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    Re: Rising angle of stars
    From: Tom Sult
    Date: 2014 Oct 27, 13:56 -0500
    This is interesting. From a practical point of view... I suppose 2 bearings and 2 alt at say 1 hour would give the rising angle but in an hour twilit would become a factor.  Or how would you find a rising angle at sea?

    Tom Sult
    Sent from my iPhone

    On Oct 27, 2014, at 13:46, Frank Reed <NoReply_FrankReed@fer3.com> wrote:

    When you're at the equator, all stars rise vertically from the eastern horizon. At other latitudes, stars due east (necessarily with declination equal to zero) rise at an angle from the vertical equal to the latitude. While driving east through dark, rural Rhode Island last night, I was enjoying the view of Orion's belt rising in front of me. I could trace Mintaka back at an angle from the vertical of 41° (near enough to my latitude) and know that this was almost exactly due east as I drove along. That got me wondering about the rising angle for other declinations or azimuths... At home I decided to sit down and work it out. It turns out that the relationship is simple. If theta, θ, is the angle with respect to the vertical for a rising (or setting) star, then 

    tan θ = tan L / sin Z

    where L is latitude and Z is azimuth. For example, if the azimuth is exactly either NE or SE as a star rises at latitude 45°, then the rising angle is 54.7° away from the vertical. This is symmetrical north and south of east, which surprised me a bit, though it makes sense in retrospect. Naturally when Z=90°, the rising angle is equal to the latitude. At all latitudes, for azimuths from 75° to 105° (within 15° of due east), the rising angle is less than one degree greater than the latitude.



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