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    Re: Polaris
    From: Frank Reed
    Date: 2009 Jun 24, 19:08 -0700

    Peter, you wrote:
    "it is possible to use a table in the Nautical Almanac to find parameters a0, 
    a1, a2 and use them to "correct" the observed altitude Ho of Polaris for a 
    more accurate value of your latitude:
    
    Lat = Ho - 1 + a0 + a1 + a2"
    
    The above formula is based on a second-order expansion of the standard ZPS 
    triangle when the polar distance is small. To start, we have
    
    sin(L+dh) = cos(p)*sin(L) + sin(p)*cos(L)*cos(HA)
    
    where L is the latitude, dh is the difference between the altitude of Polaris 
    and the altitude of the north celestial pole, p is the polar distance of 
    Polaris (which presently qualifies as a small angle, less than a degree), and 
    HA is the local hour angle of Polaris. Note that dh is what we're looking 
    for. If we have a good enough expression for this in terms of hour angle, we 
    can take an observed altitude for Polaris (after correcting for dip and 
    refraction) and add dh to it to get the latitude.
    
    First step, expand the left-hand side using the sum formula. Next, replace 
    cos(dh), sin(dh) as well as cos(p) and sin(p) by the small angle expansions: 
    six(x)=x, cos(x)=(1-(1/2)*x^2). That should give you
    
    sin(L)*(1-0.5*dh^2)+cos(L)*dh=(1-0.5*p^2)*sin(L)+p*cos(L)*cos(HA).
    
    Now collect together the terms in sin(L) on the right-hand side and divide both sides by cos(L). You get
    
    dh = p*cos(HA)+(1/2)*(dh^2-p^2)*tan(L).
    
    Dissecting this, we have a term linear in the polar distance. This is a simple 
    "rotating clock" term --as Polaris goes round the celestial pole, its 
    altitude oscillates up and down. Then there is a quadratic term, much smaller 
    since the polar distance is very small. But this term is seemingly circular 
    (in the sense of self-referential) since it depends on dh. But we can replace 
    dh in the quadratic term by its linear, first-order, expression. That is, 
    replace dh^2 by (p*cos(HA))^2. Then after applying the obvious trig identity, 
    you get
    
    dh = p*cos(HA) - (1/2)*(p*sin(HA))^2*tan(L)
    
    Now since p varies during the year, we should replace it by p(t) and introduce 
    p0 which would be the mean polar distance during the year. This is a small 
    change, so we can ignore the quadratic part here. The result is
    
    dh = p0*cos(HA) - (1/2)*(p0*sin(HA))^2*tan(L) + (p(t)-p0)*cos(HA).
    
    We're getting close. There's a very old trick for making all the corrections 
    positive: add some arbitrary constant to each term such that the constants 
    add up to one degree and then subtract one degree from the final result. We 
    then have:
    
    dh = -60' + a0 + a1 + a2
     where
    a0 = k0 + p0*cos(HA)
    a1 = k1 - (1/2)*(p0*sin(HA))^2*tan(L)/3438
    a2 = k2 + (p(t)-p0)*cos(HA)
     and 
    k0+k1+k2 = 60'
    (and I'm assuming that dh and p0 are both in minutes of arc so the quadratic 
    part of a1 has to be divided by 3438).
    
    Two last bits: HA = HA aries+SHA and the SHA of Polaris doesn't change much, 
    so we can tabulate in terms of HA aries. And also to reduce the range of a1, 
    we can add k'*(sin(HA))^2 to a0 and subtract the same from a1. This is 
    optional.
    
    So... build yourself a spreadsheet. Get the SHA and Dec of Polaris from the 
    almanac for some recent year. Then fiddle with k0,k1,k2, and maybe k' until 
    the spreadsheet matches the Polaris tables. I think that's everything, but no 
    guarantees! :-)
    
    -FER
    
    
    
    
    
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