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    Re: An equal altitude puzzle
    From: Frank Reed
    Date: 2013 May 20, 10:40 -0700

    Looks like you all worked it out much as I did.

    This got started when someone in another forum recently noted that there is a nice geometric asterism that appears in the evening sky in the latitude of Baltimore at this time of year. It's a huge trapezoid consisting of Castor and Pollux on a short side at the top and Capella and Procyon on a long side that is parallel to the horizon on the bottom side. He noted that this occurred at a particular sidereal time (for navigators, equivalent to SHA at the zenith and identical to LHA of Aries). So anyone in about the same latitude could see the pairs of stars at equal altitudes at the same sidereal time. In terms of local time this time of year, this happens in the early evening, convenient for casual viewing. But that got me thinking, if I go to a different latitude, how much would the corresponding time change? And then I switched it up and started thinking about the related problem for observations at a specific GMT. The puzzle followed.

    You can figure out that the set of places where both stars have the same altitude is a great circle by a couple of different methods. I started by thinking of circles of position, equal in size since the altitudes are equal. Varying the radii of the pair of circles of position, the crossing points, the possible fixes, move away equal distances from both stars' GPs. Of course! So I need the set of points that would be equidistant from a pair of places on a globe. And that is a great circle.

    To place some limits on the puzzle, I decided the stars really ought to be visible, so that's why I said their altitudes had to be greater than 1°. That chops off just about half of the great circle. Next it has to be late enough after sunset to see stars so I went with the traditional definition of civil twilight: the Sun is at least 6° below the horizon. The set of places where the sky is dark enough is slightly less than half the globe so that's another "nearly" great circle slicing away most of the rest of the possible locations. That would leave us with a set of points along an arc of a great circle though not a very long stretch.

    Finding these points is not difficult, and there are many ways to do it. First, we have one case which is a straight-forward celestial fix problem. That extreme end of the possible points is defined by both stars at altitudes of 1° 00' exactly. We can find that point. Like any two-body fix, there are two solutions, but these points will be separated on the globe by nearly 180° so the other point should be in daylight. Sure enough, one location is in the Atlantic east of Bermuda where it would be dark, and the other is in southwest Australia in daylight. We can rule out Australia. The position in the Atlantic anchoring one end of the great circle arc is 31° 42.5' N, 59° 53' W. To find the other end of the arc, we have to find the spot where the great circle of equal Capella-Procyon altitude crosses the LOP defined by the Sun's altitude equal to 6° 00'. I did this by fiddling around on the well-known USNO "celestial navigation data for assumed position and time" page here: https://aa.usno.navy.mil/data/docs/celnavtable.php, changing lat and lon until both stars were at the same altitude. As it turns out, that altitude was 25° 59' when the Sun was exactly 6 degrees below the horizon. As a couple of you have already posted, that location is 42° 03' N, 92° 37' W, which is in Iowa. We now have the endpoints.

    There are several ways to find intermediate points along a great circle. We can calculate them, as in Peter's spreadsheet. For something quick and visual, get out a globe. For any pair of points on a globe, if I want to see the great circle connecting them (and if it's not too close to 180° distance separating the points), I can stretch a rubber band between the two points. The minimum energy state for a stretched elastic is minimum distance so it's an analog computer of the great circle path between the two points on the globe. I also sometimes estimate the points of a great circle using a virtual globe. If you open, for example, Google Earth, and stay in a full globe view, you can find the great circle arc connecting two points by rotating the globe, usually in two dimensions, until the points in question are both aligned along a line that passes through the center of the projected image of the globe. In other words, on-screen the globe looks like a circle. If you drag the globe around until the two points you're connecting are aligned on opposite sides of the visual center of the circular image of the globe, then the straight line connecting them is guaranteed to be the great circle arc between those two places. This is a very quick way to figure out approximate intermediate points on a great circle path between two widely-separated places. Naturally, many such software tools include a simple function to draw a great circle connecting two points directly.

    In the event, I used the brute force approach with same USNO calculating web page to get the intermediate points since I wanted the stars' altitudes, too. Here are the results for every 5 degrees of longitude between the end points (lat N/lon W):
    42° 03', 92° 37', Sun -6° 00', stars 25° 59'
    41° 39', 90° 00', Sun -7° 50', stars 24° 14'
    40° 41', 85° 00', Sun -11° 24', stars 20° 48'
    39° 28', 80° 00', stars 17° 14'
    37° 59', 75° 00', stars 13° 30'
    36° 13', 70° 00', stars 9° 35'
    34° 10', 65° 00', stars 5° 28'
    31° 46', 60° 00', stars 1° 07'
    31° 42.5', 59° 53', stars 1° 00'

    And I've made a map. It's attached.

    -FER

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