NavList:

I caused Bob Crawley and Greg Rudzinski some frustration by taking a couple of bad Moon sights on our party boat. Sorry! I blame the rum. ... Actually I'm not quite sure how I screwed it up. Fortunately, there was better data available. Here's the setup again, and this time I'll list fully corrected altitudes so you don't have to worry about dip or Moon SD or refraction or parallax.

The scenario:
You know that it is late January 2018. You left NYC Friday night. Maybe today is Tuesday, maybe Wednesday. The sky is clear, and not long after sunset the stars come out. The Moon appears full, low in the east. It's nautical twilight, and Orion is in the southeast, off the bow. [other details no longer matter]

You take the following sights at approximate one-minute intervals, timed by a cheap watch pulled off the wrist of a guest from Wisconsin. When was it last set? You have no idea...

watch time: body, corrected altitude (geocentric, body center, =Ho)
4:10:05: Procyon, 15° 55' 
4:11:01: Moon, 15° 56'  
4:12:08: Sirius, 13° 17' 
4:13:04: Moon, 16° 19'
4:14:14: Procyon, 16° 46'

Ignore the Moon sights for now. You probably don't need them, right? The sights before and after Sirius are paired so that we can easily average. What would the altitude of Procyon be if made simultaneous with the Sirius sight?

Now work out your latitude. How can you do that? We discussed options in the thread related to the ice cave. They all apply here, too. It's just two simultaneous star altitudes, and they should yield one specific latitude. Or are there two latitudes? This is a calculation with many options. For example, could you estimate your latitude using only the two Procyon altitudes and the time between them? But that's not particularly accurate, is it? What is the most accurate procedure for calculating your latitude?

What do you find? Are you north or south of Bermuda's latitude?

Next, for the expert dying for an even trickier challenge: can you estimate your longitude well enough to be useful? For this, you can try to use the Moon's observed position. First, even without calculation, can you decide the date? Note that the Moon usually looks full for three days in a row, so the phase alone is not quite sufficient. But there's enough info here. Next, can you estimate GMT/UT based on the Moon's altitude?? This is similar to a lunar, but it's not as accurate. Nevertheless, you should be able to get a longitude to go along with your latitude. Can you trust it? What should you do?

....

All of that is basically as I wrote previously with the raw sextant altitudes for all objects replaced by corrected altitudes and with the Moon sights completely replaced. The main point of this is latitude by two simultaneous star altitudes at a relatively uncertain GMT. A very reasonable approach is to treat it as an ordinary two-body sight reduction. You could use Bermuda for your AP and assume that the time on the watch (above) is exact Central Standard Time (6 hours begind GMT) and see what happens. The latitude will be right regardless of the longitude, but you probably have to do one iteration to get short intercepts. Still, that's no probem. Another approach is to solve the set of simultaneous spherical triangles. There are multiple ways to do that, none requiring anything more than a common calculator. More on that in my next post.

With the latitude out of the way, how do you get the date? This is actually a trick that can come in handy, maybe not in traditional celestial navigation scenarios, but perhaps in a scenario like this where a passenger has been "promoted" to navigator, or maybe for a solo sailor after a multi-day storm (a scenario proposed by John Letcher in his fine, old book). This also can come up when we're trying to reverse-engineer observations after the fact. 

And finally, with better Moon data, can you estimate GMT and thus longitude? If so, how much can you trust it? Can you set a course to Bermuda or not? If not, is there anything useful you can do with your latitude as found? Is there anyway to get to Bermuda and save the party?!

Bermuda... Yeah, Bermuda sounds OK... Or I could just stay here with 16 inches of snow on the ground and the air temperature at 11° F (and falling) and the wind howling at 30 mph. To freeze, or not to freeze...

Frank Reed