# NavList:

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

**Re: Certaine Errors in Navigation Corrected**

**From:**Gary LaPook

**Date:**2007 Sep 21, 15:50 -0700

When computing great circle distances using Wright's method you can get better accuracy by using a modern plotting sheet such as the Radar Manoeuvering Board which has more even divisions and has a larger scale. It takes much less time to do the computation than it takes to discribe the process. gl On Sep 21, 1:29 pm, Gary LaPookwrote: > In 1980 I discovered a book in my college's library which was a reprint > of Edward Wright's "Certaine Errors in Navigation" which had been > published in 1599. I still remember how pleasurable it was to read, it > was like sitting down with an old friend and discussing navigation. He > seemed like a thoroughly modern man. I highly recommend this book if you > can find a copy. There were other navigation books reprinted as part of > "The English Experience" series and they were also worth reading. > > Edward Wright was a mathematician who turned his interest to navigation > and was the first one to devise how to compute the table of meridional > parts necessary to create charts on "Mercator's Projection" and > published such a chart covering England to the Azores based on a voyage > he took with the Earle of Cumberland in 1589. I attached a copy of that > chart to a previous post. For biographical information on Wright you > can go to: http://books.google.com/books?id=uTsJAAAAIAAJ&pg=PA101&lpg=PA101&dq=c... > > In "Certine Errors" Wright laid out a method to compute the great circle > distance between two points on the earth using only a straight edge and > a pair of dividers! I have never seen this method described in any other > navigation text. Although it is now trivial to do this computation with > a calculator it think it might be interesting to look at Wright's > method. I am attaching 13 pages of his book that I photocopied back in > 1980 but I am missing several pages which would fall between the second > page of reproduced text (ending with "places.") and the next reproduced > page starting with "foote," (they didn't number their pages.) Don't be > intimidated by the type and spelling as you quickly get the hang of it > and can read it as quickly as modern text. There also appears to be a > couple of typographical errors that I will point out. > > I will also offer my own explanation. > > Wright starts out with special cases but I think it is easier if we > start out with the general case. Wright's description of the general > case starts on the last line of the third page of reproduced text ("If > the latitudes be not both equall...") and continues with a couple of > examples using London and Jerusalem ("Hierusalem") and London and Cusco > as the examples. The coordinates he uses for London are 51� 32' north > (which is as accurate as its modern latitude and a longitude of 22� (all > of his longitudes are east). ( I do not know where he got this value > since it does not comport with his own chart (which would make it about > 16�) published in the same book, perhaps an earlier longitude using a > different prime meridian (perhaps the Cape Verde Islands) since he > couldn't be off by 6� between the Isle of Wright and London.) The > latitude of Jerusalem he gives as 32� north while its modern value is > 31� 16' north and he uses 11� south for Cusco while its correct value is > 13� 28' south. The longitude he uses for Jerusalem is 68�, making it > 46� east of London while its true value is 35� 14' east, a longitude > error of 11�. He gives the longitude of Cusco as 295� making it 273� > east of London while its modern longitude is 72� west, the same as 288� > east of Greenwich showing an error of 15� between Wright's longitude and > the modern value. > > We will use Wright's values for this illustration, 51.5 for latitude of > the London, 22 for the longitude of London; 32 and 68 for Jerusalem; > and 11 south and 295 for Cusco. > > Wright produces a diagram of the method however is is very complicated > since all the lines for his many examples are contained on one diagram. > We will draw only the lines we are interested in but will use Wright's > labels for the lines. > > Starting with the first example of figuring out the distance between > London and Jerusalem, we first draw a line across the base circle > diagram from the longitude of London, 22, "B" all the way across to the > opposite side at 202., "C." Wright would have us next do the same for > the longitude of Jerusalem but instead lets draw the line representing > the latitude of London. Add its latitude, 51.5, to 22 and place a dot on > the circle at 73.5, "E." Then taking a triangle or a plotter place its > straight edge on the line "B-C" and draw a line from "E" that is > perpendicular to "B-C," marking the intersection as "F." (Of interest is > that the length of "E-F" is the sine of 51.5 and the length from the > center of the circle (not labeled) to "F" is the cosine of 51.5.) > > Next we do the same for the position of Jerusalem, drawing a line from > 68, "M," across to 248, "J." Then we draw the perpendicular representing > Jerusalem's latitude, from 100 (68+32) "N" to intercept "M-J" at "O." > (The lengths of these lines represent the sine and cosine of 32.) > > Since London and Jerusalem are both in the northern hemisphere, we next > subtract the length of Jerusalem's latitude, "N-O," (the smaller value) > from the length of the latitude of London, "E-F," by taking a pair of > dividers and setting them to "N-O" and then placing one point on "E" and > the other point on the line "E-F" and marking this point "Q." (Note, if > you are following Wright's description at this point, there is a typo in > the text were it states "and P Q equall to N O" it should say "and E Q > equall to N O.") Next we determine the distance between the points "O" > and "F" (the intersections of the latitude lines with their respective > longitude lines) with our dividers and leaving one point on "F" we swing > the other point to the line "B-C" and mark it "P." Leaving that point on > "P" we next move the other point of the dividers to "Q," this is the end > of the chase. With the dividers set to the spacing between "P" and "Q" > we move the dividers to the scale on the circle and take out the number > of degrees between the divider points, this is the number of degrees of > the great circle joining the two places on earth. Multiply by 60 to > calculate the number of nautical miles. > > Does it work? Wright came up with "38 degrees and about 3/4 that is 775 > leagues, which are 2325 miles." Using a modern digital calculator your > end up with 38.614 degrees, 2316.8 miles a difference of 8.2 miles or a > 0.3% difference. Not bad with just a straight edge and dividers. > > Wright uses London and Cusco to illustrate the case when the two places > are on opposite sides of the equator. The only difference is that > instead of subtracting the latitude of one from the other you add them. > You do this by extending the longer latitude line till it goes outside > of the circle and then marking on this extended line the distance of the > latitude of the other location. Using the diagram we already have and > pretending that Jerusalem's latitude is south instead of north, we > extend the line "E-F' some convenient distance outside the circle. Then > placing the divider points on "N-O" we take off that spacing and placing > one point on "E" we place the other point on the extended line outside > the circle and this becomes our new "Q," lets call it "Q2." Then taking > the distance from "Q2" to our original "P" we take out the spacing and > place the dividers on the circle and find the number of degrees of the > great circle between London and a Jerusalem located in the southern > hemisphere. I come up with 92 degrees for a great circle distance of > 5520 miles. Using a computer I get 5565 miles, a difference 45 miles or > 0.8%. > > Wright uses Cusco to illustrate this case, drawing in lines "R-V" and > "S-T" in addition to the original lines for London. Taking the distance > between "S" and "F" and using it to establish "X, "(the equivalent of > "P" in the first example.) Then the line "E-F" is extended and point "Y" > is established by adding the length of "S-T" to "F-E" which takes us > outside of the circle to "Y." The distance from "Y" to "X" (the > equivalent of "P to Q" in the first example) when placed against the > degree scale on the circle gave Wright "almost 97 degrees, of a great > circle that is 1940 leagues or 5820 miles." the calculator gives 96.74 > degrees or 5804 mile, a difference of 16 miles or 0.3% difference. > > I don't know why it works and maybe somebody on the list can explain it > to me. I spent some time with the law of cosines and found a way to > reproduce this method using a calculator but you end up with a very long > and inelegant formula. > > BTW, it turns out that you never stop learning. I had always written the > law of cosines as c= square root of (a ^2 + b ^2 - 2ab cos C) but > working with it now I come up with c^2 = a^2 + b^2 -2ab cos C. I then > saw the pattern for the first time, the first part looks just like > Pythagoras' theorem. I then realized that the Pythagoras theorem is just > a special case of the law of cosines. When C=90� its cosine becomes zero > and the second part of the expression drops out leaving our old friend > Pythagoras! > -- > > ___ > > English Experience.pdf > 22KDownload > > Certaine.pdf > 392KDownload > > Illustration.pdf > 151KDownload --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---