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    Re: Derivation of the time-azimuth formula
    From: Frank Reed
    Date: 2024 Oct 29, 04:23 -0700

    Pat K, you wrote:
    "I had not heard of the "four parts formula"."

    If you just want a compendium of solution formulae, Wikipedia has a relatively good article on this: Solutions of Triangles - Solving Spherical Triangles.

    But is this really helping? There's value, unquestionably, in learning that many spherical triangle problems we deal with in navigation are just slight variations on a theme. The "great circle" equation is at the heart of almost everything. But beginning navigators with mathematical inclinations often make the mistake of thinking that they will understand it all "better" if they can dis-assemble that great circle equation and see it in terms of more basic trigonometry. This is mostly an illusion. It doesn't usually lead anywhere useful. Yes, there are various ways of "deriving" the great circle equation (in one form, known historically as the "law of cosines of spherical trigonometry"), but you gain little insight by doing so. It's good clean fun (more below), and it helps sate our mathematical hunger, but it doesn't usually make a better navigator.

    Something else to ponder: why use the second form for calculating azimuth at all? Just calculate or observe the altitude, and then determine the azimuth by the first formula. In practice, if the exact azimuth is really required (is it ever?), that's almost always sufficient, and it saves you the trouble, whether in human memory or computer code, of studying, memorizing, and debugging a quite different equation.

    Of course the simplest reason for puzzling out the details of this second azimuth, the time-azimuth formula as you're calling it, is that it's good clean fun. And if that's what you're looking for, then Lars Bergman has provided you the key clue that you can follow. Dig through that Wikipedia article, or any historical text on spherical trigonometry, and find a derivation of the "four parts formula". A related puzzle: rather than deriving this second formula, could you prove, or at least provide a satisfying plausibility argument, that it is identical to the two-step (altitude first) computation of azimuth. The simplest "plausibility" argument wouldn't satisfy any mathematical requirements, but you could set up a spreadsheet and calculate every case for whole-degree inputs and convince yourself that the two approaches yields absolutely identical azimuths out to a dozen or more digits in the final result.

    By the way, in the intercept method(s) of celestial navigation, one gets the corrected observed altitude, Ho, and then calculates the Hc, the predicted altitude at some arbitrary assumed position. The difference between them gives us a distance separating the celestial LOP at the assumed position from the LOP at our actual position. This process should be done with maximum reasonable precision. Common practice is to work this to the nearest tenth of a minute of arc, though a minute of arc is often good enough. We also need to know the direction of the celestial LOP or equivalently the direction of the celestial body (90° different from the LOP). But this azimuth is required to considerably lower precision. Common practice asks for the azimuth to the nearest tenth of a degree (note that this is 60 times reduced precision compared to the Hc calculation), and if the AP is close to the true position, then azimuth to the nearest whole degree is just fine. Because of this low precision requirement, it was possible and popular a century ago to get these azimuths using various graphical methods. No computation at all. In later decades, tables typically listed the azimuth directly, and again no computation was required, except possibly a small interpolation. So again, in practical terms, this second formula rarely showed its ugly head! :)

    Finally, the computation of azimuth for celestial lines of position is only an artifact of the intercept method. We can also calculate points on the celestial LOP by more direct methods, and azimuth isn't required at all --not even in principle. This is covered in my Modern Celestial Navigation workshop (25-26 Jan 2025, which you're already registered for :).

    Frank Reed
    Clockwork Mapping / ReedNavigation.com
    Conanicut Island USA

       
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