NavList:

Hello Bruce,

Provided I understood everything correctly (I figured out that the first UT was 16h22m02.2s and so on ....) and provided I correctly entered all ... 325 numbers on my Calculator - while knowing that just one single typo can ruin all the results - I found the following results (7th order regression) :

UT of Transit (i.e. body full south of Observer's Position) 16h44m12,8 , hence a Longitude of W070°04'3 ,

UT of Culmination occurred 8,23s earlier, i.e. at UT = 16h44m04s6 . The 8.23s difference between both comes from the Sun Declination which is decreasing. In this case, a first order value for the time difference in seconds of time is (48/Pi)* Declination change in arc minutes per hour.

I assummed that the Observer is steady on Earth.

Observed latitude N42°02'3 S

Standard Deviation of the Observations : 0.9 NM (defines quality of Observations)

Longitude Uncertainty / Latitude uncertainty (both in arc minutes) <= 3.9

In other words, the Latitude should be accurate to +/- 2.7 NM at 3 Sigma, and the Longitude should be accurate to +/- 10'5 at 3 Sigma, as long as I have correctly entered the 64 different values for these 32 Observations.

Best Regards to all

Kermit
----------------------------------------------------------------
NavList message boards and member settings: www.fer3.com/NavList
Members may optionally receive posts by email.
To cancel email delivery, send a message to NoMail[at]fer3.com
----------------------------------------------------------------