A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Jeff Gottfred
Date: 1996 Jan 18, 17:09 -0700
From: Jeff Gottfred
Date: 1996 Jan 18, 17:09 -0700
[NOTE: COPY of message originally posted on 18 Dec 1995] Why do I do lunars? The short answer is for fun. A slightly longer answer is that I am very interested in the navigational techniques of Alexander Mackenzie, David Thompson, Peter Fidler, Lewis & Clark, &c. For a hobby I do workshops and demonstrations of the techniques that these men used at various historical sites. [As a small aside, this summer, while giving a demonstration at the Lewis and Clark festival in Great Falls, I took a noon sight for the latitude of the Giant Springs of the Missouri River. To some embarrassment, I discovered that Lewis was in error by some 17 nautical miles. (Oops!). Even if you ignore refraction, it's hard to explain that kind of error. I wonder if he was just having a bad day, or if all their data was that bad? Perhaps some massive frontal system had just gone through or something...] In this message, I talk about my first successful attempts at lunars, and then describe an historical method by Young (1848). If anyone has fiddled around with lunars before, please let me know, I'd be keen to know how it worked out for you and what methods you used. If any of you are really keen to try this stuff using the methods, tables, &c used from about 1790-1820 let me know, I can provide you with them. What is a Lunar? The idea behind lunar distance observations is conceptually easy. First, you must set your watch to local time-- I use solar time because its easier, and my watch is then useful for other purposes, but you could use any convenient body. Once you have set your watch to local apparent time, you apply the equation of time correction (as found at the bottom of each page in the Nautical Almanac) to set your watch to local mean time (NB, this is NOT the same as mean zone time!!) so you can relate it to the time data in the almanac, which uses the mean sun. Next, you then measure the angular distance of the moon from some other object on the ecliptic (the sun is my fave), and then look up in a lunar distance table when the moon would be seen to be that distance by an observer in Greenwich. Voila!, you now know the relationship between Greenwich and local time, and therefore (at four minutes per degree) your longitude. Actually doing one: O.K., problem #1, just how do you correct for refraction & parallax &c when you make this observation? This is called the "clearing the distance" problem. For my first attempt (in the absence of better information-- but more on that later), I simplified the problem of parallax by making my observation at the moment of lunar transit, and ignoring refraction by making sure to keep my observation above 20 degrees apparent altitude for both bodies. To make it easy with a reflecting horizon, I used the sun as the second body. In this case, to clear the distance I had to apply corrections for the semi-diameter of both bodies (I used near limb to near limb, therefore added both corrections), and also the small correction for lunar augmentation. At the moment of lunar transit, there is no parallax in the plane perpendicular to the meridian, only the normal parallax in altitude (PA) which is derived from the horizontal parallax value in the almanac (H.P.). The PA is computed simply as: PA = HP cos A where A is the apparent altitude of the moon. I have therefore constructed a spherical triangle that looks (something) like this: Z . /| / | / | / | S/----+ Ma \. | `\| Mo Where Z is the zenith, Ma is the actual position of the moon. Mo is the observed position of the moon, and S is the sun; and where Z-Mo is the apparent zenith distance of the moon, and Z-S is the apparent zenith distance to the sun, and Ma-Mo the parallax in altitude of the moon (PA). Now for problem #2. Lunar distance tables. Lunar distance tables haven't been produced for over 80 years so I generated a little visual basic program that would crank out lunar distance solutions for every ten seconds for the hour of the observation--i.e., just enter the sun's GHA & dec, and moon's GHA and dec, and solve the problem... I then just go down the list until the find the distance that corresponds to what I measure, and Bingo! I know the time in Greenwich at the time of my obs! When I have done this using a mechanical pocket watch which gains about 1.8 seconds per hour (and the rate varies!) I can find my longitude to within about 16' or 10 nm at this latitude using this technique. The really cool thing about this is that MOST of my errors come from trying to pick the exact moment of local apparent noon! The problem with this technique is that it is so restrictive, I mean, how many times per month do you get the moon transiting at same time that the sun is more than 20 degrees above the horizon, and yet far enough from the moon so that the moon is visible, and not more than 130 degrees from the moon (so you can measure the distance with a sextant)? Well, you can count 'em on the fingers of one elbow... e.g., a couple of times a month... An Historical Technique: This summer I stumbled upon the most excellent source on this stuff: Charles H. Cotter, "A History of Nautical Astronomy", American Elsevier Publishing Company, Inc., 52 Vanderbilt Avenue, New York, New York, 10017. 1968 Here is a method for clearing the distance first published by Young, in 1856. First, construct the following spherical triangle (only better than this one!): Z ,/\ ,/ `\ ,/ `\ ,/ `\ M s,/--, ,----\ ,/ `-,----' `\ ,/ ,----' `-----, `\ m S,/----' `-------`\ ,/ `\ H /-----------------------------------`\ O Z is the zenith, M is the actual position of the moon, m is the apparent position of the moon, S is the actual position of the sun, s is the apparent position of the sun. Zs is the apparent zenith distance to the sun, ZS is the actual zenith distance to the sun, ZM is the actual zenith distance to the moon, and Zm is the apparent zenith distance to the moon. HO is the observers horizon. Zs is less than ZS due to refraction, but Zm is greater than ZM because the effect of the lunar parallax in altitude (PA) is (normally) greater then the refraction. What we see (and measure) in the sky is the observed lunar distance ms, what we want to accomplish by "clearing the distance" is to compute the actual distance MS -- this distance can then be compared to the lunar distance table/output as described above. For triangle ZSM, using the law of cosines for spherical triangles we can write: cos Z = (cos SM - cos ZS * cos ZM) / (sin ZS * sin ZM) remembering your math teacher who insisted that: sin (90-x) = cos x cos (90-x) = sin x we can write: cos Z = (cos SM - sin HS * sin OM) / (cos HS * cos OM) Next, for triangle Zsm, we can write another equation for Z: cos Z = (cos sm - cos Zs * cos Zm) / (sin Zs * sin Zm) An again, remembering our trig: cos Z = (cos sm - sin Hs * sin Om) / (cos Hs * cos Om) This gets a bit ugly, so let's simplify the notation a bit, let's define some new variables: D = SM, the true lunar distance S = HS, the true altitude of the sun's center. M = OM, the true altitude of the moon's center. d = sm, the observed lunar distance. s = Hs, the sun's apparent altitude. m = Hm, the moon's apparent altitude. we can now re-write the above equations as: cos D - sin S * sin M cos Z = --------------------- cos S * cos M and cos d - sin s * sin m cos Z = --------------------- cos s * cos m Now, add one to both sides of each equation: cos S * cos M cos D - sin S * sin M 1 + cos Z = ------------- + --------------------- cos S * cos M cos S * cos M or, cos D + cos S * cos M - sin S * sin M 1 + cos Z = ------------------------------------- cos S * cos M and likewise to the second equation: cos d + cos s * cos m - sin s * sin m 1 + cos Z = ------------------------------------- cos s * cos m Now, remebering that wonderful trig formula (that I just now had to go and look up again...) cos (x + y) = cos x * cos y - sin x * sin y we can now write: cos D + cos (M + S) 1 + cos Z = ------------------- cos M * cos S and, cos d + cos (m + s) 1 + cos Z = ------------------- cos m * cos s so, now be equating the two equations we get: cos D + cos (M + S) cos d + cos (m + s) ------------------- = ------------------- cos M * cos S cos m * cos s Multiplying both sides by cos M * cos S we get: [cos d + cos (m + s)] * cos M * cos S cos D + cos (M + S) = ------------------------------------- cos m * cos s This can be written as: cos M * cos S cos D + cos (M + S) = [cos d + cos (m + s)] * ------------- cos m * cos s Subtracting cos (M + S) from both sides we get (at last): cos M * cos S cos D = [cos d + cos (m + s)] * ------------- - cos (M + S) cos m * cos s This is Young's formula for clearing the distance. In the best case, you use three observers to measure the alitiude of the moon, the altitude of the sun, and the lunar distance at the same instant. If you are doing this alone, then you must take a few obs of the lunar and solar altitudes, then measure the distance, then a few more lunar and solar altitudes. You must then plot the altitudes and pick the altitudes at the instant of the lunar distance measurement. One of the benefits of Young's method is that slight errors in the altitudes do not have a large effect on the result. If anyone is really keen, I have more methods form Cotter's book...