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    Re: Correcting for the movement of an observer: a plausible explanation?
    From: Frank Reed
    Date: 2019 Dec 31, 15:56 -0800

    David Pike, in reply to my post, you wrote:
    "I’m not sure that it needs to be as complicated as that."

    I wrote that post for those readers who might benefit from that analogy, but that doesn't mean that everyone will! Sorry. Nonetheless, I stand by what I wrote. That's exactly how those tables work. 

    I'll try a different analogy. Suppose I fire a ballistic projectile into the air with conditions such that air resistance and other factors are minimal, and the motion is determined almost entirely by the nearly constant gravitational acceleration near ground level. Ha! That's a long-winded way of saying, "I pick up a rock and throw it". My ballistic projectile follows a simple, well-known trajectory. Suppose I have thrown this rock hard enough that it stays in the air for at least five seconds. Two seconds after leaving my hand, I measure its velocity (maybe by video, maybe by radar, details don't matter...), and I find that its speed in the vertical direction is 1 foot per second and its speed horizontally is 4 feet per second. It is presently 15.0 feet above the ground, and it is 32.0 feet "down-range" (distance measured along the ground in the direction thrown). Where will the rock be one tenth of a second later? That's easy. The rock will have climbed 0.1 feet and advanced down-range 0.4 feet. And that means its height is now 15.1 feet and down-range distance is 32.4 feet. Obvious, right? But it's a first-order linear approximation, and if we try to use that simple math to predict the location of the rock in ten seconds, our results will be radically wrong because we're ignoring the downward acceleration of gravity.

    Now switch back to the altitude of a star. If we know where a star "is" in terms of its altitude, and if we have a table (the MOB table) that tells us how the altitude changes with time, then we can predict where the star will be a few seconds later using its "velocity", which is all that the MOB table provides. Eventually that prediction will fail because it's a linear "differential" approximation.

    You also wrote:
    "If you look at tables 1 & 2 in AP3270.  All they are are plane trigometry tables.  Although they appear in celestial tables and have units recorded in minutes of arc, they’re not particularly connected with the height of the star."

    What?! Either we're talking about different tables, or there is some misunderstanding here. Let's deal with just one of the tables, the MOB table, Table 2. It tells us the change in altitude for one minute of time, but clearly this is a differential relationship. It's telling us the change in altitude for some very short period of time, and one minute just happens to be short enough that we don't have to worry any more. But it's strictly the "first derivative". It's only telling us the change in altitude to first order in time, and it does break down for longer time intervals (muchh like the motion of that "thrown rock" in the analogy above). And furthermore, the relationship is exactly and specifically tied to the height of a star, and it has no other basis. If you take the standard spherical trig expression for a celestial altitude:
      sin(h) = sin(Dec) · sin(Lat) + cos(Dec) · cos(Lat) · cos(k·T),
    where I have replaced Hour Angle, HA, with kT, where k is the normal rate for the body in question (15.04' per minute for stars), and T is time (in minutes with that choice for k), then we can directly differentiate with respect to time, T, and after some re-arranging and subtitution, we find the rate of change of altitude, which I will call "AltRate":
      AltRate = dh/dT = 15.04' · cos(Lat) · sin(Zn).
    And you can use that "AltRate" to predict changes in a celestial body's altitude --up to a point. Much as you can calculate a value on a sine curve by knowing the rate of change of the curve at that spot, call it SineRate, and then using sin(x+dx) = sin(x) + dx · SineRate (where SinRate just happens to be cos(x), nice and simple), we can similarly adjust a celestial altitude using h(T+dT) = h(T) + dT · AltRate where AltRate is given by the expression above. Using Table 1, the MOB table, you look up the value for AltRate (which is just a tabulation of the quantity above for the full range of Lat and Zn) and then you multiply by dT, the number of elapsed minutes.

    The problem here is that we're only looking at the first derivative, the first-order rate of change of the altitude, and that's only good some short period of time. So how can we determine how far off it is? It's really not difficult. Work up the altitude at some later time using the expression that generates the table, and also work up the altitude using the complete spherical triangle expression. It's easy to see why it fails, too, by going back to the analogy.

    Frank Reed
    PS: Happy UT New Year... Just five minutes to go as I finish this post.

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