End of the third Stage.CHAPTERLXXVII.PRACTICE OF THE THIRD EXAMPLE,REFERRING TO THE SECTIONS.[132]Section 410.BELOW: Barom. Inches 30, .0168:Attached Therm.60°.6;Air-ditto, 60°.2:Above: Barom.- -Inches 29, .5218:Attached Therm.56°.6;Air-ditto, 57°.Subtract thecolder——from thewarmer,and there remains4°of Heat to be addedto thecolderBarometer; to give it anequalTemperature: which is to be done bythe 1st Table, thus:Section, 356.To find the Expansionwith4° of Heat,onthecolderBarometer; (which, as before, is theupperBarometer) standing at Inches 29, .5218 Tenths.First,with4°on29 Inches =.0117:2d,with4°on.5218 Tenthsabove29 Inches:In order to obtain which, beginwith4°on29 =.0117thenwith4°on30 =.0121Subtract for the Expansionwith——4°on1 Inch above 29, and there remains.0004.Section 362.Then for the Expansionwith4°on.1 Tenth of an Inch above 29 Inches; add a Cypher and decimal Point,Section 363.viz..00004:Then for the Expansion on.5128,multiplythe two last Terms, and divide———the Product by the first Term.1: the Answer is.0002|0872Add the Expansionwith4°on29 Inches, just found,.0117to the Inches of thecolderBarometer, viz.29.5218———Answer; Inches29.5337Tenths ofthecolderBarometer, arenowexpanded equallywith thewarmer: (rejecting the Decimals as inSection 395.)Place the Barometers, thus:UpperBarometer,29.5337LowerBarometer,30.0168End of the first Stage.Section 371.411.By the 2d Table, and in the 2d Column, find theHeightof each Barometer,withtheStandard-Heat, in Feet and Tenths, corresponding to the Inches andnearestTenthaboveandbelowthe Point required: andFirst of theupper, at29.5337:The Inches andnearest Tenthis above Feet.29.5, corresp. to2119.7}Difference between .5and .6 above 29 Inches.and below 29.6, cor. to2031.5———88|.2Section 373.412.By the 3d Table,withtheDifference88Feet, find theExpansionon the remaining Decimals, above 29.5, viz. on .0337, thus:on03= 26decimated26.003= 262.60007= 62.62——Feet29.22From theHeightcorresponding to29.5viz. Feet2119.7Tenths,subtract the29.22,i. e. Height cor. to.0338and there————————remains2090.4|8,theHeightcor. to29.5338with Expansion of the Standard-Heat.413. Repeat the 4 last Steps for thelowerBarometer, at 30.0168.1st. The Inches andnearestTenth isabove30. corresp. to Feet1681.7}Difference of .1 above 30 Inches.andbelow30.1 cor.1595.0———86|.72d. Then with 86 Feet, find theExpansionon the remaining Decimals, above 30,viz. .0168, thus: on01=99.006=525.20008=69.69———Feet14.89414. (3d.) From theHeightcorresponding to30 Inches, viz. Feet1681.7Tenths,subtract the Height14.89corresp. to .0168,————and there remains1666.8|1,the Height corresp.to 30.0168, withExpansionof the Standard-Heat.4th. From theupperHeight, at2090.48Subtract thelowerHeight, at1666.81———And there remains the Height423.67in Feetand Tenths of the upper Barometer, with theStandard Temperature.End of the second Stage.Section 374.415.Detached Therm.above57°Detached ditto,below60.2——WholeHeat117.2HalfHeat58.6(0 adding aCypher)StandardHeat31.24———which being deducted, leaves27°.36,viz. Degreesof Heat more than theStandard, for each Barometer.Section 380.416. By the 4th Table, find the Expansion of Air, with 27°.36, on Feet 423.67 Tenths.Section 406.First,with27°,on423.67, thus:viz.on400as 4000 = 262.4decimated26.2420as 2000 = 131.21.3123as 3000 = 196.8.1968.6as 6000 = 393.6.03936.07as 7000 = 459.2.004592—————Expansion =27.692752Section 407.Second,with.36onthesame, thus:on 400as 4000 = 349.9decimated.349920as 2000 = 174.9.017493as 3000 = 262.4.002624.6as 6000 = 524.8.0005248.07as 7000 = 612.3.00006123—————Expansion =.37050003Add the former27.692752—————Height in Feet28.06325203417. Which Height for Expansion of Air,with more thanthe Standard Heat, beingadded[133]to the Height, for Expansion of the Barometer,withthe Standard-Heat, gives the true Height of the upper Barometer, at the given Heat.ForExpansionofAirabove Standard Heat,Height in Feet28.0ForExpansionofBarometer,with Standard: Height in Feet423.6———418. True Height of theupperBarometer451.6LowerBarometer 1 Foot above the Water1.0Height of the Top of the Cross above the Gallery50.0———Height of the Top of the Cross above the Tyber502.6Height of the same, measured the same Daygeometrically, wasFeet502.9End of the last Stage.
End of the third Stage.
CHAPTERLXXVII.
REFERRING TO THE SECTIONS.[132]
Section 410.BELOW: Barom. Inches 30, .0168:
60°.6;
- -
56°.6;
——
4°
to thecolderBarometer; to give it anequalTemperature: which is to be done bythe 1st Table, thus:
Section, 356.
To find the Expansionwith4° of Heat,onthecolderBarometer; (which, as before, is theupperBarometer) standing at Inches 29, .5218 Tenths.
First,with4°on29 Inches =
with4°on29 =
thenwith4°on30 =
Subtract for the Expansionwith
4°on1 Inch above 29, and there remains
Section 362.
Then for the Expansionwith4°on.1 Tenth of an Inch above 29 Inches; add a Cypher and decimal Point,
Section 363.
.00004
.5128,
———
.0002
.0117
29.5218
———
29.5337
29.5337
30.0168
End of the first Stage.
Section 371.
411.By the 2d Table, and in the 2d Column, find theHeightof each Barometer,withtheStandard-Heat, in Feet and Tenths, corresponding to the Inches andnearestTenthaboveandbelowthe Point required: and
29.5337:
2119.7
2031.5
———
88|.2
Section 373.
412.By the 3d Table,withtheDifference88Feet, find theExpansionon the remaining Decimals, above 29.5, viz. on .0337, thus:
on
decimated
26.
2.6
.62
——
Feet
29.22
29.5
2119.7
29.22,
.0338
————
————
2090.4|8,
29.5338
413. Repeat the 4 last Steps for thelowerBarometer, at 30.0168.
1681.7
1595.0
———
86|.7
2d. Then with 86 Feet, find theExpansionon the remaining Decimals, above 30,
=
9
9.
=
52
5.2
=
69
.69
———
Feet
14.89
414. (3d.) From theHeightcorresponding to
1681.7
14.89
————
1666.8|1,
to 30.0168, withExpansionof the Standard-Heat.
2090.48
1666.81
———
423.67
End of the second Stage.
Section 374.
57°
60.2
——
117.2
58.6
31.24
———
27°.36,
of Heat more than theStandard, for each Barometer.
Section 380.
416. By the 4th Table, find the Expansion of Air, with 27°.36, on Feet 423.67 Tenths.
Section 406.
First,with27°,on423.67, thus:
viz.on400
as 4000 = 262.4
20
as 2000 = 131.2
3
as 3000 = 196.8
.6
as 6000 = 393.6
.07
as 7000 = 459.2
Expansion =
Section 407.
Second,with.36onthesame, thus:
on 400
as 4000 = 349.9
20
as 2000 = 174.9
3
as 3000 = 262.4
.6
as 6000 = 524.8
.07
as 7000 = 612.3
Expansion =
Add the former
Height in Feet
417. Which Height for Expansion of Air,with more thanthe Standard Heat, beingadded[133]to the Height, for Expansion of the Barometer,withthe Standard-Heat, gives the true Height of the upper Barometer, at the given Heat.
Height in Feet
28.0
with Standard: Height in Feet
423.6
———
451.6
1.0
Height of the Top of the Cross above the Gallery
50.0
———
502.6
502.9
End of the last Stage.