Hay. The value of hay is said to depend upon the preservation of the green colour of the grass, and upon its being juicy, fresh, and free from all sorts of mustiness. (Mac. Com. Dict.) To measure an oblong stack, take the height from the ground to the eaves, and add one-third (some writers say one-half) of the height from the eaves to the top. Multiply the sum by the length, and the product by the breadth, all in feet. Divide the amount by 27; the quotient is the content in cubic yards. Ex.: a stack is 60 feet long, 30 broad, 12 in height from the ground to the eaves, and 9 from the eaves to the top. Then 12 + (9 ÷ 3) = 3 = 15 × 30 = 450 X 1000; being 9000 stones of old hay. (Vide 60 27000 27 = infra). Another mode of measuring a round hay-stack is to divide it into as many sections as it is yards in height, and take the circumference in yards of each section in the middle of the section. Then find the area of each section, and add them together; the sum will be the content in solid yards. The areas are found by table of circular areas (16), which may be extended to other numbers by multiplying the square of the circumference by 079577. A yard of new hay is calculated to contain 6 stones; if the stack has stood a considerable time, 8 stones; and if it be old hay, 9 stones. In London hay is sold by the truss, weighing 56 lb., so that a load is exactly a ton; except that new hay, until 4th September, weighs 60 lb. A truss of hay or clover is 3 feet long and 2 feet wide, and if well settled, 16 inches deep: this gives 113 cubic yards to a ton. If 17 inches deep, 1215 yards to a ton. If 18 inches, 13 yards to a ton. If a cubic foot be found to weigh 5 lb. 3 oz., it will require 16 yards to make a ton; 5 lb. 8 oz., 16 yards; 6 lb., 14 yards; 6 lb. 6 oz., 13 yards; 6 lb. 14 oz., 12 yards; 7 lb. 8 oz., 11 yards; 8 lb. 4 10 yards; 9 lb. 3 oz., 9 yards; and 10 lb., 8 yards. OZ., Manure.-Heaps of manure are measured by the rules for measuring timber or corn (supra, as to corn, division (i) as to timber). If the content be shown in feet, it may be reduced into yards by dividing by 27, or into loads by dividing by 40. To find the number of heaps of manure required for an acre of land, ascertain how many heaps one load may be divided into; multiply the distance of each row by the distance of each heap, and divide 4840 by the product; divide the dividend by the number of heaps into which a load is to be divided. (m) SHIPS. The value of ships and vessels is dependent on their size, fittings-up, age, wear and classification, and the probability of their being usefully and profitably employed. The mode in which the tonnage of merchant vessels is to be measured is defined by 17 & 18 Vict. c. 104; which tonnage is required to be entered in the ship's register, and carved on the main beam, so that it may be readily ascertained. (Sect. 25.) APPENDIX C. USEFUL TABLES. Introductory Remarks. TABLE 2. (Number of days between one month and another.) Where the day is not the same in both months, add or subtract the difference. Thus, from the 7th April to the 7th August is 122 days; but if required to the 19th August, add 12 days, making 134 days, and if to the 2nd August, substract five days, making 117 days. If both days are to be included, add one day, and in leap year add one day after 28th February. TABLE 3. (Income and salaries.) Multiply the sum per annum by the number of days, and the table will show the amount. Example. Salary £100 a year, amount for 10 days required. 100 x 10 = 1000; and the table shows the amount to be £2 14s. 93d. Should the result be a larger number than any in the table, the amount required will be found as in the following example. == Required amount of salary at £100 a year, for 30 days. 100 × 30 = 3000. Refer to the decimal column for 300, and take the decimal point as one figure to its right, and the amount will be 8.21918, and the value in money of the decimals will (see Table 12) make the entire amount £8 4s. 5d. TABLE 5 (Interest) is explained in a note at foot. TABLE 6 shows the amount of £1, improved at compound interest for any number of years to 70, at 3, 4, or 5 per cent. TABLE 7 is a table of the present value of £1 per annum for years certain; to pay interest at rates specified, and replace capital at 4 per cent. (See title "Annuities," in Appendix B, ante). TABLE 8 (Present value of £1 due at end of given number of years) is of use in calculating the value of deferred payments, or reversionary interests, and is explained in Appendix B. The rule is to multiply the sum by the decimal in the table. Example. Required the present value, at 4 per cent., of £100, to be received at the end of 10 years. The decimal in the table is 6756, which, multiplied by 100 (the decimal point moved 2 figures) will give 67.56 = (by Table 12) £67 11s. 24d. TABLES 9, 10 and 11 (Values of annuity on a single life) show the values for a £1 a year. The values must be multiplied by the pounds and decimals of a pound in the annuity the value of which is to be calculated. Example. Annuity of £10 on female life of 35 at 4 per cent. Table 9 gives the value of £1 a year, at 16.88; which × 10, the value required, will be £168.8 or (see Table 12) £168 16s. TABLE 14 (Square measure). With the exception of Table (c), giving the yards and feet in perches, the tables in No. 14, as well as all the tables of cubic measure in No. 15, are based on the application to all numbers of the decimals of units, by the simple alteration of the place of the decimal point. The 1st column gives the number of units, or of decimals of units, and the other columns give the multiples of the extent stated in the heading, the figures and decimals being marked in the 2nd column, and the other columns giving the number of places of figures, or the place of the decimal. Subdivision (a), showing the contents of inches in feet, gives in the 1st column the number of units: in the 2nd column the contents of the numbers multiplied by 100,000 (C M): the 3rd, the number of figures, in the numbers multiplied by 10,000 (X M): the 4th, the numbers multiplied by 1000 (M): the 5th, the numbers multiplied by 100 (C): the 6th, by 10 (X): and the 7th, units (U). Example. Required the contents in feet of 131,000 inches. It will seldom be necessary to take more than two places of decimals, adding one in the second place, should the third be 5 or more. Subdivision (b) not only gives the decimals of the first place, as shown in the table, but also the decimals of the second place (hundredths), or of the third or any other place, by altering the position of the decimal points: thus, 09 of a foot = 12.96 inches, and ⚫009 - 1.296. From this table can also be ascertained the number of inches in any number of feet, thus: required inches in 96 feet. As 9 would give 4 places of figures, 90 would give 5 places, and, the multiples being 10, every figure beyond those in the table will be 0; the 90 feet will therefore contain 12960 inches, and the 6 feet, 864 inches. Together 13824 inches. Subdivision (c) gives the number of yards, and number of feet in perches, up to 9; and will be readily made applicable to any multiple of the units. H H Subdivision (d). (Contents of yards in acres and decimals of an acre.) This table will be used in the same manner as (a). Example. Required contents of 185,400 yards, to 2 places of decimals. 100,000 (1 C M) = 20.66 80,000 (8 X M) = 16.53 5,000 (5 M) = 1.03 Subdivision (e) (Decimals of an acre) will give the value of 3 in the last example at 4 perches 24.2 yards. TABLE 15. (Squares with side 11 to 170.) The use of this table will not be limited to the ascertaining the square of a side having whole numbers, but to one having numbers and decimals. For example, required the contents of a square of which the side is 15.7. The contents will be the same as those of a side of 157, taking 2 as decimals, and will be 246-49. The table will also be applicable to any numbers, whether whole numbers or decimals, which may be double or treble or quadruple any number in the table, by simply multiplying by 4 for the double, 9 for the treble number, and 16 for the quadruple number. Also for half, a third, or fourth part, on dividing by 4, 9, or 16. For ten times the number x 100. The examples are confined to quantities within the table, that the accuracy of the rules may be at once seen; the rules apply, however, to any quantities. Example 1. Required contents of square with side of 40, from the tabular value of side of 20. The contents of square with side of 20 are, by table, 400, which x 4: = 1600, which the table shows to give side of 40. Example 2. Required contents with side of 60 in like manner from side of 20. Here 20 will give 400, which × 9 = 3600, which the table will show is the square of 60. It is not necessary to give examples of the division by 4 or 9, for contents of the half or third of side where whole numbers are concerned, but it may be of use to give an example where decimals are used. Example 3. Required contents of square with side of 805. This will be the half of 161, and consequently 259214, or 6480-25 will be the contents required. Example 4. Required contents of square with side of 52, or onethird of 156. Here 2433692704 will be the contents. A most important use of the table is in the calculation of the contents in gallons of circular areas, which can readily be ascertained by this and the two short tables Nos. 16 and 17. TABLE 16. (Circular areas.) Subdivision (a) gives the area from diameter, and Subdivision (b), from circumference. The table may be made applicable to areas of larger or smaller D. or C. than those stated; and also for fractions. 1. Thus, for double any number, take four times the area of that number; for treble the number, multiply by nine. 2. For half the number, divide by four; and for one-third of the number, divide by nine. 3. For four times, or for one-fourth, it would be requisite to multiply or divide by sixteen. 4. For numbers and decimals, the area can be ascertained by adding to the square of the number that of the decimal, and twice the number multiplied by the decimal. In giving examples, some will be taken in which the areas of the multiple or part are both given in the table. Example on Case 1.-Diameter 10, and area 785, &c. (see table). The area of 20 will be four times the extent, or 3141, &c. (see table, area of 20 D.). The area of D. 10 being as above, multiply by 9 for that of 30, and it will be found to equal the area of 30, as in the table. Case 2 is sufficiently explained by the examples before given, by commencing with the larger diameter; nor need any example be given of Case 3. Case 4. The most valuable modification of the table will be the making it available for the easy calculation of areas where the diameter or circumference contains a number and decimal beyond the direct scope of the table. The practical application of proposition 4 in the 2nd book of Euclid will serve the desired purpose, by adding to the square of the number that of the decimal, and twice the number multiplied by the decimal; this gives the square, which × 7854 gives the area. In practice, the square of the decimal will be unimportant, and the necessity for × 7854 will be avoided by using the next table. = 40.3. Example. Required the circular area, where the diameter Here 40 × 40 gives 1600, and 40 × 6 = 24 together 1624. : The square of the decimal is not added, being only 09, or nine hundredths of an inch. The following table gives the circular area, where the square with side equal to the diameter contains the 1624 inches, as being 1275 inches. The gallons are given in Table 19 (a) for each inch in depth. This table is applicable to the contents of superficial areas, and those of which the depth is an unit of measurement; thus, the measurement being in inches, the contents in inches will be given, and the solid contents will be ascertained by multiplying the area by the number of units, or the units and decimals of an unit in the depth. For example: the area is found to be 78.5398 inches, and the depth 10 inches, the contents in inches will be 785 398. If the contents in gallons are required, Table 19 (a) will give the contents. |