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Airopaidia

Chapter 148: CHAPTER LXXVII.
By Thomas Baldwin
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History - Early Modern (c. 1450-1750)Science - PhysicsTravel Writing

About This Book

The work presents a first-person account of an 18th-century balloon voyage launched from a regional town, pairing vivid descriptions of aerial scenes with practical instruction. The narrative records observations made during ascent and flight and is accompanied by engraved views; it also offers proposals for balloon design and a method of inflation by steam, precautions against descent over water, and inquiries into atmospheric conditions affecting direction. A clear, practical explanation of measuring height with the barometer is included, together with extensive tables. The material functions as an introduction to aërostatic navigation that mixes travel narrative, natural-philosophical observation, and engineering advice.

End of the third Stage.

CHAPTER  LXXVII.

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 the colder
——
 from the warmer,
and there remains
4°
 of Heat to be added

to the colder Barometer; to give it an equal Temperature: which is to be done by the 1st Table, thus:

Section, 356.

To find the Expansion with 4° of Heat, on the colder Barometer; (which, as before, is the upper Barometer) standing at Inches 29, .5218 Tenths.

First, with 4° on 29 Inches =
 .0117:
2d, with 4° on .5218 Tenths above 29 Inches:
In order to obtain which, begin
with 4° on 29 =
 .0117
then with 4° on 30 =
 .0121
Subtract for the Expansion with
 ——
4° on 1 Inch above 29, and there remains
 .0004.

Section 362.

Then for the Expansion with 4° 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,
 multiply
the two last Terms, and divide
———
the Product by the first Term
.1: the Answer is
.0002
 |0872
Add the Expansion with 4° on
29 Inches, just found,
.0117
to the Inches of the colder
Barometer,  viz.
29.5218
———
Answer; Inches
29.5337
 Tenths of
the colder Barometer, are now expanded equally
with the warmer: (rejecting the Decimals as in
Section 395.)
Place the Barometers, thus:
Upper Barometer,
29.5337
Lower Barometer,
30.0168
End of the first Stage.
Section 371.

411. By the 2d Table, and in the 2d Column, find the Height of each Barometer, with the Standard-Heat, in Feet and Tenths, corresponding to the Inches and nearest Tenth above and below the Point required: and

First of the upper, at
29.5337:
The Inches and nearest Tenth is above Feet.
29.5, corresp. to
2119.7
 Difference between .5
and .6 above 29 Inches.
and below 29.6, cor. to
2031.5
———
88|.2
Section 373.

412. By the 3d Table, with the Difference 88 Feet, find the Expansion on the remaining Decimals, above 29.5, viz. on .0337, thus:

on
 03 = 26
decimated
26.
003 = 26
2.6
0007 = 62
.62
——
Feet
29.22
From the Height corresponding to
29.5
viz. Feet
2119.7
 Tenths,
subtract the
29.22,
 i. e. Height cor. to
.0338
and there
————
————
remains
2090.4|8,
 the Height cor. to
29.5338
with Expansion of the Standard-Heat.

413. Repeat the 4 last Steps for the lower Barometer, at 30.0168.

1st. The Inches and nearest Tenth is above
30. corresp. to Feet
1681.7
 Difference of .1 above 30 Inches.
and below 30.1 cor.
1595.0
———
86|.7

2d. Then with 86 Feet, find the Expansion on the remaining Decimals, above 30,

viz. .0168, thus: on 01
=
9
9.
006
=
52
5.2
0008
=
69
.69
———
Feet
14.89

414. (3d.) From the Height corresponding to

30 Inches, viz. Feet
1681.7
 Tenths,
subtract the Height
14.89
 corresp. to .0168,
————
and there remains
1666.8|1,
 the Height corresp.

to 30.0168, with Expansion of the Standard-Heat.

4th. From the upper Height, at
2090.48
Subtract the lower Height, at
1666.81
———
And there remains the Height
423.67
 in Feet
and Tenths of the upper Barometer, with the
Standard Temperature.
End of the second Stage.
Section 374.
415. Detached Therm. above
57°
Detached ditto, below
60.2
——
Whole Heat
117.2
Half Heat
58.6
 (0 adding a
Cypher)
Standard Heat
31.24
———
which being deducted, leaves
27°.36,
 viz. Degrees
of Heat more than the Standard, 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, with 27°, on 423.67, thus:

viz. on 400
as 4000 = 262.4
 decimated 26.24
20
as 2000 = 131.2
 1.312
3
as 3000 = 196.8
 .1968
.6
as 6000 = 393.6
 .03936
.07
as 7000 = 459.2
 .004592
—————
Expansion =
 27.692752
Section 407.

Second, with .36 on the same, thus:

on 400
as 4000 = 349.9
 decimated .3499
20
as 2000 = 174.9
 .01749
3
as 3000 = 262.4
 .002624
.6
as 6000 = 524.8
 .0005248
.07
as 7000 = 612.3
 .00006123
—————
Expansion =
 .37050003
Add the former
 27.692752
—————
Height in Feet
 28.06325203

417. Which Height for Expansion of Air, with more than the Standard Heat, being added⁠[133] to the Height, for Expansion of the Barometer, with the Standard-Heat, gives the true Height of the upper Barometer, at the given Heat.

For Expansion of Air above Standard Heat,
Height in Feet
28.0
For Expansion of Barometer,
with Standard:  Height in Feet
423.6
———
418. True Height of the upper Barometer
451.6
Lower Barometer 1 Foot above the Water
1.0
Height of the Top of the Cross above the Gallery
50.0
———
Height of the Top of the Cross above the Tyber
502.6
Height of the same, measured the same Day
geometrically, was Feet
502.9
End of the last Stage.