@geeoddmike?ÿ Fichot was reference number 24 in P.D.Thomas work! Also what does the page you posted above have to do with?ÿMr. Dave Lindell's question.?ÿ ?ÿ NOTHING
@mkennedy?ÿ?ÿNO GUESSING.
It is either 42 or C. ??? ????ÿ
I don't have enough cash to buy a clue on this one.
Did everyone give up on this question?
I don't think I got an answer from here.
@dave-lindell?ÿ?ÿCongratulation are in order for you DAVE. It looks like you have stumped the board again.?ÿI think you should give them just a little more?ÿTIME, say till?ÿ midnight of 10 April 2021.?ÿIf no answer by then I guess you or I will give the?ÿ answer (how does that sound?).
JOHN NOLTON
From page 8 of the publication:
The published inverse solutions have been more varied than the direct. The series expansion for the
geodesic length in the flattening f, spherical length d ( with reference to the geodetic latitude of the vertex
of the great elliptic arc) in the form
S = a [ d - F1(d)f + F2 (d )f^2+ .. ]
was published by A.R. Forsyth in 1895 , [ 20] . (emphasis added along with the notation for f squared)
So the first distance calculated, S1, stops the formula at the second term while the second distance, S2, stops the formula at the third term. Thus, S1 is a first-order approximation and S2 is a second-order approximation.
In Appendix 3, there are many completed calculation sheets. In each case, the author first chose a beginning lat/lon, an azimuth, and a distance. Then he used his Direct calculation to calculate the end points the determined line.
The Inverse calculation on his following sheet essentially verified the accuracy of both algorithms.
Great piece of mathematics; thanks John and Dave!
?ÿ
?ÿ
?ÿ
?ÿ
@mathteacher?ÿ?ÿOutstanding MathTeacher. If you go to page 12 of P.D. Thomas work, 11 and 12 lines up from the bottom you will see "We can now compute S(sub1) (first order for comparison) or go directly to S(sub2) for the second order distance as shown in figure 5.?ÿ (Figure 5 is on page 14).
Its interesting to note that the distance P.D. Thomas computes (using only 2nd order terms) is 9,649,412.793 meters. NGS (Vincenty method) gets 9,649,412.805?ÿ ...12mm difference. The azimuth is minus 0.003 seconds for the direct and plus 0.004 seconds for the reverse.?ÿNot to bad when you are only looking for the distance to be 1/100,000 (of the true distance) and the Azimuth to be within 1 second.
JOHN NOLTON
Why here is a printed copy:
BTW, the clear scanned copy you linked and I couldn??t access was probably due to having to enter a displayed ??garbled? text. My search terms were Spheroidal geodesics thomas
Why here is a printed copy:
A little rich. I doubt it will sell at that price, with free pdf available.