Brake Tests

199 407. The work done during the stop by the retarding force F acting through the distance S. must be equal to the initial energy of the car, which is: 1M.va. Where v = initial speed in feet per second and MC= mass x W. of car = — g Therefore F.S.-J M.v2 (neglecting train resistance on the one hand, and rotative energy on the other). W.v F.S. - YEY . 2g If the velocity be expressed in miles per hour instead of feet second:— or feet per Then FcSc=.0334 W2V2 or FC= 0.0334 WV2 S. 408. If more than one car is used let N=number of the total average retarding force on the cars will be cars then NP 0.0334 X NWCV2 INTe ‘ - o---——- S. also for the locomotive F1 X S, = 0.0334 WV2, and R, = 0.0334 X W,V Si 409. But the total retarding force F, acting on the trains of N cars and locomotive = NFt+F, the trains of F,=NF.+F, also (NF.+F)S,= 0.0334 W.V. Substituting the values of NF, and F, just obtained (0.0334 X NWeV2 0.0334 W,V3 X SC + S, ) St= 0.0334 W.V2. Dividing this expression by 0.0334 V2, we have /NW,. WA S+S) S.= W.. •Solving for S, we have sW, S, S. S2W1+NS W. 410. It will be noted that by means of the above formula a., car and locomotive train stop can be computed for any initial sne twelve vided the following single car and Separate itial speed, pro- i L 8 separate locomotive test data known for the same speed. data are Weight of locomotive W1. Weight of single car Wc.