CHAPTERLXXVIII.PRACTICE OF THE FOURTH EXAMPLE,[134]FOR MEASURING SMALL HEIGHTS.By this Example,smallHeights are easily measured.Section 419.Attached Therm.below,71°.0Attached Therm.above,70 .5——Subtract, and there remains.5Tenths of a Degree of Heat to be added to thecolderBarometer (which in the present Case is theupper, but might possibly have been otherwise) by the 1st Table.First,with0°.5on29 Inches. To obtain which, beginwith1°.0on29 Inches= .002:with0°.1 above 1°,on29= .0002: thenwith0°.5 above 1°,on29= .001.Prepare it for Addition to thecolderBarometer.colder Barometer29.985Expansionwith.5 above 1°,on29.001———29.986Secondly,with.5 Tenths above 1°,on.985 Tenths above 29 Inches. To obtain which, (having already found the Height from Expansionwith.5 above 1°,on29 Inches, to be .001;) since the Expansion on .985 Tenths above 29 Inches, is somewhere above 29, yet below 30Inches; find the Expansionwith.5 above 1°,on30 Inches, thus:first,with1°,on30 = .0032d.with0°.1 above 1°,on30 = .00033d.with0°.5 above 1°,on30 = .0015Subtract the Expansionwith.5 Tenths above 1°,on29 Inches, from the Expansionwith.5 Tenths above 1°,on30 Inches:viz.on30 =.0015on29 =.001——The Answer is.0005,the Height from Expansion,with.5 Tenths above 1°,on1 Inch above 29, i. e. on the 30th Inch: Then, if 1 Inch above 29 gives .0005;.1 gives.00005:and985———multiplied00025as whole00040Numbers,00045————give.0004|925add the former Number29.986and, for the three remaining Decimals,maybe substituted 1 Decimalin the fourth Place1———colder Barometer of equal Heatwith thewarmer29.9865420.When the Quicksilver in each Barometer indicates the same Number of Inches, differingbutoneortwo Tenthsat 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 Barometerseparately,with theStandard-Heat, by the2d Columnof the 2d Table, as in Section 411;)—it will be more convenient,1st.To subtract the lower Barometer from the upper. Then,2dly.By the3d Columnof the same Table, find thedifference, (viz. ofoneortwo Tenthsat the most) belowthe Inches andnearest Tenthof thelowerBarometer.Andlastly,with thatdifference,find by the3d Table,the Height at the Standard-Heat, corresponding to the remaining DecimalsabovetheupperBarometer.421. (1st.) From the lower Barom. viz.30.082Subtract the upper29.9865———Remaining Decimalsabovethe upper.09552d.Find, by the 2d Table, the Height corresponding to the Inches, andnearestTenthaboveandbelowthe Point at which the Quicksilver rests in the lower Barometer.The Inches andnearestTenth isabove30 Inches, correspond. to Feet1681.7andbelow30.1, corresponding to1595.0———86.7which is thedifferenceof .1below30.1.Lastly.Find, by the 3d Table,withthedifference, viz. 86 Feet,onthe remaining Decimals, for the Height, in Feet, corresponding to the Standard-Heat.viz..0977=77. Feet..00543=4.3.000543=.43———Answer, Height in Feet81.73corresponding to .0955 above Inches 29.9865Tenths 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.above76°.Detached Thermom.below68.0———Whole Heat144.0Half Heat72.0(0 adding a Cypher)Standard-Heat31.24which deduct, and thereremains———40.76:with which, by the 4th Table, find the Expansion of Air on Feet 81.73:First, with40°,on81.73, thus:on80.as 8000777.6 =7.7761.as 100097.2 =.0972.7as 7000680.4 =.06804.03as 3000291.6 =.002916————7.944156Second,with.76on81.73, thus:on80.as 80001477.4 =.147741.as 1000184.6 =.001846.7as 70001292.7 =.0012927.03as 3000554.0 =.0000554————Expansion.1509341add the former Expansion7.944156————Sum of the Expansions, viz.Height in Feetfrom Excess of Heat aboveStandard,with40°.76on81.73,8.0950901addedto the Height at theStandard-Heat, in Feet81.73————gives, in Feet and Tenths, the trueHeight of the Tarpeian Rock89.8|2.
CHAPTERLXXVIII.
By this Example,smallHeights are easily measured.
Section 419.
71°.0
70 .5
——
.5
Tenths of a Degree of Heat to be added to thecolderBarometer (which in the present Case is theupper, but might possibly have been otherwise) by the 1st Table.
First,with0°.5on29 Inches. To obtain which, begin
Prepare it for Addition to thecolderBarometer.
29.985
.001
———
29.986
Secondly,with.5 Tenths above 1°,on.985 Tenths above 29 Inches. To obtain which, (having already found the Height from Expansionwith.5 above 1°,on29 Inches, to be .001;) since the Expansion on .985 Tenths above 29 Inches, is somewhere above 29, yet below 30Inches; find the Expansionwith.5 above 1°,on30 Inches, thus:
Subtract the Expansionwith.5 Tenths above 1°,on29 Inches, from the Expansionwith.5 Tenths above 1°,on30 Inches:
viz.on30 =
on29 =
the Height from Expansion,with.5 Tenths above 1°,on1 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,maybe substituted 1 Decimalin the fourth Place
1
———
colder Barometer of equal Heatwith thewarmer
29.9865
420.When the Quicksilver in each Barometer indicates the same Number of Inches, differingbutoneortwo Tenthsat 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 Barometerseparately,with theStandard-Heat, by the2d Columnof the 2d Table, as in Section 411;)—it will be more convenient,
1st.To subtract the lower Barometer from the upper. Then,
2dly.By the3d Columnof the same Table, find thedifference, (viz. ofoneortwo Tenthsat the most) belowthe Inches andnearest Tenthof thelowerBarometer.
Andlastly,with thatdifference,find by the3d Table,the Height at the Standard-Heat, corresponding to the remaining DecimalsabovetheupperBarometer.
30.082
29.9865
———
.0955
2d.Find, by the 2d Table, the Height corresponding to the Inches, andnearestTenthaboveandbelowthe Point at which the Quicksilver rests in the lower Barometer.
The Inches andnearestTenth is
1681.7
1595.0
———
86.7
which is thedifferenceof .1below30.1.
Lastly.Find, by the 3d Table,withthedifference, viz. 86 Feet,onthe remaining Decimals, for the Height, in Feet, corresponding to the Standard-Heat.
corresponding to .0955 above Inches 29.9865Tenths 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:
76°.
68.0
———
144.0
72.0
31.24
———
40.76:
with which, by the 4th Table, find the Expansion of Air on Feet 81.73:
on80.
777.6 =
1.
97.2 =
.7
680.4 =
.03
291.6 =
on80.
1477.4 =
1.
184.6 =
.7
1292.7 =
.03
554.0 =
Expansion