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Author
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Topic: Relation of d value to dec. min. when calculating Hc
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Henry Breyer
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posted February 19, 2005 07:42 AM
I'm working on 5.7 Exercises and have a question. In question 1, I'm working on the #104 workform in section 5 (Hc) I have the tab Hc as 53deg. 00',and d is 60. Table 5 does not have a line for a d of 60 (it ends at 59) so I looked at the answer to see how you interpolated the d and noticed that the d in the answer section was the same number as the Dec. min. (34.4). Can you explain this? Was this explained in the text? Does this only work when the celestial body you are observing is on your meridian? How do you figure Hc minutes when the celestial body is not on your meridian and the d=60.
Thanks,
Bill Breyer
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David Burch
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posted February 20, 2005 08:46 PM
The d-value used in sight reduction tables is defined in the online dictionary and in the course materials, when first introduced, but i am not sure if this special case comes up in any of the printed materials.
This d-value is how much the Hc changes if the declination changes by 1°. We use it to interpolate the correction to Hc based on the minutes part of the declination.
If d, for example, were 30', it would mean Hc changes by 30' when dec minutes change by 60'. In other words, in this case the Hc changes half as fast as the dec changes. If the dec minutes were 30' (and d=30' as well), it would mean the d correction was 15'.
If d=60', it means Hc changes 60' when dec changes 60', so there is a one to one correspondence. If the dec minutes are 20' the correction is 20', if dec minutes are 34.4, then the correction is 34.4 and so on.
This is purely a specific mathematical result in the tables. I cannot see offhand any reason that it should only apply to any particular set or classification of sights. In other words, it would seem you can end up with d=60 for meridian sights or for any other type of sight. It would seem to be more a unique combination of LHA, dec, and a-Lat that did it, and not any one factor alone, such as LHA = 0 corresponding to meridian passage.
With all that said, however, one can see that with LHA = 0, there is a one to one correspondence of Hc and dec(which is why we can do noon sights without sight reduction!), so this would be an expected result. Indeed, i will have to spend more time thinking on this to be certain that it could happen for other combinations of LHA, dec, and aLat. Perhaps skimming through the SR tables might be the shortest solution.
From: Starpath, Seattle, WA
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David Burch
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posted February 22, 2005 05:52 PM
Checking the Sight Reduction tables, we find that d can be 60' for several cases of low LHA, it does not have to be zero. At high latitudes, you can end up with d=60 for LHAs on the order of 25°. For "normal" sights one would not expect this d=60 unless you were actually doing a sight reduction of a noon sight or a sight very near LAN... or when Hc is near zero.
The latter strange situation comes about in emergency navigation when you simply time the sunset and call it Hs = 0° 0' and then do a normal sight reduction for a sunline. These cases will often end up with a d=60 value to work with, to further confuse how to handle the negative H values you get in these sights.
From: Starpath, Seattle, WA
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