. Electric traction and transmission engineering . ee curve is quite accurately ? or 5730 feet, 2 T and consequently the number of degrees of curvature, c,of a curve, specified according to con-vention by radius i?. Fig. 9, is c = ^^ degrees. Curve resistance is usually taken as from to pound per ton of train weight per degree of curvature, and Fig. 9. depends upon the speed of the train. Prof. E. C. Schmidt gives the following formula for curve resistance as a result of tests on a 28-ton car: C = cV pounds per ton, where V is the velocity in miles per hour. When a car moves arou
. Electric traction and transmission engineering . ee curve is quite accurately ? or 5730 feet, 2 T and consequently the number of degrees of curvature, c,of a curve, specified according to con-vention by radius i?. Fig. 9, is c = ^^ degrees. Curve resistance is usually taken as from to pound per ton of train weight per degree of curvature, and Fig. 9. depends upon the speed of the train. Prof. E. C. Schmidt gives the following formula for curve resistance as a result of tests on a 28-ton car: C = cV pounds per ton, where V is the velocity in miles per hour. When a car moves around a curve it experiences a cen-trifugal force which depends in magnitude upon the speedand mass of the car, and the degree of curvature. Thisforce tends to derail the car by rotating its center of massoutwardly around the outer rail. To neutralize this ten-dency the outer rail is raised above the inner rail to such anextent that the plane of the track is perpendicular to theresultant of the centrifugal and gravitational forces actingon the 22 TRACTION AND TRANSMISSION. Let m = mass of car in pounds, v = speed in feet persecond, g = acceleration of gravity in ^, and R =radius of curve in feet. Then R = horizontal centrifugal force, and mg = vertical gravitational force. An inspection of Fig. lo shows that the resultant of theseforces will be perpendicular to the plane of the track whenthat plane makes an angle dwith the horizontal such that e = tan~i Kg
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