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Having some unexpected "free time" for this week-end, I was able to better 
study the "exotic" LAN case which Gary published in [NavList 10009] . This 
was made possible thanks to the "solution" he gave us in [NavList 10041] as 
being LAN Time = 17h16m32s with LAN Position N22d47'7 W079d26'5 .

I wish to share some interesting results.

First, from Gary's solution I found that Culmination time is 18h14m18.05s with 
a culmination height of 46d15'715. Hence the difference between LAN and 
Culmination times is 57m29.80s .

The only Culmination Time published as a answer to [NavList 10009] was made by 
Andres in [NavList 10014]. Andres found a Culmination Time of 18h14m12.8s 
which is pretty a impressive result given the difficulty of the case 
submitted by Gary ("Exotic vessel" speeding at 300 kts due south).

Two members published results for LAN :

Andres in the same [NavList 10014] gave 17h12m45.9s , with LAN Position 
N23d05'96 and W079d42'4 . At time 17h12m45.9s actual "ship" position computed 
from Gary's results in [NavList 10014] was N23d05'84 and 79d26'5 . Therefore 
the result computed by Andres, although not accurate as regards LAN Time was 
EXCELLENT for the Latitude he finds at his indicated LAN Time.

Peter Hakel in [NavList 10012] gave 17h13m35.0s , with LAN Position N22d40'8 
and W079�42'4 . At time 17h13m35.0 s, actual airship position from Gary's 
results in [NavList 10014] was 23d01'75. This result is off by som 14' , but 
on the other hand Peter's result for LAN is fairly close from Gary's 
position.

The main cause of the differences in positions found by both Andres and Peter 
-who, under "non exotic" cases, always publish accurate results - can 
certainly be attributed to a fact which they both analyzed very well in 
[NavList 10014] and [NavList 10012], i.e. results given by Formulae such as 
the one published by Jim can no longer remain accurate under such huge 
changes in Latitude during the time span of the Observations.


And Last but not least, I discovered something I was not expecting at all 
given the very high N/S speed which in my feeling would do more than 
offset/compensate for a culmination hight quite far from zenith : the 
Altitude curve of Gary's example is ALMOST PERFECTLY SYMMETRICAL when studied 
from its Culmination time. Simple results as follows :

Culmination Time / Height :  18h14m18.05s / 46d15'715

15 minutes earlier and later :

         17h59m18.05s / 46d05'393    and    18h29m18.05s / 46d05'318


30 minutes earlier and later :

         17h44m18.05s / 45d34'684    and    18h44m18.05s / 45d34'631


60 minutes earlier and later :

         17h14m18.05s / 43d35'551    and    19h14m18.05s / 43d35'280


I am quite surprised by these results because for culminations close to zenit 
(say above 85d) the effect of a high N/S speed component and/or change in 
declination significantly skews the heights vs times curve. Given the almost 
perfect symmetry of this Altitude curve, the algorithm used by Andres - 
apparently a second degree best fit as seems to be indicated on his site - 
was very powerful to pick up almost the exact the Culmination time. Only the 
presence of higher order terms (in t**4 and maybe t**6) prevented Andres to 
do the perfect pick of the exact Culmination time.

I would conclude by thanking Gary, Andres and Peter for the interesting 
unusual/exotic case we had the opportunity to study together.

Best Regards to all


Antoine

Antoine M. "Kermit" Couette


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