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    Re: Geometry of SNO-T
    From: Alexandre Eremenko
    Date: 2004 Oct 14, 12:21 -0500

    Dear Herbert,
    I find your version of the proof OK
    (and I am glad to conclude that we are in 100% agreement now:-).
    The only little point that still leaves me somewhat uncomfortable
    is that nobody seems to have noticed the effect we discuss.
    Now I understand that Astra III (one of the most widespread sextants!)
    is constructed the same way as SNO (the axis is away from the mirror
    plane by 5 or 6 mm), and this gives a large and well visible
    effect when testing perpendicularity.
    On using vectors vs Greek math.
    On my opinion, there was some progress in math in the
    last 2200 years or so, (since the Greek geometry was invented)
    and vector notations are easier and more convenient to use.
    (Greek-style proofs always required some ingenuity.
    Vectors reduce this to a mechanical procedure, an algorithm.
    And you can compute with them, instead of making pictures)
    Unfortunately, the process of introduction of such
    innovations into curriculum (on all levels) is very slow.
    Many great math inventions that are hundreds years old
    still have not found their way to the usual math curriculum.
    But let me explain the reflection formula that I used.
    (It may be useful in other questions of sextant geometry).
    By the way, most spherical trig formulas also become
    much more transparent and easier to understand when using vectors
    rather than the ancient "Ptolemy/Arab notation".
    1. Dot product of two vectors.
    This is a NUMBER, equal to the product of the lengths
    of these vectors times cosine of the angle between them.
    If the two vectors have coordinates (x,y,z) and (a,b,c)
    there is a very nice formula for this dot product:
    Notice that no cosine is involved!! Only multiplication and
    addition. The dot product of a vector with itself is its length squared:
    2. Projection of a vector x onto the direction of a vector a.
    It is defined as a vector y, having the same direction as a,
    and minimising the length of the difference |x-y|.
    The explicit expression in terms of the dot product
    is p=a(a.x)/|a|^2=a(a.x)/(a.a)
    3. Now it is easy to derive the reflection formula.
    Let N be a normal (=perpendicular) vector to the mirror, of unit length.
    (For a given mirror, there are exactly two such vectors N, chose any of
    them, it does not matter).
    We want to find the reflection y of a vector x in this mirror.
    First, the projection p  of x onto the direction N is
    Now you have to make a picture yourself:
    Draw x and p starting from a common point O,
    and complete this drawing to a triangle, the third side
    of this triangle is the vector p-x (from the endpoint of x to the
    endpoint of p. According to the law of reflection,
    the reflected vector y should be x+2(p-x). (Draw it to see this!)
    Thus we get our final result:
    Exercise: by using only this formula, show that if y is
    the reflection of x then x is the reflection of y:-)
    On Thu, 14 Oct 2004, Herbert Prinz wrote:
    > 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

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