418 
MESSES. T. E. THOEPE AND A. W. EUCKEE ON 
Table XV. 
I. 
II. 
III. 
IY. 
I. 
II. 
III. 
IY. 
Temperature. 
Specific 
Gravity. 
Proportional 
parts for 1° C. 
Proportional 
parts for -00001 
increase in sp. gr. 
Temperature. 
Specific 
Gravity. 
Proportional 
parts for 1° C. 
Proportional 
parts for '00001 
increase in sp. gr. 
•0000 
•0000 
•0000 
•0000 
O 
0 
1-02000 
3 
1 
O 
19 
1-01740 
25 
0-944 
1 
1-01997 
4 
0-995 
20 
1-01715 
25 
0-943 
2 
1-01993 
5 
0-990 
21 
1-01690 
26 
0-941 
3 
1-01988 
6 
0-986 
22 
1-01664 
27 
0-940 
4 
1-01982 
8 
0-982 
23 
1-01637 
28 
0-938 
5 
1-01974 
9 
0-979 
24 
1-01609 
29 
0-937 
6 
1-01965 
11 
0-975 
25 
1-01580 
29 
0*935 
7 
1-01954 
12 
0-972 
26 
1-01551 
30 
0-934 
8 
1-01942 
13 
0-969 
27 
1-01521 
30 
0-932 
9 
1-01929 
14 
0-966 
28 
1-01491 
31 
0-930 
10 
1-01915 
15 
0-963 
29 
1-01460 
32 
0-928 
11 
1-01900 
17 
0-961 
30 
1-01428 
32 
0-925 
12 
1-01883 
17 
0-958 
31 
1-01396 
32 
0-922 
13 
1-01866 
19 
0-956 
32 
1-01364 
33 
0-919 
14 
1-01847 
20 
0-954 
33 
1-01331 
33 
0-915 
15 
1-01827 
21 
0-952 
34 
1-01298 
33 
0-912 
16 
1-01806 
21 
0-950 
35 
1-01265 
34 
0-908 
17 
1-01785 
22 
0-948 
36 
1-01231 
34 
0-903 
18 
1-01763 
23 
0-946 
In order to facilitate the use of the Table we subjoin directions for its application in 
the form of rules, and give a couple of examples. 
Given the specific gravity of a sample of sea-water at any temperature t, to find it 
at 0° C. : — Look out in column I. the figure giving the number of entire degrees of the 
temperature ; multiply the fraction (if any) by which the observed temperature exceeds 
that number by the corresponding number in III., and subtract the result from the corre- 
sponding number in column II. Subtract the difference from the observed specific gravity, 
and divide the number so obtained by that corresponding to the observed temperature 
in column IV. (without prefixing the ciphers at the top of the column); add the 
quotient to 1-02000, and the sum will be the specific gravity required. 
Example I. Specific gravity observed at 18 0- 5 C. = L02475. To find it at 0° C., number- 
opposite 18 in column III. is -00023, which multiplied by -5 equals *000115; and 
1-017630 — -000115=1-017515. 
Subtract this from the observed specific gravity, 
1-024750— 1-017515= -007235. 
Divide by "945, and the quotient is -00765, which added to 1-02000 gives 1-02765 as 
the specific gravity at 0° C. 
Example II. Specific gravity observed at 15° =1-02570. To find it at 0°C., 
1-02570 — 1-01827= -00743, 
and 
-00743 
•952 
=•00780. 
Therefore specific gravity at Q°=T02Q0Q + -00780=1-02780. 
