YAK
635
YAE
y
9
YARD, (from the Saxon , gerd), a well-known measure,
3 feet in length ; the square yard will, therefore, contain9 square feet, and the cubic yard 27 solid or cubic feet.
The square yard is used by artificers in measuring their'vork, and the cubic yard in finding the capacities of cavitiesdug in the earth. The solid yard is also used in measuringgreat masses of brick or stone work; of which a few examples,for practice, are here given.
Example 1.—Suppose a room 28 feet 6 inches long, and14 feet 9 inches wide; how many square yards are containedin the ceiling, or floor ? i
28
6
14
9
119
0
28
21
4 6
9 ) 420
4 6
46 6
Therefore, the superficial contents are 46 square yards,6 feet superficial, 4-twelfths of a foot superficial, and 6-twelfthsof the twelfth part of a foot superficial, which is the contentsof the floor, or ceiling, of the room.
Example 2.—Suppose a pit dug in the earth 58 feet long,39 feet broad, and 19 feet deep ; how many solid yards havebeen excavated ?
58
39
522
174
2262
19
20358
2262
27 ) 42978 (159127
159
135
247
243
48
27
21
So that the contents are 1591 solid yards, and 21 solid feet.
. Example 3.—Suppose a mound of earth, whose section18 everywhere a given segment of a circle, the breadth of
the mound 27 feet, the height 9 feet, and the length3837 feet ; what will be the expense of raising it, at 15d.per yard ?
The easiest method of finding the area of the segment isthe following, taken from the article Mensuration, viz,Multiply the chord and versed sine together, to two-thirds otthe product add the cube of the altitude, divided by twice thechord, and the sum will be the area of the section, which,being multiplied by the length of the mound, will give thesolidity in feet : see the operation.
27
9
3 ) 243
81
2
162
Again,
9
9
81
9
54 ) 729 (1354
189
162
27
So that 162 + 13 = 175 feet, the area of the segmen
Then 3837175
19185
26859
3837
27 ) 671475 ( 2486954
131
108
234
216
187
162
255
243
12