| my account | login-logout | resources | classroom help | support | catalog | home | get webcard |

Online Classroom


Post New Topic  Post A Reply
search | help desk | commons
  next oldest topic   next newest topic
» Online Classroom   »   » Public Discussion of Cel Nav   » What day is it?

   
Author Topic: What day is it?
Leslie


 - posted May 02, 2010 05:43 AM      Profile for Leslie           Edit/Delete Post 
I'm a bit flummoxed by the following problem....perhaps someone can help....

You are sailing east, not quite three days out of the Chesapeake, bound for "The Onion Patch." You get a very good noon sight and Ho of the Sun is "80-56.5." GPS (just being used as a backup for real navigation, of course) puts latitude at 32-30.5N. What's the most likely date?

I do not have a nautical almanac nor do I know how to use one. So my attempt is as follows:

For Vernal & Autumnal equinoxes the maximum altitude of the Sun is 90 deg - latitude or 57d 29.5' . For Summer solstice the maximum altitude of the Sun would be 23D 30' minutes more or
Ho = 80D 59.5' . I then assume the Suns altitude follows a Sinusoidial path throughout the astronomical year of 365.25 days. My guess then is that it is about 1 day earlier or later than Summer Solstice so June 20 or June 22nd. Is there a more analytical or refined approach?

Appreciate any help or guidance.

Thank you,
Leslie

HHEW


 - posted May 02, 2010 10:07 AM      Profile for HHEW           Edit/Delete Post 
You came pretty darn close - bracketing the date, in fact. The only glitch I see in the given data is the declination of the sun at June solstice. It's more like 23° 26.4'. (Mnemonic: the numerical series 2345 gives solstice declination in degrees - i.e., 23.45°)

The classical approach to this problem would be to subtract Ho from 90° to get the distance ( the so-called zenith distance) between the GPS latitude and the GP of the sun and then apply that distance to the GPS latitude to get the sun's declination.

89° 60.0' (90° written to facilitate subtraction of minutes)
- 80° 56.5'
9° 03.5' = zenith distance

32° 30.5' N
- 9° 03.5'
23° 27.0' = sun's declination N (.6' high - sun's decl doesn't go above N-S 23° 26.4')

So, it''s the day of the June solstice, June 21.

-Hewitt Schlereth

Leslie


 - posted May 02, 2010 10:19 PM      Profile for Leslie           Edit/Delete Post 
Many thanks !!!

I got the 23 and half degrees from one of my daughters elementary school books on astronomy describing how the earth is tilted.

Should have checked that info more carefully.

Leslie

HHEW


 - posted May 03, 2010 07:45 AM      Profile for HHEW           Edit/Delete Post 
You're entirely welcome. As a result of your question I've been brought up to date (see below) and I thank you for that.

BTW, there are several online almanacs. The one I like is at http://www.tecepe.com.br/scripts/AlmanacPagesISAPI.isa. Its output looks much like the Nautical Almanac pages. If you go there and ask for June 21, 2010, you'll get the pages for 21, 22, 23. You can plainly see the sun's declination going to 23° 26.3' at 0100 hours UT (GMT) on the 21st and hanging there thru 2200.

-Hewitt Schlereth

PS, As you see, I erred in stating the solstice declination as 23° 26.4'. That's what it was 30 years ago. (Welcome to the 21st Century, Hewitt) :-)



All times are Pacific  
Post New Topic  Post A Reply Close Topic    Move Topic    Delete Topic next oldest topic   next newest topic
Hop To:

Starpath School of Navigation

Copyright, 2003-2018, Starpath Corporation

Powered by Infopop Corporation
UBB.classicTM 6.3.1.1