NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: navigation problem
From: Frank Reed CT
Date: 2006 Feb 25, 02:45 EST
From: Frank Reed CT
Date: 2006 Feb 25, 02:45 EST
Dick W wrote: "The moon was barely visible above the fog. They computed the way back. For any meridian the moon will cross just 49 minutes after the sun crosses and the moon falls back 49 minutes every day. By counting the days after new moon, multiplying that number by 49 and dividing by 60, one arrives at the time in hours and minutes when the moon is due south. With south noted and knowing the reciprocal bearing back the ship they could safely return." This will get you a rough bearing. Because, as George H has already pointed out, the Moon's motion in hour angle is not uniform (both because its orbital speed is variable and because the path through the sky is tilted with respect to the celestial equator), it will not cross the meridian with regular timing. That said, as long as the Moon is not too high in the sky, this will still yield a decent azimuth, accurate within 10 to 15 degrees or so. Also note that this would be a true bearing. If they kept the bearing outbound with a magnetic compass, that can be a significant correction, but I think we can assume that anyone smart to think about the Moon's bearing is smart enough to know the magnetic variation. By the way, I don't think it's true that pocket watches aboard ship were uncommon, especially at the end of the century. But there's another wrinkle here. Suppose I work out the time of the Moon's meridian passage and discover that it's due at 02:30. But my watch reads midnight. Do I sit around and wait? With yet a little more knowledge of astronomy and a good estimate of your latitude, you could convert the Moon's hour angle to azimuth even if it's not on the meridian. So this method will work --up to a point. But is it plausible? It depends on someone knowing the date of New Moon as well as the 49 minute per day rule. I would suggest that any sailor or young officer with that knowledge would also be likely to have one other bit of knowledge that's more immediately useful: the Moon's position in the sky from the previous night. If you spend any amount of time on deck at all when the Moon is visible, you'll probably have a pretty good idea of its compass bearing at various times. And it's easy to estimate where it would be one day later. By the way, for the mathematically inclined, there's a nice way to think about the delay of the Moon's arrival at the meridian. The 49 minute rule tracks a fictitious "Mean Moon" in the same way that an ordinary clock tracks the "Mean Sun". The delay of the Sun's arrival at the meridian every day is the "Equation of Time" and it depends on the inclination of the Sun's path to the celestial equator and the eccentricity of the Earth's orbit in about equal measures. We can define a similar "Moon Equation of Time" and calculate it based on the inclination of the Moon's path through the sky and the eccentricity of the Moon's orbit. Since the Moon's orbit is about three times more eccentric than the Earth's, this will be a bigger, dominant contribution for the Moon Eq.T (about 1600 seconds in amplitude vs 350-900 seconds for the inclination term). Additionally, there is no fixed relationship between the phase of the inclination contribution and the eccentricity contribution. But it can all be worked out, and if I've done it right, the result is that a "typical" Moon Equation of Time value is about 20 minutes with extreme values around 40 minutes. In other words, that's how early or late the Moon will arrive on the meridian compared to the fictitious Mean Moon. Using that typical "20 minute" value translates to a typical azimuth error of about 5 degrees. Add in other sources of error, like the uncertainty in the exact time of New Moon, and the average azimuth error will climb to 10 or 15 degrees. -FER 42.0N 87.7W, or 41.4N 72.1W. www.HistoricalAtlas.com/lunars