\fdO'^ - Old Forge Coal Mines

160 GEOMETRY AND TRIGONOMETRY.
(341) (a) Area of piston = 19' X .7854 = 283.529 sq.
in., or 1.9689 square feet.
Length of stroke plus the clearance = 1.14 X 2 ft. (24
in. =2 ft.) = 2.28 ft.
1.9689 X 2.28 = 4.489 cubic feet, or the volume of steam
in the small cylinder. Ans.
(d) Area of piston = 31* X .7854 = 754.7694 sq. in., or
5. 2414 square feet.
Length of stroke plus the clearance = 1.08 X 2 = 2.16 ft.
5.2414 X 2.16 = 11.321 cubic feet, or the volume of steam
in the large cylinder. Ans.
(c)
1 1 QOI
Ratio = , T, ,
4. 4o9
or 2. 522 : 1. Ans.
• (342) {a) Area of cross-section of pipe = 8* X . 7854 =
50.2656 sq. in.
• ir 1 * 50.2656 X 7 _ .
,,_ ^, Volume of pipe = — = 2.443 cu. ft. Ans.
144
(d) Ratio of volume of pipe to volume of small cylinder
= 47489' '''^•^^•^- ^''^-
1 R
(343) (a) In Fig. 15, given 0B= -^ or 8 inches, and
13 1 .
O A = — or 6— inches, to find the volume, area and weight :
Radius of center circle equals — ^^^-^ —
jj or 7 J inches. Length
of center line =
2 X 3.1416 X 7^ = 45.5532 inches.
4
Fig 15. '^^^ radius of the inner circle is 6 —
inches, and of the outer circle 8 inches ; therefore, the dia
meter of the cross-section on the line A B is 1 — inches.
Then, the area of the ring is 1 ^ X 3.1416 X 45.553 = 214.665
square inches. Ans.
*