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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: dip, dip short, distance off with buildings, etc.
From: Frank Reed CT
Date: 2006 Jan 15, 21:22 EST

```Bill you wrote:
"OK with treating ABCA as a right triangle given the  ratio
of R to H and h, but semi-clearly an oblique triangle in almost all  cases
once the horizon "dips" due to height of eye/curvature/refraction and B  is
raised above the horizon."

The triangle isn't a right triangle (and  you don't need to assume it is),
although it will be nearly so in many practical  cases.

But let's continue with the development of the equation behind  Table 15.

To recap, you have point C at the center of the
Earth,  point A at height h above the Earth's surface, and point B some
distance
away at height H above the Earth's surface.Those three points make a big
triangle ABC. Our GOAL is to find the angle in the triangle at point C
(since, if
we multiply that by the Earth's radius, we get the distance). Now  what do we
measure with our sextant or theodolite? Well, we're at point A,  and we
measure
altitudes above the horizontal which is, by definition, the  plane
perpendicular to side AC of the triangle. Now how is that related to  the
angle at point
A in the big triangle (the angle "CAB")? Clearly that's  just 90 degrees plus
the measured altitude. So we KNOW the angle CAB. This  is equivalent to the
measured parameter (or at minimum, it's related to the  measured parameter by
a
simple relationship). So we know the corner angle at  point A (the observer),
which I named "gamma", and we are seeking the corner  angle at point C (the
center of the Earth), which I named "phi". But, uh-oh,  we're stuck with
another
angle --the one at point B. Wait... no we're  not. Since it's a simple plane
triangle, all three angles must add up to 180  degrees. That means that the
angle in the big triangle at point B is  NECESSARILY equal to
180-(gamma+phi).
Ok so far?? If you haven't drawn a  picture of this yet, you can't possibly
be
ok here  so please make  sure you've got a picture of this. And if you
don't want to draw your own  picture, see image 488 in the archive (see
below).

Now we are in a  position where we can apply the law of sines (the ordinary
plane trig law of  sines for oblique triangles...) to the big triangle ABC.
Set it up as
sin("angle at A")/(R+H)=sin("angle at B")/(R+h).
And work from  there.

The angle at B is 180-(gamma+phi) so the the sine of that is the  same as the
sine of (gamma+phi). We now  have

sin(gamma)/(R+H)=sin(gamma+phi)/(R+h)
which is equivalent  to
(R+h)/(R+H)=[sin(gamma)*cos(phi)+sin(phi)*cos(gamma)]/sin(gamma)
or
(R+h)/(R+H)=cos(phi)+sin(phi)*cot(gamma)

Next  we replace gamma by using its relationship with the measured altitude
(assumed  to be corrected for dip) which is gamma=90+alt. For any angle x, it
is  always true that cot(90+x)=-tan(x), so that gives us

(R+h)/(R+H)=cos(phi)-sin(phi)*tan(alt).

We could stop here since  this equation can be solved iteratively (glorified
"trial and error") to get the  value of phi for any combination of h,H, and
alt. With phi in hand, you then get  distance from d=R*phi. And if you want to
incorporate refraction, just replace R  by R'=R/(1-beta) where beta is
typically somewhere between 0.13 and 0.4  depending on the lapse rate.

But if we don't want to stop here we can start throwing in some assumptions
about the relative "smallness" of certain angles and ratios. For starters, it
is  certainly true that h and H are very much smaller than the radius of the
Earth  in cases of practical interest. So it is an excellent approximation to
replace  (R+h)/(R+H) by the expression 1+h/R-H/R. That gives us

1+h/R-H/R=cos(phi)-sin(phi)*tan(alt).

Next, we go to the "small  angle" approximation for the trig functions of the
angle phi. This is a very  good approximation whenever the angle phi is less
than about 0.1 (a tenth of a  radian or roughly 5.7 degrees). In our case, the
angle phi will rarely exceed  even 1 degrees, so we can replace cos(phi) by
1-(1/2)*phi^2 and sin(phi) by phi  (if you don't believe this, try calculating
the sine and cosine of 1 degree or 2  degrees using these equations... don't
forget to convert to radians first).  Replacing cos(phi) and sin(phi) now
yields
1+h/R-H/R=1-(1/2)*phi^2-phi*tan(alt).

Let's cancel 1 from both sides and introduce some temporary names for
tan(alt) and (H-h)/R. We'll set f=tan(alt) and g=(H-h)/R. Then we  have
(1/2)*phi^2 + f*phi - g = 0.

Aha! (say it out loud)
This is just a plain old quadratic equation which we learned how to solve  in
first year algebra. The solution of a quadratic equation of the form a*x^2 +
b*x + c =0 is x=[-b+sqrt(b^2-4*a*c)]/(2*a) or the same thing with a negative
sign in front of the square root. So we can now solve "our" quadratic
equation.  The solution is
phi = -f+sqrt(f^2+2*g).

Bringing back the original  constants (replacing f and g), we have
phi =  -tan(alt)+sqrt[(tan(alt))^2+2*(H-h)/R]
And of course, distance = R*phi, so  the distance from the base of A to the
base of B  is
d=sqrt[(R*tan(alt))^2+2*R*(H-h)] - R*tan(alt).

As I mentioned  above, you can incorporate refraction simply by replacing R
in this equation by  R'. Finally converting everything to minutes of arc and
nautical miles and  picking one very specific value for the terrestrial
refraction reproduces the  formula in Bowditch exactly. You can go beyond the Bowditch
equation by  experimenting with different values for the refraction.

By the way, there  is another way of deriving this result. You can start with
refraction explicitly  at the beginning of it all by working from the
so-called "refraction invariant".  More work, same result (which is re-assuring!).

-FER
42.0N 87.7W, or  41.4N 72.1W.
www.HistoricalAtlas.com/lunars

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