Suppose the whole Distance to be any Line, as A. B. to be divided in C. Then, as 7 is the whole Line, and 4 the greater Part; say, as the whole 7 is to the greater Part 4, so is the whole Distance to a fourth Term proportional, which will be equal to the greater Distance sought:

Whole Distance
in Yards.		 Greater Distance in Yards.
Thus 7, : 4 ::

2332

: 1332⁠4⁄7 Ans.

4

———

7)9328
2332 the whole.

1332⁠4⁄7

1332⁠4⁄7 being the greater Distance found; take the greater from the whole, and then will remain the lesser Distance wanted, viz. 999⁠3⁄7: the 1332⁠4⁄7 = the greater Distance, and 999⁠3⁄7 = the lesser Distance: and adding the Fractions 4⁄7 3⁄7 = 1 to the 999; we have 1332 Yards for the greater Distance, or Height of the Balloon above the Summit of the superior Clouds: and 1000 Yards for the less Distance, or Height from the Earth to the Summit of the superior Clouds.

Note. The Line A. B. here selected is the famous Measure of (half) a mathematical Rhinland and Roman Foot, according to Snellius. (See Geographia Generalis of Varenius, published by Newton. Lib. 1. Cap. 2. De variis Mensuris.)

[18]

PROBLEM.

To find the circular Boundary of the celestial Prospect over the Tops of the superior Clouds, from the Balloon at the Height of near a Mile and half above the Surface of the Earth, viz. 2332 Yards. The Height from the Earth to the upper Surface or Floor of Clouds being 1000 Yards; and the Height above the Floor to the Balloon being 1332 Yards.

On the Curvature of the Earth and Clouds, and Elevation of the Eye above their circular Horizon.

Rule. To the Earth’s Diameter, equal to 7940 geographical Miles, add the Height of the Eye above its Surface: multiply the Sum by that Height: then the square Root of the Product gives the Distance at which an Object on the Surface of the Earth can be seen by an Eye so elevated. Note the Diameter of the Earth, in Feet, is 41798117, according to Newton. (See Practical Navigator, by J. Moore, 7th Ed. Page 251.)

FIRST.