# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Moonrise over Connecticut, 19 October 2021
From: Antoine Couëtte
Date: 2021 Oct 23, 08:35 -0700

Hello to all,

I know, Frank, that you subsequently came up with a different questionnaire, but I am simply answering this initial one, i.e. not knowing (yet) the Observer's Position.

Connecticut Moon rise published on 20 th October has to be Moon rise on 19 th of October, which is confirmed by both your title and the picture info. Then it has to happen by Sun set, which I am guessing to happen roughly by 19th Oct 12:00 UT.

For the UT Time, hit the button and get for the Moon SD = 15.03' (HP = 55.15') and Dec N 04°16' .

I printed your picture. Getting Diameter = 83 mm = 30.06'

Estimating the height of the 3 level house bottom right of Lady Moon (with 2 triangular upwards tops) at 8 meters (for 3 levels) for 12 mm, i.e. at 8 meters for 4.3' .

Distance of this house is equal in meters to 8 / tan(4.3'), i.e. close to 6,400 m, i.e. close to 3.45 NM

This house is beyond the apparent horizon, otherwise we would see the beach.

Hence HOE is low enough so that we do not see a beach at 3.4 NM at the most (no dip short).

Looking up my depression tables, I feel confident that HOE should be no more than 3 m (10 ft) probably a little less, but certainly it should not exceed 12ft-13ft. Let's go for 8-9 ft.

Next : sailboats hulls are probably 4.5 m long (15 ft). The rightmost one shows 42mm on my printed picture, which gives an angle of 0.2535°, hence at a distance ranging from 0.5 NM to 0.6 NM at least for this one.

One of the sailboat under the water because of a short swell.

Latitude can be roughly - probably within 1° - determined through noticing that "Mare Crisium" on the Moon is about 33° right of the vertical. I do not have the applicable system/software to derive it.

From the picture, I can determine 2 more environmental parameters :

- Since the Moon Dec is close from 4°16', then its Azimuth is close from 095°. If we know the place showing on the picture, we can get the spot from which the pic was taken.

- And finally through extremely intense calculations and reflexion - I am really worn out now - I have determined the horizontal visibility to be at least 400,000 km, i.e. in excess of 215,00 NM. Any taker here ?

Fun ... thanks a lot.

Now, where do I stand in terms of grade / mark ... ?  :-)

Kermit

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