minutes, if any, in I to decimals of a degree; then divide by the degree of curvature. Ecample 1. 1 20° 40), 1) = 3o. 20° 40' is equivalent to 20.67 degrees. Dividing by 3, we have 6.89 chord lengths for the length of the If the chords, as is usual, are each 100 feet long, the length of the curve in this case will be 689 feet. If the chord lengths were 50 feet each, the length of the curve would be half this number of feet. 14. If the degree of curvature is fractional, the more convenient method of effecting the division is, first, to reduce both I and D to minutes; then divide the former by the latter. curve. Example 2. I - 300 22', D = 2° 45'. These are equivalent, respectively, 10 1,822 and 165 minutes. Dividing the former by the latter, we have 1,104 feet for the length of the curve. 15. The ingenious assistant who will attentively consider the preceding figures cannot fail to detect other obvious analogies which it has not been thought necessary to include in this compendium. 16. In railroad field practice it is usually sufficient to determine angles to the nearest minute, and distances to the nearest foot. The nicety of seconds and tenths appears generally to be quite superfluous; the time consumed on them were better employed in;"pushing ahead. H Н If the chords are 100 feet long, as is usual in railroad prác 5,730, the radius of a 1o curve, by the deflection angle, or degree of curvature. Thus, in the foregoing example, 5,730 • 4 1,432.5. 18. GIVEN ANY RADIUS R, AND CHORD C, TO FIND THE DE FLECTION ANGLE D. From the preceding equation and example: Sin ; D=1C; R= 50 - 1,432.7=.0349 = sin 2° = ID ..D=49. 19. GIVEN ANY RADIUS R, AND CHORD C, TO FIND THE DE FLECTION DISTANCE d. First find the deflection angle by above method (18). Then, angle H AB in the figure being made equal to D, and H A. BA, BH will be the deflection distance. Draw, A K to the middle point of H B. Then d=HB=2 KB=2 A B X sin KAB= 2 x sin ID. Example. Let R=1,146 feet, C= 100 feet. By (18) D will be found 5o. Then d 2 C X sin #D= 200 x .0436 = 8.72 feet. 20. If the chords are 100 feet long, as is usual in field measurement, divide the constant number 10,000 by the radius in feet: the quotient will be the deflection distance. The deflection distance with radius of 10,000 feet and chord of 100 feet is one foot: this rule is based upon the principle that deflection distances, the chord length being fixed, will vary inversely as: the radii. Thus, in the foregoing example, 10,000+ 1,146 = 8.72. 21. GIVEN ANY RADIUS R, AND CHORD C, TO FIND THE TAN GENTIAL ANGLE T. The angle T is equal to 1 D by construction; for mode of 22. GIVEN ANY RADIUS R, AND CHORD C, TO FIND THE TAN. GENTIAL DISTANCE 1. First find the tangential angle, as above directed. Then, angle B A E in the figure being made equal to T, and AE : A B, BE will be the tangential distance. Draw AN to the middle point of B E. Then t =EB=2 NB= 2 A B X sin NAB = 2C x zin 1 T. E.crumple. Let R=1,146 feet, C = = 100 feet. By sect. 1, T will be found = 2° 30'. Thent 2 C X sin 1 T 200 x .0218 = 4.36 feet. 23. In ordinary railroad practice the tangential distance may be considered equal to half the deflection distance. XIX. ORDINATES. 1. GIVEN ANY RADIUS R, AND CHORD C, TO FIND THE MID DLE ORDINATE M. In the annexed figure, HN=M, HG=R, AB=C. NG=NA G2 – A N= VR– +C2; HN=HG-NG, Example. M=819 — 1670761 — 2500= 1.53. 2. Angle HAN=4 HGB; HGB= } AGB, .. HAN = * AGB HN AN X tan. HAN; i.e., M C X tan. į D; D being the central angle subtended by the chord. Example. D=70, С M= X tan. & D= 50 X 0.03055 1.528. 3. GIVEN THE RADIUS R, CHORD C, AND MIDDLE ORDINATE M, TO FIND ANY OTHER ORDINATE E K M', DISTANT d E L=NG E – K=R? -đỏ; N G (1) = R3 - C. Then E K=M'= VR—-VR— À C2. 4. It is a property of the parabola, that ordinates vary as the products of their abscissas. This property may be assigned to the circle in cases where the arc encloses a small angle. Applying it here we have — HN: EK :: ANX NB: AK X K B. Call any segments A K, K B, of the chord, a and b. Then M:M:: C? : ab, .: M=M X 4 ab • 02. Example. M' =1.528 X 9600 • 10000 = 1.528 x 0.96 = 1.467. 5. Multiply the corresponding ordinate of a 1o curve from the annexed table by the degree of curvature: the product will ORDINATES OF A 1o CURVE, CHORD 100 FEET. DISTANCES OF THE ORDINATES FROM THE END OF THE 100-FEET CHORD. Example. What is the ordinate of a 6o curve, 30 feet from the end of the 100-feet chord? The corresponding tabular ordinate of a 1o curve is .183; which, multiplied by 1.098, the required ordinate. 6. A quick way of laying off ordinates on the ground, and one sufficiently exact for the field, is, after fixing the point H by means of the middle ordinate H N, to stretch a line from I to A, and make the middle ordinate F0=$HN; then from F to A and F to H, making the middle ordinates = {F0; and SO on. 7. A good track-layer will seldom require points at shorter intervals than 25 feet. |