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I had looked my atlas wrong when I came up with the latitude for
Jerusalem of 31� 16' north. It is actually at 31� 46' north, only 14
miles south of the latitude of 32� used by Wright.


gl

On Sep 21, 1:29 pm, Gary LaPook  wrote:
> 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


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