Handling Floating-Point Precision Problems

When converting to a JD value during calculation — for example, for 00:00:00 on January 1, 2000, the JD value is 2451545.4166707597 — restoration yields December 31, 1999, 23:59:59.999986588954926, due to floating-point arithmetic issues.

After a manual correction adding one second (1/86400), the seconds become 0.999981164932251111; in other words, the stable floating-point value here is around 0.99998. Since 1000 milliseconds make one second and astronomical calculations cannot be that precise anyway, the computation can be simplified.

Therefore, a floating-point correction is needed:

var xxx = Math.Round(v - Math.Floor(v), 6);
if (xxx < 0.50001)
{
 jd = v - xxx - 0.5 + 0.00001;
}
if (xxx > 0.50001)
{
 jd = Math.Floor(v) + 0.50001;
}

return jd;