tourse of bricks, (4 of which usually make a foot in depth) and this is but 4 inches, or half a brick thick ; what will this piece of work come to at $32 per statute rod ? Ans. : $119.8576. TO CALCULATE FREIGH ON PACKAGES OF MERCHA DIZE. RULE--Find the cubic feet aš in case II. of Solids, which divide by 40 for tons, then the amount of freight may be found by the Rule of Three, or Practice. EXAMPLES. 1. What is the freight of a bale that is 7 feet 9 inches in length, 6 feet in breadth, and 3 feet 6 inches thick, at 20 cents per cubic foot ? Ans. $32.55. 2. Required the freight at 18d. per cubic foot on a trunk that is 3 feet 6 inches in length, 2 feet 2 inches in breadth, and 1 foot 8 inches in depth. Ans. 188. 10.3. 3. I demand the freight of that case whose length is 7 feet 6 inches, breadth 5 feet 9 inches, and depth 5 feet, at 814 Ans. $75.46875. per ton ? TO FIND THE BURTHEN OF SHIPS. RULE-Multiply the length of the keel in feet, by the breadth of the mid-ship-beam in feet, and that product again multiplied by the depth of the hold also in feet, which last product divide by 95 for merchants ships, and by 100 for ships of war, and the quotient is the tons. EXAMPLES. 1. If the keel of a ship be 95 feet in length, and the breadth of the mid-ship-beam 32, and the depth of the hold 16 feet: What is the burthen? $ 512 tons as a merchants ship. Ans. 486.4 tons as a ship of war. 2. The proportions -of Noah's Ark were as follows, viz: 300 cubit the length of the keel, 50 feet, the breadth of the culita mid-ship-beam, and 30 feet the depth of the hold : Required its burthion allowing the tubit as it is found by modern tray. ollers, to be 22 inches ? $ 2918834 tons as a merchant ship. 227729 tons as a ship.of war. t-cutits. o Ans. 16.30 GAUGING. I. To gauge any vessel, or to find what quantity of liquor it can contain. RULE.-Find how many solid inches will fill the cavity of the vessel, and divide these by the number of solid inches contained in any given measure, and the quotient is the con. tent in that measure. Table of cubic inches in several measures. ditto, = 1 wine gallon. 268.8 ditto, = 1 corn gallon. 2150.42 ditto, 1 corn or malt bushel. II. To find multipliers and gauge points for right lined surfaces. RULE-Divide 1 by the cubic inches in each measure, and the quotient will be equivalent multipliers, and their square roots, will be the gauge points. EXAMPLE 1. divisors. multipliers. gauge points. A.G. 282)1.000(.003546 A.G. 28216.79 A. G. W.G. 231)1.000/.004329 W.G. 231=15.19 W.G. C.G. 268.8)1.000( .00372 C. G. 268.8=16.39 C. G. M.B.2150.42)1.000(.000465 M. B. 2150.42=46.37 M. B. III. To find the area of any rectangular tun, back, or cooler, &c. in ale and wine gallons, &c. RULE-Multiply the length by the breadth) both being in inches,) and divide the product hy the divisors, or multiply it by the multipliers, the result will be the area in the mcas. ure required. EXAMPLES. 2. Required the area of a square cooler whose side is 1243 inches, in ale and wine gallons, &c. 124.5 x 124.5=15500.25 by Division. by Multiplication. 282)15500.25(54.96 A. G.|15500.25%.003546=54.96 A.G. 231)15500.25(67.10 W.G. 15500.25%.004329=67.10 w.g. 2150.42)15500.25(7.208 M. B. 15500.257.000465=7.208 m. B. By the Sliding Rule. Set the divisor upon B. to the side of the square on A. against the side of the square on B. you have the area on A. in ale and wine gallons, &c. If the tun, back, &c. be a rectangular oblong, set the proper divisor on B. to the breadth on A. then against the length on B. is the area on A. These areas multiplied by the depth, produce the content. 3. A vessel in the form of a rectangle 235 inches in length areils. and 68 inches in breadth : Required its area in ale and wine gallons, &c. Ans. 56.66 A. G. 69.17 W. G. 59.44 C. G. and 7.43 M. B. IV. To find divisors, multipliers, and gauge-points for circular RULE-Divide the cubic inches in measures by .785398 or .7854 (the area of a circle whose diameter is 1) the quotients will be divisors, and divide .785398 or .7854 by the culo bic inches in measures, will quote multipliers. The square roots of those divisors will give gauge-points. EXAMPLE 4. Multipliers. Gauge-points. 359.05 A. G. .002785 A. G. 18.95 A. G. 294.12 W.G. .0034 W.G. 17.15 W. G. 342.24 C. G. .00292 C. G. 18.5 C. G. 2738 M. B. .000365 M. B. 52.32 M. B. V. To find the area of a circle in ale and wine gallons, &c. RULE-Divide the square of the diameter by the divisors, or multiply the same square by the multipliers, the result will be the area in ale and wine gallons, &c. EXAMPLE. 5. Required the area of a circle whose diameter is 80 in. ches, in ale and wine gallons, &c. Ans. } 17:13 A. G: 21.76 W.6.2.336 M. B. by multiplication. G. By the Sliding Rule. Set the divisor on B, to the diameter on A. and against the diameter on B. is the area on A. VI. To find the content of any cube, parallelopiped, prism, or cylinder, in ale and wine gallons, &c. Rule 1–Find the area of the base in ale and wine gallons, &c. and multiply by the depth produces the content, or 2. Find the solidity in inches as taught in Solids, this content, divided by the divisors, or multiplied by the multiplia ers, will give the content in ale and wine gallons, &c. EXAMPLES 6. How many ale, wine gallons, and malt bushels will a vessel in the form of a parallelopipedon contain, the length being 72, the breadth and depth 82 inches ? 690.89A.G. 843.42W.G. 90.60M.B. by the 1st rule divis. Ans. 690.89 A.G. 843.42W.G.90.60 M.B by division. 2nd. (790.87 A.G. 843.42W.G. 90.59M.B. by multipli.) rule. BY THE SLIDING RULE. 690.89 A. G. Set 82 on G. to 15.19 mean proportional be 843.42 W.G. 46.37 Štween 72 & 33 on D. you will find the contenton C. 90.6 M. B. 7. A cylinder, whose diameter is 72 inches, and depth 48, it is required to find the contents in ale and wine galons ? Ans. 693 A. G. 846 W. G. $18.95 2 on D, and against 72 on D. 2693 A. G. Set 48 on C. to 17.15S is the content on C. $ 846 W.G., VII. To find the content of a pyramid, cone or their frustums is ale and wine gallons, &c. Rule-Compute the solidity in inches by the rules given in solids, which divided by the divisors, or multiplied by the multipliers, the result will give the gallons or bushels required. EXAMPLES. 8. Suppose the side of the base of a square pyramid be 35 inches, and the altitude 57, required its contents in ale and wine gallons ? Ans. 82.53 A. G. 100.75 W. G. 10.82 M. B. BY THE SLIDING RULE. on the line D, and 82.53 A. G. Set on C. to 15.19 against 35 on D, is 100.75 W. G. 246.375 the content on c. S 10.82 M. B. 9. Required the content in ale and wine gallons and malt bushels of a conical vessel whose base diameter is 40 inches and altitude 60 ? Ans. 89.12 A. G. 108.8 W. G. 11.68 M. B. BY THE SLIDING RULE. $18.957 on D.then against 40(the 789.12 A. G. Set on C. to 17.15 diameter of the base) on 108.8 W.G. (51.32J D.the con. are foundonc ) 11.68 M.B. 10. Required the content in ale, wine gallons and malt bushels of a vessel, whose bases are rectangles, the greater 90 by 60 inches, and the lesser 69 by 46, and depth 40 inches ? Ans. 601.11 A. G.733.85 W. G: 78.82 M. B. by multiplication 11. Required the content of the lower frustum of a cone, the diameter of the greater base being 32 inches, that of the lesser 24, and depth 20 in ale and wine gallons ? Ans. 43.96 A. G. 53.67 W. G. VIII. The divisors for ale and wine gallons, &-c. for a cylinder being ziven, to find divisors for any of the following solids ; namely, the globe =j, spheroid=, parabolic conoid ), hyperbolic conoide, parabolic spindle 15, and cone = of the cir. oumscribing cylinder, Rule-Observe what part each is of the circumscribing cylinder; then say, As the numerator : is to the denominator : : so is the ale, wine and malt divisors of a cylinder : to the like divisors for the required figures. 5 8 EXAMPLES. 12. Required ale, wine, and malt divisors for the cone, those of the cylinders being 359.05 for ale, 294.12 for wine, and 2738 for malt ? Ans. 1077.15 ale div. 882.36 wine div. 8214 malt divisor. · And so on for any other. 13. A cone whose diameter is 40 inches, and altitude 60, required its content in ale and wine gallons, &c. 40 X 40 X 60=96000 8214)96000(11.68 malt bushels. What is here she wn of the cone is sufficient to inform the student how to proceed with the other solids mentioned in the problem. QUESTIONS FOR EXERCISE. 14. How many ale, wine gallons, and malt bushels will a Fessel in the form of a spheroid, contain whose fixed axis is 100, and revolving 60 inches ? Ans. 668.4 A. G. 816.4 W. G. and 87.6 malt bushels. 15. Required the content of a parabolic conoid in ale and wine gallons, the height being 30, and diameter of its base. 20 inches ? Ans. 16.7 A. G. and 20.4 wine gallons. 16. Required the content of the hyperbolic conoid, the base being 100, and altitude 60 inches ? Ans. 696.2 ale, and 850 wine gallons. 17. Required the content of a parabolic spindle, whose length is 60, and greatest diameter 64 inches, in ale and wine gallons ? Ans. 103.02 ale gallons, and 125.7 wine gallons.. CASK-GAUGING. Cask are distinguished into the following four varieties : 1st. The middle frustum of a spheroid. 2nd. The middle frustum of a parabolic spindle. 3rd. The middle frustums of two parabolic conoids. 4th. The middle frustums of two cones. IX. To find the content of a cask. RULE 1-If the staves are very much curved, the cask is of the first variety, then to the square of the head diameter, add twice the square of the bung diameter, multiply the sum by the length, and divide the product by 1077.15 for ale, and by 882.36 for wine gallon. 2. If the staves are less curved than was supposed in the last, the cask is taken for the second variety: then to 9 times the square of the bung diameter, add 6 times the square of : a |