Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

parately, and add these contents, and the liquor required to ver the crown: the sum will be the content.

2. Suppose the depth 36 inches, and the greatest bulge 15 ches from the top; the diameters at the top 80.5 and 80.8, id at the bulge 89.0 and 89.5, and in the middle between ese 85.5 and 860; also, the diameters at the top of the own 83.0 and 83.5, and half-way between it and the greatest ilge 86·5 and 87·0; the liquor required to cover the crown 3.5 gallons. Required the content.

Ans. 775-36435 imperial gallons. 3. Let the depth of a still be 42-8 inches, and the height the greatest bulge from the bottom 20·5; and let the diaeters at the top be 21.0, and at the bulge 47.8 and 47.3, id half-way between them 454 and 460; also, the diamers at the bottom 43.5 and 44.0, and half-way between the ttom and the bulge 470 inches. Required the content in perial gallons, supposing 7 gallons to cover the crown.

Ans. 249-31137 imperial gallons. Stills are generally measured by taking cross diameters at e middle of every six inches, and finding the content of each art as if it were a cylinder; and the top is calculated like a ustum or zone of a sphere.

4. Suppose the top of the still to be 7.3 inches, and its reatest diameters 41·5 and 40·8, and its least 210; the body the still 35.5 inches deep, and the cross diameters in the iddle of every six inches from the top to be, first, 43.9 and 3.2; second, 47·0 and 46.2; third, 47.8 and 47.3; fourth, 7·6 and 47·4; fifth, 46.5 and 46·5; and in the middle of he undermost 5 inches, 450 and 45.2 inches. Required he content in imperial gallons, supposing 7.5 gallons to cover Ans. 248-226245 imperial gallons.

he crown.

CASK GAUGING.

THE easiest way of finding the contents of casks is by the iagonal-rod.

OF THE DIAGONAL-ROD.

This rod is 4 feet long and 4 of an inch square. It is di-ided into 4 equal parts by joints. The principal line on it the diagonal line for imperial gallons, which may be made

hus:

It is found by experiment, that a cask containing 144 mperial gallons has a diagonal of 40 inches: therefore 144 is placed at 40 inches; and, since the contents are as the cubes of the diagonals, 144 : 403 64000 :: 114: 152000, the cube root of which is 37, therefore 114 is put at 37; and

in the same manner any other name if gis may be par op the rod. A line of inces is as on the same side of the rd.

Upon another side of the rod is a Le marked Seg. St. for fus ng the usage of a standing Cask

On a third side are tatus fe Lacing he cab, is theme of ha for whole bogsheads, ď s. 18 12 and 13) and wine gains. The depth is then in inches, and the re is given in gabus

The fourth side contains lines for claring asks of kon dimensions, as a half-anker, a frkin, a re, a bgshead, i puncheon, &c. either lying or standing. Put into the ban that end of the rod from which the divisios for the given casi are numbered, until it rests upon the opposite stave; and the division on the rod intersected by the surface will be the ullage

PROB. XIV. To find the content of a cask by the rod.

Put in the end covered with brass at the bung, and extend it to the opposite corner of the head, and mark the gallts and parts at the middle of the bung; then extend it to the other head of the cask, and mark the gallons and parts. Hi the sum of these two, if they do not agree, will be the content NOTE. The contents on the rod are made for the most common forms of casks.

1. Suppose a cask to be 21 inches long, the bung-diameter 19, and the head-diameter 16. Required the content in ale gallons.

If the rod be extended from the bung to the opposite corner of the head, it will give 19-3 imperial gallons nearly.

PROB. XV. To find the content of a cask by the pen.

In common casks, the cube of the diagonal divided by 44! will give the content in imperial gallons. Therefore, to the square of half the length add the square of half the sum of the diameters, to get the square of the diagonal: this multiplied by its square root, and divided by 4444, gives the content.

Half the sum of the diameters in last example is 175 therefore 17.52+10.52 = 416·5, the square root of which is 204, and 416.5 x 20.4444319-12 imperial gallons the

content.

OF THE VARIETIES OF CASKS.

Casks are commonly divided into four varieties, according to the degree of their curvature.

. The middle zone of a spheroid, measured by Prob. XX. MENSURATION OF SOLIDS.

ii. The middle zone of a parabolic spindle, gauged by ob. XXX. of the same.

iii. Two equal frustums of a parabolic conoid, by Prob. XIV. of the same.

iv. Two equal frustums of a cone, by Prob. X. of the same. 2. Required the content, in imperial gallons, of a cask of e first variety, of which the length is 40, the bung-diameter 2, and the head-diameter 24 inches.

(2 × 322 +242) × 40 = 104960

⚫0009442

99-1032 imperial gallons.

By the Sliding-Rule.

et 32.544 on D to 40 on C; and at 24 on D is 21.75 on C,

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

3. Suppose the cask to be of the second variety, and the dimensions the same as in the last.

(2 × 322 +2424x82) × 40=103936

⚫0009442

98.1364 imperial gallons.

Set 32.544 on D to 40 on C; then at 8 on D is 2.417 on C, which, multiplied by 4, gives 9668 of an imperial gallon to be taken from the content found in the last example, and leaves 98.13643 imperial gallons.

4. Let the cask be of the third variety, and the dimensions as before.

Ans. (322+242) X 20 X 0028326-90-6432 imp. gals. Set 18.95 on D to 20 on C; then at 32 on D is 58.01 on C. . . 18.95 . 20

[ocr errors]

24 . 32.63

[ocr errors]
[ocr errors]

90.64 imp. gal.

5. Let the cask be of the fourth variety, and the dimensions still the same.

Ans.

(56232 x 24) x 40 x 0009442

rial gallons.

89.4346 impe

Set 24 on D to 24 on C; and at 32 on C is 27·7 on D, the mean proportional.

Set 29.7 on D to 40 on C; and at 56 on D is 118.4 on C. 27.7.. 29.0.

. . 29.7 . . 40

[ocr errors]
[ocr errors]
[ocr errors]

89.4 imp. gal.

6. Let the length be 20, and the diameters 16 and 12 inches. Required the contents in imperial gallons, according to all the varieties.

Ans. First var. 12.3879, second var. 12.267, third var 11.3304, fourth var. 11·1793 imperial gallons.

7. Let the length be 40, and the diameters 32 and 26 inches. Required the content, according to all the varieties. Ans. First var. 102-88, second var. 102.3362, third var. 96.3084, fourth var. 95-6286 imperial gallons.

8. Let the length be 45 inches, and the diameters 36 and 30 inches. Required the content, according to all the varieties, in imperial gallons.

Ans. First var. 148.3716, second var. 147·7597, third var. 139.9588, fourth var. 139·194 imperial gallons.

9. Let the length be 48 inches, and the diameters 40 and 32 inches. Required the content, according to all the varieties. Ans. First var. 191-4384, second var. 190·2782, third var. 178.3858, fourth var. 176·9355 imperial gallons.

NOTE. The second variety comes nearer to the form of common casks than any of the others, but it does not entirely agree with them.

PROB. XVI. To gauge a cask by reducing it to a cylinder.

on the

RULE. Divide the head by the bung diameter, and find the quotient in the column titled Quot. in the following table. In the column answering to the variety of the cask, same line with the quotient, will be found a number, which, multiplied by the difference between the bung and head diameters, and the product added to the head diameter, will give the mean diameter, or that of a cylinder equal to the cask. Then multiply the square of the mean diameter by the length of the cask, and by 0028326, for the content in imperial gallons.

By the Sliding-Rule.

Find the difference between the head and bung diameters on the edge of the rule, and against it, in the proper line, is he number to be added to the head, to get the diameter of he cylinder, called the mean diameter.

Quot.

1st 2d 3d 4th Var. Var. Var. Var.

Quot.

1st 2d 3d 4th Var. Var. Var. Var.

76 695678 534 511

[blocks in formation]

532 510

530 510

522 508

507

506

77 694 677 78 693 677 79 691676 529 510 80690 676 527 509 81 689 675 526 508 82 688 675 524 508 83 686 674 84 685 674 521 85 684 673 520 86683 673 519 506 87 682 672.517 505 88680671.516.505 89 679 671515 504 90 678 671.513 504 91 677 670 511 503 92675 670 510 503 93 674 669 509 503 94 673 668 507 502 95672668.506 501 96670667.505.500 97 670 667 | ·503 | ·500 98 667 666 501 99 666 666 .500 .500

71702 681 541
72 701 680 540 513
73 699 680539513
74 698 679 537 512 1.00
75697678 | ·535 | 512

500

1. Suppose the length 40, and the diameters 32 and 26 inches. Required the mean diameter and content in imperial gallons, according to all the varieties.

26÷3281, opposite to which, in the table, are '689, 675, 526, 508. Then,

6×689+26=30134 mean diameter, and 30-1342 × 40 X0028326102-8866 imperial gallons in first variety. 6×675+26=30.05 mean diameter, and 30.052 × 40 X 0028326102-3138 imperial gallons in second variety.

« ΠροηγούμενηΣυνέχεια »