Chapter LXXXIII: Section 419: Attached Therm. below, 71°.0

Attached Therm. _above_, 70 .5 ————— Subtract, and there remains .5 Tenths of a Degree of Heat to be added to the _colder_ Barometer (which in the present Case is the _upper_, but might possibly have been otherwise) by the 1st Table.

_First_, _with_ 0°.5 _on_ 29 Inches. To obtain which, begin
_with_ 1°.0 _on_ 29 Inches = .002:
_with_ 0°.1 above 1°, _on_ 29 = .0002: then
_with_ 0°.5 above 1°, _on_ 29 = .001.

Prepare it for Addition to the _colder_ Barometer.
colder Barometer 29.985
Expansion _with_ .5 above 1°, _on_ 29 .001
——————
29.986

_Secondly_, _with_ .5 Tenths above 1°, _on_ .985 Tenths above 29 Inches. To obtain which, (having already found the Height from Expansion _with_ .5 above 1°, _on_ 29 Inches, to be .001;) since the Expansion on .985 Tenths above 29 Inches, is somewhere above 29, yet below 30 Inches; find the Expansion _with_ .5 above 1°, _on_ 30 Inches, thus:

first, _with_ 1°, _on_ 30 = .003
2d. _with_ 0°.1 above 1°, _on_ 30 = .0003
3d. _with_ 0°.5 above 1°, _on_ 30 = .0015

Subtract the Expansion _with_ .5 Tenths above 1°, _on_ 29 Inches, from the Expansion _with_ .5 Tenths above 1°, _on_ 30 Inches:

viz. _on_ 30 = .0015
_on_ 29 = .001
—————
The Answer is .0005, the Height from Expansion, _with_ .5 Tenths
above 1°, _on_ 1 Inch above 29, i.e. on the 30th Inch: Then, if 1 Inch
above 29 gives .0005; .1 gives .00005:
and 985
———————
multiplied 00025
as whole 00040
Numbers, 00045
————————
give .0004|925
add the former Number 29.986
and, for the three remaining Decimals,
_may_ be substituted 1 Decimal
in the fourth Place 1
———————
colder Barometer of equal Heat } 29.9865
with the _warmer_ }

420. _When the Quicksilver in each Barometer indicates the same Number of Inches_, differing _but_ one _or_ two Tenths _at the most; (which will frequently be the Case, in levelling flat Countries, or measuring small Heights;—instead of the usual Method, (to find the Height of each Barometer_ separately, _with the Standard-Heat, by the_ 2d Column _of the 2d Table, as in Section 411;)—it will be more convenient,_

1st. _To subtract the lower Barometer from the upper. Then,_

2dly. _By the_ 3d Column _of the same Table, find the_ +difference+, (_viz. of_ one _or_ two Tenths _at the most_) below _the Inches and_ nearest Tenth _of the_ lower _Barometer_.

_And_ lastly, _with that_ +difference+, _find by the_ 3d Table, _the Height at the Standard-Heat, corresponding to the remaining Decimals_ above _the_ upper _Barometer_.

421.(_1st._) From the lower Barom. viz. 30.082
Subtract the upper 29.9865
———————
Remaining Decimals _above_ the upper .0955

_2d._ Find, by the 2d Table, the Height corresponding to the Inches, and _nearest_ Tenth _above_ and _below_ the Point at which the Quicksilver rests in the lower Barometer.

The Inches and _nearest_ Tenth is
_above_ 30 Inches, correspond. to Feet 1681.7
and _below_ 30.1, corresponding to 1595.0
——————
86.7
which is the +difference+ of .1 _below_ 30.1.

_Lastly._ Find, by the 3d Table, _with_ the +difference+, viz. 86 Feet,
_on_ the remaining Decimals, for the Height, in Feet, corresponding to
the Standard-Heat.
viz. .09 77 = 77. Feet.
.005 43 = 4.3
.0005 43 = .43
—————
Answer, Height in Feet 81.73
corresponding to .0955 above Inches 29.9865 Tenths of an Inch, of
Quicksilver in the upper Barometer thus brought to the Standard-Heat.

422. Prepare for Expansion of Air from Excess above Standard-Heat, on
the same Number of Feet:
Detached Thermom. _above_ 76°.
Detached Thermom. _below_ 68.0
————
Whole Heat 144.0
Half Heat 72.0(0 adding a Cypher)
Standard-Heat 31.24
—————
which deduct, and there remains 40.76: with which, by the 4th Table,
find the Expansion of Air on Feet 81.73:
_First, with_ 40°, _on_ 81.73, thus:
_on_ 80. as 8000 – 777.6 = 7.776
1. as 1000 – 97.2 = .0972
.7 as 7000 – 680.4 = .06804
.03 as 3000 – 291.6 = .002916
————————
7.944156
_Second_, _with_ .76 _on_ 81.73, thus:
_on_ 80. as 8000 – 1477.4 = .14774
1. as 1000 – 184.6 = .001846
.7 as 7000 – 1292.7 = .0012927
.03 as 3000 – 554.0 = .0000554
————————
Expansion .1509341
add the former Expansion 7.944156
—————————
Sum of the Expansions, viz. }
Height in Feet } 8.0950901
from Excess of Heat above Standard,
_with_ 40°.76 _on_ 81.73,
+added+ to the Height at the Standard-Heat, }
in Feet } 81.73
gives, in Feet and Tenths, the true ——————
Height of the Tarpeian Rock 89.8|2.