CHAPTER LXXVI.

PRACTICE OF THE FIRST TABLE IN THE SECOND EXAMPLE.

_4th Step applied in the 2d Example._

[Sidenote: _4th Step applied._]

Section 391. The _Order_ to be observed in finding the Expansion _with_ 4°.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. _With_ 4° _on_ 24 Inches.

2d. Part. _With_ 4° _on_ .178 Tenths of an Inch above 24 Inches.

_Case the 2d._

1st. Part. _With_ .6 Tenths of a Degree, _on_ 24 Inches.

2d. Part. _With_ .6 Tenths of a Degree, _on_ .178 Tenths above 24 Inches.

+specifically+, _thus_:

1st. Part of _Case the 1st._ To find the Expansion,

_With_ 4° _on_ 24 Inches.

2d. Part of _Case the 1st._

_With_ 4°, _on_ .178 Tenths of an Inch above 24 Inches; begin thus:

_With_ 4°, _on_ 24 Inches: then,

_With_ 4°, _on_ 25: then,

_With_ 4°, _on_ 1 Inch above 24, i.e. _on_ the 25th Inch: then,

_With_ 4°, _on_ .1 Tenth above 24: then,

_With_ 4°, _on_ .178 Tenths above 24.

1st Part of _Case the 2d._ To find the Expansion,

_With_ .6 above 4° _on_ 24; begin thus:

_With_ 4° _on_ 24 Inches: then,

_With_ 5° _on_ 24: then,

_With_ 1° above 4°, _on_ 24, i.e. the 5th°: then,

_With_ .1 Tenth above 4°, _on_ 24: then

_With_ .6 Tenths above 4°, _on_ 24.

2d Part of _Case the 2d._ To find the Expansion,

_With_ .6 Tenths above 4° of Heat _on_ .178 Tenths above 24 Inches: to be done thus:

_The_ +expansion+ _with 4°, on .178 Tenths above 24 Inches, being once found; divide_ +it+ _by 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 the_ +part+ _of 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, the_ +part+ _is 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. +practice+ of the first Part of _Case the 1st._

For the Expansion _with_ 4°, _on_ 24 Inches; look, in the first Table, (Sect. 363) and in the left vertical Column, _with_ 4 Degrees of the Thermometer; and along the upper horizontal Line, _on_ 24 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

+practice+ of the 2d Part of _Case the first_.

393. In order to obtain the Expansion, _with_ 4°, of Heat _on_ .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, _with_ 4 Degrees of Heat, _on_ 24 Inches; then _with_ 4° _on_ 25 Inches: and the respective Numbers are .0097 and .0101.

And by taking the Expansion _with_ 4° _on_ 24 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:

} 25 = .0101 from; _With_ 4° _on_ } } 24 = .0097 subtract: ————— .0004:

This therefore is the Expansion _with_ 4°, _on_ 1 Inch above 24 Inches.

Then _with_ 4°, _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 Expansion _with_ 4°, _on_ .1 Tenth of an Inch above 24 Inches.

Then for the Expansion _with_ 4°, _on_ .178 Tenths, say,

If the Expansion _with_ 4°, _on_ .1 Tenth above 24 Inches gives .00004 Part of an Inch, what will the Expansion _with_ 4°, _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 Expansion _with_ 4°, _on_ .178 Tenths of an Inch above 24 Inches: prepare it for _Addition_, as the former, 24.178 .0097 .0000712

+practice+ of the first Part of _Case the 2d._

394. For the Expansion of .6 Tenths of a Degree of Heat, (more than the 4 Degrees) on 24 Inches of the _coldest_ Barometer; it shoud be 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); first _with_ 4 Degrees of Heat, _on_ 24 Inches, and then _with_ 5 Degrees of Heat _on_ 24 Inches; and the respective Numbers are .0097 and .0121: and by taking the Expansion _with_ 4 Degrees _on_ 24 Inches, from the Expansion _with_ 5 Degrees _on_ the same 24 Inches; the Remainder will be the Expansion _with_ 1 Degree above 4° _on_ 24 Inches: viz.

_with_ {5° = .0121} _on_ 24 Inches, as in whole {4° = .0097} Numbers. ————— Remainder, .0024

This therefore is the Expansion _with_ 1 Degree of Heat, above 4, viz. _with_ the 5th Degree, _on_ 24 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, _on_ 24 Inches of the Barometer; what Height will 6 Degrees give? Answer 6 Times _more_.

Multiply the 2d and 3d Terms, and divide by the first, thus;

1 : .0024 :: 6? 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 this _Height_ for Addition to the Numbers already found.

+practice+ of the 2d Part of _Case the 2d._

395. To find the Expansion of .6 above 4° on .178 above 24 Inches.

The Expansion _with_ 4° _on_ .178 is already found to be .0000712: divide it by 4, and the Answer is .0000178, viz. the Expansion _with_ 1° _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 .178 Tenths 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.

[Sidenote: 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 Expansion _with_ 4°, .6 _on_ 24, .178:

1st. _with_ 4°, _on_ 24, .0097 2d. _with_ .6 _on_ 24, .00144 3d. _with_ 4°, _on_ .178 .0000712 4th. _with_ .6 _on_ .178 .00001068 ————————— The Expansions with 4°,.6 added = .01122188 To the Sum add the Height of the _colder_ Barometer 24.178 ——————————— 24.1892|

The Answer is Height of the _colder_ Barometer, now equal in Temperature to the _warmer_: (rejecting all but the four first Decimals.)

[Sidenote: 6th Step.]

397. 6th Step. Place the Barometers _now_ of the same Temperature, i.e. _equal_ to the warmer, in one View, thus:

1st. the _upper_ Barometer, 24.1892 2d. the _lower_ Barometer, 28.1328

END OF THE FIRST STAGE.

_The 7th Step applied in the second Example._

[Sidenote: _7th Step._]

398. Find the Height, in Feet, in the 2d Column of the 2d Table, corresponding to Inches and Tenths of the _upper_ barometric 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 to _such_ Inches and Tenths can lie: and the Answer is, somewhere _above_ 24 Inches .1 Tenth, but not so high as 24 Inches .2 Tenths: 24 Inches .1892 Tenths, being _more_ than 24 Inches .1 Tenth, but _less_ than 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, is _only_ 7280.1.

[Sidenote: 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 Question _then_ arises, viz. what are the Heights in Feet and Tenths, corresponding to the remaining Tenths or Decimals of an Inch above Inches 24, .1 Tenth, viz. .08 .009 .0002? which is to be resolved, by Application of the 3d Table, or Table for _Tenths_, which see, (Section 373.)

[Sidenote: _9th Step._]

400. _9th Step applied in the 2d. Example._

First for the _upper_ Barometer.

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 Answer _gradually_, 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 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.

[Sidenote: _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 Remainder 7292.1 gives the real, viz. the _less_ Height of the _upper_ Barometer, at 24.1892 with the Standard Temperature.

Repeat the same Process, viz. the 9th. and 10th. Steps, for the _lower_ Barometer, thus:

For the lower Barometer in the 2d. Example.

First, Find the Height, in _Feet_, 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 not so 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 for 28.1, and the corresponding Height, in Feet, is 3386.6: but the Height for 28.2, is only 3294.0: —————— subtracting the less from the greater; the Remainder is 92.6, the same as in the 3d. Column, viz. the Height, in Feet and Tenths, corresponding to _one Tenth only_ above 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 which _sole_ Purpose, 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 3 answering to .03, which gives 28 = Feet 28. 1 to .001, 9 = .9 8 to .0008, 74 = .74 —————— 29.6|4

Which 29.6 is the _Height_ in Feet and Tenths corresponding to .0318 Tenths above Inches 28.1 Tenth: and Inches 28.1 Tenth gave Feet 3386.6 Tenths 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, a _less_ Height of the _lower_ Barometer, elevated above the _imaginary_ Level indicated by the Quicksilver resting in the Tube at 32 Inches.[130]

402. Therefore subtract the _Height_, in Feet, corresponding to the _Expansion on_ .0318 Tenths of an Inch (_less_ than Inches 28.2 Tenths of the _lower_ barometric Tube,) from the Height, in Feet, corresponding to the _Expansion on_ 28.1 Tenth of the same Barometer, viz. 3386.6 29.6 —————— and the Remainder - 3357.0, gives the _real_ Height in Feet of the lower Barometer, at 28.1318 when above the _imaginary_ Level, and with the Temperature of _freezing_ by the second Table.

403. Then, by taking the Number of Feet and Tenths _above_ the imaginary Level, (indicated by the Quicksilver, in both Tubes, resting at 32 Inches) answering to the _Expansion on_ Inches and Tenths of the _lower_ Tube, from the Number of Feet, &c. by the former Process, answering to that of the _upper_ Tube; viz. _upper_ 7292.1 _lower_ 3357.0 —————— the remaining Feet 3935.1 Tenth is the _Height_, by which the _Station_ of the _upper_ Barometer exceeds the _Station_ of the _lower_; both being at the Temperature of 31°.24 on Farenheit’s Scale. See Section 371.

END OF THE SECOND STAGE

* * * * *

[Sidenote: 11th Step.]

Section 404. 11th Step.

(See the Practice in the 1st Example, Sect. 376.)

_Air_-Thermom. +above+ was 56°. _Air_-Thermom. +below+ was 63.9 ————— Whole Heat 119.9(0 adding a Cypher) Half Heat 59.95 Standard-Heat 31.24 which deduct; and there —————— remains each Moiety, 28.71 above the Standard-Heat.

[Sidenote: _12th Step._]

405. 12th Step. (See the Practice in the first Example, Section 377.)

By the fourth Table, find the Expansion of Air, _with_ 28.71, (more than the Standard-Temperature) _on_ Feet 3935, .1 Tenth, gradually, thus:

406.

_First_ _with_ 28° _on_ Feet 3000 = 204.1[131] 900 as 9000 = 612.3 30 3000 = 204.1 5 5000 = 340.1 .1 1000 = 68.0

Note: 1st. The decimal Point in the Answer corresponding to the Place of _Thousands_, in the Question, is to remain, as taken from the Table calculated for thousand Feet, thus: 204.1.

2d. For _Hundreds_ in the Question, remove the decimal Point _one Place_ in the Answer, thus: 612.3 becomes 61.23:

3d. For _Tens_, _two_ Places, thus: 204.1 becomes 2.041:

4th. For _Units_, _three_ Places, thus: 340.1 becomes .3401:

5th. And for each _Decimal_, 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 same 204.1 612.3 = becomes 61.23 204.1 = 2.041 340.1 = .3401 68.0 = .00680 ————————— viz. Expansion of Air _with_} 267.7|179 28° _on_ 3935.1 }

408. _Second_, _with_ .71° _on_ Feet 3000 = 517.5 900 as 9000 = 1552.7 30 3000 = 517.5 5 5000 = 862.6 .1 1000 = 172.5

In order to decimate these Answers, it must be observed that the Expansion was not _with_ 71 Degrees, but with .71 _Tenths_ of a Degree of Heat; therefore the decimal Point corresponding to 3000 Feet in the Question, must in the Answer be removed _two_ Places to the left, thus: 517.5 becomes 5.175: for the 100, three Places: for 1.5527 the 10, _four_ Places: and so .05175 on. .008626 .0001725 —————————— 6.7|882485

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 Answer.

Which _Height_ in Feet and Tenths, corresponding to the _Expansion_ of Air with 28°.71 Tenths of a Degree of Heat more than the Standard 31°.24, being added to the _Height_ in Feet and Tenths, corresponding to the _Expansion on Inches_ of the Quicksilver in the _upper_ Barometer, with the Standard-Heat, already found, viz. 3935.1 gives the _real Height_ of the _Mountain_, 274.4 —————— or _upper Station_, sought. 4209.5

END OF THE THIRD STAGE.

* * * * *

_The second Example_ briefly _stated: referring to the Sections._

[Sidenote: 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 the _colder_ Barometer at Inches 24.178 Tenths, by the first Table, viz. .0112 Parts of an Inch of the Quicksilver in the Barometer, raised by 4°.6 of Heat. ——————— The Sum 24.1892

is the +point+, in Inches and Tenths of an Inch, at which the upper Barometer _now_ rests, being of _equal_ Heat with the lower.

_End of the first Stage._

[Sidenote: Section, 399.]

By the 2d. Table, find the _Height_, in Feet and Tenths, corresponding to the _said_ +point+ when at the Standard-Heat; gradually, thus: the _Height_ corresponding to Feet 24.1 is 7388.0: then with the Difference 107.9, (rejecting the .9).

[Sidenote: Section, 400.]

Find the Height by the 3d. Table corresponding to .08 86.0 } .009 9.7 } = Feet 95.9 Tenths. .0002 .2 }

Which Height subtract from 7388.0 95.9 —————— And there remains, in Feet, 7292.1

The Height corresponding to Inches 24.1892 Tenths of the _upper_ Barometer, with the Standard Temperature of 31.24; for which sole Purpose the 2d. Table is calculated.

Repeat the last Process with the _lower_ Barometer, resting at 28.1318, gradually, thus:

[Sidenote: Section, 401.]

By the 2d. Table, find the _Height_ corresponding to 28.1, which is 3386.61; then with the Difference 92.6 (rejecting the .6) find the corresponding _Height_, by the 3d. Table for the remaining Tenths or Decimals of an Inch, above 28.1, viz. .03 28.0 } .001 .9 } = Feet 29.6 Tenths. .0008 .7 }

[Sidenote: Section 402.]

Which _Height_ subtract from 3386.6 29.6 —————— And there remains, 3357.0 viz. the _Height_ in 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.

[Sidenote: Section 403.]

Subtract the _Height_ in Feet, corresponding to Inches of Quicksilver in the upper Barometer, viz. 7292.1 from ditto in _lower_ Barometer, viz. 3357.0 and there remains the _Height_ in Feet —————— of the upper Barometer at the Standard-Temperature viz. 3935.1 of 31.24.

_End of the second Stage._

[Sidenote: Section, 404.]

On which Number of Feet, viz. 3935.1, by the 4th Table, find the _Height_, with 28°.71 of Heat: _With_ 28°. _on_ Feet 3935.1 = 267.7 and _With_ .71 _on_ the same = 6.7 ————— Sum 274.4: which Height, more than the Standard-Heat, being _added_ to 3935.1 the Height, with the Standard, —————— gives the true Height, viz. 4209.5.

_End of the third Stage._