The Commissioners appointed by the British Government to construct standard weights and measures, decided that the normal boiling-point, 212°, on the thermometer should represent the temperature of steam generated under an atmospheric pressure equal in inches of mercury, at the temperature of freezing water, to 29·922 + (cos. 2 latitude × ·0766) + (·00000179 × height in feet above the sea-level). Hence, at London, lat. 51°30´ N., we deduce 29·905 as the barometric pressure representing the normal boiling point of water,—the trifling correction due to height being neglected. If then, in the latitude of London, the barometric pressure, at the time of fixing the boiling point, be not 29·905 inches, that point will be higher or lower, according to the difference of the pressure from the normal. Near the sea-level about 0·59 inch of such difference is equivalent to 1° Fahrenheit in the boiling point.
Suppose, then, the atmospheric pressure at London to be 30·785 inches, the following calculation gives the corresponding boiling temperature for Fahrenheit’s scale:—
Observed | pressure |  | 30·785 |
Normal | " |  | 29·905 |
Difference |  | ·880 |
As 0·59 is to 0·88, so is 1° to 1°·5.
That is, the water boils at 1°·5 above its normal temperature; so that, in this case, the normal temperature to be placed on the scale, viz. 212°, must be 1°·5 lower than the mark made on the tube at the height at which the mercury stood under the influence of the boiling water.
The temperature of the vapour of boiling water may be found, at any time and place, as follows:—Multiply the atmospheric pressure by the factor due to the latitude, given in the annexed Table V., and with the result seek the temperature in Table VI.
Table V. | Table VI. | ||||
Latitude. | Factor. | Temperature of Vapour. | Tension. | Temperature of Vapour. | Tension. |
Degrees. | Â | Degrees. | Inches. | Degrees. | Inches. |
0 | 0·99735 | 179 | 14·934 | 197 | 22·036 |
5 | 0·99739 | 180 | 15·271 | 198 | 22·501 |
10 | 0·99751 | 181 | 15·614 | 199 | 22·974 |
15 | 0·99770 | 182 | 15·963 | 200 | 23·456 |
20 | 0·99797 | 183 | 16·318 | 201 | 23·946 |
25 | 0·99830 | 184 | 16·680 | 202 | 24·445 |
30 | 0·99868 | 185 | 17·049 | 203 | 24·952 |
35 | 0·99910 | 186 | 17·425 | 204 | 25·468 |
40 | 0·99954 | 187 | 17·808 | 205 | 25·993 |
45 | 1·00000 | 188 | 18·197 | 206 | 26·527 |
50 | 1·00046 | 189 | 18·594 | 207 | 27·070 |
55 | 1·00090 | 190 | 18·998 | 208 | 27·623 |
60 | 1·00132 | 191 | 19·409 | 209 | 28·185 |
65 | 1·00170 | 192 | 19·828 | 210 | 28·756 |
70 | 1·00203 | 193 | 20·254 | 211 | 29·335 |
75 | 1·00230 | 194 | 20·688 | 212 | 29·922 |
80 | 1·00249 | 195 | 21·129 | 213 | 30·515 |
 |  | 196 | 21·578 | 214 | 31·115 |
How to use the Tables.—When the temperature is known to decimals of a degree, take out the tension for the degree, and multiply the difference between it and the next tension by the decimals of the temperature, and add the product to the tension, for the degree.
Required the tension corresponding to 197°·84.
° | |||||||
197 |  | = | 22·036 |  | ·465 × ·84 | = | ·391 |
198 |  | = | 22·501 |  | 197° | = | 22·036 |
 | Difference |  | ·465 |  | 197·84 | = | 22·427 |
When the tension is given, take the difference between it and the next less tension in the Table, and divide this difference by the difference between the next less and next greater tensions. The quotient will be the decimals to add to the degree opposite the next less tension.
Thus, for 23·214 inches, required the temperature.
Given | 23·214 |  |  |  | Next | greater |  | 23·456 |
 | 22·974 |  |  |  | Next | less |  | 22·974 |
 | ·240 |  | Difference |  | ·482 | |||
 | And  | ·240 |  | = | ·5 | |||
·482 | ||||||||
Temperature opposite next less |  | 199·0 | ||||||
Temperature required |  | 199·5 |
A similar method of interpolation in taking out numerical quantities is applicable to almost all tables; and should be practised with all those given in this work.
Example.—Thus, in Liverpool, lat. 53° 30´ N., the barometer reading 29·876 inches, its attached thermometer 55°, and the correction of the instrument being + ·015 (including index error, capillarity and capacity), what temperature should be assigned for the boiling point marked on the thermometer?
Observed barometer |  | 29·876 |
Correction |  | + ·015 |
 |  | 29·891 |
Correction for temperature |  | - ·074 |
Reduced reading |  | 29·817 |
Factor from Table V. |  | 1·00077 |
 |  | 208719 |
 |  | 208719 |
 |  | 29817 |
Equivalent for lat. 45° |  | 29·83995909 |
In Table VI., 29·84 gives temperature 211°·86.