C 6^ ] 
« Example. 
In 74212 mean lunations. Qu. How many days, 
hours, minutes, and feconds ? 
Lun. 
Days Dccim, of a Day. 
70000 
2067141.3595756 
4000 
1 18 122.36340432 
200 
5906.1 18170216 
10 
295 ' 3 ° 59 '= 85 io 8 
59.061 18170216 
2 
74212 
1 2191524.20824034896 
Anf, 2191524 days and .20824034896 
decimal parts of a day. 
And by redudion, 2191 524 days contain 6ooo 
Julian years, 24 days, and .20824034896 decimal 
parts of a day, contain 4 hours, 59 minutes, 51 
feconds, 57 But in pradice, it is fuf- 
hcient to take in four or five of the decimal figures. 
Having got this hint from Mr. Rivet, I reverfed 
it into a way of finding the number of mean luna- 
tions contained in any given quantity of time j for 
which purpofe I calculated the following table, upon 
the above length of a mean lunation j which comes 
the neareft to the truth of any length I have yet 
found, when carried back from the prefent times 
to the recorded times of antient eclipfes, if the proper 
equations depending upon the anomalies of the fun 
and moon are applyed to the mean times of new and 
full moons. 
A Table 
