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CALCULATION.

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Here 36.8x18.3=673.44, the area of the base; and 44.6 x 25.9= 1155.14, the area of the section at the eaves.

Also, (36.8+44.6)^-2=81.4-^2=40.7, the length of the middle section; and (18.3 +25.9)-s-2=44.2-f-2=:22.1, the breadth of the middle section; then 40.7x22.1 x4 = 3597.88, four times the area of the middle section; whence (673.44 +1155.14 +3597.88) x 1 8.6h-6 = 5426.46 x 3.1 = 16822.026feet, the solidity of the prismoid ABCDEF.

Again, (25.9 x 15.5)-=-2=401.45-s-2=200.725, the area of the ewrfEDG; and 200.725 x 43.7=8771.6825 feet, the solidity of the prism FEDGH.

Lastly, (16822.026+8771.6825)-s-27=25593.7085 H-27 = 947.9151 cubic yards, the answer required.

2. How many cubic yards are contained in a hay-stack of the following dimensions; viz. the length at the bottom 49.4, and the breadth 24.6; the length at the eaves 58.8, and the breadth 34.2; the perpendicular height of the lower part 25.8; the perpendicular height of the upper part 21.7 feet, and its length the same as that at the eaves? Ans. 2335.143 yards.

CASE n.

When a stack is bulged from the bottom to the eaves, and from the eaves to the top, as in the following figure ; recourse must be had to the equi-distant ordinate method.

RULE.

1. Find the areas of as many equi-distant, parallel sections as you think sufficient, with which proceed as if they were the equi-distant ordinates; and the result will be the solidity of the stack, from the bottom to the uppermost section.

2. Multiply the area of the end of the remaining piece, at the top, which may be considered as a triangular prism, by its length; and the product will be its solidity.

3. Add these two solidities together, and the sum will be the content of the whole stack.

Note 1. The weight of a stack may be ascertained to a considerable degree of accuracy, in the following manner: Cut out a portion extending to the centre of the stack, from the top to the bottom, and weigh it: and also measure the vacuity from which it is taken: then say, as the content of this part is to its weight, so is. the content of the whole stack, to its weight.

2. If the stack have a rectangular base, the portion to be weighed should not be cut off the end, but taken out about half way between the end and the middle; where it may be supposed the hay is of a medium density.

EXAMPLES.

1.. Let the annexed figure represent a hay-stack, the dimensions of which are as follow; viz. the length AB 38 feet, and breadth BC 13 feet; the length DE 41 feet, and breadth EF 18 feet; the length GH 45 feet, and breadth HI 22 feet; the length KL 44 feet, and breadth LM 19 feet; the length NO 42 feet, and breadth OP 12 feet; the mean length of the triangular prism, at the top, 41 feet,- and the perpendicular height of its end 5 feet; how many cubic yards are contained in the stack; the perpendicular distance between each parallel section being 6 feet ?

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CALCULATION.

Here 38x13 = 494, the area of the first section ; 41 X 18 = 738, the area of the second section; 45x22 = 990, the area of the third section; 44x 19 = 836, the area of the fourth section; and42 x 12=504, the area of the fifth or last section ; then by proceeding according to the Rule for equi-distant ordinates, we have A=494+504=998, B=738+836= 1574, C=990, and D=6; consequently, (A+4B+2C)+3 x D=(998 + 6296 +1980) x f=9274 x 2=18548feet, the solidity of the part ABCPON.

Again, (12 x 5)-^2=60-r-2=30, the area of the end OPR; and 41 x 30=1230feet, the solidity of tlie prism NOPRS.

Lastly, (18548 +1230)-^27=19778h-27=732 yards, 14 feet, the answer required,

2. The dimensions of a hay-stack are as follow; viz. the length of the bottom or first section 70.8, and its breadth 20.2 ; the length of the second 75.4, and its breadth 28.7; the length of the third 81.2, and its breadth 32.4; the length of the fourth 86.7, and its breadth 36.8; the length of the fifth 84.9, and its breadth 346 ; the length of the sixth 83.4, and its breadth 30.9 ; the length of the seventh 82.7, and its breadth 16.5 ; the mean length of the triangular prism, at the top, 81.5, and the perpendicular height of its end 6.4 feet; how many cubic yards are contained in the stack; the perpendicular distance between each section being 8 feet ? Ans. 4668.8 yards.

3. What is the weight of a hay-stack which measures 425 cubic yards 18 feet; the weight of a piece cut out of the stack from the top to the bottom, being 1 ton, 3 cwt. 2 qrs. 15 lb., and the vacuity from which it is taken, 16 cubic yards 12 feet ? Ans. 30 tons, 11 cwt. 3 qrs. 1 lb.

REMAKE.

Sometimes hay is stored up in bams or shades. When this is the case, the lower part of the mow, from the bottom to the eaves, will be a parallelopipedon, the content of which may be found by Problem 2 ; and the upper part, from the eaves to the ridge of the building, will be a triangular prism, the content of which may be obtained by Problem 3, Part IV.; then the sum of these two contents will be the whole content of the mow.

If the hay does not extend to the ridge, the upper part will be a trapezoidal prism, the content of which may be obtained by finding the area of the end by Problem 8, Part II.; and multiplying this area by the length of the prism. (See Problem 18, Part IV., on the method of measuring compound and irregular solids.)

Note. When hay-mows are irregular, mean dimensions must be taken.

DRAINS AND CANALS.

A Drain is an artificial channel made to convey water from marshes, bogs, and other low grounds, for the purpose of making improvements in agriculture, &c.

A Canal is an artificial, navigable river, most commonly made for the purpose of transporting goods from one place to another, by water-carriage.

Note 1. Drains and Canals are made with sloping sides, and are almost invariably dug by the cubic yard ; hence, it is of the utmost moment to ascertain their coo. tents with accuracy.

The common method of measuring them is to take the breadths of the top and bottom, in different places, and their sum being divided by their number, the quotient is considered as a mean breadth. Several depths are likewise measured, in various places, and their sum divided by their number, is taken for a mean depth ; then the length, breadth, and depth being multiplied continually together, the last product is taken for the content; but it is evident that this process must lead to very erroneous results.

2. In this country, many thousands of acres of bogs, marshes, and fens have of late been made fit for the purposes of agriculture, by means of drains, particularly in the counties of York and Lincoln; and improvements of this kind- are still carried on in the latter county, with great spirit.

PROBLEM III .

To find the number of cubic yards which have been dug out of a drain or canal.

RULE.

Find the areas of as many equi-distant perpendicular transverse sections, as you judge sufficient, with which proceed as if they were equi-distant ordinates; and the result will be the content between the first and last sections. (See Problem 23, Part II.)

Note 1. The distance between the sections must be measured along the middle of the bottom of the drain or canal; and when the ground is very uneven, it will be necessary to take the sections nearer to each other, than when it is pretty level.

2. When a drain or canal is very long, divide it into several parts or lengths, and find the content of each separately by the above Rule ; and the sum of these contents will be the content of the whole.

3. Let the annexed figure represent a perpendicular, transverse section of -a drain or canal, where the ground is even; then it is evident that if a line AB, be stretched across the top, it will be parallel to the bottom DC, admitting it to be level; and the perpendicular Dm will be equal to the perpendicular Cn, which perpendiculars may be easily found by erecting a straight stafF perpendicularly to the bottom DC, and so as to touch the line AB, at m and n.

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In this case, the section ABCD is a trapezoid, the area of which may be found by Problem 8, Part II.

4. Let the subjoined figure denote a perpendicular, transverse section of a drain or canal, where the ground is uneven; then it is evident that the section is a trapezium, because AB is not parallel to DC.

Now, as two perpendiculars cannot be taken upon either of the diagonals, the figure must be divided into two triangles in the following manner: Measure the diagonal BD, and at right-angles to it, the perpendicular Cn; measure also the line AB, and at right-angles to it, the perpendicular Dm; hence, the area of the section may be found by Problem 7, Part II.

Or, the section may be divided » n into two right-angled triangles, and a trapezoid, as in the annexed figure, by measuring the line AB, and at right-angles to it, the two perpendiculars Dm and Cn; and hence $he area may be found.

Or, the area of the section may be obtained by dividing it into the two triangles ADm, BCn, and the trapezoid DmnC.

In this case, Dm and Cn, which
are at right-angles to DC, are the
parallel sides of the trapezoid, and
also the bases of the two triangles; and their perpendiculars are Ar
and Ba.

5. If the bottom of a drain or
canal be a curve line, as in the sub-
joined section; then the area of the
quadrilateral figure ABCD, must
be found by some of the foregoing
methods; and the area of the part ^

DEC, by the method of equi-distant ordinates, described in Pro-
blem 23, Part II.

Note. The Rule given in Problem 23, Part II., being expressed in an algebraic form, is seldom perfectly comprehended by learners; but the following one may be easily understood, and committed to memory.

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KTJLE.

To the sum of the first and last ordinates, add four times the sum of all the even ordinates, and twice the sum of all the odd ordinates, not including the first and last; multiply this sum by the common distance of the ordinates, divide the product by 3, and the quotient will be the area required.

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