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. | |
Reducing then the Numbers 66 and 72, to the lowest Denomination, thus 6)66⁄72 = 11⁄12 the Proportion which the Length of the Shadow bears to the Height of the Object is thereby obtained: that is
[26] If the Length of the Shadow be divided into 12 Parts, the Height of the Object would be 11 of those Parts. See Moore’s Practical Navigator.
PROBLEM.
An easy Way to find the Proportion which the Length of the Shadow bears to the Height of an Object is, at any time when the sun shines, to fix a Plummet Line and frame upright in the Ground; measure the Length of its Shadow, and compare it with the Height of the frame.
[27] Equal to 3 Quarters of a Mile and 121 Yards.
[28] i. e. When the Barometer below is at 30 Inches, and Thermometer below at 60° viz. about 1000 Yards high in fine Weather, and 500 in changeable.
[29] Being 1083 Yards, i. e. half a Mile, and 203 Yards.