Navigaion -determining position with ref. to the Sun/Moon
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Navigation -determining position with ref. to the Sun/Moon
Any tricks on how one can determine direction with reference to the position of the Sun or Moon.
I was told this one which workd (try it out).
Around Midday if you point point the hour hand on your watch in the direction of the sun the midpoint between 12 on the watch and the hour hand is due south.
Early morning /late afternoon evening -- the direction of the Sun can help you determine East/West.
Any others?
I was told this one which workd (try it out).
Around Midday if you point point the hour hand on your watch in the direction of the sun the midpoint between 12 on the watch and the hour hand is due south.
Early morning /late afternoon evening -- the direction of the Sun can help you determine East/West.
Any others?
Last edited by shon7; 27th January 2006 at 14:00.
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Without trying to make it too complicated, try this one. as taught by the RAF nav school.
Take the time using the 24 hour clock (whole hours), multiply it by 1.5, add a zero and that's the sun direction in degrees.
Example: 4pm is 16hrs = 24+0=240 degrees. You've got to interpolate the in between bits and remember about any local daylight saving time but it was accurate enough to tell me how lost I was on a 3000ft hill.
Take the time using the 24 hour clock (whole hours), multiply it by 1.5, add a zero and that's the sun direction in degrees.
Example: 4pm is 16hrs = 24+0=240 degrees. You've got to interpolate the in between bits and remember about any local daylight saving time but it was accurate enough to tell me how lost I was on a 3000ft hill.
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From: Estonia
Around Midday if you point point the hour hand on your watch in the direction of the sun the midpoint between 12 on the watch and the hour hand is due south.
Early morning /late afternoon evening -- the direction of the Sun can help you determine East/West.
Early morning /late afternoon evening -- the direction of the Sun can help you determine East/West.
ECON cruise, LR cruise...
Joined: Apr 2002
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From: MIRSI hold - give or take...
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.
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).
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
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.
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).
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
Joined: Feb 2000
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From: 500 miles from Chaikhosi, Yogistan
Re: Navigaion -determining position with ref. to the Sun/Moon
I fly with the GPS in true as the variations in my part of the world are greater than 60 degrees. As there are few ground features (infact often completely absent) as a gross error check on the GPS, I do the following on the back of the flight log.
First, the sun moves 360 degrees in 23 hours 56 minutes and 4 seconds. That's 15.04 deg/hr.
Now I am lon E078 in the southern hemisphere, it's 0430 UTC, what brg is the sun? There are two stages to work this out. First is to work out where the sun is now w.r.t. an observer in my hemisphere at 0deg lon.
At 0 deg lon 0000 UTC the sun should be due south (southern hemisphere) (180T). So now at 0 long it should be 4.5*15.04deg/hr east of south = brg 112T.
The second part of the calculation is to correct for my longitude. That is to work out what fraction of 24 hours I am ahead or behind the observer at Greenwich. I am Lon East, so I am ahead. 78/360*24=5.2 hours.
5.2 hours @15.04deg/hr= 78 degrees.
The sun should be 78 degrees east of the previously calculated point as I am to the east.
So the sun is BRG 034True at this point in time.
There are other formula to work it out, but as I am hopeless at remembering them, I calculate it practically by thinking out the above, and there are a couple of different ways to do the above. Remember to convert the times/direction for your hemisphere, then add on your local variation.
First, the sun moves 360 degrees in 23 hours 56 minutes and 4 seconds. That's 15.04 deg/hr.
Now I am lon E078 in the southern hemisphere, it's 0430 UTC, what brg is the sun? There are two stages to work this out. First is to work out where the sun is now w.r.t. an observer in my hemisphere at 0deg lon.
At 0 deg lon 0000 UTC the sun should be due south (southern hemisphere) (180T). So now at 0 long it should be 4.5*15.04deg/hr east of south = brg 112T.
The second part of the calculation is to correct for my longitude. That is to work out what fraction of 24 hours I am ahead or behind the observer at Greenwich. I am Lon East, so I am ahead. 78/360*24=5.2 hours.
5.2 hours @15.04deg/hr= 78 degrees.
The sun should be 78 degrees east of the previously calculated point as I am to the east.
So the sun is BRG 034True at this point in time.
There are other formula to work it out, but as I am hopeless at remembering them, I calculate it practically by thinking out the above, and there are a couple of different ways to do the above. Remember to convert the times/direction for your hemisphere, then add on your local variation.
Joined: Mar 2000
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From: Oakland CA USA
Originally Posted by compressor stall
the sun moves 360 degrees in 23 hours 56 minutes and 4 seconds. That's 15.04 deg/hr.
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From: Australia
1 Mean Solar Day = 24:03:56.55536 of mean siderial time,
1 Mean Solar Day = 1.00273791 earth rotations with respect to the Mean Vernal Equinox,
1 Mean Solar Day = 1.0027378118868 earth rotations with respect to the Mean position of Stars,
1 Mean Solar Day = 1 earth rotation with respect to the Mean Sun position - EXACTLY!
I like the last one best.
1 mean siderial day = 23:56:04.09054 of mean solar time.
Tim Zukas nailed it, the very small difference (approximately 4 minutes) between the apparent solar day and the siderial day is that the earth moves along it's orbit of the sun by about the equivalent of 4 minutes by the time the sun returns to the same Longitude. 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.
Regards,
Old Smokey
1 Mean Solar Day = 1.00273791 earth rotations with respect to the Mean Vernal Equinox,
1 Mean Solar Day = 1.0027378118868 earth rotations with respect to the Mean position of Stars,
1 Mean Solar Day = 1 earth rotation with respect to the Mean Sun position - EXACTLY!
I like the last one best.
1 mean siderial day = 23:56:04.09054 of mean solar time.
Tim Zukas nailed it, the very small difference (approximately 4 minutes) between the apparent solar day and the siderial day is that the earth moves along it's orbit of the sun by about the equivalent of 4 minutes by the time the sun returns to the same Longitude. 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.
Regards,
Old Smokey
Joined: Aug 2005
Posts: 834
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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.
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.
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).
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
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
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.
