NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Navigation exercise
From: Peter Fogg
Date: 2008 May 17, 22:03 +1000
From: Peter Fogg
Date: 2008 May 17, 22:03 +1000
Thanks for the detail on how you worked this out, Jeremy. Here is how I do it.
The time of LAN, both as GMT and LZT: Start from Local Apparent Noon, which happens at 12h 00m 00s by definition. Then apply the equation of time, giving Local Mean Time. Adding or subtracting the DR Longitude, expressed as time, converts this to GMT. Adding or subtracting the hours of difference between GMT and the time zone brings this back to the Local Zone Time, or the standard time in that part of the world. Thus:
LAT 12 00 00
EoT -3 40
LMT 11 56 20
Lon -9 42 40 +W, -E
GMT 02 13 40
Zone +10 00 00 -W, +E
LZT 12 13 40
This might be what you call the "Equation of Time method". I can see your "GHA method" gives a more accurate result. Although I have a couple of versions of analemmas, the best only indicates a result to about the nearest third of a minute of time. Do you know a better way to derive the Equation of Time? Not that it matters in this case; there being no practical difference in altitude, or information derived from an almanac, over a few seconds of time at meridian passage.
The next step is to take the sextant altitude, apply Dip, Index Error and the correction for refraction and limb to derive the observed altitude, which is subtracted from 90d 00' to give the Zenith Distance. We use a slightly different Dip, as mine comes from the formula: Dip = [square root of height of eye]*0.97 (for feet) which in the case of 106ft (big ship!) is 9.987, rounded to nearest tenth: 10.0.
Hs 85 59.0
Dip -10.0
IC + 0.8
Ha 85 49.8
Cor +15.8 (for the sun's lower limb)
Ho 86d 05.6'
Zenith distance = 90d 00' minus Ho
89 59.10
-86 05.6
Zd 03d 54.4'
Next we need the sun's Declination at meridian passage, which is N19d 09.6'. Since we're about 15d to the north of the equator, and the sun is about another 4 degrees further to the north, we are observing to the north, so the azimuth will be 000d 00.0' at meridian passage.
Our latitude is either the sum or the difference between Zenith Distance and Declination. Since we are already north of the equator, and the sun is to the north of us, it is the difference.
Dec 19 09.6
Zd -03 54.4
Lat 15d 15.2'
Our latitude by observation at Local Apparent Noon on 16 May 08 is N15d 15.2'.
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The time of LAN, both as GMT and LZT: Start from Local Apparent Noon, which happens at 12h 00m 00s by definition. Then apply the equation of time, giving Local Mean Time. Adding or subtracting the DR Longitude, expressed as time, converts this to GMT. Adding or subtracting the hours of difference between GMT and the time zone brings this back to the Local Zone Time, or the standard time in that part of the world. Thus:
LAT 12 00 00
EoT -3 40
LMT 11 56 20
Lon -9 42 40 +W, -E
GMT 02 13 40
Zone +10 00 00 -W, +E
LZT 12 13 40
This might be what you call the "Equation of Time method". I can see your "GHA method" gives a more accurate result. Although I have a couple of versions of analemmas, the best only indicates a result to about the nearest third of a minute of time. Do you know a better way to derive the Equation of Time? Not that it matters in this case; there being no practical difference in altitude, or information derived from an almanac, over a few seconds of time at meridian passage.
The next step is to take the sextant altitude, apply Dip, Index Error and the correction for refraction and limb to derive the observed altitude, which is subtracted from 90d 00' to give the Zenith Distance. We use a slightly different Dip, as mine comes from the formula: Dip = [square root of height of eye]*0.97 (for feet) which in the case of 106ft (big ship!) is 9.987, rounded to nearest tenth: 10.0.
Hs 85 59.0
Dip -10.0
IC + 0.8
Ha 85 49.8
Cor +15.8 (for the sun's lower limb)
Ho 86d 05.6'
Zenith distance = 90d 00' minus Ho
89 59.10
-86 05.6
Zd 03d 54.4'
Next we need the sun's Declination at meridian passage, which is N19d 09.6'. Since we're about 15d to the north of the equator, and the sun is about another 4 degrees further to the north, we are observing to the north, so the azimuth will be 000d 00.0' at meridian passage.
Our latitude is either the sum or the difference between Zenith Distance and Declination. Since we are already north of the equator, and the sun is to the north of us, it is the difference.
Dec 19 09.6
Zd -03 54.4
Lat 15d 15.2'
Our latitude by observation at Local Apparent Noon on 16 May 08 is N15d 15.2'.
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Navigation List archive: www.fer3.com/arc
To post, email NavList@fer3.com
To , email NavList-@fer3.com
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