There being no attached Thermometers; the first Table is useless: the Barometer below is therefore supposed to be of the same Temperature as when above; the detached Thermometer remaining at the same Degree, viz. 65°.

State the Barometer, thus: when below, at29.8
when above, at23.25.

End of the first Stage.

424. Find the Height (at the Standard-Heat) corresponding to the Inches and nearest Tenth above and below 23.25: i. e. above 23.2, and below 23.3: by the 2d Table.

Now 23.2 corresponds to 8379.7: and the Difference of .1 above, i. e. to 23.3, is in Feet = 112|.1: by the 3d Column of the same Table.

With this Difference, consult the 3d Table: i. e. with 112, (omitting the .1 as too minute) on the remaining Decimals above 23.2, viz. on 05, as on 5, or 5⁄10; and the Answer is 56 Feet: which Number being subtracted from 8379.7, the Remainder 8323.7, is the Height in Feet of the Barometer in the Car, at the Standard-Heat.

Repeat the last Process for the Barometer on the Ground.

Now 29.8, by the 2d Table, corresponds to 1856.0; and there being no Parts or Decimals more minute than a Tenth, viz. .8, there is no Occasion for the 3d Table.

Subtract the Barometer in the Car, from the same when on the Ground; and, by the 2d Table,

upper Barom. 23.25,

corresp. to

 8323.7, and the
lower Barom. 29.8,

to

 1856.0: the
Remainder is the Height in Feet ——— of the
Barometer in the Car

viz.

 6467.7, with the Standard-Heat.