A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Antoine Couëtte
Date: 2020 May 8, 05:30 -0700
Thanks for you interesting question.
Here is one manner - among certainly very many - to solve it.
With (O) being the unknown Observer's position, with "Ht(O)" (54.9317°) being the Sun Center Topocentric unrefracted Altitude, and finally with "SZ(O)" (120.4650°) being the Sun Azimuth seen from (O) :
A - Solution on a spherical Earth
A1 - From Ephemeris Data, compute position of Subsolar point (S) on Earth
A2 - From Ht(O) compute Sun Center Geocentric Height "Hg(O)"
A3 - From Hg(O) compute Great Circle Distance "GCD" from (S) to Observer (O) : GCD = 90° - Hg
A4 - From currently known elements, compute approximate Meridian Convergence Correction #1 "MCC1" from (S) to (O)
A5 - ITERATE
Bold Characters here-under are the back bone for "iteration support" since they do not change.
A.5.1 - Determine Observer's Azimuth "OZi" as seen from (S). OZi= SZ + 180° + MCCi
A.5.2 - From (S) determine a point (Pi) on Azimuth OZi and at a Distance GCD
A.5.3 - From (Pi) compute Azimuth to Sun "SZi"
A.5.4 - Compute [ SZ - SZi ] and if smaller than 0.1', GOTO B
A.5.5 - Compute Updated MCCi+1 = MMCi + [ SZ - SZi ] , then GOTO A.5.1
B - Further Checks : Solution on WGS 84 Ellipsoid
(Pi) should now be very close from unknown (O).
B.1 - From (Pi) compute Ht(Pi) and SZ(Pi)
B.2 - Compare Ht(Pi) to Ht(O) and SZ(Pi) to SZ(0)
B.3.1 - If B.2 differences outside 0.1', fine-tune (Pi) and GOTO B.1
B.3.2. - If B.2 differences less than 0.1', GOTO B4
B4 - The position thus derived is the requested Observer's Postion (O) .
With (O) being the unknown Observer's position, and "Ht(O)" (54.9317°) being the Sun Center Topocentric unrefracted Altitude, and finally "SZ(O)" (120.4650°) being the Sun Azimuth seen from (O)
Here are my results (4 digits published only to avoid possible accumulation of round-off errors):
A.1 - (S) N16.7797° W045.8520° Sun Distance = 1.0090 UA
A.2 - Only required correction here is parallax equal to 8.8" / (Sun Distance) = 8.7218" = 0.0024°
Hence Hg(O) = Ht(O) + 0.0024° = 54.9317° + 0.0024° = 54.9341°
A.3 - GCD = 90° - 54.9341° = 35.0659°= 2103.9546 NM
A.4 - From (S), start a Great Circle at Azimuth SZ(O) + 180° at a Distance of 2103.9546 NM.
Such Great Circle covers approximately :
35.0659 ° x sin (SZ + 180°) = 35.0659 ° * Sin (320.4650°) = 17.7788 ° in Latitude.
Hence get an approximate mid-Latitude of 16.7797° + (17.7788° / 2) = 25.6691°
Such Great Circle also covers approximately :
35.0659° x cos (SZ + 180°) / Cos (Mid Lat) = 35.0659° cos (320.4650°) / Cos (25.6691°) = 33.5341° in Longitude.
Approximate Meridian Convergence (1 st Order) : Longitude Difference * Sin (Mid-Latitude) = 33.5341 * sin 25.6691° = 14.5261 ° .
We will therefore start up with MCC1 = 15° and with, as indicated here-above, SZ(O) = 120.4650°
A.5 – ITERATE
A.5.1 - OZ1 = SZ(O) + 180°+ MCC1 = 315.4650°
A.5.2 - (P1) N38.9311° W077.0487°
A.5.3 - SZ1 = 120.3226°
A.5.4 - SZ(O) - SZ1 = + 0.1424°
A.5.5 - MCC2 = MCC1 + (SZ - SZ1) = 15.1424°
A.5.1 - OZ2 = SZ(O) + 180° + MCC2 = 315.6074°
A.5.2 - (P2) N38.99788° W076.99828°
A5.3 - SZ2 = 120.46458°
A5.4 - SZ(O) - SZ2 = - 0.00042° = 0.02491’ which is smaller than 0.1’ , then we can exit after this 2nd loop.
B.1 - (P2) N38.99788° W076.99828° , i.e. N38°59.87’ W076°59.90’
Fine-tuning on WGS84 Ellipsoid :
Get Ht(P2) = 54°55’961° (vs. 54°55.9’) and SZ(P2) = 120.46445° = 120.27’87 (vs. 120°27.9’)
Azimuth is fine, and we just need to decrease the height by 0.061’ .
From position here-above, proceed 0.061 NM in the direction opposite to the Azimuth
Hence Δ Lat = - 0.061’ cos Az = + 0.031’ Lat (O) = N38°59.99’
Δ Lon = 0.061’ sin Az / Cos Lat = 0.068’ Lon (O) = W076°59.97’
The Observer’s position is thought to be at N39°00.0’ W077°00.0’ within the limits of the data accuracy (0.1').
Antoine M. "Kermit" Couëtte