C—3a

32

Area _ 2fi Log. 25 = 1-3979400 Perimeter ~** „ 20 = 1-3010300 2)0-0969100 = 1-25, mean depth. 0-0484550 = 1-118 y mean depth. Area, 25 ft. Log. 1-3979400 V mean depth =1-118 „ 0-0484550 1-4463950 = 27-951 square feet. Area 43-528 Log. 1-6387640 Perimeter 26-3904 „ 1-4214394 2)0-2173246 = 16494 mean depth. 0-1086623 = 1-2843 J mean depth. Area 43-528 Log. 1-6387640 V mean depth 1-2843 „ 0-1086623 1-7474263 = 55-902 square feet. Now, as 27-951 : 55-902 : : 10,000 : 20,000. So that both the above rules show that if an airway 5 ft. x 5 ft. = 25 square feet sectional area, and 6,000 ft. long circulates 10,000 cubic feet of air,' an airway 6-5976 ft. x 6-5976 ft. = 43-528 square feet sectional area, and 6,000 ft. long will circulate double the quantity or 20,000 cubic feet, always provided that the circulating-pressure is the same in both cases. The formula 113 A yRS will also show that the larger of the two airways referred to will carry twice the quantity of the others. Airway, 5 ft. x 5 ft. =25 square feet, 6,000 ft. long, with 2 in. on water-gauge (or 10-4 lb. per square foot). Formula No. 2. Area, 5 ft. x 5 ft. = 25 square feet = A. Area _25 __ i-25 =B. * Perimeter = a/1-26 X 0-0017333 = 0-046547 7BS = 0-046547 Log. = 2-6678960 113 „ 2-0530784 Area, 25 „ 1-3979400 Seconds in a minute, 60 „ 1-7781513 3-8970657 = 7,889-9 cubic feet. Say, 7,890 cubic feet per minute. Formula No. 2. Airway, 6-5976 ft. x 65976 ft. = 43-528 square feet, 6,000 ft. long, with 2 in. on water-gauge (or 10-4lb. per square foot). Area, 6-5976 x 6-5976 = 43528 square feet = A. Area 43-528 _ Perimeter 263904 04y4 =W* = 0-0017333 „ =S. Length 6,000 _ = v'l-6494 x 0-001733 = 0-053469 v/RS = 0-053469 Log. 2-7281033 113 „ 2-0530784 Area, 43528 „ 1-6387640 Seconds in minute, 60 „ 1-7781513 4-1980970 = 15,780 cubic feet. 15,780 cubic feet per minute. , The carrying-capacity of the larger airway is twice the carrying-capacity of the other. At page 142 of the " Miners' Guide " a rule is given to find the size of a rectangular airway equal to square airways (that is, equal in carrying-capacity). The question worked out is, What would be the width of a rectangular airway 7 ft. in height equal in carrying-capacity to a square