. Railway mechanical engineer . D = 15 ft. 0 in., £ = S ft. 6 in. and f = 4 ft. 9 in. Fromthe table the angle ^ = 8 deg., 38 min. and cosineA = Using formula (4) (100 X ) — Cot. B = —B = 12 deg., 15 + inin.; Ian. B :- 0.; Losing formula (5) II = [1100 X X ] - (15 + ) = — = ft. It should be noted that formula (5) for the distance Hdoes not consider the swing of the coupler sidewise, whichvaries to suit the detail of the striking plate. Fig. 2 is used to find the distance of the highest point ofthe wheel flange from the cent
. Railway mechanical engineer . D = 15 ft. 0 in., £ = S ft. 6 in. and f = 4 ft. 9 in. Fromthe table the angle ^ = 8 deg., 38 min. and cosineA = Using formula (4) (100 X ) — Cot. B = —B = 12 deg., 15 + inin.; Ian. B :- 0.; Losing formula (5) II = [1100 X X ] - (15 + ) = — = ft. It should be noted that formula (5) for the distance Hdoes not consider the swing of the coupler sidewise, whichvaries to suit the detail of the striking plate. Fig. 2 is used to find the distance of the highest point ofthe wheel flange from the center line of the car and also thetravel of the truck side bearing. Referring now to Fig. 2: A =: The angle between tlie center line of the car and thecenter line of the truck, also between the center lines of thebody and truck bolsters, and the axles. B = The distance from the center line of the car to thecenter line of the side bearing. C = Travel of the truck side Fig. 2 D = The distance from the center line of the truck tothe center line of the axle. E =: The distance from tlie center line of the truck to thehighest point on the wheel flange. F and H --= The distance from the highest point on thewheel flange to the center line of the car. G = The distance from the center line of the car to thecenter line of the truck at the axle. The angle A is the same as in Fig. 1 and is found by for-mula (1) or from the table. With these values the formulae for Fig. 2 are: (6) C = B X A (7) G = D X sine A(81 F = (E X COS. A)(9) H = (E X COS. A) ■ G : ■ G (E X COS. A) — D X sine A)(Ex COS. A) + (I) X sine A) The following example will illustrate the use of the for-mulae for Fig 2. Let .4 = 8 deg., 38 min.; B ^ 2 ft. 1 in.;D = 2 ft. 9 in., and £ = 2 ft. i]/^ in. From the table, sineA =; COS. .4 = Using formula (6) C = 2 ft. 1 in. X = in. Using formula (8) F = (2 f-. i\/i in. X ) — (2 ft. 9 in. X ) = i
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