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
[20] Ullòa in his voyage to South-America relates, that in passing over
the +Deserts+, Írides are frequently seen by Travellers round _their
own Heads_ as the Center of the _Iris_; and visible only to themselves.
But what Analogy the _Balloon Iris_ bears to them, Time and future
Experiments may discover. See his “Voyage to South America, Vol. 1. Pa.
442.”
[21] As Sound travels 1142 Feet in a
Second, it must have moved in 30 Seconds
————
Feet in a Yard 3 )34260 = Feet
Yards in a Mile 1760 )11420( 6 Miles
10560
—————
Yards in a Quarter of a Mile 440)860( 1 Quarter
440
—————
Answer 6 Miles, 1 Quarter, and 420 Yards.
[22] Equal to 2085 Yards; or 1 Mile, 325 Yards.
[23] Long’s Astronomy. Pages 227, 229.
[24] Also called the _Horsham Stone_, from a Place so named, in Surrey,
where great Quantities are found.
[25] PROBLEM.
To find the _Length_ of the _Shadow_ from a Person of _middle_ Stature,
(five Feet and a half High) viz. at XII o’Clock, on the 8th Day of
September, 1785, at Chester, whose North Latitude is 53° 12′; (and 3°
11′ West Longitude from London.)
FIRST,
To find the Sun’s Altitude at XII.
From 90°. 00′ Subtract
The Latitude 53. 12
———————
The Remain. 36. 48 is the Complement of Latitude,
to which add (from the Tables)
Sun’s N. Decl. 5. 29
———————
The Remain. 42. 17 is the Sun’s Altitude (viz. at XII.)
SECOND,
For the Shadow say,
As the Sine of the Sun’s Altitude 42° 17′
To the Person’s Height, viz. 66 Inches,
So is the Co-Sine of the Sun’s Altitude,
To the Length of the Shadow.
For the Sine of the Sun’s Altitude 42° 17′ in the Table
of artificial Sines, is the Logarithm 9.82788, which, subtracted
from the arithmetic Complement, viz. 9.99999 (supposing
the last Figure a 10) becomes, .17212
Then for the Person’s Height, viz. 66 Inches:
in the Table of Logarithms is the corresponding
Number, 1.81254
And for the Co-Sine (had by subtracting the Altitude
42.17 from 90.00) viz. 47.43: among the artificial
Sines is the Logarithm, 9.86913
————————
The above Sums added, are 11.86079
which logarithmic Number (deducting the _Initial_ 1 as useless) viz.
1.86079, in the Table of Logarithms, corresponds to 72.57, equal to 72
Inches, for the Length of the Shadow at XII.
Public-domain text, read in full here on John Shaqi.
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