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Re: Math behind horizontal sextant angles
From: Mike Wescott
Date: 1997 Mar 04, 15:46 EST

```> I've known since I was a beginner that if you measure the horizontal
> angle between two objects, it gives a circle of position.  I even
> plotted a few points on a sheet of paper once to convince myself.  But
> every time I sit down to try to generate a mathematical proof of this,
> I come up smack against my relatively poor memory of trig identities.
[...]
> Can anybody point me to a reference work where I could look up a
> proof or derivation?
Can't come up with a pointer but I can sketch out the proof.  We can,
WLOG*, put the two objects on the cartesian plane at A=(0,0) and
B=(1,0). Now consider a point X=(x,y) such that the angle AXB is T.
Drop a perpendicular from X to the x-axis and decompose Tinto
T1 and T2:
(x,y)
.
|
/    T1 | T2    \
/          |           \
/             |               \
.               .                 .
(0,0)           (x,0)             (1,0)
(Sorry for the poor diagram).
We get:
T = T1 + T2
tan(T1) = x/y
tan(T2) = (1-x)/y
tan(T) = tan(T1 + T2) = (tan(T1) + tan(T2))/(1 - tan(T1)tan(T2))
substituting and simplifying we get
tan(T) = y/(y^2 - x + x^2)
let tan(T) be the constant k and we can rearrange into
y^2 - y/k + x^2 - x = 1
which can, in turn, be arranged:
(x - 1/2)^2 + (y - 1/(2k))^2 = 5/4 + 1/(4k^2)
i.e. (x,y) lies on a circle. A little more work shows that all such
points on the circle and above the x-axis have the same angle. A
symmetry argument shows that there is another circle centered below
the x-axis with the same property.
*WLOG - Without Loss of Generality - a mathematical term meaning that
I've figured out that I can make this simplification but I'm not going
to explain it to you. :-)
<PRE>
--
-Mike Wescott
mike.wescott{at}XXX.XXX
</PRE>
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