3d. With 107, and 2, (as a whole Number,) answering to 0002: which, in the Place of Meeting, gives 21.

Place them in View, and add, and bring them back again into Decimals, thus:

With 107 and 8, answering to .08 giving 86. Feet
and 9, to .009 9.7
and 2, to .0002 .21
———
95.9|1

(Next: with the 9, if required; which was before rejected:) but there being no .9 Tenths in the left Vertical, call it 90, and allow for it in each Answer by moving the decimal Point two Places to the left, thus:

with 90, and 8, answering to .08 giving 72 = .72
and 9, to .009 81 = .081
and 2, to .0002 18 = .0018
———

To

 .8|00|28

Add the former Sum

 95.9|
———

Total =

 96.7)

Which 95.9 is the Height in Feet and Tenths corresponding to .0892 Decimals of an Inch above Inches 24 .1 Tenth: and 24 .1 gave Feet 7388.0 in Height; therefore an additional Height, of so many Tenths of an Inch of Quicksilver in the Tube of the Barometer, must give in Feet, a less Height of the Barometer elevated above the imaginary Level indicated at 32 Inches.

10th. Step.

401. 10th. Step. Subtract the Height in Feet, corresponding to the Expansion on .0892 Tenths of an Inch, (less than Inches 24.2 Tenths, of the upper barometric Tube,) from the Height, in Feet, corresponding to the Expansion on Inches 24.1 Tenth of the same barometric Tube, continuing at the Standard Heat,⁠[129]

viz.7388.0
95.9
———
The Remainder7292.1

gives the real, viz. the less Height of the upper Barometer, at 24.1892 with the Standard Temperature.