110 Method of finding the Heights of Mountains. [Fes. 
Amounts of il. at five per Cent. Compound Interest. 
Years, Amounts, Years. Amounts, 
i 1°05 ll 1:71034 
2 1°1025 12 1°79586 
3 1°15762 15 1°8S565 
4 hoo) 14 1°97993 
5 1:27628 15 2:07893 
6 1°34010 16. 2°18287 
7 1:40710 17 229202 
8 1:47746 18 2°40662 
9) Lop 1s3 19 2°52695 
10 1°62895 20 2°65330 
‘To make the example as simple as possible, it may be proper, in 
the first place, to suppose a case in which corresponding parts of the 
numbers expressing the heights of the barometrical columns, are 
found among the amounts exactly. Thus, let the height of the 
barometer at the lower station be 28°142 in., and at the upper 21 
in.; if each be divided by 20, we obtain 1°4071 and 1:05, which 
are the exact amounts for seven years and one year ; but these latter 
numbers being the logarithms of the former, 
Hh >= 212[/.D —1.d)= 212 x7 — 1 = 1272 fathoms, 
If formula (2) be used, which, in most cases, will be more con- 
venient, 
H = 212 x 1(2) = 212 x 1(75*) = 212 x 7. (13401) 
= 212 x G= 1272 fathoms. ‘The result by Brigg’s logarithms is 
1271°357 fathoms. 
Next, let a case be taken in which neither the barometrical 
heights themselves, nor corresponding parts of them, can be found 
in the table exactly: thus, employing the data of the above 
example, and taking ~!.th instead of th of the heights of the 
barometrical columns, we obtain 2°34517 and 1°75. As neither of 
these numbers is found among the amounts, we must take the years 
corresponding to the next lowest for both, and then find a propor- 
tional part for the excess of each. The next lowest among the 
amounts to 2°34517 is 2:29202, the amount for 17 years, and the 
difference is *05315; also the excess of the next highest, viz. 
240662, above 2:29202 is ‘1146, 
And *1146 ; 05315 :: 1: °4638. 
Hence 2°34517 of amount corresponds to 17°46378 years. 
Tn Ike manner, it will be found that 1°75 amount corresponds to 
11°46374 years. 
