A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Bennett's '...Celestial Navigator' --An improved Zn calculation
From: George Huxtable
Date: 2005 Nov 26, 00:30 -0000
From: George Huxtable
Date: 2005 Nov 26, 00:30 -0000
D. Walden wrote- > With some trepidation, I raise again the question of using '...Celestial > Navigator' to obtain Azimuth. Wow, that WAS a long time ago! It was a matter that seemed to excite great passion in the breasts of Bennett's adherents; or in one such breast, at least. =========================== D.Walden doesn't quote my original posting in detail, or give its date, and I no longer have easy access to it, so I will have to go a bit by memory. For new readers: it concerned the azimuth tables in "The complete on-board celestial navigator" 2003-2007 edition, by George G Bennett, about which it was claimed- "No interpolation is required, and it is one of the simplest techniques for finding azimuth with an accuracy of one or two degrees." I decided to put that claim to the test, and put forward what I thought to be a worst-case set of inputs for which the azimuth tables gave a result that was all of 15 degrees different from the true azimuth. These values are what Walden refers to when he writes- >Using the infamous Huxtable example: > Dec=55-30 > LHA=54-30 > Alt=61-30 To keep the record straight, I don't recognise those numbers as exactly the same worst-case as I suggested. Just to ensure that the worst-case choice was made between adjacent rows and columns, when finding the nearest value without interpolation, I think I actually proposed- > Dec=55-29 > LHA=54-31 > Alt=61-31 When those values are plugged in to the Bennett azimuth table as nearest whole-number degrees, we get dec = 55, lat = 55, which gives a resultant value for X of 469. Combining that with an alt = 62, the azimuth table gave a resulting value of 270 degrees, or due West. (In my initial mailing, I had in error stated that to be 90 degrees, but that's by the way). This result of 270 is nearly 15 degrees out from the azimuth you get if you make a calculation with spherical trig., using- Az = arcsin ( - sin LHA * cos dec / cos alt), which gives Az as 255deg 21' or 284deg 39' That said, I'm quite happy for Walden to round off those inputs to the nearest half-degree, as he did. ========================== Walden continued- > George H. didn't give the corresponding Lat, but it can be found to be: > Lat=60-18 ============================ Let's pause again here. How, I wonder, has he arrived at that value, which I do not dispute. He doesn't explain. Actually, there are two possible such latitudes. There's a position circle, centred on the geographical position of the body given by dec, with radius (90 - alt) degrees, and any meridian that crosses it, defined by LHA, must cross it twice. In this case, one such latitude is at 60 deg 18', as Walden states, and the other is at 76deg 11'. As for me, if I needed lat given those quantities, I would work it out using lat = arctan(tan dec / cos LHA) + arccos ( sin ( arctan ( tan dec / cos LHA) )*sin alt /sin dec) Because arc cos has two possible values, this gives two solutions, as above. There are other ways of expressing it, which may be simpler. How did Walden obtain that value for lat, which he needed to use in his proposed different way of using the Bennett table, "backwards"?. Walden then abandons the Bennett azimuth table, the thing that causes all the trouble. Instead, he takes the procedure and tables that Bennett provided to calculate the altitude, and uses them to obtain the azimuth instead, by cleverly twisting the astronomical triangle round. In those tables, quantities can be entered to the nearest minute, unlike Bennett's azimuth tables, in which they are only to the nearest degree. So it's no surprise that he ends up with a much better result. .> First going through the altitude calculation using the Bennett work form, on page 168, to generate altitude from given Dec, LHA, and Lat. > > line 13 Local Hour Angle 54-30 -> 8841 > line 14 DR Latitude N 60-18 -> 3974 > line 15 Declination N 55-30 -> 3217 > ______ > line 16 (theta=28-04) SUM 16032 -> RES 11760 > line 17 Latitude ~ Declination > (ABS(Lat-Dec)) 4-48 ------------> 351 > ______ > line 18 Computed Altitude 61-30.5 <------ ALT 12111 > > Now for the new method to calculate Zn. In a sentence, use Bennett's > table 'backwards' substituting Alt for Dec, and Dec for Alt. The final Z > will be the LHA value. Continuing with the infamous example from above: > > remember, substitute Dec for Alt > line 18 Computed Altitude 55-30 --------> ALT 17587 > now, Lat~Dec becomes Lat~Alt ______ > line 17 Latitude ~ Declination 1-12 --------> 22 > now, calculate what RES must be for sum to equal ALT, (17587-22) > line 16 (theta=34-29) SUM 13763 <- RES 17565 > ______ > now, we have the sum of three, we know two, so we can solve for the > third. > remember, substitute Alt for Dec > line 15 Declination 61-30.5 -> 4187 > note, line 14 is the same as above > line 14 DR Lat N 60-18 -> 3974 > for SUM to be correct, line 13 must be 13763-4187-3974 > note use top of column LHA value as Z > line 13 Local Hour Angle 75-07 <- 5602 > now, we apply our one rule, if LHA(the real LHA)<180, Zn=360-Z, else Zn=Z > So, Zn=360-(75-07)=284-53 Exactly the ATAN2 formula result! > > Note, there is a typo, which I don't recall seeing mentioned before, in > Bennett's response to Huxtable: "If, however, the Tables are interpolated > (X=460) the azimuth is found to be 255 or 285 (not 075 or 105) which > compares favourably with the results from direct calculation of 255.3 and > 254.8." The last number should be 284.8, as above. I agree, that must be a typo. Bennett's argument, that if you interpolate you get a much better answer, though true, was beside the point, because his book specifically stated that interpolation wasn't needed. He accepted that at some angles large errors could result, and promised to put a note to that effect in future printings of his book. That was what I was after. I presume that he has done so. > Pretty slick, eh? I can only agree. At first sight, pretty slick! But it needs a bit more than that single example to fully convince us if the virtues of Walden's procedure. The next sentence is a bit ominous... > (Some adjustments of signs for special cases are left to the reader as an > exercise.) D. Walden should be congratulated on a useful piece of lateral thinking. George.