon with only Scotland V Ireland on the box....

Next?

 

[nial]

common sense

This is a trick question. If someone is stupid enough to find himself on an island that far North and furthermore not even know which damn island it is then he doesn't have any common sense.
nial

 

[macd]

If someone is stupid enough to find himself on an island that far North

The question gives no suggestion of 'far north', simply somewhere north of the equator. 24 hours of clear sky is not 24 hours of daylight. For all we know the island could be Sheppy.

 

[ARO]

... attempts to find his position using nothing but common sense:

Some questions:

1. How does one measure the hight of Polaris using "nothing but common sense"?
2. How does one measure time using "nothing but common sense"?
3. How does one find GMT using "nothing but common sense"?
4. How does one find local noon using "nothing but common sense"?

Even if a timepiece is available (much more than just common sense):
Which time does it give? Local time or GMT?

The minimum equipment needed to fix the position is a timepiece adjusted to GMT and some instrument to measure an angle. Common sense is simply not enough to solve the problem.

Using only the data provided in the question there is no answer to it.