# NavList:

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

**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@XXX.XXX </PRE>