Intermediate Time of Phenomenon (Sunrise/Sunset and Moonrise/Moonset)

An more moderately paced and clearer explanation of using table 1 would be useful. Neither do the instructions in NP314/23 (UK version) help at all. It lacks the usual worked example too. Simply slowing down the video does not help. The guidance is absent. Overall impression on this occasion is one of indecent haste.

This must be Algebra the high level 🎚️ type right 😂😂 lol 😮 this is interesting 🤔 what is that book 📚📖 title name

About the table 1 how to find tabular interval

Very good instructional video, although I miss the Pacific Seacraft.

បានទិកប្រាក់សុំឃើញបានទេលោកិមឲនាផ្សេងចូលមកក្នុងទំព៍នេះលើយ

ទិកប្រាក់រកឃើញត្រូវបញ្ចូលទៅធានាគាអេធីអេលេខបញ្ចីមាដូចនេះ ០០២៤៣០៥៥៤

For instance you have in the last problem +8 uniform and 125 (-8) degrees west, shouldn't it be subtracting if we apply basic navigation rules, as West equals "-" Negative? And where did you get those names from (I know they're found in the CIQ Code)?

Can you explain where the 1.69 is derived from?

Another method uses sight reduction tables, an almanac, and a Visibility Table.  The expected marine-sextant height (Hs) of the sea horizon is 0°. The dip angle---the correction for height of the eye--tells the apparent height (Ha). From the observer to the horizon, the terrestrial refraction is found by comparing the dip angle in minutes with the NM distance to the horizon; (the distance to the horizon is tabulated in he American Practical Navigator and in H.O. No. 208. It is about 1.15 NM times the square root of the altitude in feet). According to the almanac, the astronomical refraction (beyond the horizon) is about 34 minutes. These corrections and the semidiameter correction give the expected observed height (Ho) of the center of the Sun disk. If the sight reduction tables don't produce such a low height, then force a higher height by assuming a position far enough out in the direction of the Sun. Draw a perpendicular Sun Line through that point. What is the new, higher height---the precomputed height (Hp)? Find the LHA of the Sun in the sight reduction tables. Using arithmetic find the GHA of the Sun. In the almanac fInd the time at which the Sun has that GHA. Look at where the plotted Sun Line crosses your parallel of latitude. How many arcminutes of longitude is it from your position? Adjust the time of sunrise by 4 seconds of time for each arcminute of longitudinal difference.

hello sir chris nolan thank you for this video .im a little bit confused about this
1749H -1840H =diff. 69 mins?
or 1749H - 1840H = diff. 51?
or you have add some equation.

What is meaning of +2 ( Oscar ) or +4 ( Quebec ) ? Is it time zone ?

i dont know where the 2 oscar came from and the 4 quebec, can u please explain where those nos. came from? and the time different, what are your reference there?

your conversion of the latitude pos. must 41.59. corect me if im wrong? a little bit confused with that

may i ask you a question?you calculated sunrise or sunset for middle date of page.if i want calculate for other date of page.how do i do it?

Please make more videos! Your explanation is even better than my teacher's. ;)