PART VI.

THE METHOD OF

MEASURING HAY-STACKS,

DRAINS, CANALS, MARL-PITS, PONDS, MILL-DAMS, EMBANKMENTS, QUARRIES, COAL-HEAPS, AND CLAY-HEAPS.

HAY-STACKS.

THE contents of hay-stacks are found in order to ascertain their weights; which must, of course, vary according to the density of the hay.

4

Some stacks will not weigh more than 8 or 10 stones per cubic yard, and others will weigh 15 or 16 stones; it is not, however, the measurer's province to determine the weight, but only the number of cubic yards which the stack contains; and leave the buyer and seller to settle about the weight as they think proper.

PROBLEM I.

To measure a hay-stack, having a circular base.

CASE I.

When the stack is straight from the bottom to the eaves, and from the eaves to the top, as in the following figure; the upper part may be taken as a cone, and the lower part as a conical frustum.

RULE.

Multiply the square of the circumference at the bottom AB, by .07958, or for general practice, by .08; and the product will be the area of the base.

manner.

Find the area of a section at the eaves DC, in the same To the sum of these areas, add the square root of their product; multiply this sum by the perpendicular height GF HD, and of the product will be the solidity of the frustum ABCD.

Multiply the area of the section at the eaves, by the perpendicular height FE=DK, and of the product will be the solidity of the cone DCE.

To the solidity of the frustum add that of the cone; and the sum will be the content of the whole solid ABCEDA.

Note 1. Some measurers take the dimensions of hay-stacks, canals, marl-pits, &c. with a tape divided into yards, tenths, and hundredths; but one divided into feet and tenths, is considered, by most practitioners, to be much preferable.

2. When the dimensions are taken in feet, the content must be divided by 27, in order to reduce it to cubic yards.

EXAMPLES.

1. The circumference at the base AB, of the following figure, is 40, the circumference at the eaves DC 60, the perpendicular height GF=HD 15, and FE: = DK 16 feet; how many solid yards does the stack contain ?

CALCULATION.

C

Here 40 x 40 x.08=128, the area of the base; and 60 x 60 x .08-288, H A the area of a section at the eaves;

also, ✓ (128 × 288)=/36864-192, the square root of their product; then (128+288+192) × 15÷3=608 x 5 =3040 cubic feet, the content of the frustum ABCD.

Again, (288 x 16)÷3=4608÷3=1536 cubic feet, the content of the cone CDE.

Lastly, (3040+1536)+27=4576+27=169 yards, 13 feet, the answer required.

2. The circumference of the base of a hay-stack is 52 feet 9 inches, the circumference at the eaves 75 feet 6 inches, the perpendicular height of the lower part 16 feet 3 inches, and that of the upper part 18 feet 6 inches; how many tons are contained in the stack; admitting each cubic yard to weigh 15 stones or 210 pounds?

Ans. 28 tons, 10 cwt. 1 qr. 17 lb.

CASE II.

When the stack is bulged from the bottom to the top, as in the following figure; the equidistant ordinate method, described in Problem 23, Part II., must be adopted.

RULE

1. Find the areas of as many circular sections, taken at equal perpendicular distances from the bottom, as you judge sufficient, by multiplying the square of the circumference of each section by .08. Proceed with these areas in the same manner as if they were equidistant ordinates; and the result will be the solidity of the stack, from the bottom to the uppermost or last section.

2. Multiply the area of the base of the remaining part, at the top, which may be considered as a cone, by its perpendicular height; and of the product will be the solidity.

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

Note 1. Always make choice of an odd number of sections, in order that the number of parts into which the solid is divided, may be equal. Five or seven will, in general, be sufficient; of which one must be at the bottom, and another at the eaves, or as near to them as possible.

2. Great care must be taken to obtain the dimensions of the sections at equal perpendicular distances; for if the slanting distances be taken, it is evident that the content will be made too much.

SCHOLIUM.

The method of finding the areas of curvilineal figures, by means of equi-distant, perpendicular ordinates, was first demonstrated by the illustrious Sir Isaac Newton.

Mr. Robert Shirtcliffe, in his Theory and Practice of Gauging, appears to have been the first who applied it to finding the areas of curvilineal vessels used by Brewers, Distillers, &c.; and after him Mr. Samuel Farrer, in the Appendix to Overley's Gauging. Their Rules, however, were extremely tedious; and, though true to demonstration, were not general, but particular, according to the number of ordinates used.

To obviate this inconvenience, the general rule given in Problem 23, Part II. of this Work, was deduced from SIMPSON'S Dissertations, page 109, by Mr. Thomas Moss; and demonstrated in his valuable Treatise of Gauging, page 235.

Dr. Hutton, in his Mensuration, Proposition I., Section II., Part IV., has also given an elegant demonstration of the same Rule; and adds, in a Corollary, that it will obtain for the contents of all solids, by using the areas of the sections perpendicular to the axe, instead of the ordinates.

The Doctor particularly recommends it for the purpose of gauging and ullaging casks; hence, it is evident that it may be applied with propriety and success to the Mensuration of Hay-stacks, Canals, Marl-pits, and other irregular figures, as being the best approximation that has yet been, or perhaps ever can be given; for by taking an indefinite number of sections, the content of an irregular solid may be obtained to any degree of accuracy.

Here 36 × 36 ×.08=103.68 the area of the bottom or first section; 54 x 54 x .08=233.28, the area of the second section; 66 x 66 x.08=348.48, the area of the third section; 58 x 58 x.08=269.12, the area of the fourth section; and 37 x 37 x.08=109.52, the area of the fifth or last section; then by proceeding according to the Rule for equi-distant ordinates, we have A=103.68+109.52=213.2, B=233.28 +269.12=502.4, C=348.48, and D=5; consequently (A+4B+ 2C)÷3xD=(213.2+2009.6+696.96)÷3 × 5 (2919.76 x 5)÷3=14598.8÷3=4866.26 feet, the solidity of the part ABLK.

Again, (109.52 x 4.5)÷3-492.84÷3-164.28 solidity of the conical part KLM.

feet, the

Lastly, (4866.26+164.28)+27=5030.54-27=186.31 cubic yards, the content required.

2. The dimensions of a hay-stack are as follow; viz. the girt of the bottom or first section = 127.2, the girt of the second = 145.4, the girt of the third = 156.5, the girt of the fourth 168.7, the girt of the fifth 148.3, the girt of the sixth = 121.8, and the girt of the seventh or last section 68.6 feet. The perpendicular distance between each section is 8 feet, and the perpendicular height of the conical part, at the top, 7.4 feet; how many cubic yards are contained in the stack?

PROBLEM II.

To measure a hay-stack, having a rectangular base.

CASE I.

When the stack is straight from the bottom to the eaves, and from the eaves to the top, as in the following figure; the lower part may be taken as a prismoid, and the upper part as a triangular prism.

RULE.

1. Multiply the mean length of the bottom by the mear breadth, and the product will be the area of the bottom. Find the area of a section at the eaves in the same manner. Multiply half the sum of the lengths of the bottom and eaves, by half the sum of the breadths; and the product will be the area of a section equally distant from the bottom and eaves.. To the area of the bottom add the area of the section at the eaves, and four times the area of the middle section; multiply this sum by the perpendiculan height, from the bottom to the eaves; and of the product will be the solidity of the lower part.

2. Multiply the breadth at the eaves by the perpendicular height from the eaves to the top; and half the product will be the area of the end; which being multiplied by the mean length, will give the solidity of the upper part.

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

Note 1. Sometimes stacks are longer on one side than the other, and broader at one end than the other; in such cases, take half the sum of the lengths for a mean length, and half the sum of the breadths for a mean breadth.

2. Some stacks are higher and broader at the ends than in the middle; when this is the case, a proper allowance must be made in taking the dimensions. Allowance also ought to be made for the thatch.

3. If the ends of the upper part be not equal, find the area of both ends; and take half their sum for a mean area, which multiply by the length for the solidity. 4. If the length of the top or ridge GH, be more or less than the length of the base or eave EF, it is evident that the upper part of the stack is in the form of a cuneus or wedge; hence its truc content may be found by Problem 9, Section I. Part IV.

EXAMPLES.

1. Let the annexed figure represent a hay-stack, the dimensions of which are as follow; viz. the mean length at the bottom 36.8, and the mean breadth 18.3; the mean. length at the eaves 44.6, and the mean breadth 25.9; the perpendicular height from the bottom to the eaves 18.6, and from the eaves to the top 15.5 feet; how many cubic yards does the stack contain; the mean length of the upper part being 43.7 feet?