Elementary arithmetic, with brief notices of its history... by Robert Potts.

30
2. 851208LW A- gallons. 3. 16-26 per cent.
5. If 3'141592659 times the square of the radius be taken for the area of a circle,
and 5 miles be employed instead of 26253 feet, the volume of the atmosphere will be
found to be the difference of two spheres whose radii are respectively 4000 and 4005
miles, and equal to 100656681 cubic miles nearly. Next finding what weight of
mercury is equivalent to a cubic mile of air, the whole weight of the atmosphere may
be determined according to the suppositions made.
6. 12'61208cwt.
XXXI.
1. 5205 384... grains Troy. 2. 2150-428... cubic inches. 3. 2'225... gallons.
4. 69~1bs. Avoirdupois.
XXXII.
1. 23-25305 degrees. 2. 147617283950 of a right angle. 3. 57~ 17' 446". 4.
40-5 French degrees.
XXXIII.
1. '550115740 of 1 day. 2. '4844421... of a day of 12 hours: 25 days l7hrs.
8min. 26'88sec. 3. '6634534... of a Julian year. 4. 5 0376... an abstract number.
5. The decimals of the month and the week must first be reduced to days before
performing the operation. 6. 49-00302083 days. 7. '142850i42857 of a week is
the exact difference: any less decimal will fulfil the condition. 8. 4 hours.
XXXIV.
2. 1 hour= 0001140791... of a Julian year. 3. 1 hour, 33min. 51- sec.
4. 1037 133... miles an hour.
5. 88~ 20' longitude correspond to 5hrs. 52~min., and 59~ 50' to 3hrs. 59'min.
Suppose the time to be noon at London, then the clocks of Calcutta are 3hrs. 52'min.
if'ter 12, and the clocks of Barbadoes are 3hrs. 59Lmin. before 12: or the clocks of
Calcutta indicate 7min. 40sec. past 6 o'clock p.m., and those of Barbadoes 40sec. to
8 o'clock a.m. The clocks of Calcutta are 9hrs. 512min. behind the clocks of Barbadoes.
6. Divide the number of days in 19 years by the number of days in one lunar
month.
7. Divide the whole period by the interval between two successive full moons.
X.XXV.
1. See "On the Divisions and Measures of Time," pp. 14, 17. 2. The product
of the correct number of days in one year, and the number of years in the interval,
will give the required number. 3. The error in one year is '007736 of a day, which
in 1257 years amounted to 9*724152 days, or 10 days nearly, the number of days
omitted in 1582 by Pope Gregory in his adjustment of the calendar. 4. It may be
shewn that the Gregorian intercalation is more accurate than the Persian. 5. The
error in 52 years is '34728 of I day, and this error amounts to one day in
52x2 8795=149 731, or nearly 150 years.