Airopaidia

by Thomas Baldwin

PRACTICE OF THE FIRST TABLE IN THE SECOND EXAMPLE.

4th Step applied in the 2d Example.

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:

With 4° on		25 =	.0101	from;
		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 Numbers.
		4° = .0097
		——
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.

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.)

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

The 7th Step applied in the second Example.

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.

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].)

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.

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
	———
and 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

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, above the Standard-Heat.	———
	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, 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 28° on 3935.1		267.7|179

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 the 10, four Places: and so on.
	1.5527
	.05175
	.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, or upper Station, sought.	274.4
	———
	4209.5

END OF THE THIRD STAGE.

The second Example briefly stated: 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

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.

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).

Section, 400.

Find the Height by the 3d. Table corresponding to

.08	86.0		= Feet 95.9 Tenths.
.009	9.7
.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:

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		= Feet 29.6 Tenths.
.001	.9
.0008	.7

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.

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.

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.