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Part V., (80 x 80 x 25)

359.05

160000 ÷

359.05 = 445.62, the content in ale gallons.

EXAM. 38.

By Rule I., Prob. VIII., Part V., we have 126,3 × 158.6 x 3 = 20031.18 x 3 60093.54, three times the product of the diameter.

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Also 158.6 126.3 = 32.3, the difference of the diameters; and 32.3 x 32.3 = 1043.29, the square of the difference; then (60093.54 + 1043.29) x 132.761136.83 × 132.7 = 8112857.341; and 8112857.341 882.36 = 9194.398, the content in wine gallons,

EXAM. 39.

By Prob. IX., Part V., we have 82.4 x 42.6 = 3510.24, the area of the bottom; and 104.2 × 54.8 = 5710,16, the area of the top.

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48.7, the breadth of the middle section; then 93.3 × 48.7 X 4 = 4543.71 × 4 18174.84, four times the area of the middle section; whence (3510.24 + 5710.1618174.84) X

112.2
6

= 27395.24 x 18.7

=512290.988, the content in cubic inches; and 512290.9882150.42 238.228, the content in

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116.8, the transverse diameter of the middle

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Now 99.2 x 75.6 × 7854 = 7499.52 x .7854 = 5890.123008, the area of the bottom; 116.8 x 88.2 x 7854 X 4 10301.76 x .7854 × 4 = 8091.002304 x 4 = 32364.009216, four times the area of the middle section; and 134.4 × 100.8 × -785413547.52 X 7854 10640.222208, the area of the top; then (5890.123008+ 32364.009216 93.6 +10640.222208) X 48894.354432 × 15.6

6

-762751.9291392, the content in cubic inches; hence 762751.929 ÷ 282 = 2704.794, the content in ale gallons.

EXAM. 41.

By Rule II., Prob. X., Part V., we have 52.4 × 52.4 × 52.4 x .00227 = 2745.76 × 52.4 × .00227 = 143877.824 x .00227 326.60266048, the content in wine gallons.

EXAM. 42.

By Rule I., Prob. XI., Part V., we have 17.8 × 17.8 × 3 = 316.84 × 3 = 950.52, three times the square of half the top diameter; also 15.8 x 15.8 = 249.64, the square of the depth; then (950.52 + 249.64) x 15.8 x .5236 = 1200.16 x 15.8 x .523618962.528 × .5236 = 9928.7796608, the content in cubic inches; hence 9928.779 282 = 35.208, the content in ale gallons.

EXAM. 43.

By Rule II., Prob. XI., Part V., we have 53.8 × 53.8 2894.44, the square of the diameter of the base; and 32.7 X 32.7 = 1069.29, the square of the 1069.29 depth; also = 356.43, one-third of the

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square of the depth; then (2894.44+ 1069.29 + 356.43) × 16.35 = 4320.16 × 16.35 = 70634.616; and 70634,616294.12 = 240.155, the content in wine gallons.

EXAM. 44.

722 + 84%

=

By Prob. XII., Part V., we have 2

5184 +7056

2

=

12240

2

= 6120, half the sum of

the squares of the two diameters; and 60% x }

= 3600 × =

7200
3

= 2400, two-thirds of the

square of the depth; then (6120 + 2400) x 60 = 8520 × 60 = 511200; and 511200 ÷ 2738 = 186.705, the content in malt bushels.

EXAM. 45.

By Prob XIII., Part V., we have 82 × 82 × 96 = 6724 × 96 = 645504; and 645504 ÷ 538.58 1198.529, the content in ale gallons.

EXAM. 46.

=

By the Remark at the end of Prob. XIII., Part V., we have (40 x 40 x 28) 441.18 (1600 × 28) 441.18 = 44800 441.18 101.545, the content in wine gallons.

EXAM. 47.

By Prob. XIV., Part V., we have 48 × 48 × 2 = 2304 x 2 = 4608, twice the square of the middle diameter, 40 x 40 = 1600, the square of the end diameter; and 4608 1600 × 50 = 6208 x 50 = 310400; hence 3104001077.15288.167, the content in ale gallons.

EXAM. 48.

By the Remark at the end of Prob. XIV., Part V., we have 72 × 72 × 2 = 5184 x 2 = 10368, twice the square of the greater diameter; and 54 × 54 =2916, the square of the less diameter; then 10368 + 296 × 80 = 13284 x 80 = 1062720 ; hence 1062720 882.36 1204.406, the content in wine gallons.

EXAM. 49.

By Prob. XV., Part V., we have 50 x 50 x 80 = 2500 × 80 = 200000, the square of the middle diameter, multiplied by the length; then 200000 → 673.22 297.079, the content in ale gallons.

EXAM. 50.

By Prob. XVI., Part V., we have 50 x 50 x 8 = 2500 × 8 = 20000, eight times the square of the middle diameter; 40 x 40 x 3 = 1600 × 3 = 4800, three times the square of the end diameter; and 50 x 40 x 4 2000 × 4 = 8000, four times the product of the diameters; then (20000 + 4800 + 8000) × 54 = 32800 X 54 =. 1771200; and 1771200 4411.8 401.468, the content in wine gallons.

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EXAM. 51.

By Prob. XVII., Part V., we have 54 x 54 x 32 = 2916 × 32 = 93312, the square of the diameter of the top, multiplied by half the depth; then 93312 359.05 = 259.885, the content in ale gallons.

EXAM. 52.

By the Remark at the end of Prob. XVII., Part V., we have 30+ 422 = 900 + 1764 2664, the square of the radius of the top, added to the square of the middle diameter; then 2664 x 42 x .5236 =

12288 X 32 = 785845568, the antent in cubic inches; and 383845508 – 231=ines, the consent at write gaults

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By Prize XVII. Part V. we have 794 + 48 x 30 = (5244 + 9804; x 30 = 7488 x 30=224640 the sum of the squares of the diameters, multiplied by the depth; and BRR÷TIN1 = $12.895, the contet ne gius

Exex 34

By the Remark at the end of Prob. XVIII., Part V., we have 244 + 40° + 682 = 570 + 1000 + 4024 = 6800, the sum of the squares of the semi-diameters of the top and bottom, added to the square of the midde Cameter; then 6800 x 90 x 5236 = 652800 × 236 = 341806.08, the content in cubic inches; and $41806.08 ÷ 231 = 1479.68, the content in wine galets.

EXAM. 55.

By Prob. XIX., Part V., we have 49° + 45a = 1704 + 2023 = 3789, the sum of the squares of the top and bottom diameters; and 47a × 4 = 2209 × 4 = 8836, four times the square of the middle diameter; then (3789 + 8836) x 48 = 12625 × 48 = 600000; and 606000 ÷ 2154.3 = 281.297, the content in ale gallons.

EXAM. 56.

By Prob. XX., Part V., we have 692 + 78 = 4701 + 6084 = 10845, the sum of the squares of the top and bottom diameters; 95a × 2 = 9025 × 2 = 18050, twice the square of the middle diameter; and 879 +90° x 4 = (7569 + 8100) × 4 = 15669 × 4 = 62676, four times the sum of the squares of the diameters taken at one-fourth, and at three-fourths

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