PPRuNe Forums - View Single Post - Navigaion -determining position with ref. to the Sun/Moon
Old 29th January 2006 | 08:28
  #8 (permalink)  
chornedsnorkack
 
Joined: Aug 2005
Posts: 834
Likes: 0
From: Estonia
Originally Posted by Empty Cruise
A crude sextant can be constructed using a protractor & a piece of string with a bob @ the end. When approaching midday, don (several of) your sunglasses and aim the protractor at the lower edge of the sun disc, then take note of the angle between the protrctor & local vertical (the piece of string & its bob).
Next, correct for the suns declination. At summer solstice, subtract 23,5 deg. in the nortern hemisphere, winter solstice add 23,5 deg, and at equinox, the correction is zero. If an almanac is at hand, it'll give you the exact declination, otherwise a sine curve drawn to connect these points in a grid will do nicely. This procedure corrects for the earths axis being offset 23,5 deg. from the ecliptic plane.
The value you arrive at is called the "co-lattitude" - subtract it from 90 to get your present lattitude.
But you cannot determine if it is midday unless you already know your present longitude!
Originally Posted by Empty Cruise
If you have a watch that runs to any known timezone, you can also work out your longtitude by noting exactly at what time you measured the maximum sum elevation and appplying a correction of 15 deg. lat per hour difference (all time zones start at the longtitude that is divisible by 15, apart from the Z time zone, which obviously is zero).
But if you are moving at a significant speed, the elevation of the Sun is changing both because it passes culmination AND because your latitude is changing. This would throw off the estimate of midday.
Originally Posted by Empty Cruise
And if you have an almanac, a bubble sextant & a decent calculator (plus a high frustration threshold) - you can get a position line from the moon But if you're so well equipped, I'd stay with shooting 5-7 stars to get intersecting position lines, the stars are much nicer to deal with, and - unless you have a lot of practice in using a sextant (not to mention the calculator!) - are far more likely to get you a decent result
Empty
True, stars are nicer if you can find several of them. Especially if they are near zenith. Can you find the zenith and nadir in a plane, though?

Anyway, twilight is even more a nuisance. Sun is below horizon and its location cannot be measured exactly, but the sky is too bright to see stars, or enough stars to be able to identify them. So flying west in high latitudes has great risk of getting lost.

Originally Posted by Old Smokey
The very small variation in the actual solar day (which is why we have to refer to the MEAN day) that Tim alludes to is because the earth's speed around the sun varies slightly because it follows an ellipse, not a perfect circular orbit.

There are two members to the equation of time - the other comes from the inclination of the equator.
chornedsnorkack is offline