Airopaidia : $b Containing the narrative of a balloon excursion from Chester, the eighth of September, 1785, taken from minutes made during the voyage; hints on the improvement of balloons ... To which is subjoined, mensuration of heights by the barometer, made plain; with extensive tables. The whole serving as an introduction to aërial navigation.Baldwin, Thomas
History
Airopaidia : $b Containing the narrative of a balloon excursion from Chester, the eighth of September, 1785, taken from minutes made during the voyage; hints on the improvement of balloons ... To which is subjoined, mensuration of heights by the barometer, made plain; with extensive tables. The whole serving as an introduction to aërial navigation.
Baldwin, Thomas
Aeronautics -- Early works to 1900; Balloons -- Early works to 1800
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:
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