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    Re: Rising angle of stars
    From: Steve E. Bryant
    Date: 2014 Oct 28, 17:08 -0500

    Would it be correct to assume that Mintaka's declination was zero?
    
    Sent from my iPhone
    
    > On Oct 27, 2014, at 1:47 PM, Frank Reed  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.
    > -FER
    >  
    > 
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    > : 
    > http://fer3.com/arc/m2.aspx/Rising-angle-stars-FrankReed-oct-2014-g29075
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    > 
    

       
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