Note 25 — Chapter XIV

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.

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