Welcome to the NavList Message Boards.

NavList:

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

Compose Your Message

Message:αβγ
Message:abc
Add Images & Files
    or...
       
    Reply
    Re: Dip and the refractive invariant
    From: Paul Hirose
    Date: 2013 Apr 22, 12:16 -0700

    Bruce J. Pennino wrote:
    > Paul,  I have one set of dip data that indicates a value of dip 
    significantly higher than would be predicted in the NA or that I've measured 
    elsewhere.  Interestingly, it is  a rushed data set at one of my lowest HoE.  
     The wind was blowing smog in my face from Long Island and NYC. I was 
    sighting over Long Island Sound. Considering Frank's pictures and your 
    analysis, how does smog influence refraction.  I would imagine I was in the 
    same smog as the horizon.... your thoughts?
    
    If it was not a false horizon, a dip greater than expected suggests the
    indices of refraction at the observer and sea level were abnormally
    similar. In fact, if they are identical, refracted dip equals the
    geometric dip.
    
    I don't know how smog would affect refraction.
    
    
    I wrote:
    > Combining all the above, dip (radians) = √(2 * (dn + h))
    >
    > where dn is the difference in the refractive indices of the air at the
    > sea surface and at the observer (negative if the latter is less), and h
    > is the height of eye in earth radii.
    
    For a more convenient formula, change h (height of eye) to meters
    instead of Earth radii. The value of dn must be multiplied by the same
    factor to keep the sum (dn + h) valid. Let dn′ = dn * 6371000, and h
    = height of eye in meters. Then:
    
    dip (radians) = √(2 * (dn′ / 6371000 + h / 6371000))
    = √(dn′ / 3185500 + h / 3185500)
    = √(1 / 3185500) * √(dn′ + h)
    
    therefore,
    
    dip (arcmin) = 180 / π * 60 * √(1 / 3185500) * √(dn′ + h)
    dip (arcmin) = 1.926 * √(dn′ + h)
    
    where dn′ = 6371000 * ((refractive index at eye) - (refractive index at
    sea))
    h = height of eye (meters)
    
    Test the formula with my previous example at 10 meters height of eye.
    The value of dn was -2.65e-7, so dn′ = -1.69. Dip = 1.926 * √(-1.69 +
    10) = 5.55′, which agrees with the almanac and the computation in my
    previous message.
    
    In other words, the decrease in refractive index of the air between
    sea level and the eye is equivalent to a reduction in height of eye by
    that many Earth radii.
    
    --
    I filter out messages with attachments or HTML.
    

       
    Reply
    Browse Files

    Drop Files

    NavList

    What is NavList?

    Join NavList

    Name:
    (please, no nicknames or handles)
    Email:
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
    Email:

    Email Settings

    Posting Code:

    Custom Index

    Subject:
    Author:
    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site