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

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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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