. Electric railway journal . h of the brake beams or shoescontrolled by the cylinder in question, and finally bydividing the sum of all of these last mentioned forcesby the assumed force on the push rod. Suppose now that the car, on which the brake rigging shown in Fig. 1, is installed, weighs 35,000 lb. and letit be required to find the percentage of braking poweron this car when a brake cylinder pressure of 50 square inch is developed. As shown in the preceding discussion the total brak-ing force (or braking power) on all of the specifiedwheels (, for the whole car in this case) i
. Electric railway journal . h of the brake beams or shoescontrolled by the cylinder in question, and finally bydividing the sum of all of these last mentioned forcesby the assumed force on the push rod. Suppose now that the car, on which the brake rigging shown in Fig. 1, is installed, weighs 35,000 lb. and letit be required to find the percentage of braking poweron this car when a brake cylinder pressure of 50 square inch is developed. As shown in the preceding discussion the total brak-ing force (or braking power) on all of the specifiedwheels (, for the whole car in this case) is equal to26,388 lb. when an 8-in. cylinder is used and the airpressure is 50 lb. per square inch. Consequently,Percentage of braking power = 100 x total braking force on specified wheelstotal weight on specified wheels26 388 Percentage of braking power = 100 X So, UUO — per cent (based on 50-lb. cylinder pressure)As an additional example consider the Americanequalized driver brake rigging illustrated in Fig. 4 Beam D. , „ . .- RodG laSOJLb.^ -r ^ i/i6 RodM12724Lb. Rod R 6731 Lb. ,FIG 5 FIG, 4—LEVERAGE SYSTEM FOR AMERICAN EQUALIZEDDRIVER BRAKE. PIG. 5—AMERICAN EQUALIZED BRAKEAPPLIED TO ATLANTIC TYPE LOCOMOTIVE and let it be required to find the total leverage and thepercentage of braking power when two 16-in. diametercylinders are used and the air pressure is 50 lb. persquare inch, the total weight on the drivers being107,000 lb. The total braking force on each shoe is13,404 lb. and therefore the total braking force on allsix shoes is 6 X 13,404 = 80,424 lb. Consequently, 80,424 Percentage of braking power = 100 X qqq = per cent (based on 50-lb. cylinder pressure)Now in determining the total leverage, for this case,where two brake cylinders are used, it is merely neces-sary to consider each side of the engine as a separatesystem, which it really is, and to find the total shoepressure developed by each cylinder alone. By referringto Fig. 4 it is seen that each
Size: 1496px × 1671px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1900, bookpublishernewyorkmcgrawhillp
