# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: finding your latitude through double altitudes (and elapsed time).**

**From:**George Huxtable

**Date:**2007 Apr 1, 14:53 +0100

Joel asked, in [NavList 2512] , about finding your latitude through double altitudes (and elapsed time). I don't recall previous contributions from Joel to this list, and wonder if this is his first? If so, may I welcome him, and his interesting and well-expressed question, and hope we may get more from him along similar lines. I'm not going to answer that question here, but instead point him to where he can find the answer. If such mathematical questions interest him, he should certainly invest in a copy of William Chauvenet, Manual of spherical and practical astronomy, 1863 (in two volumes). It's become a Dover reprint, which I think was squeezed into a single volume. It may, by now, be available in digital form. Before he took up his chair at Washington University, St. Louis, Chauvenet was astronomer to the US Navy, so he knew what he was talking about. Chauvenet approach is rigorous to say the least, and often hard going. He loves to work on the most general example possible, and then simplify that general case when approximations are practical. His chapter 6, on "Finding the latitude by astronomical observations", runs in vol.1 of my edition from pages 223 to 316, and includes 11 methods (!) for use on land, and 8 (!) for use at sea. I give the relevant page numbers from my edition in case that same edition exists as a digital scan. The relevant method is the 5th on-land "by two altitudes of the same star, or different stars, and the elapsed time between the observations", which runs from page 257 to 277, articles 178 to 183. And you then have to combine that information with the 8th method for finding the latitude at sea, "by two altitudes with the elapsed time between them", pages 313 - 316, articles 209 to 212. Because of the way Chauvenet keeps referring back to an earlier diagram (on page 257) and scattered definitions and equations, I am unable to distil a succinct answer from it to Joel's question. Chauvenet deals specifically with the way that Bowditch tackled the matter (on page 269), giving a numerical example. All the information that Joel might seek is there, but it's rather too much to send by email. If Joel passes me his postal address, I'll be happy to scan the relevant pages and post them. They are well out of copyright! I have not detected any errors in all I have read from Chavenet's work, and regard him as completely trustworthy, if somewhat pernickety about the detail. He works with the precision that's required of an astronomer, rather than a navigator. ==================== The double altitude problem is also discussed at some length in Charles H Cotter's "A History of Nautical Astronomy", 1968, and Joel might find it useful to refer to pages 152-155. I have a serious distrust of Cotter when he gets into anything mathematical, and a long list of perceived errors in that book, collected by Jan Kalivoda and me, can be found at www.huxtable.u-net.com/cotter01.htm . However, he has delved into many obscure texts, and provided information which exists nowhere else. It's in interpreting them that he seems to get out of his depth. It's a really useful book to have, but now hard to find and expensive. He looks into Ivory's method of 1821, and Ainsley's of 1867, both of which seem to correspond with the Bowditch procedure that Joel describes so well. His diagram, figure 7, corresponds exactly in its lettering with Joel's descripion, but to understand what's going on, some more construction is called for. The initial presumption is that changes in the Sun's declination during the observation are small enought to be neglected, as long as a mid-value is used for declination. First, Joel needs to join points X and Y by a great circle (not by the path of the Sun, which is a great circle only at the dates of the equinoxes). Because X and Y have the same declination, that great circle arc XY is symmetrical about its vertex, the mid point where it comes closest to the pole P. That mid-point is to be marked M. The resulting two triangles PXM and PYM are both right-angled (at M). The arc MX = MY is what Joel has taken as A. The latitude of M then corresponds to B. And his first two equations then follow from the rules for right-angled spherical triangles. Next, some more construction. Join M by an arc to the pole P. Where that arc passes closest to the observer's zenith Z, mark another point R, and join ZR. MRZ and PRZ are now two more right-angled triangles (at R). As I see it, Joel's C now corresponds to the arc MR, his Z to the arc ZR, and E to the arc PR. Now, applying more spherical trig to the right-angled triangles ZMX, ZMY, and ZMR, brings him to his third equation (for C) and from that to his fourth equation (for Z). I haven't even tried to follow the details of those manipulations, but they are all in Cotter. Then, if that's correct, I get a bit puzzled about his fifth equation, which he states to be E = Z +/- B, because as I see it, those angles are not in the same plane. It seems more likely to me that the intention is to derive the lat (or perhaps colat) of R from the lat (or perhaps colat) of M, using the difference angle C. So I think that fifth equation should be combining B and C to obtain E, and suggest that Joel might take another look at it. Now we know E (= PR) and Z (= ZR) in the right-angled triangle PZR, so we can compute the third side PZ, which is the colat (and from that the lat) of the observer. which was what was wanted. Note that I have completely ignored any subtleties in the complex sign-rules that Bowditch presents, in aiming just to give a general picture of how that complex calculation was meant to work. Hope I've got it right, but not completely confident... Anyway, it may give Joel a start-off in the right direction. George. contact George Huxtable at george@huxtable.u-net.com or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. ==================== Joel Silverman wrote in Navlist 2512- | I am having difficulty figuring out the logic behind the calculations | for finding the latitude using the method of double altitudes. I am | working from an 1833 edition of Bowditch. I do not know how far back | this particular method goes ... I | am sure it is earlier than 1833. This edition of Bowditch gives three | methods. The first method is the one I am asking about. The second | method is identified as basically the method of Douwes and the third | method takes into account the possibility that the declination or | polar distance of the object at the second observation may differ from | that of the first observation. | | I'll summarize what the method does, and how the algorithm proceeds | and then I'll ask my question(s). | | The method is as follows: determine the corrected altitude of the | sun at two points in time, noting the elapsed time between the two | observations. You do not need to know the actual time of either | observation, simply how much time has passed between observation #1 | and observation #2. A series of logarithmic calculations lead | eventually to your latitude. | | The method in its simplest form assumes that the ship is stationary | during the interval, but it is easily modified to take into account | movement of the ship between the observations using a dead reckonning | of the change in position between them. | | The details, as described in Bowditch are as follows: | | The sum of the log-csc of the elapsed time and the log-sec of the | declination is the log-csc of arc or angle A. | (The values of numerical examples in Bowditch indicate that he is not | really using the the log-csc of the elapsed time as he claims, but the | log-csc of half the elapsed time). The sum of the log-csc of the | declination and the log-cos of A determines the log-csc of B. The | sum of the log-cos of half the sum of the two altitudes and the log- | cos of half the difference in the two altitudes is the log-sin of | C. The sum of the log-cos of A, the log-csc of the half sum of the | alititudes, the log-sec of half the difference in altitudes, and the | log-cos of C determines the log-sec of azimuth angle Z. | | Z is named north if the zenith and the north pole lie on the same side | of the great circle between the two observed points and is named | south if the zenith and north pole lie on opposite sides of that great | circle. B has the same name as the declination. E is the sum of B | and Z if B and Z have like names and is the differnce if B and Z have | contrary names. Finally the sum of the log-cos of C and the log-sine | of E is the log-sin of the latitude. | | I am struggling to see what this elaborate rule does in terms of the | spherical triangle P-X-Y , the spherical triangle Z-X-Y, and the arc | P-Z (where X and Y are the positions of the sun at the time of the | two observations, P is the north celestial pole, and Z is the | zenith. (P-Z is the complement of the Latitude, P-X =P-Y = | complement of the declination, Z-X is the complement of the first | altitude, Z-Y the complement of the second altitude, and X-Y is the | arc of the great circle between the two solar positions. | | With the availability of computers and calculators, it is no longer | essential to view things in their logarithmic form. | Direct solution of the trigonometric problem is now easy and it is far | easier to see what the method is actually doing from a geometric or | trigonometric viewpoint without the logs in the way. I assume that | the calculations involving half the sum of the altitudes and half the | difference in the altitudes derive from the trigonometric identity for | converting sums and differences of trig functions to products of trig | functions (so that logarithms may be used). | | If I reexpress Bowditch's method without using logarithms or the sum/ | difference to product identity, the calculations are equivalent to | solving the set of equations below. | | Let PD (polar distance) be the complement of the solar declination and | let theta be the hour angle corresponding to the elapsed time between | observations. Let H1 be the altitude of the first solar observation | and let H2 be the altitude of the second solar observation. | | The rule is then captured by solving the following six equations: | | sin(A) = sin(theta/2) * sin(PD) [solve this equation for | A] | | cos(PD) = sin(B) * cos(A) [solve this equation for | B] | | 2 sin(C) * sin(A) = sin (H2) - sin(H1) [solve this equation for | C] | | 2 cos(A) * cos(C) * cos(Z) = sin(H1) + sin(H2) [solve this | equation for Z] | | E = Z + B (or E = Z - B as per rule) [solve this equation | for E] | | sin(Latitude) = sin(E) * cos(C) [solve this equation | for Latitude]. | | Can anyone help me understand what these angles and arcs represent. | Where in the diagram are A, B, C, Z, and E ?? How do these | calculations derive from the geometry of the situation ? It seems to | me that A is the value of half the length of the great circle segment | connecting the geographic position of the sun and the time of the two | observations. Why he wants half of the segment I do not understand. | I have been staring at these equations for weeks and can not see what | B, C, Z , and E are. If there are auxilliary lines that have been | added to the original triangles to convert the oblique spherical | triangles to right triangles, then what are those lines, which | triangles are being considered and how to A,B,C,Z, and E relate to | these auxilliary figures? If anyone can untangle even some parts of | this rule, perhaps it will put me back on track for figuring out the | rest. | | I have seen a direct method explained in several sources, where the | declination and elapsed time are used to compute the length of XY, | then the law of sines is used to calculate one of the base angles in | PXY. Now the three sides of ZXY are known and the law of cosines is | used to find the angles. Subtracting the base angles gives enough | information to find an angle in PZY (or PZX) and PY and ZY being | known, the side PZ can be found by the law of cosines and the | complement of PZ is the latitude. But I have not been able to see a | connection between Bowditch rule and this direct method. I don't | know if that is because there is no connection, but rather a totally | different way of solving the problem, or if I'm just not seeing it. | | I hope someone can shed some light.... this is driving me crazy. | | | | | --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@googlegroups.com To unsubscribe, send email to NavList-unsubscribe@googlegroups.com -~----------~----~----~----~------~----~------~--~---