A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Navigation exercise
From: Gary LaPook
Date: 2008 May 25, 22:07 -0700
From: Gary LaPook
Date: 2008 May 25, 22:07 -0700
Gary LaPook writes: I have found this discussion interesting in exploring ways of achieving the highest possible accuracy from an LAN sight for latitude. But there is another way to work such a sight, simply treat it as you would any other celestial sight with an LHA of zero degrees and use your normal sight reduction tables or formulas and draw the usual LOP. Doing it this way obviates the criticality of timing since the LHA of zero lasts for a period of four minutes, the azimuth is 180 (or 360 depending on latitude and declination) so an assumed position within 30 minutes east or west of your DR will not affect the LOP. Since this is the way you would work any other sight it must produce an LOP accurate enough for practical navigation at sea (or in the air.) gl On May 25, 12:02 pm, "George Huxtable"
wrote: > Bill is worrying away at this question of the shape of the altitude curve of > the Sun, around noon, and the effect on it of the speed and direction of the > vessel. > > Let me take, first, his final question, which was- > > "What about east/west displacement if trying to determine longitude from > LAN?", and will assume, for that purpose, that the ship has no motion in the > North-South direction, but is travelling East-West only, and the Sun's > decllination is unchanging, at a solstice.. > > The moment of LAN (Local Apparent Noon) is when the Sun and the ship are on > the same meridian. The Sun will always be crossing meridians, going > Westward, at almost exactly 15 degrees per hour. The ship will be crossing > meridians travelling much more slowly, at a few tens of minutes in an hour, > perhaps in the same direction as the Sun, perhaps the opposite direction. > The moment when they cross the same meridian will depend on that speed, so > to calculate that moment you must take the speed into account, but the > maximum height of the Sun will occur at that moment and will not depend on > the speed of the vessel. The only effect you would see, if you examined the > curve of altitude against time more closely, was that it would be very > slightly peakier, changing slightly faster with time if the vessel was > travelling Eastward than vice versa. However, it would still have the same > maximum value, at the same moment, of LAN. There would be no surprises. > > North-South travel, around noon, has a very different effect, the details of > which Bill questions. Judging by his words, he finds it hard to accept that > it can result in a symmetrical curve of altitude against time: symmetrical, > not about noon, but about a moment displaced, earlier or later, in time. > It's because of the way two quite different changes combine together. > > The speed, Northward or Southward, of a vessel, has a very similar but > greater effect than the changing declination of the Sun (which can be up to > 1 knot at the equinoxes) So just let's consider a vessel moving Southward at > say 10 knots, at Winter solstice, at a position of say 56 deg North. And > let's make an enormous supposition, that at the moment of LAN, we could stop > the rotation of the Earth. The Sun would then really "hang " in the sky, > absolutely stationary. But to our observer on the Ship, not quite > stationary. He would still be travelling Southward, toward the Sun, at 10 > arcminutes per hour, so to him the Sun would be slowly increasing in > altitude, at that rate. > > Now put that picture to one side, and set the Earth rotating again. And now, > take the vessel to be stationary. Now our observer sees a familiar curve: > the Sun climbs, reaches a peak, then falls, quite symmetrically about LAN. > Actually, near the peak, that's as close as you like to a parabolic curve, > where the drop, from the peak value, changes with the square of the time > from the peak. > > And now, in real life, we have to combine those two motions together, the > symmetrical parabola with the steadily changing altitude, increasing at 10 > minutes an hour. And I suspect that the only way to convince Bill will be to > persuade him to take a bit of graph paper and combine them for himself. > > As an example, let's take the ship approaching the Clyde in Winter, that I > tried to get Frank to consider as a noon-longitude exercise (but he ducked). > The noon Sun (neglecting corrections) would be at 10 deg 30' altitude. > Before and after that moment, if the ship is stationary, it will be somewhat > less. Here are some numbers. > > First column, time in minutes before and after LAN. > Second column, Sun altitude with stationary observer. > Third column, change in altitude due to observer's motion. > Fourth column, combination of the two. > > time (min) alt1 shift sum > LAN-20 10d 23.1 -3.3 10d 19.7 > LAN-15 10d 26.2 -2.5 10d 23.7 > LAN-10 10d 28.3 -1.7 10d 26.6 > LAN-5 10d 29.6 -0.8 10d 28.8 > LAN 10d 30.0 0.0 10d 30.0 > LAN+5 10d 29.6 +0.8 10d 30.4 > LAN+10 10d 28.3 +1.7 10d 30.0 > LAN+15 10d 26.2 +2.5 10d 28.7 > LAN+20 10d 23.1 +3.3 10d 26.4 > > Anyone who cares to plot out the fourth column will find it to be just > another parabola, perfectly symmetrical in itself, but with its centre of > symmetry displaced to be about 5 minutes later than LAN. It also has a > slightly higher peak value but not very significantly so, which allows an > observer to take either the altitude at LAN, or the maximum altitude, > without bothering much about the difference. > > But not so the difference in the timing of that central value, to anyone > trying to use Frank's proposed method to find his longitude. The correction > that has to be made for the ship's south-going speed of 10 knots is all of 5 > minutes of time, or 1.25 degrees of longitude, which has to accommodate the > speed of the ship, any tidal current and any declination changes in the Sun > position. With modern shipping commonly travelling at over 25 knots, it's > obvious what an enormous correction this must be, and how precisely it would > have to be made. > > And yet Frank is claiming that he can derive the central moment of that > altitude curve, and then make that correction, to provide an overall error > in the whole process of no more than 5 miles in longitude, which corresponds > to 35 seconds of time. No wonder he is reluctant to disclose the details. > > George. > > contact George Huxtable at geo...@huxtable.u-net.com > or at +44 1865 820222 (from UK, 01865 820222) > or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavListemail@example.com -~----------~----~----~----~------~----~------~--~---