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07162019, 06:59 PM  #16 
Seaman
Join Date: Apr 2016
Location: AN72
Posts: 39
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Thanks, I did see that save button before but didnt know it would reload nav marks or super marks on the map

07192019, 04:53 PM  #17 
Seaman
Join Date: Apr 2016
Location: AN72
Posts: 39
Downloads: 21
Uploads: 0

oops

07202019, 12:34 AM  #18  
Navigator

Sorry I didn't respond earlier. I somehow missed your posts. If you're still looking for some answers ...
Quote:
This is just a way of simplifying the calculation. 89°60' is equal to 90°00' (there are 60 minutes in a degree). When subtracting 42°54' from 90°00', you would have to do this anyway  you can't subtract the 54' from 00', so you would borrow 60' from 90°. Another practice that was common during the time celestial navigation was used was to apply a simple correction to a sun sight of +12' for a lower limb sight or 20' for an upper limb sight. This was a "combined correction" including refraction, semidiameter and dip of the horizon. In reality, these values vary with different conditions (altitude of the sun, height of eye of the observer, etc.), but these simple corrections were considered "close enough" for the average sight. Quote:
I do not know enough about that to provide a good answer, sorry. Quote:
The generally accepted definitions of "sunrise" and "sunset" are when the upper limb (the "top") of the sun meets the horizon. This is the definition that the "official" Nautical Almanac uses. However, it is interesting to note that when you see the sun very near the horizon, it is actually well below it. This is due to the effect of refraction of the atmosphere. The positions of the sun tabulated in the Nautical Almanac are for the apparent center of the body. The average semidiameter of the sun is 16'. Refraction can be calculated using the following formula: 0.0167°/tan(H+7.32/(H+4.32)) ... where "H" is the altitude of the body. At 0° altitude, we find that the value for refraction equals 0.0167°/tan(0°+7.32/(0°+4.32)) = 0.5645...° or 0°34'. Add the 16' of semidiameter to this (to account for the difference between the apparent center and the upper limb) and we find that, at the "moment of sunrise/set", the sun is actually 0°50' below the horizon. That's nearly two apparent sun diameters. The time of sunrise/set for a particular latitude and longitude can be calculated using the formula for a "time sight": cos(LHA)=(sin(Ho)sin(Dec.)·sin(Lat.))/(cos(Dec.)·cos(Lat.)) Once the LHA (local hour angle) is determined, it can be corrected for longitude by using the standard 15°/hour "movement" of the sun (in 24 hours the Sun appears to move 360° around the Earth  360° ÷ 24hr. = 15° per hr.), and for the "equation of time" [EoT]  the difference between the "apparent" and "mean" places of the sun. The EoT is due to the differing speed of the Earth in its orbit around the Sun and varies from about 14 minutes of time to about +16 minutes of time*. These corrections are necessary because clocks generally keep track of "mean zone time" as opposed to "apparent local solar time". If you would like me to expand on any of these concepts, just ask. I would be more than happy to. *The EoT varies in a somewhat irregular way due to the [slightly] elliptical orbit of the Earth around the Sun. This is the source of the "analemma" diagram sometimes found on globes and sun dials.
__________________
If you have a question about celestial navigation ... ask me! Last edited by Nathaniel B.; 07202019 at 01:10 AM. 

07202019, 03:05 AM  #19 
Seaman
Join Date: Apr 2016
Location: AN72
Posts: 39
Downloads: 21
Uploads: 0

Hi thanks for responding and answering my questions. I already figured the first question out myself from reading things on external sites but haven't got round to looking sunrises up yet. thanks you have saved me a lot of time I think.
I will shortly send you a PM if you dont mind taking a peak. 
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