A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Frank Reed
Date: 2021 Jan 12, 09:06 -0800
That's an interesting --and clever!-- approach. If I'm interpreting this correctly, you're using the Sun's daily angular rate as a surrogate for its distance. And then distance determines SD. When you say "determine how much the sun has moved over 1 day", what do you mean specifically? What calculation are you doing for that distance moved? Also, I suspect there's a way to dispense with the square root in your method, right? Something binomial: sqrt(1 + x) ~ 1 + x/2 when x is small.
Another source to compare against: my almanac data web app which includes the Sun's semi-diameter to 0.01' at the foot of the data table. The SD numbers agree with your other resources to +/-0.01' (I aim for one arcsecond in these web apps, so that's consistent).
The simplest quick calculation for the Sun semi-diameter is probably:
SD = 16' + 0.27' · cos(0.986° · dDays)
where dDays is the number of days since perihelion (ten days ago). And in fact, you can do quite well by sketching a simple graph of a "sine" curve (corresponding to this short equation).