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Q2 Z2,(Q2) Z2(Q2) - Institute for Water Resources - U.S. Army

Q2 Z2,(Q2) Z2(Q2) - Institute for Water Resources - U.S. Army

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and (5.26), and 2-miles-per-hour speed intervals, Hay calcul'ates it<br />

would require 49,295 feet and 694.29 seconds to achieve the cruis-<br />

speod of 60 miles per hour. While these appear to be large -<br />

figures, it is interesting to calculate the error resulting from<br />

ignoring acceleration. Suppose it is determined the train operates<br />

at a cruising speed of 60 miles per hour. At this speed the time<br />

required to cover 49,295 feet is 560 seconds. There<strong>for</strong>e, the error<br />

in ignoring acceleration time is an underestimate by the amount 134<br />

seconds or slightly over 2 minutes. Distance traveled would be<br />

overestimated by 10,777 feet or less than 2 miles. It would seem<br />

that unless one were concerned with short trips and/or those in-<br />

volving frequent stops, acceleration time and distance can be ig-<br />

nored with only small resulting error.<br />

Deceleration. There exist in the railroad engineering litera-<br />

ture <strong>for</strong>mulas <strong>for</strong> obtaining the time required and the distance trav-<br />

eled in the deceleration of trains. For example, Hay derives quite<br />

complicated deceleration relationships by equating the kinetic en-<br />

ergy of the train with the work done in stopping the vehicle. 11 In<br />

this derivation, consideration is made of such things as the time<br />

required <strong>for</strong> brakes to act, weight of train, brake cylinder pressure,<br />

etc.<br />

An alternative, and much simpler, procedure <strong>for</strong> obtaining esti-<br />

mates of deceleration time and distance was outlined in Chapter III.<br />

Assume a constant rate of deceleration d; e.g., 0.75 miles per hour<br />

per second, suggested in Car and Locomotive Cyclopedia. Then the

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