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Well, I was hoping one of you ‘locked-down’ mathematicians’ would explain this AP3270/HO249 reducing of minus d with increasing LHA and eventually going plus when lat<dec in the same hemisphere problem for me, but no luck so far, so I had a go at it myself.  I started with a drawing of a sphere including Zenith Distances from the observer to the star for two successive lines of equal declination (isodecs, not great circles) for increasing LHAs e.g. 0, 20, 40, 60.  It soon became obvious in side elevation that as LHA was increased the difference between the ZDs began reducing from -60.  I even managed to show in plan view that d would eventually change sign.  You’d probably never come across this in real life, because by the time it happens, the star would be so low in the sky, you’d have to be really desperate to use it.  However, my drawings ended up so messy that I’d be embarrassed to show them, maybe next time.

Instead, I thought it was about time I revised my spherical trig, so I had a go at it in Excel.  Well you've got to do something to keep the little grey cells active during the lockdown.

Getting Zenith Distance is a side-angle-side cosine rule problem.
CosZD=CosCoLatxCosCoDec+SinCoLatxSinCodecxCosLHA

I decide to use an Observer’s Latitude of Five Degrees North and a declination between 29 and 30 North, the limit of declination in Vol 2.  Once I had the ZDs for 30N and 29N, I subtracted one from the other to get d.  The result is in the attached spreadsheet.  d changes from minus to plus somewhere around 80 degrees LHA.  Perhaps one of you could check it for me although I’m fairly confident it’s OK, because the results tie in with Vol 2 within a minute.  Any difference is probably because 3270/249 and Excel round decimals differently.

My next homework is to learn how to make Excel draw graphs.  DaveP