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Looks like the equinox this year will be at about 09:37 UT on March 20. For at least two weeks before and after the equinox, you can work out the Sun's Declination in your head by knowing that it's gliding North at a speed of one knot. It's not quite one knot so take back a mile every 3 days, but nearly so. For example, suppose I want to know the Sun's Dec at 11:30 UT today, 18 Mar 2021 (11:30 UT is 07:30 Eastern US time =EDT). That's 46 hours before the equinox so the Dec of the Sun should be 0°46' S. The USNOclone app says 0°45.6' S. Close enough! What do you get assuming one knot north and minus 1' of Dec every three days a week from now? Two weeks?? Test it out. How long can you go with this simple rule as a "useful approximation"? It all depends on what we mean by useful, but pick a definition and test it out.

One knot northbound, 900 knots westbound. The Sun is flying!

Where will this happen? At the time of the equinox, it looks like the subSun point (where the Sun is at the zenith) will be in Kenya not far from Mount Kenya Park. About 20 minutes later with the Sun now a fraction of a mile above the equator (what fraction??), the Sun will be over Lake Victoria about a mile off the end of the runway at Entebbe, Uganda.  

Frank Reed