# NavList:

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

**Re: Geometry of SNO-T**

**From:**Herbert Prinz

**Date:**2004 Oct 14, 08:48 -0400

Alexandre Thank you for posting your proof and doing the tedious work of typing it in. Unless you find a flaw in the following, we are in 100% agreement now. I have been trying to give a purely geometrical explanation, but could not see how to do it without a drawing. Your proof has shown the way. Most list members will be able to follow your proof up to the point where you introduce the general formula for reflection. This and the dot product might scare a few people away. So, if you don't mind, here is your proof again, after a small cosmetic change in the form of a coordinate transformation. I will still use cartesian coordinates, because I can't draw, but the argument is entirely based on good old proportions, Greek style, avoids complicated vector algebra and can therefore be drawn on paper if the reader cares to. Let O = (0,0,0) be the front edge of the mirror at which we reflect. Let the mirror intersect the x-y-plane in the y-axis. Let M be a point on the y-axis. Let A be the left visor that we see directly. Let B = (p,q,0) be the right visor that we see in the mirror. Because angles AOM = MOB, the right triangles that form the coordinates of points A and B are similar and symmetric, and the coordinates are proportional. So A can be written as k*(p,-q,0) or A = (k*p, k*-q , 0) Let B' be the reflection of B, and therefore B' = (p, -q, 0) The rest goes like you say: After the observer lifts his eye, the rays go through O' = (0,0, h). O'A = (k*p, k*-q , -h) O'B = (p, -q, -h) These vectors can only be parallel if h = 0 or k = 1. If h = 0, then k is arbitrary, meaning that the mirror does not have to be in the symmetry plane of the two visors. But since the pivotal axis has to be in it, we know that the mirror may be outside of the pivotal axis. Therefore the test is always valid if we look in the x-y-plane. If h > 0, then k = 1. Inserting this into the definition of A, above, we get A = (p, -q , 0) This means that A and B' can be seen in the same direction from a point O' outside the x-y-plane only if A and B are symmetric with respect to the mirror. In other words the mirror is the symmetry plane. But the pivotal axis is always in the symmetry plane, and thus in the mirror. *** (Since line AB is perpendicular to the mirror if and only if k = 1, this is also criterion for whether the visors will coincide for every h. We don't need this here, I just mention it for completeness, since I brought it up earlier.) Herbert Prinz