NavList:

Gentlemen

I undertook to find “Ohama Reno Harbour”.  This apparently is an old name, since there is no reference to it on the web.

However, using Kermit’s derived lat, long, we can find that the location is near to Tahiti, yet it is still out in the Pacific Ocean.

This is not in a harbor.  It is shown by the red marker “A”.

Since we do know that it is a harbor and it must be in French Polynesia, I poked around, until I found a location called

Cook’s Bay.  It is on the middle of the three islands pictured above.  The latitude was given as S17.49

and the longitude given as W149.82.  In investigating this, a website indicated that this is not where he

actually anchored ( http://www.pacific-travel-guides.com/tahiti-islands/travel-guide/moorea.html )

That same website indicated that Cook actually anchored in “Opunohu Bay”, French Polynesia. 

Latitude S17.49 W149.857.  It is the next bay over from Cook’s Bay.  There are two bays on the north shore,

this is the western bay.

I believe we have just been handed the keys to the kingdom.  Firstly, we can fiddle with the altitude given until

it provides us with an appropriate Latitude.  There are only so many corrections that can apply.  It’s a simple

combinatorial problem.   One caution is that the bay is somewhat long in the N/S direction, from S17.48 to S17.51

With the altitude problem resolved, we can move on to the longitude.

Best Regards

Brad

Dear George and Brad,

Please find hereunder the positions derived from 3 of the 4 Lunars taken in Ohama Reno Harbour. As for ALL previous results which I have published here I keep assuming that all 1773 published Heights were corrected for ONLY instrument errors. For the time being this still remains a most reasonable assumption.

*******

07 Sep 1773 , delta T = + 16.4 s

IF WE ASSUME HEIGHT OF EYE IS 0 FT, then get the following :

Lunar Distance Observation Time UT = 17h05m47s3, and
Observed position : S 16°22°02" / E 208°10'51", and

IF WE ASSUME HEIGHT OF EYE is 12 FT, then get the following results :

Lunar Distance Observation Time UT = 17h05m37s6, and
Observed position : S 16°29'19" / E 208°11'01"", and

IF WE ASSUME HEIGHT OF EYE is 24 FT, then get the following results :

Lunar Distance Observation Time UT = 17h05m33s6, and
Observed position : S 16°32'19" / E 208°11'05"

*******

09 Sep 1773 , delta T = + 16.4 s

IF WE ASSUME HEIGHT OF EYE IS 0 FT, then get the following :

Lunar Distance Observation Time UT = 18h05m38s0, and
Observed position : S 16°21'22" / E 208°38'53", and

IF WE ASSUME HEIGHT OF EYE is 12 FT, then get the following results :

Lunar Distance Observation Time UT = 18h05m32s7, and
Observed position : S 16°27'04" / E 208°38'17", and

IF WE ASSUME HEIGHT OF EYE is 24 FT, then get the following results :

Lunar Distance Observation Time UT = 18h05m30s3, and
Observed position : S 16°29'26" / E 208°38'10"

*******

11 Sep 1773, , delta T = + 16.4 s, with results here-below already published in [NavList 12623] earlier to-day.

IF WE ASSUME HEIGHT OF EYE IS 0 FT, then get the following :

Lunar Distance Observation Time UT = 19h01m07s0, and
Observed position : S 16°31'37" / E 208°23'51", and

IF WE ASSUME HEIGHT OF EYE is 12 FT, then get the following results :

Lunar Distance Observation Time UT = 19h01m03s3, and
Observed position : S 16°36'09" / E 208°22'48", and

IF WE ASSUME HEIGHT OF EYE is 24 FT, then get the following results :

Lunar Distance Observation Time UT = 19h01m01s8, and
Observed position : S 16°38'02" / E 208°22'21"

*******

Comments :

1 - Why no position computed for Sep 08 th ?

Simply because there is no Sun height published for that date. I could easily compute one, and fully process the rest of the data, but I prefer not since I would "play on data" which I do not wish to.

2 - Interesting set of observations in which the derived Longitudes are almost insensitive to the various HoE's. So average observed Longitudes are close from E 208°24' . What does Google Earth say here ?

3 - Again, on these 3 observations, the Latitudes closest from the published one (16°45'30") are obtained for HoE's equal to 24' . Should we tweak this 23 ft value up ? What do you think, George ?

4 - The mysterious "Printer dagger signs" do not seem in any way related to HoE, at least in these examples. :-) !!!

Best Regards, and thank you for your comments,

Antoine M. "Kermit" Couëtte
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