BY THE SLIDING BULE.

CASE III.

When the warm work is in a circular by-tač.

Given the top and bottom diameters, and the perper dicular depth of a vessel in the form of the frus tum of a cone, to determine the intermediate dia meters, at any assigned depth.

EXAM. 2.

58-42 16

Here

35

=== .457, the common factor or 35

multiplier; then 58

5 X .457 = 58 — 2.285 = 55.715, the diameter at 5 inches from the bottom; 58 10 x .457 = 58 4.570 = 53.430, the diameter at 10 inches from the bottom; 58 15 x 457 = 58 - 6.855 51.145, the diameter at 15 inches from the bottom; 58 - 20 x .457 = 58 — 9.140 = 48.860, the diameter at 20 inches from the bottom; 58-25 X.457 = 58 – 11.425 = 46.575, the dia meter at 25 inches from the bottom; and 58 - 30 × .457 = 58 — 13.710 = 44.290, the diameter at 30 inches from the bottom.

PROBLEM XV.

Given the top and bottom diameters, and the perpendicular depth of a vessel in the form of the frustum of a semi-sphere, to determine the intermediate diameters, at any assigned depth from the top of the vessel.

es, the distance of the bottom diameter of the frustum, from the centre of the sphere.

Again, 2√7.25 + 182 = 2 √52.5625 + 824 = 2√376.5625 = 19.4 × 238.8 inches, the dia meter of the sphere.

To find the height of the segment.

By Rule II., we have 19.4 (8 + 7.25) = 19.4 4.15 inches, the height of the segment.

15.25

To find the intermediate diameters.

By Rule III., we have 2√(38.8 - 5.15) × 5.15 2√33.65 × 5.15 = 2 √ 173.297513.16×226.32 inches, the diameter at 1 inch from the top of the given vessel.

Again, 2 (38.8 6.15) x 6.152√32.65 x 6.15 =2√200.7975=14.17 × 2 = 28.34 inches, the diameter at 2 inches from the top of the vessel.

૨૩

14.07 inches, the distance of the bottom diamete i the fiscum from the centre of the sphere.

the diameter of the sphere.

To fut the beagit of the segment.

By Edle II. we have 53.13

(15.08 + 12) = 55.13 — 30.35 = 5.1 inches, the height of the se

To End the intermediate diameters.

By Rule 1.1, we have 2,766.26 — 4.1; × 41. = i

√6206 × 4.1 = v£54.856 = 15,95 × 2=$1.98 inthes, the diameter at 1 inch from the top of the given vessel.

e

Again, 206.26 - 5.1) x 5.1 = 261.16 x 5.1 = 2 311916 17.66 x 2 = $5.39 inches,

=

the diameter at 2 inches from top.

Again, 2 (66.26-6.1) × 6.1 = 2√60.16 × 6.1 = ✔366.976 = 19.15 x 2 = 38.30 inches, the diameter at 3 inches from the top.

By a similar process we find the fourth intermediate diameter to be 40.98, the fifth 43.40, the sixth 45.60, the seventh 47.62, the eighth 49.48, the ninth 51.18, the tenth 52.76, and the eleventh 54.22 inches.

SECTION II.

THE METHOD OF GAUGING AND INCHING COMMON

BREWERS' UTENSILS, AS PRACTISED

EXCISE.

IN THE

PROBLEM I.

To guage and inch a mash-tun in the form of the frustum of a cone, standing upon its greater base.

EXAM. 2.

BY THE PEN.

To find the area.

Inches. Inches. Divisor. M. T. gallons.

90.4 x 90.428928.277 area of the 1st section. 88.2 X 88.228926.917 area of the 2nd section. 86.1 x 86.128925.651 area of the 3rd section. 83.8 X 83.8289 81.6 x 81.6289

24.299 area of the 4th section. 23.040 area of the 5th section.

From the question, and from the fregung ara Di comers, we form the Dimension Book, as below.

AZ & Bount Masi Tun, Na 2. gorged Jvig 20,1821.

56 23:40 276.46 0.2704) 4 244 26.200 242.99 0.3 0.299 3 62.99 96.1 25.651 256.51 0.3 1.651 4 0051 $2.2 26.917 269-17 a 3 2917 4 1317 96.4 25.377 282.77 0.34.277 4 3277