The length of the antediluvian month and year

If the earth's spin increases then the days in a year increases. The year—as measured in a stable time metric, such as atomic-seconds—is unchanged because it is dependant on the distance from the sun, but the year as measured in days does change because the length of the day is dependant on the spin. The same can be said about the month.

If we define the year length to be y in unchanging units; the day to be d_a for the antediluvian day and d_p for the postdiluvian day; n_a is number of days in an antediluvian year and n_p the number of days in a postdiluvian year; then

d_a × n_a = y = d_p × n_p

d_a = d_p × n_p/n_a

d_p/d_a = n_a/n_p

Many ancient calendars use 360 days for a year. Perhaps these are stylised, especially as these calendars are postdiluvian; but what if they were based on memory of an antediluvian year of such a length? In such a case the length of an antediluvian day would have been

d_a = 24 hours × 365.25/360 = 24 hours 21 minutes.

We will define a month to be m in unchanging units and the number of days in an antediluvian month k_a; then

d_a × k_a = m = d_p × k_p

k_a = k_p × d_p/d_a = k_p × n_a/n_p

d_p/d_a = k_a/k_p = n_a/n_p

If a (synodic) month is now 29.53 days (new moon to new moon)

d_p = 24 hours, k_p = 29.53 days

then the antediluvian month (assuming a 360-day year) was

k_a = 29.5 days × 360/365.25 = 29.1 days

Now consider instead if the antediluvian month was 30 days in length

k_a = 30 days, k_p = 29.53 days

then the antediluvian day was

d_a = d_p × k_p/k_a = 24 × 29.53/30 = 23 hours 38 minutes

and the antediluvian year was

n_a = n_p × k_a/k_p = 365 × 30/29.53 = 371 days

These calculations assume no change in distance from the earth to the sun or the moon.