dobsonian telescope - Cancelling out earth rotation speed, Altazimuth mount

Given ra, dec, lat, lon in radians, and d in number of fractional days
since '1970-01-01 00:00:00 UTC', the azimuth of a star is:

$-cot ^{-1}(tan (text{dec}) cos (text{lat}) sec (text{d1}+text{lon}-text{ra})+sin (text{lat}) tan (text{d1}+text{lon}-text{ra}))$

and the elevation is:

$tan ^{-1}left(frac{sin (text{dec}) sin (text{lat})-cos (text{dec}) cos (text{lat}) sin (text{d1}+text{lon}-text{ra})}{sqrt{(cos (text{dec}) sin (text{lat}) sin (text{d1}+text{lon}-text{ra})+sin (text{dec}) cos (text{lat}))^2+cos ^2(text{dec}) cos ^2(text{d1}+text{lon}-text{ra})}}right)$

where d1 is: $frac{pi (401095163740318 d+11366224765515)}{200000000000000}$

(you will need to resolve the ambiguity in the arc(co)tangent, but
this isn't difficult).

These formulas aren't as daunting as they seem, since, for you, lat,
lon, ra, and dec will be fixed, and the only thing that changes is d.

Hope this helps, but I'm worried that it just demonstrates how complicated these formulas are.