NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: running fix + lunar
From: Paul Hirose
Date: 2010 Nov 22, 21:31 -0800
From: Paul Hirose
Date: 2010 Nov 22, 21:31 -0800
Andres Ruiz wrote: > The "practical circumstance", is: one is sailing with known course and > speed and to obtain a fix uses two altitudes with a sextant and a > lunar distance at different times ant the chronometer or on-board > watch has an error. Please Paul, correct me if I have not > misunderstood your example. You understand me perfectly. The idea of celestial sight reduction with a set of linear equations is nothing new. An especially elaborate example is the Kaplan paper, referenced in your other message. The only new idea I've introduced is that time is only approximately known, and must be found by a lunar. In reality this is an improbable scenario, but I thought a few readers might find it interesting. Here is another example. This time we'll shoot an afternoon lunar with the Sun, followed by three altitudes at evening twilight. Normally, Sun and Moon altitudes would be observed with the lunar distance, but to make the problem more difficult let's say the horizon is bad at that time. So the lunar distance must be combined with three altitudes several hours later. Here are the observations on 2010-12-01 and -02, with UTC from the chronometer, and estimated positions. 21:36:02 at N25°57.7' W142°00.5' Moon-Sun 49°53.1' 02:27:11 at N26°35.3' W140°05.7' Jupiter 51°28.1' 02:32:56 at N26°36.1' W140°03.5' Schedar 51°11.7' 02:38:24 at N26°36.8' W140°01.4' Altair 50°21.1' Those times and positions have considerable error. As before, assume the same error affects all observations. The observed angles are accurate. The lunar is near limb to near limb. Once again I'll try to describe the method with simple language. Begin by computing altitude (or lunar distance) at the given times and locations. Compare these to the observed angles. Moon - Sun computed separation angle is 1.7' greater than observed. One degree north changes computed angle -.7'. One degree east changes it +.6'. One minute later changes it -.4'. I don't know a way to gauge the effect of latitude and longitude on lunar distance, except to recompute from scratch for a different position. However, I have a program which outputs the rate of change per minute. The "partial derivatives", as these are called, are much easier to compute for the altitude observations. The increase in altitude equals the (north) increase in latitude, times the cosine of azimuth. It also equals the (east) increase in longitude, times the sine of azimuth, times the cosine of latitude. And since the passage of time has the effect of moving east in longitude, its partial derivative is simply some fraction of the longitude partial derivative. E.g., one minute of time is equivalent to 1/4 degree of longitude, so if 1 degree of longitude changes the computed altitude by 40', 1 minute of time changes it 10'. Getting back to our observations, Jupiter's computed altitude is 1°41' less than observed. Moving one degree north changes computed altitude -42'. Moving one degree east changes it +38'. One minute later changes it +9.5'. Schedar computed altitude is 3°09' less than observed. Moving one degree north changes computed altitude +50'. One degree east changes it +29'. One minute later changes it +7.3'. Altair computed altitude is 4°05' greater than observed. One degree north changes computed altitude -24'. One degree east changes it -49'. One minute later changes it -12.3'. Set up the equations as described in my previous message: -1.7 = -.7∆φ + .6∆λ - .4∆t 101 = -42∆φ + 38∆λ + 9.5∆t 189 = 50∆φ + 29∆λ + 7.3∆t -245 = -24∆φ - 49∆λ - 12.3∆t Unlike before, we have three unknowns and four (not three) equations. A solution may be found by the method of least squares. Results: ∆φ = +1.38° = N 1°23' ∆λ = +2.75° = E 2°45' ∆t = +5.99 m = +5m 59s Add those corrections to the times and positions. The observed angles are the same as before. 21:42:01 at N27 20.7 W139 15.5 Moon 49 53.1 02:33:10 at N27 58.3 W137 20.7 Jupiter 51 28.1 02:38:55 at N27 59.1 W137 18.5 Schedar 51 11.7 02:44:23 at N27 59.8 W137 16.4 Altair 50 21.1 Now repeat the process of determining (computed - observed), and the partial derivatives. Set up new equations: .6 = -.7∆φ + .4∆λ - .4∆t 11.4 = -46.6∆φ + 33.3∆λ + 8.3∆t -2.8 = 50.9∆φ + 28.1∆λ + 7.0∆t -4.3 = -22.4∆φ - 49.1∆λ - 12.3∆t Note the values on the left side are much smaller than before. It shows the first set of corrections were a big improvement. New solution: ∆φ = -.138° = S 8.2' ∆λ = +.369° = E 22.2' ∆t = -.878 m = -53 s In the upper line of each pair below I show the corrected time and position, and the corresponding computed lunar distance or altitude. The lower line has the true values, which are unknown to the navigator (except for the observed angles). 21:41:08 at N27 12.5 W138 53.3 Moon-Sun 49 53.1 21:41:18 at N27 11.8 W138 53.5 Moon-Sun 49 53.1 02:32:17 at N27 50.1 W136 58.5 Jupiter 51 28.1 02:32:27 at N27 50.0 W137 01.0 Jupiter 51 28.1 02:38:02 at N27 50.9 W136 56.3 Schedar 51 11.7 02:38:12 at N27 50.8 W136 58.8 Schedar 51 11.7 02:43:30 at N27 51.6 W136 54.2 Altair 50 21.1 02:43:40 at N27 51.5 W136 56.7 Altair 50 21.1 We now have a perfect match between computed and observed angles. However, some error remains in the time and position. I believe this is mainly due to error in the dead reckoning between the afternoon lunar and the evening sights. The navigator used 22 knots on true course 069 to come up with the estimated positions at dusk. I introduced an error in both course and distance made good, equivalent to 3% of the distance run. Thus, the same set of time and position corrections cannot be valid for both the afternoon and evening observations. The results were still pretty good. I could have injected errors in the observed angles too, but there are limits to how hard I want to work! Fortunately my HP 49G calculator makes it easy to solve simultaneous linear equations. It automatically uses the method of least squares if the solution is "overdetermined" (the number of equations exceeds the number of unknowns). This is not a very modern calculator, either. It's about 10 years old. The problem in my first message, with three equations in three unknowns, could actually be solved with a simple calculator, pencil, and paper. Some readers may have solved similar sets of equations that way in high school. Unfortunately, a least squares solution is not so simple. I'd hate to work it with nothing but a simple calculator. It wouldn't be too bad with something like an early 1980s HP-15C, which can invert and multiply matrices. That's the hard part. One refinement of the least squares solution is to put additional weight on a lunar distance, because usually that angle is more accurate than an altitude. In my example, that would be accomplished by multiplying the coefficients of the first equation by say 1.5 or 2.0. The same thing can be accomplished by expressing the residuals (observed - computed value) and the partial derivatives in "sigmas", where one sigma is the standard deviation of the measurement. This could simply be the navigator's estimate. --