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    Re: Lunars by series, quadratic correction
    From: Frank Reed
    Date: 2019 May 8, 08:55 -0700

    Sean, you asked:
    "So (just to make sure I have this straight), if you had written this "new" correction in your original essay it might've looked something like:
    Q = (dh_Moon - dh_Star)^2 / 2 * cot(dist) * (1-A^2) / 3438 "

    Yes. Just so. :)

    During the 15% of a century that has elapsed since I wrote the original essay, my taste in organizing expressions like this has changed. Back then, I was happy using the cotangent function in expressions, but I assume these days that relatively fewer readers are comfortable with that function, and, after all, it is identical to 1/tangent. In addition, various steady improvements in online typography have made it easier to write equations like this more attractively. And finally to remove one possible confusion, I usually put the "/2" out front as a factor of "0.5" or sometimes a factor of "(1/2)" depending on how the wind is blowing that day. So I would write this today as:

    Q = 0.5 (dhm - dhs)2 (1 - A2) / tan(dist) / 3438. 

    It's assumed that the dh's are in minutes of arc and also that the result will be in minutes of arc, which is why we divide by 3438 at the end. If instead the dh's and Q are pure angles (also known as radians), then dividing by 3438 is un-necessary. It's important to remember that the difference in the altitude corrections assumes that they normally have opposite signs. It is, therefore, equivalent to the sum of the unsigned magnitudes (absolute values) of the altitude corrections. In simple terms, the number (dhm - dhs) should be larger in magnitude than dhm alone by as much as 5-10%.

    I should emphasize again that using the series expansion (as in "Easy Lunars") is useful for various historical and practical reasons, but if you're clearing lunars on a modern computing device (where the details are not relevant and are hidden from the end user and where dozens of digits of accuracy are normal), then there's no point. If the details are hidden, then just use the standard direct triangle solution which depends on the constancy of the difference in azimuth between the two bodies. The series expansion should be used when you're working with lower precision calculations (like a handheld calculator possibly or certainly on a slide rule or similar --you can work a series method with no more than four digits past the decimal point at all steps), or it should be used when you're trying to gain insight into the numerous historical methods which were based on the series expansion (for example, all of the methods published in Bowditch's Navigator during the heyday of lunars).

    Frank Reed

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