Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROBLEM XI

T: fad the scatent of a vessel, in the form of the segment of a sphere.

Exix. 2.

By Rule I.

Here 241 × 3 = 76 × 3 = 1728, three times the square of the radius of the top; also 21 = 441, the square of the depth; then (1728 + 441) × 21 x 5236 = 2160 x 21 x 5236 = 45549 × 5236 = 238404564, the content in cubic inches; hence 238444504 — 231 = 103.244, the content in wine

[ocr errors][merged small]

Here 38a = 1444, the square of the diameter of the base; and 152 = 225, the square of the height; also

225

= 15, one third of the square of the height; then 3 (1444 + 225 + 73) × 7-5 = 1744 x 7.5 = 13080; and 13080 ÷ 9738 = 4.777, the content in malt bashels.

PROBLEM XII.

To find the content of a vessel in the form of the frustum or zone of a sphere.

[merged small][ocr errors][merged small]

Here

[blocks in formation]

= 1802,

half the sum of the squares of the two diameters; and

[blocks in formation]

square of the depth; then 1802 + 216 x 18 = 2018 x 18 = 36324; and 36324 ÷ 294.12 = 123.500, the content in wine gallons.

[blocks in formation]

half the sum of the squares of the two diameters; and

800

202 × = 400 = × } = =266.66, two thirds

3

of the square of the altitude; then 600.5 + 266.66 × 20 = 867.16 × 20 = 17343.20; and 17343.20 → 2738 = 6.334, the content in malt bushels.

PROBLEM XIII.

To find the content of a vessel, in the form of a prolate spheroid.

EXAM. 2.

Here 56' x 72.5 = 3136 × 72.5 = 227360; and

227360

441.18

515.345, the content in wine gallons.

EXAM. 3.

Here 422 × 48 = 1764 × 48 = 84672; and

84672

20.616, the content in malt bushels.

PROBLEM XIV.

To find the content of a vessel, in the form of the middle frustum of a prolate spheroid.

EXAM. 2.

Middle Diameter.

24

24

96

48

576 square.

2

1152 twice the square of the middle diameter. 400=20×20, the square of the end diameter. 1552 sum.

25 length.

7760

3104

1077.15)38800.00(36.020 content in ale gallons.

323145

648550
646290

226000
215430

105700

EXAM. 3.

Here 35 x 35 x 2 = 1225 x 2 = 2450, twice the square of the middle diameter; and 30 × 30 = 900, the square of the end diameter; then 2450 + 900 × 38 = 3350 × 38 = 127300; hence 127300 ÷ 882.36 = 144.272, the content in wine gallons.

EXAM. 4.

Here 52 x 52 x 2 = 2704 x 2 5408, twice the square of the middle diameter; and 40 × 40 = 1600, the square of the end diameter; then 5408 + 1600 × 587008 × 58 = 406464; hence 406464 ÷ 8214 49.484, the content in malt bushels.

PROBLEM XV.

To find the content of a vessel, in the form of a parabolic spindle.

EXAM. 2.

=

Here 24 X 24 = 576, the square of the middle diameter; and 576 × 65 37440, the square of the diameter multiplied by the length; then 37440 ÷ 551.48 67.890, the content in wine gallons.

EXAM. 3.

Here 56 x 56 = 3136, the square of the middle diameter; and 3136 × 150 = 470400, the square of the diameter multiplied by the length; then 470400 5133.75 91.628, the content in malt bushels.

PROBLEM XVI.

To find the content of a vessel, in the form of the middle frustum of a parabolic spindle.

EXAM. 2.

Here 27 × 27 × 8 = 729 × 8 = 5832, eight times the square of the middle diameter; 22 × 22 × 3 =

484 × 3 = 1452, three times the square of the head diameter; and 27 × 22 × 4 = 594 × 4 = 2376, four times the product of the two diameters; then (5832 +1452 +2376) × 30 = 9660 × 30 = 289800; and 2898005385.79 = 53.808, the content in ale gallons; and 2898004411.76 65.688, the con tent in wine gallons.

EXAM. 3.

Here 25 x 25 x 8 = 625 × 8 = 5000, eight times the square of the middle diameter; 20 × 20 × 3 = 400 × 3 = 1200, three times the square of the top or bottom diameters; and 25 × 20 × 4 = 500 × 4 = 2000, four times the product of the two diameters; then (5000 + 1200 + 2000) × 27 = 8200 × 27 = 221400; and 221400 ÷ 41069.92 = 5.390, the content in malt bushels.

PROBLEM XVII.

To find the content of a vessel, in the form of a para bolic conoid.

EXAM. 2.

Here 24 × 24 = 576, the square of the diameter of the base; and 576 × 21 = 12096, the square of the diameter multiplied by half the altitude; then 12096

359.05 = 33.688, the content in ale gallons; and 12096294.12 = 41.126, the content in wine gallons.

EXAM. 3.

Here 24 × 24 = 576, the square of the diameter of the base; and 576 × 9 = 5184, the square of the diameter of the base, multiplied by half the depth;

« ΠροηγούμενηΣυνέχεια »