. Railroad construction. Theory and practice . ed. Such a spiralwould require a length of 375 feetto run on to an 8° curve, which isinconveniently long, but it mightbe used to run on to a 4° curve,where its length would be only 175feet. Three spirals have been de-veloped in Table IV, each withchords of 25 feet, the rate of in-crease in the degree of curvaturebeing 0° 30, 1° and 2° per of these will be suitable forany curvature found on ordinarysteam-railroads. 47. To find the ordinates of ai°-per-25-feet spiral. Since thefirst chord subtends a 1° curve, its central angle is 0° 15 and
. Railroad construction. Theory and practice . ed. Such a spiralwould require a length of 375 feetto run on to an 8° curve, which isinconveniently long, but it mightbe used to run on to a 4° curve,where its length would be only 175feet. Three spirals have been de-veloped in Table IV, each withchords of 25 feet, the rate of in-crease in the degree of curvaturebeing 0° 30, 1° and 2° per of these will be suitable forany curvature found on ordinarysteam-railroads. 47. To find the ordinates of ai°-per-25-feet spiral. Since thefirst chord subtends a 1° curve, its central angle is 0° 15 and theangle aQV (Fig. 31) is 7 30. The tangent at a makes anangle of 15 with VQ. The angle between the chord ba and thetangent at a is i(30)=15, and the angle 6a?) = J(30) +15= 30. Similarly the angle chc = 4(45) + 30 +15 = 67 30 = 1° 07 30,and the angle dcd= 2° 0. The ordinate aa=25 sin 7 30, andQa=25 cos + ab = 25 (cos 730 +cos 30).66=66 + 66 = 25 (sin 730 +sin 30) „Similarly the ordinates of c, d, etc., may be Fig. 31. 50 RAILROAD CONSTRUCTION. §48. 48. To find the deflections from any point of the = 7S0\ Tan hQV=h¥---Qh\ tan cQV=; we are enabled to find the deflection angles from the tan-gent at Q to any point of the spiral. The tangent to the curve at c (Fig. 32) makes an angle of
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