CHAPTERLXXVI.PRACTICE OF THE FIRST TABLE IN THE SECOND EXAMPLE.4th Step applied in the 2d Example.4th Step applied.Section 391.THEOrderto be observed in finding the Expansionwith4°.6, i. e. with 4 Degrees, .6 Tenths of Heat, on 24.178, i. e. 24 Inches, .178 Tenths of the coldest Barometer.Find the Expansion required, thus:Case the 1st.1st. Part.With4°on24 Inches.2d. Part.With4°on.178 Tenths of an Inch above 24 Inches.Case the 2d.1st. Part.With.6 Tenths of a Degree,on24 Inches.2d. Part.With.6 Tenths of a Degree,on.178 Tenths above 24 Inches.specifically,thus:1st. Part ofCase the 1st.To find the Expansion,With4°on24 Inches.2d. Part ofCase the 1st.With4°,on.178 Tenths of an Inch above 24 Inches; begin thus:With4°,on24 Inches: then,With4°,on25: then,With4°,on1 Inch above 24, i. e.onthe 25th Inch: then,With4°,on.1 Tenth above 24: then,With4°,on.178 Tenths above 24.1st Part ofCase the 2d.To find the Expansion,With.6 above 4°on24; begin thus:With4°on24 Inches: then,With5°on24: then,With1° above 4°,on24, i. e. the 5th°: then,With.1 Tenth above 4°,on24: thenWith.6 Tenths above 4°,on24.2d Part ofCase the 2d.To find the Expansion,With.6 Tenths above 4° of Heaton.178 Tenths above 24 Inches: to be done thus:Theexpansionwith 4°, on .178 Tenths above 24 Inches, being once found; divideitby 4: and the Quotient is the Expansion with 1° above 4°, on .178 Tenths of an Inch above 24 Inches.Then for the Expansion with .1 Tenth above 4°, on .178 Tenths above 24 Inches; add a Cypher and decimal Point to the left of the same Quotient.Then for the Expansion with .6; multiply that Sum into .6, and add a Cypher and decimal Point.The Answer is thepartof an Inch, to which .6 Tenths of a Degree above 4° of Heat, on .178 Tenths of an Inch above 24 Inches, raises the Barometer.It is true, thepartis so minute as to be rejected: yet the Mode of Proceeding, in order to investigate the Expansion with Precision, is proper to be retained.392.practiceof the first Part ofCase the 1st.For the Expansionwith4°,on24 Inches;look, in the first Table, (Sect. 363) and in the left vertical Column,with4 Degrees of the Thermometer; and along the upper horizontal Line,on24 Inches of Quicksilver in the Tube of the Barometer: the Point of Meeting gives the Expansion .0097;[128]which, preparatory to Addition,is to be placed under the 24, .178 thus,.0097practiceof the 2d Part ofCase the first.393. In order to obtain the Expansion,with4°, of Heaton.178 Tenths of an Inch above 24 Inches of the Barometer; let it be considered where it ought to be found in the Table: for, Tenths of 1 Inch above 24 Inches, are at some intermediate Point between 24 and 25; that is, above 24, yet not so high as 25: or more than 24, yet less than 25.Look therefore in the Table,with4 Degrees of Heat,on24 Inches; thenwith4°on25 Inches: and the respective Numbers are .0097 and .0101.And by taking the Expansionwith4°on24 Inches, from 4° on 25; the Remainder will be the Expansion with 4° on 1 Inch above 24 Inches, viz. on the 25th Inch, thus:With4°on}25 =.0101from;24 =.0097subtract:——.0004:This therefore is the Expansionwith4°,on1 Inch above 24 Inches.Thenwith4°,on.1 Tenth of an Inch above 24 Inches.The Answer is the same as the former, viz. .0004, with the Addition of a Cypher and decimal Point to the left, thus; .0004 becomes .00004, viz. the Expansionwith4°,on.1 Tenth of an Inch above 24 Inches.Then for the Expansionwith4°,on.178 Tenths, say,If the Expansionwith4°,on.1 Tenth above 24 Inches gives .00004 Part of an Inch, what will the Expansionwith4°,on.178 give?Thus; .1 : .00004 :: .178?Multiply the two last Terms, thus:.00004.178————000320002800004————0000712:and, as in Multiplication of Decimals, the Product must have as many decimal Places, as are in the Factors; a Cypher must be added to the left Hand, thus: .00000712: but having divided that Product by the first Term .1, viz. a Decimal, the Answer is a Cypher less; viz. .0000712.This Answer is the Expansionwith4°,on.178 Tenths of an Inch above 24 Inches: prepare it forAddition, as the former,24.178.0097.0000712practiceof the first Part ofCase the 2d.394. For the Expansion of .6 Tenths of a Degree of Heat, (more than the 4 Degrees) on 24 Inches of thecoldestBarometer; it shoudbe considered where such Tenths can lie in the Table.Now .6 Tenths of 1 Degree, (more than the 4°) are at some intermediate Point of the Thermometer between 1 and 2 Degrees: above 1; yet not so high as 2: or more than 1; yet less than 2.Therefore .6 Tenths of 1 Degree above 4 Degrees, are somewhere between the 4th and 5th Degree: above 4; yet not so high as 5: or more than 4; yet less than 5.Look in the Table (Section 363); firstwith4 Degrees of Heat,on24 Inches, and thenwith5 Degrees of Heaton24 Inches; and the respective Numbers are .0097 and .0121: and by taking the Expansionwith4 Degreeson24 Inches, from the Expansionwith5 Degreesonthe same 24 Inches; the Remainder will be the Expansionwith1 Degree above 4°on24 Inches: viz.with{5° = .0121}on24 Inches, as in whole Numbers.4° = .0097——Remainder, .0024This therefore is the Expansionwith1 Degree of Heat, above 4, viz.withthe 5th Degree,on24 Inches of the Barometer.Then say, if 1 Degree of the Thermometer (above 4, viz. the 5th Degree) gives by Expansion, a certain additional Height, or Part of an Inch, viz. .0024,on24 Inches of the Barometer; what Height will 6 Degrees give? Answer 6 Timesmore.Multiply the 2d and 3d Terms, and divide by the first, thus;1 :.0024:: 6?6——.0144is the Expansion, or Height, in Parts of an Inch, for 6 Degrees.And farther, to proportion for the Decimal; say as .1 Tenth of a Degree gives a certain Tenth of the former .0024, in additional Height, viz. .00024; what Height will .6 Tenths give? Answer, .00144.Prepare thisHeightfor Addition to the Numbers already found.practiceof the 2d Part ofCase the 2d.395. To find the Expansion of .6 above 4° on .178 above 24 Inches.The Expansionwith4°on.178 is already found to be .0000712: divide it by 4, and the Answer is .0000178, viz. the Expansionwith1°on.178 above 24 Inches:And, for the Expansion with .1 Tenth; the Answer, with the Addition of a Cypher and decimal Point to the left, becomes .00000178.Lastly, for the Expansion with .6, say,If .1 : .00000178 :: .6?Multiply the 2d and 3d Terms, and divide by first:.00000178.6—————.000001068.The Answer is a Decimal less, viz. .00001068; i. e. the Decimal of an Inch, to which .6 Tenths of a Degree above 4 Degrees of Heat, on .178Tenths of an Inch above 24 Inches, raises the Barometer: which, after all, is so inconsiderable, that it may be fairly rejected.Yet the Rules by which these Deductions are made, may be useful in other Cases.Prepare for Addition, as before.The Decimals, in the Answers, may be omitted, when they exceed four Places.5th Step.396. 5th Step. To proceed with the second Example.Place the different Expansions now found, above each other, Units, Tens, &c. under Units, Tens, &c. preparatory to Addition, thus;For the Expansionwith4°, .6on24, .178:1st.with4°,on24,.00972d.with.6on24,.001443d.with4°,on.178.00007124th.with.6on.178.00001068—————The Expansions with 4°,.6 added =.01122188To the Sum add the Height of thecolderBarometer24.178———24.1892|The Answer is Height of thecolderBarometer, now equal in Temperature to thewarmer: (rejecting all but the four first Decimals.)6th Step.397. 6th Step. Place the Barometersnowof the same Temperature, i. e.equalto the warmer, in one View, thus:1st. theupperBarometer,24.18922d. thelowerBarometer,28.1328The 7th Step applied in the second Example.7th Step.398. Find the Height, in Feet, in the 2d Columnof the 2d Table, corresponding to Inches and Tenths of theupperbarometric Tube, in the 1st. Column of the same Table, thus: (Sect. 371.)The Barometer standing at 24.1892; it must be considered where, in the 2d Column of the 2d Table, a Height corresponding tosuchInches and Tenths can lie: and the Answer is, somewhereabove24 Inches .1 Tenth, but not so high as 24 Inches .2 Tenths: 24 Inches .1892 Tenths, beingmorethan 24 Inches .1 Tenth, butlessthan 24 Inches .2 Tenths.First then, look in the 1st Column for Inches 24, .1 Tenth; and the corresponding Height in Feet is 7388.0: but the Height for 24, .2, in the 2d Column, beneath the former Number, isonly7280.1.8th Step.399. 8th Step. Subtract the latter from the former and the Remainder is 107.9, the same as in the 3d Column: viz. the Height, in Feet and Tenths, corresponding to one Tenth only, namely, the ist Tenth above Inches 24, .1 Tenth: with the Temperature of 31.24 of Farenheit, for which sole Purpose the 2d Table is calculated.A new Questionthenarises, viz. what are the Heights in Feet and Tenths, corresponding to the remaining Tenths or Decimals of an Inch aboveInches 24,.1 Tenth,viz..08.009.0002?which is to be resolved, by Application of the 3d Table, or Table forTenths, which see, (Section 373.)9th Step.400.9th Step applied in the 2d. Example.First for theupperBarometer.Look in the Table for Tenths, in the left vertical Column with 107, (rejecting the .9, as too minute;) and along the horizontal Line at the top, with 8: and find the Answergradually, thus:1st. With 107, and 8, (as a whole Number,) answering to .08: which, in the Place of Meeting, gives 86 Feet.2d. With 107, and 9, (as a whole Number,) answering to .009: which, in the Place of Meeting, gives 97.3d. 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 107and 8,answeringto .08giving86. Feetand 9,to .0099.7and 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,answeringto .08giving72 = .72and 9,to .00981 = .081and 2,to .000218 = .0018———To.8|00|28Add the former Sum95.9|———Total =96.7)Which 95.9 is theHeightin Feet and Tenths corresponding to .0892 Decimals of an Inch above Inches 24 .1 Tenth: and 24 .1 gave Feet 7388.0 inHeight; therefore an additionalHeight, of so many Tenths of an Inch of Quicksilver in the Tube of the Barometer, must give in Feet, alessHeight of the Barometer elevated above theimaginaryLevel indicated at 32 Inches.10th. Step.401. 10th. Step. Subtract theHeightin Feet, corresponding to theExpansionon .0892 Tenths of an Inch, (lessthan Inches 24.2 Tenths, of theupperbarometric Tube,) from theHeight, in Feet, corresponding tothe Expansion onInches 24.1 Tenth of the same barometric Tube, continuing at the Standard Heat,[129]viz.7388.095.9———The Remainder7292.1gives the real, viz. thelessHeight of theupperBarometer, at 24.1892 with the Standard Temperature.Repeat the same Process, viz. the 9th. and 10th. Steps, for thelowerBarometer, thus:For the lower Barometer in the 2d. Example.First, Find the Height, inFeet, of the lower Barometer, standing at Inches 28.1318 Tenths, in the 2d. Column of the 2d. Table, corresponding to Inches and Tenths of the Quicksilver in the barometric Tube, in the first Column of the same Table, thus:The lower Barometer standing at 28.1318; it must be considered, where in the 2d. Column of the 2d. Table, a Height corresponding to such Inches and Tenths can lye: and the Answer is, somewhere above 28 Inches, .1 Tenth, but notso high as 28 Inches .2 Tenths: 28.1318 Tenths being more than 28 Inches .1 Tenth, yet less than 28 Inches .2 Tenths.First, then, look, in the first Column for28.1,and the corresponding Height, in Feet, is3386.6:but the Height for 28.2, is only3294.0:———subtracting the less from the greater; the Remainder is92.6,the same as in the 3d. Column, viz. the Height, in Feet and Tenths, corresponding toone Tenth onlyabove 28.1.Having therefore found that Feet 92.6 Tenths, are the Height, corresponding to one Tenth only above Inches 28.1 Tenth, of the lower Barometer, with the Temperature of freezing; for whichsolePurpose, the 2d Table is calculated;—a new Question arises, viz. what are the Heights, in Feet and Tenths, corresponding to the remaining Decimals above 28.1, viz..03.001.0008; to be resolved by Application of the third Table, or Table for Tenths, which see, (in Section 373.)Look in the 3d. Table, with 92, (omitting the .6 as too minute) and with3answering to.03,which gives28 =Feet28.1to.001,9 =.98to.0008,74 =.74——29.6|4Which 29.6 is theHeightin Feet and Tenths corresponding to .0318 Tenths above Inches 28.1 Tenth: and Inches 28.1 Tenth gave Feet 3386.6Tenths in Height: therefore an additional Height of so many Tenths or Decimals of an Inch of Quicksilver in the Tube of the Barometer, must give in Feet, alessHeight of thelowerBarometer, elevated above theimaginaryLevel indicated by the Quicksilver resting in the Tube at 32 Inches.[130]402. Therefore subtract theHeight, in Feet, corresponding to theExpansion on.0318 Tenths of an Inch (lessthan Inches 28.2 Tenths of thelowerbarometric Tube,) from the Height, in Feet, corresponding to theExpansion on28.1 Tenth of the same Barometer, viz.3386.629.6———and the Remainder -3357.0,gives therealHeight in Feet of the lower Barometer, at 28.1318 when above theimaginaryLevel, and with the Temperature offreezingby the second Table.403. Then, by taking the Number of Feet and Tenthsabovethe imaginary Level, (indicated by the Quicksilver, in both Tubes, resting at 32 Inches) answering to theExpansion onInches and Tenths of thelowerTube, from the Number of Feet, &c. by the former Process, answering to that of theupperTube; viz.upper7292.1lower3357.0———and the remaining Feet3935.1Tenth is theHeight, by which theStationof theupperBarometer exceeds theStationof thelower; both beingat the Temperature of 31°.24 on Farenheit’s Scale. SeeSection 371.END OF THE SECOND STAGE11th Step.Section 404. 11th Step.(See the Practice in the 1st Example, Sect. 376.)Air-Thermom.abovewas56°.Air-Thermom.belowwas63.9———Whole Heat119.9(0 adding a Cypher)Half Heat59.95Standard-Heat31.24which deduct; and thereremains each Moiety,above the Standard-Heat.———28.7112th Step.405. 12th Step. (See the Practice in the first Example, Section 377.)By the fourth Table, find the Expansion of Air,with28.71, (more than the Standard-Temperature)onFeet 3935, .1 Tenth, gradually, thus:406.Firstwith28°onFeet 3000 =204.1[131]900 as 9000 =612.3303000 =204.155000 =340.1.11000 =68.0Note: 1st. The decimal Point in the Answer corresponding to the Place ofThousands, in the Question, is to remain, as taken from the Table calculated for thousand Feet, thus: 204.1.2d. ForHundredsin the Question, remove the decimal Pointone Placein the Answer, thus: 612.3 becomes 61.23:3d. ForTens,twoPlaces, thus: 204.1 becomes 2.041:4th. ForUnits,threePlaces, thus: 340.1 becomes .3401:5th. And for eachDecimal, a Place more, by adding Cyphers to the left, if wanted, thus: 68.0 becomes .00680.407. Place the plain and decimated Answers, in one View, and add the latter together, thus:204.1 =the same204.1612.3 =becomes61.23204.1 =2.041340.1 =.340168.0 =.00680—————viz. Expansion of Airwith28°on3935.1267.7|179408.Second,with.71°onFeet 3000 =517.5900 as 9000 =1552.7303000 =517.555000 =862.6.11000 =172.5In order to decimate these Answers, it must be observed that the Expansion was notwith71 Degrees, but with .71Tenthsof a Degree of Heat; therefore the decimal Point corresponding to 3000 Feet in the Question, must in the Answer be removedtwoPlaces to the left, thus: 517.5becomes5.175:for the 100, three Places: for the 10,fourPlaces: and so on.1.5527.05175.008626.0001725—————6.7|882485The Expansion with .71 being found, viz. Feet 6.7 Tenths; add it to the Expansion on 28 Feet already found, viz.267.7———274.4Answer.WhichHeightin Feet and Tenths, corresponding to theExpansionof Air with 28°.71 Tenths of a Degree of Heat more than the Standard 31°.24, being added to theHeightin Feet and Tenths, corresponding to theExpansion on Inchesof the Quicksilver in theupperBarometer, with the Standard-Heat, already found, viz.3935.1gives thereal Heightof theMountain, orupper Station, sought.274.4———4209.5END OF THE THIRD STAGE.The second Examplebrieflystated: referring to the Sections.Section, 391.409. Below: Barometer 28.1318.Attached Thermometer 61°.8; Air ditto 63.9.Above: Barom. 24.178.Attached Thermometer 57°.2; Air ditto 56°. Degrees of Heat, viz. 4°.6 to be added tothecolderBarometer at Inches24.178Tenths,by the first Table, viz..0112Parts of an Inch of the Quicksilver in the Barometer, raised by 4°.6 of Heat.———The Sum24.1892is thepoint, in Inches and Tenths of an Inch, at which the upper Barometernowrests, being ofequalHeat with the lower.End of the first Stage.Section, 399.By the 2d. Table, find theHeight, in Feet and Tenths, corresponding to thesaidpointwhen at the Standard-Heat; gradually, thus: theHeightcorresponding to Feet 24.1 is 7388.0: then with the Difference 107.9, (rejecting the .9).Section, 400.Find the Height by the 3d. Table corresponding to.0886.0}= Feet 95.9 Tenths..0099.7.0002.2Which Height subtract from7388.095.9———And there remains, in Feet,7292.1The Height corresponding to Inches 24.1892 Tenths of theupperBarometer, with the Standard Temperature of 31.24; for which sole Purpose the 2d. Table is calculated.Repeat the last Process with thelowerBarometer, resting at 28.1318, gradually, thus:Section, 401.By the 2d. Table, find theHeightcorresponding to 28.1, which is 3386.61; then with the Difference 92.6 (rejecting the .6) find the correspondingHeight, by the 3d. Table for the remaining Tenths or Decimals of an Inch, above 28.1, viz..0328.0}= Feet 29.6 Tenths..001.9.0008.7Section 402.Which Height subtract from3386.629.6———And there remains,3357.0viz. theHeightin Feet corresponding to Inches 28.1318 Tenths of the lower Barometer, with the Standard Temperature of 31.24, for which sole Purpose the 2d. Table is calculated.Section 403.Subtract theHeightin Feet, corresponding to Inches of Quicksilver in the upper Barometer,viz.7292.1from ditto inlowerBarometer,viz.3357.0and there remains theHeightin Feet———of the upper Barometer at the Standard-Temperatureviz.3935.1of 31.24.End of the second Stage.Section, 404.On which Number of Feet, viz.3935.1,by the4th Table, find theHeight, with28°.71of Heat:With28°.onFeet 3935.1 =267.7andWith.71onthe same =6.7———Sum274.4: whichHeight, more than the Standard-Heat,beingaddedto3935.1the Height, with the Standard,———gives the true Height, viz.4209.5.
CHAPTERLXXVI.
4th Step applied in the 2d Example.
4th Step applied.
Section 391.THEOrderto be observed in finding the Expansionwith4°.6, i. e. with 4 Degrees, .6 Tenths of Heat, on 24.178, i. e. 24 Inches, .178 Tenths of the coldest Barometer.
Find the Expansion required, thus:
Case the 1st.
1st. Part.With4°on24 Inches.
2d. Part.With4°on.178 Tenths of an Inch above 24 Inches.
Case the 2d.
1st. Part.With.6 Tenths of a Degree,on24 Inches.
2d. Part.With.6 Tenths of a Degree,on.178 Tenths above 24 Inches.
specifically,thus:
1st. Part ofCase the 1st.To find the Expansion,
With4°on24 Inches.
2d. Part ofCase the 1st.
With4°,on.178 Tenths of an Inch above 24 Inches; begin thus:
With4°,on24 Inches: then,
With4°,on25: then,
With4°,on1 Inch above 24, i. e.onthe 25th Inch: then,
With4°,on.1 Tenth above 24: then,
With4°,on.178 Tenths above 24.
1st Part ofCase the 2d.To find the Expansion,
With.6 above 4°on24; begin thus:
With4°on24 Inches: then,
With5°on24: then,
With1° above 4°,on24, i. e. the 5th°: then,
With.1 Tenth above 4°,on24: then
With.6 Tenths above 4°,on24.
2d Part ofCase the 2d.To find the Expansion,
With.6 Tenths above 4° of Heaton.178 Tenths above 24 Inches: to be done thus:
Theexpansionwith 4°, on .178 Tenths above 24 Inches, being once found; divideitby 4: and the Quotient is the Expansion with 1° above 4°, on .178 Tenths of an Inch above 24 Inches.
Then for the Expansion with .1 Tenth above 4°, on .178 Tenths above 24 Inches; add a Cypher and decimal Point to the left of the same Quotient.
Then for the Expansion with .6; multiply that Sum into .6, and add a Cypher and decimal Point.
The Answer is thepartof an Inch, to which .6 Tenths of a Degree above 4° of Heat, on .178 Tenths of an Inch above 24 Inches, raises the Barometer.
It is true, thepartis so minute as to be rejected: yet the Mode of Proceeding, in order to investigate the Expansion with Precision, is proper to be retained.
392.practiceof the first Part ofCase the 1st.
For the Expansionwith4°,on24 Inches;look, in the first Table, (Sect. 363) and in the left vertical Column,with4 Degrees of the Thermometer; and along the upper horizontal Line,on24 Inches of Quicksilver in the Tube of the Barometer: the Point of Meeting gives the Expansion .0097;[128]which, preparatory to Addition,is to be placed under the 24, .178 thus,.0097
practiceof the 2d Part ofCase the first.
393. In order to obtain the Expansion,with4°, of Heaton.178 Tenths of an Inch above 24 Inches of the Barometer; let it be considered where it ought to be found in the Table: for, Tenths of 1 Inch above 24 Inches, are at some intermediate Point between 24 and 25; that is, above 24, yet not so high as 25: or more than 24, yet less than 25.
Look therefore in the Table,with4 Degrees of Heat,on24 Inches; thenwith4°on25 Inches: and the respective Numbers are .0097 and .0101.
And by taking the Expansionwith4°on24 Inches, from 4° on 25; the Remainder will be the Expansion with 4° on 1 Inch above 24 Inches, viz. on the 25th Inch, thus:
.0101
.0097
——
.0004
This therefore is the Expansionwith4°,on1 Inch above 24 Inches.
Thenwith4°,on.1 Tenth of an Inch above 24 Inches.
The Answer is the same as the former, viz. .0004, with the Addition of a Cypher and decimal Point to the left, thus; .0004 becomes .00004, viz. the Expansionwith4°,on.1 Tenth of an Inch above 24 Inches.
Then for the Expansionwith4°,on.178 Tenths, say,
If the Expansionwith4°,on.1 Tenth above 24 Inches gives .00004 Part of an Inch, what will the Expansionwith4°,on.178 give?
Thus; .1 : .00004 :: .178?
Multiply the two last Terms, thus:
.00004
.178
————
00032
00028
00004
————
0000712:
and, as in Multiplication of Decimals, the Product must have as many decimal Places, as are in the Factors; a Cypher must be added to the left Hand, thus: .00000712: but having divided that Product by the first Term .1, viz. a Decimal, the Answer is a Cypher less; viz. .0000712.
This Answer is the Expansionwith4°,on.178 Tenths of an Inch above 24 Inches: prepare it forAddition, as the former,
24.178
.0097
.0000712
practiceof the first Part ofCase the 2d.
394. For the Expansion of .6 Tenths of a Degree of Heat, (more than the 4 Degrees) on 24 Inches of thecoldestBarometer; it shoudbe considered where such Tenths can lie in the Table.
Now .6 Tenths of 1 Degree, (more than the 4°) are at some intermediate Point of the Thermometer between 1 and 2 Degrees: above 1; yet not so high as 2: or more than 1; yet less than 2.
Therefore .6 Tenths of 1 Degree above 4 Degrees, are somewhere between the 4th and 5th Degree: above 4; yet not so high as 5: or more than 4; yet less than 5.
Look in the Table (Section 363); firstwith4 Degrees of Heat,on24 Inches, and thenwith5 Degrees of Heaton24 Inches; and the respective Numbers are .0097 and .0121: and by taking the Expansionwith4 Degreeson24 Inches, from the Expansionwith5 Degreesonthe same 24 Inches; the Remainder will be the Expansionwith1 Degree above 4°on24 Inches: viz.
5° = .0121
4° = .0097
——
Remainder, .0024
This therefore is the Expansionwith1 Degree of Heat, above 4, viz.withthe 5th Degree,on24 Inches of the Barometer.
Then say, if 1 Degree of the Thermometer (above 4, viz. the 5th Degree) gives by Expansion, a certain additional Height, or Part of an Inch, viz. .0024,on24 Inches of the Barometer; what Height will 6 Degrees give? Answer 6 Timesmore.
Multiply the 2d and 3d Terms, and divide by the first, thus;
.0024
6
——
.0144
is the Expansion, or Height, in Parts of an Inch, for 6 Degrees.
And farther, to proportion for the Decimal; say as .1 Tenth of a Degree gives a certain Tenth of the former .0024, in additional Height, viz. .00024; what Height will .6 Tenths give? Answer, .00144.
Prepare thisHeightfor Addition to the Numbers already found.
practiceof the 2d Part ofCase the 2d.
395. To find the Expansion of .6 above 4° on .178 above 24 Inches.
The Expansionwith4°on.178 is already found to be .0000712: divide it by 4, and the Answer is .0000178, viz. the Expansionwith1°on.178 above 24 Inches:
And, for the Expansion with .1 Tenth; the Answer, with the Addition of a Cypher and decimal Point to the left, becomes .00000178.
Lastly, for the Expansion with .6, say,
If .1 : .00000178 :: .6?
Multiply the 2d and 3d Terms, and divide by first:
.00000178
.6
—————
.000001068.
The Answer is a Decimal less, viz. .00001068; i. e. the Decimal of an Inch, to which .6 Tenths of a Degree above 4 Degrees of Heat, on .178Tenths of an Inch above 24 Inches, raises the Barometer: which, after all, is so inconsiderable, that it may be fairly rejected.
Yet the Rules by which these Deductions are made, may be useful in other Cases.
Prepare for Addition, as before.
The Decimals, in the Answers, may be omitted, when they exceed four Places.
5th Step.
396. 5th Step. To proceed with the second Example.
Place the different Expansions now found, above each other, Units, Tens, &c. under Units, Tens, &c. preparatory to Addition, thus;
For the Expansionwith4°, .6on24, .178:
To the Sum add the Height of thecolderBarometer
The Answer is Height of thecolderBarometer, now equal in Temperature to thewarmer: (rejecting all but the four first Decimals.)
6th Step.
397. 6th Step. Place the Barometersnowof the same Temperature, i. e.equalto the warmer, in one View, thus:
The 7th Step applied in the second Example.
7th Step.
398. Find the Height, in Feet, in the 2d Columnof the 2d Table, corresponding to Inches and Tenths of theupperbarometric Tube, in the 1st. Column of the same Table, thus: (Sect. 371.)
The Barometer standing at 24.1892; it must be considered where, in the 2d Column of the 2d Table, a Height corresponding tosuchInches and Tenths can lie: and the Answer is, somewhereabove24 Inches .1 Tenth, but not so high as 24 Inches .2 Tenths: 24 Inches .1892 Tenths, beingmorethan 24 Inches .1 Tenth, butlessthan 24 Inches .2 Tenths.
First then, look in the 1st Column for Inches 24, .1 Tenth; and the corresponding Height in Feet is 7388.0: but the Height for 24, .2, in the 2d Column, beneath the former Number, isonly7280.1.
8th Step.
399. 8th Step. Subtract the latter from the former and the Remainder is 107.9, the same as in the 3d Column: viz. the Height, in Feet and Tenths, corresponding to one Tenth only, namely, the ist Tenth above Inches 24, .1 Tenth: with the Temperature of 31.24 of Farenheit, for which sole Purpose the 2d Table is calculated.
A new Questionthenarises, viz. what are the Heights in Feet and Tenths, corresponding to the remaining Tenths or Decimals of an Inch above
which is to be resolved, by Application of the 3d Table, or Table forTenths, which see, (Section 373.)
9th Step.
400.9th Step applied in the 2d. Example.
First for theupperBarometer.
Look in the Table for Tenths, in the left vertical Column with 107, (rejecting the .9, as too minute;) and along the horizontal Line at the top, with 8: and find the Answergradually, thus:
1st. With 107, and 8, (as a whole Number,) answering to .08: which, in the Place of Meeting, gives 86 Feet.
2d. With 107, and 9, (as a whole Number,) answering to .009: which, in the Place of Meeting, gives 97.
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:
(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:
To
Add the former Sum
Total =
Which 95.9 is theHeightin Feet and Tenths corresponding to .0892 Decimals of an Inch above Inches 24 .1 Tenth: and 24 .1 gave Feet 7388.0 inHeight; therefore an additionalHeight, of so many Tenths of an Inch of Quicksilver in the Tube of the Barometer, must give in Feet, alessHeight of the Barometer elevated above theimaginaryLevel indicated at 32 Inches.
10th. Step.
401. 10th. Step. Subtract theHeightin Feet, corresponding to theExpansionon .0892 Tenths of an Inch, (lessthan Inches 24.2 Tenths, of theupperbarometric Tube,) from theHeight, in Feet, corresponding tothe Expansion onInches 24.1 Tenth of the same barometric Tube, continuing at the Standard Heat,[129]
viz.
7388.0
95.9
———
The Remainder
7292.1
gives the real, viz. thelessHeight of theupperBarometer, at 24.1892 with the Standard Temperature.
Repeat the same Process, viz. the 9th. and 10th. Steps, for thelowerBarometer, thus:
For the lower Barometer in the 2d. Example.
First, Find the Height, inFeet, of the lower Barometer, standing at Inches 28.1318 Tenths, in the 2d. Column of the 2d. Table, corresponding to Inches and Tenths of the Quicksilver in the barometric Tube, in the first Column of the same Table, thus:
The lower Barometer standing at 28.1318; it must be considered, where in the 2d. Column of the 2d. Table, a Height corresponding to such Inches and Tenths can lye: and the Answer is, somewhere above 28 Inches, .1 Tenth, but notso high as 28 Inches .2 Tenths: 28.1318 Tenths being more than 28 Inches .1 Tenth, yet less than 28 Inches .2 Tenths.
28.1,
3386.6:
3294.0:
———
92.6,
the same as in the 3d. Column, viz. the Height, in Feet and Tenths, corresponding toone Tenth onlyabove 28.1.
Having therefore found that Feet 92.6 Tenths, are the Height, corresponding to one Tenth only above Inches 28.1 Tenth, of the lower Barometer, with the Temperature of freezing; for whichsolePurpose, the 2d Table is calculated;—a new Question arises, viz. what are the Heights, in Feet and Tenths, corresponding to the remaining Decimals above 28.1, viz.
.03.001.0008; to be resolved by Application of the third Table, or Table for Tenths, which see, (in Section 373.)
Look in the 3d. Table, with 92, (omitting the .6 as too minute) and with
28 =
28.
9 =
.9
74 =
.74
——
29.6|4
Which 29.6 is theHeightin Feet and Tenths corresponding to .0318 Tenths above Inches 28.1 Tenth: and Inches 28.1 Tenth gave Feet 3386.6Tenths in Height: therefore an additional Height of so many Tenths or Decimals of an Inch of Quicksilver in the Tube of the Barometer, must give in Feet, alessHeight of thelowerBarometer, elevated above theimaginaryLevel indicated by the Quicksilver resting in the Tube at 32 Inches.[130]
402. Therefore subtract theHeight, in Feet, corresponding to theExpansion on.0318 Tenths of an Inch (lessthan Inches 28.2 Tenths of thelowerbarometric Tube,) from the Height, in Feet, corresponding to theExpansion on28.1 Tenth of the same Barometer, viz.
3386.6
29.6
———
3357.0,
gives therealHeight in Feet of the lower Barometer, at 28.1318 when above theimaginaryLevel, and with the Temperature offreezingby the second Table.
403. Then, by taking the Number of Feet and Tenthsabovethe imaginary Level, (indicated by the Quicksilver, in both Tubes, resting at 32 Inches) answering to theExpansion onInches and Tenths of thelowerTube, from the Number of Feet, &c. by the former Process, answering to that of theupperTube; viz.
7292.1
3357.0
———
3935.1
Tenth is theHeight, by which theStationof theupperBarometer exceeds theStationof thelower; both beingat the Temperature of 31°.24 on Farenheit’s Scale. SeeSection 371.
END OF THE SECOND STAGE
11th Step.
Section 404. 11th Step.
(See the Practice in the 1st Example, Sect. 376.)
56°.
63.9
———
119.9
59.95
31.24
———
28.71
12th Step.
405. 12th Step. (See the Practice in the first Example, Section 377.)
By the fourth Table, find the Expansion of Air,with28.71, (more than the Standard-Temperature)onFeet 3935, .1 Tenth, gradually, thus:
406.
Firstwith28°onFeet 3000 =
900 as 9000 =
303000 =
55000 =
.11000 =
Note: 1st. The decimal Point in the Answer corresponding to the Place ofThousands, in the Question, is to remain, as taken from the Table calculated for thousand Feet, thus: 204.1.
2d. ForHundredsin the Question, remove the decimal Pointone Placein the Answer, thus: 612.3 becomes 61.23:
3d. ForTens,twoPlaces, thus: 204.1 becomes 2.041:
4th. ForUnits,threePlaces, thus: 340.1 becomes .3401:
5th. And for eachDecimal, a Place more, by adding Cyphers to the left, if wanted, thus: 68.0 becomes .00680.
407. Place the plain and decimated Answers, in one View, and add the latter together, thus:
204.1 =
612.3 =
204.1 =
340.1 =
68.0 =
408.
Second,with.71°onFeet 3000 =
900 as 9000 =
303000 =
55000 =
.11000 =
In order to decimate these Answers, it must be observed that the Expansion was notwith71 Degrees, but with .71Tenthsof a Degree of Heat; therefore the decimal Point corresponding to 3000 Feet in the Question, must in the Answer be removedtwoPlaces to the left, thus: 517.5
The Expansion with .71 being found, viz. Feet 6.7 Tenths; add it to the Expansion on 28 Feet already found, viz.
267.7
———
274.4
WhichHeightin Feet and Tenths, corresponding to theExpansionof Air with 28°.71 Tenths of a Degree of Heat more than the Standard 31°.24, being added to theHeightin Feet and Tenths, corresponding to theExpansion on Inchesof the Quicksilver in theupperBarometer, with the Standard-Heat, already found, viz.
3935.1
274.4
———
4209.5
END OF THE THIRD STAGE.
The second Examplebrieflystated: referring to the Sections.
Section, 391.
409. Below: Barometer 28.1318.
Attached Thermometer 61°.8; Air ditto 63.9.
Above: Barom. 24.178.
Attached Thermometer 57°.2; Air ditto 56°. Degrees of Heat, viz. 4°.6 to be added to
24.178
.0112
———
24.1892
is thepoint, in Inches and Tenths of an Inch, at which the upper Barometernowrests, being ofequalHeat with the lower.
End of the first Stage.
Section, 399.
By the 2d. Table, find theHeight, in Feet and Tenths, corresponding to thesaidpointwhen at the Standard-Heat; gradually, thus: theHeightcorresponding to Feet 24.1 is 7388.0: then with the Difference 107.9, (rejecting the .9).
Section, 400.
Find the Height by the 3d. Table corresponding to
86.0
9.7
.2
7388.0
95.9
———
And there remains, in Feet,
7292.1
The Height corresponding to Inches 24.1892 Tenths of theupperBarometer, with the Standard Temperature of 31.24; for which sole Purpose the 2d. Table is calculated.
Repeat the last Process with thelowerBarometer, resting at 28.1318, gradually, thus:
Section, 401.
By the 2d. Table, find theHeightcorresponding to 28.1, which is 3386.61; then with the Difference 92.6 (rejecting the .6) find the correspondingHeight, by the 3d. Table for the remaining Tenths or Decimals of an Inch, above 28.1, viz.
28.0
.9
.7
Section 402.
3386.6
29.6
———
And there remains,
3357.0
viz. theHeightin Feet corresponding to Inches 28.1318 Tenths of the lower Barometer, with the Standard Temperature of 31.24, for which sole Purpose the 2d. Table is calculated.
Section 403.
Subtract theHeightin Feet, corresponding to Inches of Quicksilver in the upper Barometer,
7292.1
3357.0
———
3935.1
End of the second Stage.
Section, 404.
On which Number of Feet, viz.
3935.1,
28°.71
onFeet 3935.1 =
267.7
onthe same =
6.7
———
Sum
274.4
3935.1
———
4209.5.