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Frank.
I guess I still don’t quite understand the topic at hand. Like you say, the question can be cast as a multi-variable differential calculus problem. When I do that (as in Exercises 4.8 and 4.9 in my book), I get results very similar to what I’ve seen you present. The one-body clearing correction to the lunar distance is (my Eq 13.9):

- A dH + 0.5 cot LD (1 – A^2) dH^2 /3438 + …

where dH is the altitude refraction correction (in minutes), and A is cos RBA (relative bearing angle). The second term, the quadratic one, is what you’ve called “Q”. Well almost -- your Q has the factor 0.55 in front, while I get 0.50. My calculation expands the exact clearing equation keeping the RBA constant, while calculating the change in the LD. That’s what happens in the clearing process.

In your example you’re changing the moon’ altitude, while keeping LD constant, and letting the RBA change. But if LD has been kept constant, how can its change be calculated?

JK

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