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    Re: Single LOP is enough?
    From: Frank Reed
    Date: 2020 Sep 17, 12:46 -0700

    Bill Lionheart, you wrote:
    "Of course if you could measure the azimuth accurately (eg on land) as well as the altitude you would be on (one of) the intersection(s) of the isoazimuth and the circle of position."

    Yes, thanks for mentioning that. It really is nowhere near as simple as I made it sound in my post on this topic. Even if you could measure azimuth with a "super-compass", there are plenty of real-world cases where that information would not give you a fix when combined with a measured altitude. 

    A special case that highlights the problem:

    Next week at 13:30:00 UT, the Sun will be directly over the equator at longitude 24°22' W (thanks to David C for reminding us that the equinox is coming and that it will occur over the Atlantic!). I'm going to pick a time for my example such that I can keep my imaginary observers in one ocean (Pacific). So we'll go to 15:00 UT, an hour and a half after the exact equinox. By that time, the Sun is a mere 1.4' below the equator. Its longitude by then would be 46° 52'. Now suppose you are located 45 degrees of longitude due west from that spot, right on the equator. That puts you a bit less than 20 miles northwest of Isla Isabela, the big island in the Galapagos Islands. You measure the altitude of the Sun, correct it, and you get 45°00'. You observe the Sun's azimuth with your super-compass and find that it is 90°02'.  With that information, you determine that you are on the equator at latitude 0°00' and your longitude is 91°52'. All fine so far.

    Now suppose you were not at that exact location. Suppose your observations from your vessel at 15:00:00 UT yielded an altitude of 44°30', and your observed azimuth was 89°32'. What is your position? Well, the altitude and the azimuth are each half a degree, 30', lower than before. That means we must be further west by 30', and, in this specific case, we are also south by 30'. These observations would be correct at latitude 0°30' S and longitude 92°22' W. In this specific case, there is a nearly exact one-to-one relationship between the change of azimuth and the change of position perpendicular to the direction to the Sun (which in this case is latitude). That's so easy. Even the math is trivial: it's a one-to-one change in latitude for azimuth difference. But don't fall for it --it's a magic trick.

    Now let's move west...

    Suppose our vessel is located another 45° further west on the equator near 136°52' W. At this location the Sun is just rising. We measure its altitude, correct it, and get 0°00'. It's right on the true horizon. We also measure its azimuth with our super-compass and get 90°01'. But what happens when we change our position? As before, let's move 30' in longitude (this time east) and 30' in longitude (south, as before). From here we find that the observed altitude of the Sun (after correction) is 0°30'. No surprise. We've moved half a degree towards the Sun so its corrected altitude is exactly that much higher. But its azimuth? Unchanged. It's 90°01'. In fact, if we move even 10° or 20° away from the equator, the azimuth will be the same. It's a big contrast. This makes good sense, of course. If you imagine the circle of position for this sunrise sight, everyone along it sees the Sun rising exactly due east on the equinox. Azimuth is the same for every observer along that circle of position. 

    Near the Galapogos, we had a simple, very promising one-to-one relationship: the azimuth changed by some angle, and the latitude changed by the same angle. But out in mid-Pacific with the Sun on the horizon, the azimuth is completely insensitive to a change in latitude. This is in the nature of the geometry. If the Sun is directly above the equator, then the azimuth will be due east for all points located exactly along the equator within that 90° band of longitude due west of it. But the azimuth will also be due west for all points located along that meridian of longitude 90 west of the Sun. Two perpendicular arcs, and the azimuth is exactly 90° along both of them. It's not an observational problem. My super-compass cannot help.

    Frank Reed

       
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