Airopaidia

Note 46 — An alphabetical index of the contents:

Reading edition

As therefore it may be supposed that the Peak of St. George, in

receding

from it, woud

vanish

at the Distance of 150 Miles; its Height may

easily

be ascertained geometrically thus:

Illustration from Airopaidia: 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.

See the Figure annexed.

Let M be the Summit of the Mountain: and let the Line M T drawn to the Circumference of the Circle at T, be the evanescent Distance of the Mountain in the Horizon, viz. 150 Miles.

Join T C, viz. a Line drawn from the Tangent to the Center of the Circle, which Line will therefore represent the Semidiameter of the Earth, viz. 3958 Miles, according to Newton.

Draw a Line from C to M, which will pass throu’ some Point of the Circumference as H, the Base of the Mountain.

Then, in the Triangle M T C, as the Angle at T is a right Angle (Euclid’s Elements, Book 3, Proposition 18;) and the Sides M T, and T C, containing the right Angle, are known; the third Side C M is readily found: (being a Corollary to the 47th Prop. 1st Book Euclid:) viz. having the two Sides of a right Angle Triangle given to find the third. Therefore

RULE.

Multiply the Sides containing the right Angle, each into itself: viz. 150 and 3958: add the Products into one Sum: from which extract the square Root; equal to the Length in Miles, of the third Side required.

From the third Side, subtract that Part, viz. C H, which is equal to the Semidiameter T C already found: and the Remainder H M is the Height of the Mountain.

Thus: 150 Miles. 3958 Miles in the Semidiameter of

150 3958 the Earth

—— ———

7500 31664

15 19790

——— 35622

22500 11874

Square of the ————

greatest visible 15665764 Square of the Semidiameter

Distance. add 22500 of the Earth.

————

Extract the sq. Root, 15688264 (3960.84 Square Root.

9 3958 subtract.

— ——

69) 668 Rem. 2.84 Answer in Miles.

621

——

786) 478.2

471 6

———

79208) 6664.00 continued to 2 Decimals.

6336 64

————

792164) 32736.00 ditto.

31686 56

————

104944

To find the .84 Part of a Mile; multiply

1760 Yards in a Mile,

Decimal Parts of a Mile to be reduced .84 into Yards.

——

7040

14080

———

1760)1478.40 (0

Subtract 1478

——

282

Answer: the Height of the Mountain is 2 Miles 282 Yards.

Return to the book