# NavList:

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

**Geometry of SNO-T**

**From:**Alexandre Eremenko

**Date:**2004 Oct 13, 18:32 -0500

Here is some simple math behind the perpendicularity test and Frank Reed's experiment on his kitchen floor. Suppose that A and B are two points on the upper edges of the visors, and C is the point where your eye is. These three points define a plane P. Let A be the point seen directly and B the point seen by reflection in the index mirror. Let O be the point of intersection of this plane P with the right-hand edge of the index mirror. Suppose that the mirror is perpendicular to P. Then (by the laws of reflection) the light rays from A to C and from B to O to C are in the plane P. Let L be the line of intersection of the mirror (reflecting surface) with the plane P. It follows from the law of reflection that L is the bissector of the angle AOB. In general, |AO| is different from |OB| (Here || denotes the length). For example, this is the case for SNO-T sextant. But apparently for many sextants we have |AO|=|OB|. (It is easy to see that this is the case for those sextants in which the axis of rotation of the index arm passes through the reflecting surface of the mirror. Then the intersextion of this axis with the plane P belongs to our line L). Let us consider this case first, that is |AO|=|OB|. Then the triangle AOB is isoceles, thus L is perpendicular to AB. And thus the index mirror is perpendicular to AB. (This was the point in one of the Herbert Prinz messages, which I misunderstood because my sextant does not have this property |AO|=|OB|). Now, as the observer moves her eye to any point near O, say to the point O', the new plane P' containing A,B and O' will be still perpendicular to the index mirror. This is because P' contains the line AB which is perpendicular to the index mirror. So the test will give positive result again, because now all light rays are in the plane P'. This explains why the usual version of the test works for such sextants (whose axis of rotation of the index lies in the silvered plane of the index mirror). (For complete mathematical justification of the test, it remains to show that if the mirror is not perpendicular to P' then the rays will never be in a same plane, so the observer will never see the edges of the visors aligned. But I skip this part). Now we pass to the general case (which I had in mind from the very beginning when I started to question this test). So suppose that |AO| is not equal |OB| .................................(1) Suppose again that the mirror is perpendicular to the plane AOB, where O is the observer's eye. Then again all the light rays we consider will be in the plane P, and the observer sees the visor edges aligned. Suppose now that the observer "lifts" her eye a little bit, that is moves it to the point C' So that CC' is perpendicular to P. Then the ray from A to C will pass through some point O' on the right edge of the mirror, so that OO' is perpendicular to P and |OO'|=h, where h is some non-zero number. I will show that with non-zero h, the edges of the visor, as seen by the observer, are never aligned. To do this I will use co-ordinates (x,y,z) in 3-space. Suppose that P coinsides with x-y plane, and the points have the following coordinates: O (0,0,0) O' (0,0,h) A (0,1,0) B (p,q,0) Let N=(t,s,0) be a normal vector to the mirror. (Its z-coordinate is zero because the mirror is assumed perpendicular to P). For the edges top be aligned, the ray from B to C' through the mirror has to pass through the point O'. Let us compute vector Y which is the reflection of the vector X=O'B=(p,q,-h) in the mirror. The general formula for reflection is (in vector notation) Y=2N(N.X)-X, (where (.) is the "dot-product"). So the coordinates of Y are 2t(tp+sq)-p, 2s((tp+sq)-q, h ....................(2) Notice that |Y|=|X| (reflection does not change length!) To have the visors edges aligned, this vector Y in (2) has to be parallel (proportional) to AO'=(0,-1,h). But (2) can be parallel to (0,-1,h) only if these two vectors Y and AO' are equal (look at their z-coordinates!) We conclude that |AO'|=|Y|=|X|=|O'B|, thus |AO|=|OB|, contradicting our assumption (1). This proves that the alignment of visors can persist dutring the vertical eye motion ONLY is the condition |AO|=|OB| is satisfied, that is only for those sextants whose silvered surface of the index mirror contains the axis of the index arm. I suggested one way of doing this adjustment for SNO-T in the end of my message of Wed Oct 13 2004 - 17:07:08 EDT. Alex.