Fhe D. Van No^and Company Lend this book to be sold to the Pubhc rt the advertised price, and supply to t Tde on terms wHch will not allow of reduction. SOLUTION OF RAILROAD PROBLEMS BY THE SLIDE RULE BY E. R. CARY, C.E. Professor of Railroad Engineering and Geodesy, Rens- selaer Polytechnic Institute, Troy, N.Y. Member American Society Civil Engineers NEW YORK D. VAN NOSTRAND COMPANY 25 PARK PLACE 1913 COPYRIGHT, 1913, BY D. VAN NOSTRAND COMPANY Stanbope iprcss F. H.GILSON COMPANY BOSTON, U.S.A. PREFACE THE ease and rapidity of solving problems in Railroad Curves by the use of the slide rule led the author to develop a set of problems for class-room work. The object of this book is to present similar problems for the convenience of students who have studied Railroad Curves and the Theory of the Slide Rule. Where it is possible the solution of the problems should be made by the use of the C and D scales of the Mannheim slide rule on account of their greater precision. The notation used in Allen's " Railroad Curves and Earthwork" is used thruout this discussion. The discussion of the slide rule is from Professor C. W. Crockett's article in the Polytechnic of May 2, 1910. The discussion of the slide rule, the develop- ment of the equations used and the discussion of the easement curve have been added to make this book of more general interest. E. R. GARY. TROY, N. Y. March 1, 1913. iii 288606 TABLE OF CONTENTS CHAPTER I THE SLIDE RULE AET. PAGE 1. The Construction and Setting of the Slide Rule 1 2. The Mannheim Slide Rule 4 CHAPTER II SIMPLE CURVES 3. The Center Line 5 4. The Degree of a Curve 5 5. The Decimal Point 7 6. The Functions of a Curve 7 7. The Middle Ordinate 10 8. The Tangent Offset 12 9. A Curve laid out by a Tape 13 10. Offsets from a Chord 15 11. Deflection Angles 16 12. Offsets from a Tangent 19 13. Special Problems 21 14. Other Special Problems 28 CHAPTER III COMPOUND CURVES 16. Compound and Reversed Curves 31 17. Method of Solution by Coordinates 33 v VI TABLE OF CONTENTS CHAPTER IV VERTICAL CURVES ART. PAGE 18. The Vertical Curve ... 48 CHAPTER V TURNOUTS 19. The Split Switch Turnout 51 20. The Turnout from a Curve 54 21. The Stub Switch Turnout 57 22. The Approximate Method for a Turnout from a Curve 58 23. The Connecting Curve 62 24. A Connection between Parallel Tracks 63 25. A Connection between Concentric Curves ... 64 26. The Y " Curve 70 CHAPTER VI THE EASEMENT CURVE 27. The Easement Curve 75 28. The True Spiral 78 29. The Spiral Angle 79 30. The Deflection Angles for a Spiral 80 31. The Offset from the Tangent 81 32. The Deflections from the Tangent thru the P.S 82 33. The Value of p 83 34. The Value of i 83 35. The Value of q 85 36. The Value of T a 87 37. To find the Tangent at any Point 87 TABLE OF CONTENTS Vll 38. The Deflections from any Point on the Spiral 88 39. The Methods of Laying out, Spirals 89 40. The Offset Method 91 41. The Spiral between the Branches of a Com- pound Curve 93 42. The Spiral Tables 94 44. Spirals not given in the Tables 95 46. The Solution of a Spiral by the Slide Rule. . 104 47. The Equation for Deflections, using the Slide Rule 112 48. The Diagram for e 113 49. The Diagram for l c 115 CHAPTER VII EARTHWORK 50. Cross-sections 117 51. Form of Notes for Cross-sections 118 52. Change from Cut to Fill 119 53. A L^vel Section 120 54. A Three-level Section 121 55. The Methods of Computation 123 56. The Prismoidal Correction 123 57. The Formulas for Slide Rule Computations. 126 PROBLEMS PBOB. 1. To find the Radius 6 2. To find the Tangent Distance 8 3. To find the External Distance 9 4. To find the Middle Ordinate 10 5. To find the Long Chord 10 Vlll TABLE OF CONTENTS 6. To find the Middle Ordinate 12 7. To find the Tangent Offset 12 8. To find the Offset from any Point on a Chord 16 9. To find the Deflections 17 10. To locate a Curve by Offsets from the Tan- gent 19 11. To change the Ending Tangent 24 12. To change the Ending Tangent 25 13. To change the Ending Tangent 26 14. Special Problem for a Simple Curve 28 15. Tangent Distances for a Compound Curve. . 32 16. Compound Curve 33 17. Compound Curve 36 18. Compound Curve 39 19. Changing a Simple to a Compound Curve. . . 41 20. Three-centered Compound Curve 43 21. A Reversed Curve 45 22. A Reversed Curve . . . 46 23. A Vertical Curve 48 24. A Split Switch Turnout 53 25. A Turnout from a Curve 60 26. A Connecting Curve 62 27. A Connection between Parallel Tracks 64 28. A Turnout and Connecting Curve for Con- centric Curved Tracks 66 29. Similar to Prob. 28 68 30. A "Y" Curve 70 31. A "Y" Curve 72 32. Deflections for a Spiral, using the Tables 97 33. Deflections for a Spiral, using the Slide Rule. 104 34. Earthwork, using the Slide Rule 128 35. Earthwork, using the Slide Rule 128 TABLE OF CONTENTS IX PROS. PAGE 36. Earthwork, using the Slide Rule. . 129 37. Earthwork, using the Slide Rule 130 38. Earthwork, using the Slide Rule 130 39. Earthwork, using the Slide Rule 131 DIAGRAMS For e 114 For l c 116 TABLES For Deflection Angles for Spirals .......... 100-103 Constants 132 Formulas for Triangles 132 Trigonometric Formulas " 133 Trigonometric Series 134 Simple Curve Formulas 134 Vertical Curve Formulas 135 Turnout Formulas 135 Spiral Formulas 136 Earthwork Formulas 136 THE SOLUTION OF RAILROAD PROBLEMS BY THE SLIDE RULE THIS discussion will be given under the following heads: The Slide Rule, Problems in Simple Curves, Problems in Compound Curves, The Vertical Curve, Turnouts, The Easement Curve, and Problems in Earthwork. CHAPTER I THE SLIDE RULE i. The Mannheim and Carpenter rules are the most common types of the slide rule. The face of either bears two kinds of scales, on one of which, for the ordinary ten-inch rule, the distance from 1 to 1 is about 5 inches, while on the other the distance from 1 to 1 is about 10 inches, the latter distance being exactly double the former. Let us call the shorter scale the single scale and the longer scale the double scale; on the 1 % THE- SLIDE RULE Mannheim rule, the single scale is called A or B and the double scale is C or Z>; on the Car- penter rule, the single scale is called A or B, and the double scale is called C. Note that the double scale is the long scale. Let a and 6 be constant numbers and x a third number to which different values are to be assigned, and suppose we wish to find the values of y in any one of the following expressions: ax ax 2 A fax t /ax 2 y = j>y = -v> y= -\T' y== \-v' ab o6 2 . fab . /ah* y =lt> y= -^' y= -\^' y = \l?' or of any similar expression containing two factors in the numerator and one in the de- nominator, any one or more being squared, a being a constant factor in the numerator. The method of setting the instrument is as follows: 1. If is in the numerator, use the slide direct. If x is in the denominator, use the slide inverted. 2. The numbers a, &, x and y are found on the instrument in the order named; a on THE SLIDE RULE O the rule, 6 on the slide, x on the slide, y on the rule; that is, we look on the rule and slide in the following order: Rule, slide, slide, rule. Or we could find a on the slide, b on the rule, x on the rule, y on the slide; that is, we look on the rule and slide in this order, Slide, rule, rule, slide. 3. If a, b or x is raised to the first power, we must find it on a single scale; if squared, on a double scale. 4. If y is a first power, we must find it on a single scale; if a square root, on a double scale. ax 2 Thus to find the value of y = -p- with the Mannheim rule, we may proceed in two ways, using the slide direct, since x is in the numerator: (a) Find a on the single scale on the rule, and opposite it set b on the double scale on the slide; find x on the double scale on the slide, and opposite it read the result y on the single scale on the rule. Or: (b) Find a on the single scale on the slide, and opposite it set b, found on the double scale on the rule; find x on the double scale on the rule, and opposite it read the result y on the single scale on the slide. 4 THE SLIDE RULE The same problem may be solved with the Carpenter or Thacher rule, but only by the second method, since the slide does not bear a double scale; this limitation is not objection- able, however, except in continued operations, where it is desirable to use the runner. /ax 2 Had we wished to find the value of y = y -j^> the settings on the Mannheim rule would have been the same as before, but the result y would have been on the double scale instead of on the single scale. The Carpenter or Thacher rule cannot solve this problem without first reading ax 2 on the slide the value of -r^-, and then deter- mining its square root. 2. The Mannheim slide rule is used in the solution of the problems given herein. The scales of this rule are as follows: The upper scale of the rule, A; the upper scale of the slide, E\ the lower scale of the slide, C; the lower scale of the rule, D. On the back of the slide are the sine and tangent scales, and when these are to be used the slide must be reversed, i.e., the slide changed in the rule so that the back face is brought to the front. CHAPTER II SIMPLE CURVES 3. The center line of a railroad track con- sists of tangents and curves (circular arcs), and, in modern practice, also of some form of easement curve connecting them. 4. In this country a curve is designated by its degree. The degree of a curve is the cen- tral angle subtended by a chord of one, hundred feet. If the metric system is used, the degree of a curve may be defined as the central angle sub- tended by a chord of ten meters. In almost all other countries a curve is designated by its radius. From Fig. 1, R sin \ D = 50; 50 R = sin J D 5 6 THE SLIDE RULE As D is usually a small angle, 50 R = iZ>sml' looJL, -4-, = 3438; sinl' 100 X 3438 = -p- - (1) In Equation (1), D is expressed in minutes and R in feet. For metric curves R in meters _ 10 X 3438 D' Problem 1. Find the radius of a 6 30' curve. To solve by the slide rule, opposite 3438 on the D scale set D in minutes on the C scale, opposite the index of the C scale read the result on the D scale for R in feet. By inverting the slide and setting 3438 on the C scale opposite the index of the D scale, the radius of any degree of curvature may be found on the D scale opposite the degree of the curve in minutes on the inverted C scale. For the above problem the setting is as follows: Opposite 3438 on the D scale set 390 on the SIMPLE CURVES 7 C scale and opposite the right index of the C scale read 882 on the D scale, or R = 882 feet. Except for odd minute curves the approximate , , D 5730 . , ,. , formula, R = ~- , gives an easy solution by the slide rule. In this formula D is in degrees. If this formula is figured exactly and T o of D is added to the value found for R, the result is practically a precise value of R. 5. To find the decimal point, always figure the result roughly by mental arithmetic, e.g., in the above problem use 3900 in place of 3438 and the result is 1000, showing that the correct result is about 1000. Hence, when the figures 882 are obtained by the slide rule, the decimal point is placed after the third figure for the value of R. 6. The functions of a curve are: T 7 , the tangent distance; E, the external distance; M, the middle ordinate; (7, the long chord; P.O., the beginning of the curve; P.T., the ending of the curve; and / the central angle of the curve. In Fig. 2, T = AV = VB. E = VF. M = FG. C = AB. I = ACB = BVX. 8 THE SLIDE RULE From Fig. 2 the following formulas are readily derived: r = #tani7 ........ (2) E = R (sec I I - 1) = R exsec i 7. (3) exsec J 7 = tan J 7 tan } 7, or # = R tan 7 tan i 7 . . . . (4) Af = JB-72.cosf J = Bversi/ . (5) vers i 7 = 2 sin 2 i 7 = sin ^ 7 tan J 7, or Af = jRsini7tani7 ..... (6) C = 2 E sin i 7 ....... (7) Fig. 2 Problem 2. Find the tangent distance for a 6 12' curve with a central angle of 30 24'. T = R tan i 7 = E tan 15 12'. fi, found as in Problem 1, = 925 feet. T = 925 tan 15 12' = 251.5 feet. SIMPLE CURVES 9 The setting for T is as follows: Using the slide reversed, opposite 925 on the D scale set the right index of the T scale and opposite 15 12' on the T scale read 251.5 on the D scale. In figuring roughly for the decimal point use the sin or tan of 1 = 0.0175 and assume that the tangents and sines vary as the angles, i.e., the sin of 15 is 15 times the sin of 1. This is only approximately true for angles less than 30. Problem 3. Find the external distance for a 6 12' curve with a central angle of 30 24'. E = R tan 7, tan J / = R tan 15 12', tan 7 36'. R, the same as in Problem 2, is 925 feet. E = 925 tan 15 12', tan 7 36' = 33.5 feet. The setting for E is as follows: Using the slide reversed, set the right index of the T scale opposite 925 on the D scale, bring the runner to 15 12' on the T scale, then bring the left index of the T scale under the runner, and opposite 7 36' on the T scale read the result 33.5 on the D scale. The position of the decimal point is found as indicated in Prob- lem 2. 10 THE SLIDE RULE Problem 4. Find the middle ordinate for a 6 12' curve with a central angle of 30 24'. M = R sin ^ 7 tan J /. j?, the same as in Problem 2, is 925 feet. M = 925 sin 15 12' tan 7 36' = 32.4 feet. The setting for M is as follows: Using the slide reversed, opposite 925 on the D scale, set the right index of the T scale and opposite 7 36' on the T scale read 1235 on the D scale. Opposite 1235 on the A scal6 set the right index of the S scale and opposite 15 12' on the S scale read 32.4 on the A scale. The position of the decimal point is found as indicated in Problem 2. Problem 5. Find the length of the long chord of a 6 12' curve with a central angle of 30 24'. jffi, found as above, = 925 feet. C = 2 R sin^ = 2 R sin 15 12' = 2 X 242 = 484 feet. Setting for 925 sin 15 12' is similar to that given in Problem 4. 7. In bending rails it is convenient to use the middle ordinate for a given length of chord to determine the degree of the curve to which the rail is bent. The middle ordinate SIMPLE CURVES 11 may be expressed in terms of the length of the chord and the radius or in terms of the length of the chord and the degree of the curve. These formulas are derived as follows: In Fig. 3, AB is the chord, c, and EF is its middle ordinate. G is at the middle of AE. AE = AFj approximately. O~E> (8) Then EF=AG \c = R M ic or R = or 5730 D ' 8 X 5730 45,840 12 THE SLIDE RULE Problem 6. Find the middle ordinate of a chord 84 feet long on a 6 12' curve. _ c*D _ 6.2 ~ 45,840 ~ * 45,840 = The setting for M is as follows: Opposite 84 on the D scale set 4584 on the right B scale, E opposite 62 on the left B scale read 953 on the A scale. Figure roughly for the decimal point. 8. In Fig. 4, EB is the tangent offset, a, and its value in terms of chord, c, c and the radius is found as Fi s- 4 follows: Triangles AEB and ACF are similar, hence EB _ AF AB~AC* but AC = R, AF = J c. Then - = ^ or a = ~ . . , f . (9) Problem 7. Find the tangent offset for a 6 12' curve for a chord of 110 feet. c 2 SIMPLE CURVES 13 a = R, found as above, = 925 feet. HO 2 0.5 X HO 2 = 6.55 feet. 2 X 925 925 Setting for A is: Opposite 5 on the A scale set 925 on the B scale and opposite 11 on the C scale read 655 on the A scale. 9. The tangent offset may be used to lay out a curve by the use of a tape only. In Fig. 5, if A is at station 16 + 40, then 60 2 D A 17 = 60 feet, and E 17 = = 11,460* All - Ell 2A17 From this AE may be found. Fig. 5 It is assumed that the direction of the line TA is established on the ground and that A 14 THE SLIDE RULE is located. Then AE may be measured out and E located. Then measuring off A 17 from A and E 17 from E } their intersection may be found and 17 located as a point on the curve. Point 16' may be located in a manner similar to locating 17. Both 16' and 17 are on the curve. By prolonging 16', 17, 100 feet from 17 to (?, measuring G 18 = 2 aioo (where aioo is the tangent offset for a chord of 100 feet), and measuring 100 feet from 17, the intersection of these measurements locates station 18. In a similar manner station 19 may be located from the prolongation of 17, 18. 18 H is aioo. 18,19 - 19 # = 200 From this 19 H may be found. Then by meas- uring 18 H from 18 and 19 ff from 19, H may be located and the line 19 H is tangent to the curve at station 19. Then station 19 + 30 may be located from this tangent in a manner similar to locating station 17 from the tangent thru A. 30 2 19 K = 77^5 . 30 minus the distance from Zn, 19 + 30 to K equals . From these dis- SIMPLE CURVES 15 tances K may be located, and K joined with station 19 + 30 gives the direction of the tangent ahead of the curve. 10. A curve may be located by offsets from a long chord. The method of finding the off- sets is as follows: F K Fig. 6 In Fig. 6, JK = LK - LJ. LK = EF. Let AB = c, and W = 6. Then it = It P tan J / tan / = (-#') tan 7. From Equation (11), (R -fi')tan7 = : tan^7 Then A A' = ^ y - tanf / Add to the station of the old P.O. AA' for the station of the new P.O. From R f the new degree of the curve can be found. From the new degree of the curve and the central angle the length of the new curve can be found, and from the new station of the P.O. and the length of the new curve the new station of the P.T. can be found. These equations are better adapted to the slide-rule solutions than the ordinary ones for these problems. If the new ending tangent is nearer the center, a change in the signs will result in the above equations. Problem 11. Find the stations of the new P.O. and P.T. for ending a 3 30' curve in a SIMPLE CURVES 25 parallel tangent 13 feet further from the center, keeping the degree of the curve the same. The central angle is 22 24' and old P.O. is at station 126 + 40. 13 ~ oro CkA sin 22 24 Setting: Opposite 13 on the A scale set 22 24' on the S scale of the slide reversed and opposite the right index of the S scale read 34.1 on the A scale. 22 24' 1344 Old P.O. at station 126 + 40 34.1 New P.O. at station 126 + 74.1 L = 6 + 40 New P.T. at station 133 + 14.1 Problem 12. Using the same curve as in Problem 11, it is desired to move ending tan- gent 13 feet nearer the center, keeping the same P.O. R' =fl- P I . T tan ~ sin / 26 THE SLIDE RULE p 100 X 3438 . R = ^ph = 1"35 feet. ZiL\J P = 13 tan i / sin / tan 11 12' sin 22 24' #' = 1635 - 172.1 = 1462.9 feet. 13 Setting for tan 11 12' sin 22 24' is: PP - site 13 on the A scale set 22 24' on the S scale of the slide reversed and opposite the right index of the S scale read 341 on the A scale; opposite 341 on the D scale set 11 12' on the T scale and opposite the left index of the T scale read 172.1 on the D scale. 100 X 3438 _ 1463 22 24' _ 1344' _ = WW~~" 234 r = P.O. at 126 + 40 P.T. at 132 + 15 Problem 13. Using the same curve as in No. 11 and ending new curve directly opposite old P.T. in a new parallel tangent 13 feet further from the center, find new degree of curve and new stations of the P.O. and the P.T. SIMPLE CURVES 27 R> = R -- - _ = 1635 tan J / tan / 10 _ i KQ ___ _ _ tan 11 12' tan 22 24' R f = 1476 feet. Setting for tanllol2 ^ an22024/ is: Oppo- site 13 on the D scale set 11 12' on the T scale of the slide reversed, bring the runner to the right index of the T scale, bring 22 24' on the T scale under the runner and opposite the left index of the T scale read 159 on the D scale. zy _ 100X3438 1476 ., 22 24' 1344 L = - AA' = (B- = 159 tan 11 12' + sin 22 24' = 31.5 + 34.1 = 65.6 feet. Old P.O. at 126 + 40 65.6 New P.O. at 127+ 5.6 L' = 5 + 78 New P.T. at 132 + 83.6 28 THE SLIDE RULE 14. Other special problems in simple curves may be solved with fair precision by the use of the slide rule. To get good results it is necessary that the angles whose sines are used shall be between and 70, and those whose cosines are used shall be between 20 and 90. The following is a sample of such special problems. Problem 14. Find the degree of the simple curve that will pass thru a given point and will join two tangents having a given inter- section angle. The values are given in Fig. 11. A 16*30' V In Fig. 11, Fig. 11 = 90 - 34 45' = 55 15 OV= SIMPLE CUEVES 29 R. R cos 18 15' In the triangle 07P, R sin 55 15' R sin OPV cos 18 15' . v sin 55 15' _ sin 55 15' cos 18 15' sin 72 45'' Solving this by the slide rule, sin OPV = sin 60. The angle OPV is greater than 90, as seen by the construction of Fig. 11. Hence OPV = 180 - 60 = 120. The setting of the slide rule for finding the sin OPV is as follows: Reverse the slide in the rule. Make the indices of the A scale and the sine scale to coincide, move the runner over 55 15' on the sine scale, bring 71 45' on the sine scale under the runner, move the runner to the right index of the sine scale and then set the indices of the sine scale to agree with the indices of the A scale, reading the result 60 under the runner. 30 THE SLIDE RULE POV = 180 - (120 + 55 15') = 4 45'. sin 55 15' A0 . t sm 4 45 This solved by the slide rule gives R = 1190 feet. D, in minutes, = - = 289' = 4 49'. ri It is probable that a 5 curve would be used under the conditions given. 15. In the remaining problems the settings of the slide rule will not be given, except in a few cases, as those already given show the methods that usually may be applied. CHAPTER III COMPOUND CURVES 1 6. It is sometimes necessary, for the best location, to combine curves of different radii. When these curves are on the same side of their common tangent, they form a compound A F V Fig. 12 curve and when on opposite sides of their common tangent, they form a reversed curve. A reversed curve may have branches of equal or unequal radii. In a compound curve the radii of the branches are always unequal. 31 32 THE SLIDE RULE Fig. 12 shows a compound curve, the tangent distances, TI and T 8 , for which may be found as follows: Ti = AV. T s = VB. Ri = AO. R s = CB. FE = FN + NE = Ri tan J Ii + R s tan J I 8 . FV and VE may be found by solving the triangle FVE. Then T l = Ei tan 7, + **7. T 8 = 8 tan J 7 a + 7#. Problem 15. Find the tangent distances for the compound curve consisting of a 3 curve with a central angle of 14 20' beginning at A and compounding with a 6 curve with a central angle of 40 40' ending at B. p 100 X 3438 f = = 191 feet - 100 X 3438 n _ f =955feet ' -860 = 1910 tan 7 10' + 955 tan 20 20' = 240 + 354 = 594. F V = o sin 40 40' = 473. sin 55 COMPOUND CURVES 33 VE = . * sin 14 20' = 179.5. sin 55 r, = 240 + 473 = 713 feet. T 8 = 354 + 179.5 = 533.5 feet. 17. Nearly all problems in compound curves may be solved by the method of coordinates, the application of which is given in the follow- ing three problems. Problem 16. The deflection angle of the tangents of a compound curve is 52. The curve begins at station 372 and the first branch is a 6 curve to the left, 5 stations long. The T 8 is to be 550 feet. Find the radius and the degree of the second branch and the stations of the P.C.C. and the P.T. In Fig. 12, the P.O. of the curve, to fulfil the condition given in the problem, is at B. I 8 = 6 X 5 = 30. Ii = I - I 3 = 52 - 30 = 22. = 955'. The lengths and directions of the lines NC, CB and BV are known, and, using NC as a meridian, their Z/'s and M's are found as follows: 34 THE SLIDE RULE Azimuth L M + - f - NC 000' Feet. 955 955 CB... 210 00' 955 827 477 BV 120 00' 550 275 476 VN 955 147 1102 476 1 477 F V Fig. 12 k cos 210 = - 955 cos 30 = - 827. (cos 30 = sin 60.) 1 2 sin 210 = - 955 sin 30 = - 477. k cos 120 = 550 cos 60 = 275. (cos 60 = sin 30.) COMPOUND CURVES 35 Z 3 sin 120 = 550 sin 60 = 476. Tan of azimuth of VN = T i T = 0.0068, or azimuth of VN = 23'. Length of VN = 147 + 2 ^^ = 147.03, practically 147 feet. In Fig. 12, the azimuth of VA is 120 - 52 = 68. The azimuth of VN is 23' and of NA is the angle ANO = \ (180 - 22) = Fig. 13 79. To find the lines AV and NA, the tri- angle shown in Fig. 13 is solved: AV = , 'sin 101 23' = 755 feet. 147 sin 11 147 sin 11 NA 711 67 37' = 711 feet. = 1865 feet. \Ii 2 sin X 100 = 184.5' = 3 ( 36 THE SLIDE RULE In practice a 3 curve would probably be used. 22 Li = -30 = 7 + 33.3 P.C.C. is at 377 P.T. is at 384 + 33.3 T l = AV = 755 feet. A 755' V Fig. 14 Problem 17. Given the data shown in Fig. 14, find the degree of the second branch of the compound curve and the stations of the P.C.C. and the P.T., beginning the compound curve with a 3 (the long radius branch) curve at station 78. Use OA as a meridian. COMPOUND CURVES 37 T in/ T AncrtVi L M + - + -. OA. (TOO' Feet. 1910 1910 AV 90 00' 755 755 VB 142 00' 550 434 338 BO 1910 434 1476 1093 1693 L 3 = 550 cos 142 = - 550 sin 52 = - 434. M 3 = 550 sin 142 = 550 sin 38 = 338. For the line OB the coordinates would be positive. Tan of the azimuth of OB = 1093 = 0.741. 1476 Az. of OB = 36 30'. Az. of BO = 216 30'. Az. of EC = 142 + 90 = 232. Angle a = 232 - 216 30' = 15 30'. In the triangle CBO, EC =R 8y CO = R t - R 9 , OB = . sin 36 30' = 1838 feet. s = ^=919+^2=1874. 38 THE SLIDE RULE i = (s - ft) (s - 1838) (1874 - R.) 36 1838ft 1838ft = 1874 - ft 51ft Sin 2 J (15 30') = sin 2 7 45' = 0.01815. The setting for sin 2 7 45' is as follows: Reverse the slide in the rule, set the indices of the sine scale and the A scale coincident and read on the A scale 1348 opposite 7 45' on the sine scale. Then set the runner over 1348 on the D scale and read 1815 on the A scale under the runner. 0.01815 X 51 X ft = 1874 - ft, or ft = 974 feet. To find 51 X 0.01815, keep the runner in the position stated above, bring the numbered face of the slide to the front, set the left index of the B scale under the runner and read 925 on the A scale opposite 51 on the B scale. Use a 6 curve. Sinl5 30 ' I a = 31 40'. COMPOUND CURVES 39 It = 52 -31 40' = 20 20'. Lj- L,= U P.C. at P.C.C. at L 8 P.T. = 5 + 27.7 stations. 86 + 00 6 + 77.8 92 + 77.8 5 + 27.7 at 98 + 05.5 Problem 18. Begin a compound curve with a 4 (long radius) curve at station 378. Find the degree of the second branch and stations of 878 Fig. 15 P.C.C. and P.T. of the compound curve to connect A and B under the conditions given in Fig. 15. THE SLIDE RULE Length L If + - + - OA 000' Feet 1432 5 1432 5 AB 116 00' 1200 525 1079 BO ... 907.5 1079 or Using OA as a meridian. 1200 cos 116 = - 525. 1200 sin 116 = 1079. 1079 907.5' . azimuth OB = 49 56'. Azimuth BC = 240 00' Azimuth BO = 229 56' Tan azimuth OB = a = 10 04' 1079 B0 = r, = 1408.* sin 49 56' s = | (1432.5 - R a + R s + 1408) = 1420.25. gin* 1 ^ (s~Rs)(s-BO) R a W (1420.25 - R.) 12.25 1408 E, * To get a good result care must be taken to read this value on the slide rule very carefully. COMPOUND CURVES 41 (1*20-25 -ft) 12-2) 00077 1408 fl, or R 3 = 755 feet. 3438 X 100 Use a 7 30' curve. l ^R s = 1432.5 - 755 = 677.5. Sin COB = j= sin 10 04', 677.5 or COB = 11 12'. 7 8 = 10 04' + 11 12' = 21 16'. /, = 60 - 21 16' = 38 44'. 3 44' i = F- = 9 + 68.3. 21 16' L< yo P.O. P.C.C. L, P.T. 30' at at at - A T OO.il. 378 9 + 68.3 387 + 68.3 2 + 83.5 390 + 51.8 Problem 19. A 4 curve begins at station 395 + 30 and ends at station 408 + 40. Find the station of the P.C.C., where a 6 curve 42 THE SLIDE RULE will compound so as to end in a parallel tangent 12 feet nearer the center. r> T> 77'D = KI K s = xI/jD. BE' = (R t - R s ) vers. I 8 = (Ri R a )(l cos I 8 ). Ri - R s = 1432.5 - 955 = 477.5. 12 =0.9749. Fig. 16 Sin (90 - 7.) = 0.9749, or 90 - I s = 77.* L = 13. 7 = 13.10 X 4 = 52 24'. I l = 52 24' - 13 = 39 24'. * Reading for this must be carefully made. COMPOUND CURVES 43 100 L s = ^ = 2 + 16.7. Li = 39^24' = g + 85 . P.O. at 395 + 30 P.C.C. at 405 + 15 L 3 2+ 16.7 P.T. at 407 + 31.7 Problem 20. An 8 curve begins at station 372 and ends at station 378 + 50. Find the stations of the P.C.C.'s and the P.T. for a three-centered compound curve, with one sta- tion of a 1 curve at each end and passing thru the same P.C. and P.T. as the 8 curve. sin ^7 Ri - R s Oi0 3 sin i /*"#*-#' / = 6.5 X 8 = 52. 7 S = 52 - 2 = 50. R L - R = 5729.7 - 716.8 = 5012.9. R l -R 8 = 5012.9 = 5200. R s = 5729.7 - 5200 = 529.7 feet. 44 THE SLIDE RULE L - 5 -4 + 61 ** ^' ~ P.O. L L P.C.C.i L 8 P.C.C.2 L t P.T. Fig. at at at at 17 372 1 + 00 + 00 373 4 + 00 + 61 377 1 + 61 + 00 378 + 61 COMPOUND CURVES 45 Problem 21. Given two straight parallel tracks with center lines 13 feet apart and with a long chord of 186.4 feet, find the degree of the equal branches of a reversed curve to con- '0 Fig. 18 nect the parallel tracks. A ED and AOD are similar triangles, then 46 THE SLIDE RULE 13 4 X 13 R r = 667 feet. 3438 X 100 = 667. 667 = 515' = 8 35'. Problem 22. Given the conditions shown in Fig. 19, find the degree of the reversed curve, with branches of equal radii, connecting the Fig. 19 tangents AB and CD, and having EC as the common tangent of the two branches, and also find the stations of the P.C., P.R.C. and P.T. 940 = R r tan 18 + R r tan 14. 940 tan 18 + tan 14' 940 0.325 + 0.2495 = 1640 feet COMPOUND CURVES 47 _ 3438 X 100 1640 Use 3 30'. R for 3 30' curve = 1637 feet. T l = 1637 tan 18 = 532 feet. 376 + 00 5 + 32 36 P.O. at 370 + 68 L x = ~^ = 10 + 28.6. 10 + 28.6 P.R.C. at 380 + 96.6 L 2 = f- = 8 + 00. o.O 8 + 00.0 P.T. at 388 + 96.6 CHAPTER IV THE VERTICAL CURVE 18. The vertical curves, used at sags and summits to connect adjacent grades, or used elsewhere to connect grades of the same kind, are usually parabolas. The following problem shows how the elevations on such a curve may be found. Problem 23. Find the elevations of the sta- tions on a ten-station vertical curve to connect the grades shown in Fig. 20. Fig. 20 The offsets from the tangent to the parabola vary as the squares of the distances from the beginning of the curve to the points where the offsets are made. The beginning of the ver- tical curve is at station 225 and the end is at THE VERTICAL CURVE 49 station 235. The elevation of station 225 is 515 and of station 235 is 517.5. The vertical offset from a point on the prolongation of the 1-per cent grade to station 235 is 7.5 feet. By the above rule the offset at station 226 is T ^ of 7.5 or 0.075. The other offsets may be found from this by the above rule. Station. Elevation on the straight grade. Offset . Elevation on the curve. 225 515.00 515 000 226 516.00 0.075 515.925 227 517.00 0.300 516.700 228 518.00 0.675 517.325 229 519.00 1.200 517.800 230 520.00 1.875 518.125 231 521.00 2.700 518.300 232 522.00 3.675 518.325 233 523.00 4.800 518.200 234 524.00 6.075 517.925 235 525.00 7.500 517.500 The setting for the offsets is as follows: Set the right index of the slide opposite 75 on the right-hand A scale, and opposite 2, 3, 4, etc., on the C scale read 3, 675, 12, etc., on the A scale. By rough figuring obtain the posi- tions of the decimal points. A check on the elevation of station 230 on the curve may be found from the principle 50 THE SLIDE RULE that the parabola is midway between its chord and its vertex at its middle point. The eleva- tion of the chord at its middle point is the mean of the end elevations The elevation of the middle of the curve is the mean of the elevation of the middle of the chord and the elevation of the vertex. Applying this to the above problem, the elevation of the middle of the chord is - ^ - = 516.25 and the ele- . 516.25+520 vat ion of the middle of the curve is - = 518.125, which checks the elevation found. By the extension of this principle, the eleva- tions of points on the vertical curve may be found or a parabola may be laid out. CHAPTER V TURNOUTS 19. Fig. 21 is a line diagram representing a split switch turnout, the one now most commonly used. AH and LK are the movable A * Fig. 21 rails. The rails A G and BP are continuous. F is the point of the frog. The rails HF and QP are curved and the rail AH, when the turnout is set for the branch or siding, is tan- 51 52 THE SLIDE RULE gent to the rail HF. AH is the switch rail, I, and HE is equal to the amount of its throw, t. EAHj the switch angle, = 8. a . e EH t = AS == r AB, the gage of the track, = g. The standard gage is 4' 8J". FOD = F, the frog angle. 0V Rj the radius of the turnout curve. (R + iflf) cos AS - (R + g)cosF = g -t. R + ^ g== cosS-cosF' or # = 1 ^ \ g. cos cos F The distance BF is called the lead, L. KF. Angle #/^C = F - f F0# = F - J (F- r_7 tan 1 = 2(R+ig)sm%(F-S) . (13) TURNOUTS 53 Fig. 22 shows the form of the frog. The , ,, , . FR ... FN . number of the frog is -= , altho is some- times used. Fig. 22 Where n is the number of the frog, f = A. . (15) Problem 24. Find the radius and the lead of a split switch turnout using a No. 9 frog, 18-foot switch rail, a S^-inch throw and standard gage track. o- a _ * _ * 5 ' 5 12X8 S = 1 28'. 5.5 216 = 0.0254. 9^ = VQ = ili = 0-0556. A 71 /\ y J.O For small angles sines and tangents are practically equal, and for all angles less than 54 THE SLIDE RULE 5 45' the sine in place of the tangent scale of the slide rule should be used for tangents of such angles. By use of the S scale of the slide, -| F = 3 11' or F = 6 22'. Setting for these is: Opposite right and left indices of the A scale set the indices of the slide reversed and opposite 0.0254 on the A scale read 1 27' on the S scale, also op- posite 0.0556 on the A scale read 3 11' on the S scale. r tan $(F + 8) 12 tan 3 55' L = 18 + 62.2 = 80.2 feet. R + ^ ~ 24 sin 3 5?' sin 2 27' = ?28 72 = 725.65 feet. 20. Where the turnout leaves the main track on a curve, the radius and the lead may be found by the precise or approximate method following. TURNOUTS 55 Precise method: 1st. When the turnout is on the inside of the main track curve. In Fig. 23, is the center of the main curve, C is the center of the turnout curve, F is the frog point and AB is the gage of the track. Fig. 23 In the triangle AFO, 2' "2/_ g Rm 56 THE SLIDE RULE tan^ ff = g cot | 2 R m ' i ^ 9 cot i F = taniO = f^ ...... (16) Km In the triangle BCF, BC = R t - | z and F(7 = ^ + , nr r, . g cot j F _ g^ , . ~ ~ ' l ; 2nd. When the turnout is on the outside of the main track curve. In Fig. 24, solving the triangle AFO by the method used above, Then solving the triangle BCF, tan i (F - 0) TURNOUTS 57 In either case, L t = 2 R m sin \ 0, or L t = 2 R t sin % (F d= 0). Fig. 24 21. Fig. 25 shows a turnout from a straight track where a stub switch is used. BF is the lead and AB is the gage. L s = BF = AB cot i F = 2 0n. . (18) 58 THE SLIDE RULE Make BC = g. Then the triangles AFC and AOF are similar and AO:AF ::AF:AC ACXAO = I/. 2g(R,+ g ^='A~f, or 2 gR, = Af-g\ ? 8 = 20n 2 . (19) (20) or gn Fig. 25 22. The approximate method for turnout from curved main track. Let R 8 = radius of turnout from a straight track using a given number of frog. TURNOUTS 59 R t = radius of turnout from curved track using same number of frog. R m = radius of curved main track. From Equations (16), (17) and (20), for the condition where the turnout is on the inside of the curved main track, p . P . p . 9 n 9 n 9 n 'tan %F ' tan J 'tan \ (F + 0) As F and are small angles, the tangents may be taken equal to the arcs of unit radius, or jp-O'F+0 The degrees of curves are practically inversely as their radii, or D 8 :D m :D t ::F : : F + 0. By composition (D s + D m ) : D t : : F + : F + 0. Hence D t = D 8 + D m . Where the turnout is on the outside of the curved main track, a similar proportion applies. D s :D m :D t ::F:0:F -0. By division D 8 -D m :D t ::F-0:F-0. Hence D t = D 3 D m . 60 THE SLIDE RULE The condition of D t = D m D 8 may arise as shown in Fig. 26, where is the center of the main track curve arid C is the center of the turnout curve. A B Fig. 26 In either case, L t = 2 R t sin \ (F =b 0), and as F and are small angles, the tangents and sines are about equal or L t = 2 R t tan \ (F 0), gn Rt Hence = 2 gn. Problem 25. Find the radius and the lead of a turnout on the inside of a 4 main track of standard gage, using a No. 9 frog. Both TURNOUTS 61 the precise and approximate methods of solu- tion follow. Precise method: 4.708 X 9 3438 X 100 ~ m = 240 tan J = 0.0296. Solving by the slide rule using the sine scale as described in Problem 24, \ = 1 42' and = 3 24'. F = 6 22' as in Problem 24. 4.709 X 9 leet. >t i i /ET , Ty: 7 - ,0 P0/ tan i (F + 0) tan 4 53 (Method of solution as given in Problem 24.) 343000 L t = 2R t sm% (F + 0) = 2 X 496 X sin 4 53' = 84.6 feet. Approximate method: D t = D m + D s . 5730 2865 ooo 4.708 X 9 2 62 THE SLIDE RULE An = 4 0' A =11 29' L, = 20n = 2 X 4.708 X 9 = 84.6 feet. The use of the slide rule for the solution has been given for similar formulas. 23. Fig. 27 shows in outline the connection between a parallel siding and the main track by a turnout and a connecting curve to which the following problem applies. Fig. 27 Problem 26. Find R^ D% and Z/ 2 , where the distance between track centers is 13 feet and a No. 9 frog is used. FA and EB may be considered as the rails of a track. Then the curve EF may be con- sidered as the curve part of a turnout. The gage for this is p g. TURNOUTS 63 Then # 2 -f = (p-g)2n*. R 2 - 6.5 = (13 - 4.71) 2 X 9 2 = 8.29 X 2 X 9 2 = 1340. R 2 = 1346.5 feet. 3438 X 100 D 2 = = 255' = 4 15'. 1346.5 F for a No. 9 frog may be found from = = = 0.055. Using tangent = sine, = 3 11', or F = 622'. L 2 = # 2 - sin ^ = 1344.3 sin 6 22' = 149 feet. 24. Fig. 28 shows a connection between two parallel tracks by the use of two similar turn- Fig. 28 outs and a piece of straight track whose center line is a common tangent to the center lines 64 THE SLIDE RULE of the curves of the turnouts. The following problem applies to this form of a connection. Problem 27. Find the distances a, b and c for the connection between two parallel straight tracks of standard gage, with 13 feet between track centers and with No. 9 frogs used in the turnouts. F = 6 22' (from Problem 26). a = c = 2 gn = 9.42 X 9 = 84.7 feet. b = p-g-gsecF = 8.292 - 4.71 sec 6 22' tan F tan F 4 ' 71 82Q2- cos 6 22' 8.292-4.73 tan F tan 6 22' = 31.9 feet. 25. Fig. 29 shows the connection between the parallel siding, on the outside of the curved main track, and the main track by a turnout and a connecting curve. Solving the triangle OBG, 7? 4-, ^-9- *" P 2 j i"?^ ^ T-V ; 7: cot0 2R m + p 2 A/w + p 2 + ^ + . 1 _ (p - flf) n or oan 7: L/ . TURNOUTS Fig. 29 Solving the triangle PBH, Rt _ +Rt _ p+ z * tan |F p - g 66 THE SLIDE RULE r> _ P _ -n Kt ~ Problem 28. By both the precise and ap- proximate methods find the radius and lead of the turnout and the radius (R 2 ) and length (L 2 ) of the connecting curve to connect a 4 30' main track of standard gage with a parallel siding using a No. 9 frog. The siding is on the outside of the main track, with 13 feet between the track centers. Precise method: n gn 4.708 X 9 "" p 3438X100 107n , -- 270~~ = Solving by the slide rule using the sine scale, J = 1 54' and = 3 48' F = 6 22' F - = 2 34' i (F - 0) = 1 17' 4.708X9 == 189 feet > 3438X100 A= ~1890~ D t = 3 02', TURNOUTS 67 _ 100 X 3 48' _ 100 X 228 _ , 4 30' 270 tan J0' = l^ = *^ = 0.0584. P p 1276.5 Solving by the slide rule using the sine scale, 1 = 3 20' 0=6 40' F = 6 22' F + = 13 2' i(F + 0)= 6 31' p _ (p - g) n . 8.29 X 9 _ ~ ~'~ and #2 = 659.5 feet. _ 3438 X 100 _ 22 _ R0 , 659.5 42 ' 100 X 13 02' 100 X 782 . , Approximate method: By Problem 25, D 8 = 7 29' D m = 4 30' D t = 2 59' = 84.6 feet. 68 THE SLIDE RULE Let R s = radius of the connecting curve if the tracks were straight. fi a ~ | = 2 (p -g) n* = 2 X 8.29 X 9 2 = 1340 feet. Rz = 1346.5 feet and 3438 X 100 1346.5 Z> 3 = 4 15' D m = 4 30' = 8 45 r 3438 X 100 L 2 = 2(p-g)n = 149 feet. Problem 29. By the precise and approxi- mate methods find the radius and the length of the connecting curve for a parallel siding on the inside of a 4 main track of standard gage, with 13 feet between the track centers and using a No. 9 frog. Precise method: TURNOUTS 69 Using the sine for the tangent i = 3 00', or = 6 00'. F as found in previous problem = 6 22', F - = 22'. _p_ (p-g)n 8.29 X 9 2 2~tani(^-0) tan 11'' tan 11' = approximately 11 X 0.00029 = 0.00319. p _ 8.29 X 9 " 2 ~ 0.00319 23 ' 39 ' or # 2 = 23,384 feet. 3438 X 100 , c , , 2 = 23384 = (approximately). 99' ,_!_ 147 feet. 1Q Approximate method: R s - = 2( p - g) n * = 2 X 8.29 X 9 2 - 1340 feet. Zt R s = 1346.5 feet. _ 3438 X 100 _ , _ , 1346.5 D 2 = 4 15' - 4 = 15'. L 2 = 2(p-g)n= 149 feet. 70 THE SLIDE RULE 26. "Y" curves are used to connect the main tracks and branch tracks, so that the trains coming in either direction may go from the main tracks to the branch tracks without backing. A "Y" and a branch track may also be used for reversing the direction of the train on the main track. Problem 30. A branch track leaves a straight main track at station 322 + 40. The radius of the branch curve is 762.7 feet. Find the P.O. on the main track and the P.T. on the branch track of a " Y" curve, of 573.7 feet radius, that will connect the main and branch tracks. A is at station 322 + 40. is the center of the branch curve. C is the center, L is the P.O., and M is the P.T. of the "Y" curve. Let a = central angle of the branch track from its P.O. to the P.T. of the "Y," B = central angle of the " Y" curve, R b = the radius of the branch curve and R y = radius of the "Y" curve. = R b - R y = 762.7 - 573.7 " R b + R y " 762.7 + 573.7 189 1336.4 = 0.142. TURNOUTS 71 By the use of the sine scale of the slide rule, sine of (90 -*)'- 0.142; 90 - a = 8 10' a = 81 50'. B = 180 - a = 180 - 81 50' = 98 10'. I = distance from P.O. of the branch to the P.O. of the "Y" measured along the main track. A 7 B Fig. 30 I = 1336.4 - 189 2 P.C. is at 2 X 1336.4 322 + 40 13 + 22.9 335 + 62.9 -. = 1322.9 feet. 72 THE SLIDE RULE 3438 X 100 762.7 81 50' X 100 730 / = 450' = 7 30', = 1090 feet, or P.T.y is at station 10 + 90 on the branch track. Problem 31. Using the same curves for the branch and " Y" tracks as in No. 30, but having Fig. 31 a 60 central angle for the branch curve and the P.T. of "Y" coming beyond the P.T. of the branch and on the tangent to the branch curve at its P.T., find the P.C. on the main track and the P.T. on the branch track for the "Y" curve. TURNOUTS 73 Using letters which represent the same things as in Problem 30, fin 1 90 I = 762.7 tan ^- + 573.7 tan ~ , 1 Z I = 762.7 tan 30 + 573.7 tan 60, ^70 7 Z = 762.7 tan 30' + ^, I = 441 + 994 = 1435 feet. Settings. Reverse the slide, set right index of T scale of the slide opposite 762.7 on the D scale and opposite 30 on the T scale read 441 on the D scale. Set 30 on the T scale opposite 573.7 on the D scale, and opposite the right index of the T scale read 994 on the D scale. Let d = the distance from the P.T. of the branch curve to the P.T. of the "Y" curve measured along the tangent to the branch curve thru its P.T. 120 _ 60 d = R y tan =- R b tan -_- - 762.7 tan 30 = 994 - 441 tan 30' 553 feet. 74 THE SLIDE RULE 322 + 40 I = 14 + 35 P.C.y at 336 + 75 on the main track. 3438 X 100 ._, ?0 ory 762.7 L b = ^5^, X 100 = 800 feet, 8 + 00 d = 5 + 53 P.T.^at 13 + 53 on the branch track. CHAPTER VI THE EASEMENT CURVE 27. The purpose of the easement curve, connecting the tangent and the circular arc, is to produce a gradually increasing centrifugal force that may be balanced by a gradually increasing centripetal force produced by the gradual elevation of the outer rail of the ease- ment curve part of the track. The ordinary form of the easement curve is the spiral or a similar curve. From the above it is seen that the length of the spiral is the distance in which the total amount of the elevation of the outer rail is gained. In the best practice the rate, by which the elevation of the outer rail is gained, is a function of the speed of the train. It has been found that a gain of 1 inches per second is not felt by a passenger in the train. Even a gain of two inches per second does not produce a disagreeable effect. Let e = the elevation of the outer rail, in inches, necessary for the cir- cular curve. 75 76 THE SLIDE RULE Let v = the velocity of the train in feet per second. S = the velocity of train in miles per hour. l c = the length of the spiral. DC and R c = the degree and the radius of the circular curve. r = the rate at which e is gained in inches per second. C = the centrifugal force of a car on the circular curve. W = the weight of the car. G = the gage of the track. g = the acceleration due to gravity. Then from mechanics, gR c ' From Fig. 32, C ED ED Wv* gR c e Gv 2 -W = G or e = W c ' ' * v = 1.47 S, G = 4.71, g = 32.2, and R c = THE EASEMENT CURVE 77 Substituting these values in Equation (21), reducing and expressing the result in inches, e = 0.00067 S 2 D C . At a rate of gain, for e, of r inches per second, I = e Xv = 'WW7S*D Cv c r r Substituting for v its value in terms of S, (22) , 0.00067 S*D C w 1 _ S 3 D C . . l c = - f- - X 1.47 AS = 1QQO r (approx.). For a rate of 1 inches per second, For a rate of 2 inches per second, 7 _^ ~ 2000' 78 THE SLIDE RULE 28. The equation for a true spiral is derived as follows: If I is the distance from the beginning of the spiral, P.S., to any point, P, on the spiral where its radius is R and its degree of curva- ture is D, e P = 0.00067 S*D and 0.00067 S 2 X 5730 v 5.735 ~ r In a similar way it is readily shown that Hence El =- R c l c or y = ^- c . . (23) ' 'c This is the equation for a true spiral and should be used for very long curves which partake more of the nature of a curve compounded many times than of an easement curve which connects a tangent and a circular arc. This equation shows that the radius of the spiral varies inversely as the distance from the beginning of the spiral, or its degree of curva- THE EASEMENT CURVE 79 ture varies uniformly with the distance from the beginning of the spiral. B H Fig. 33 In Fig. 33, ADE represents part of an ease- ment curve or spiral, beginning at A, the point of spiral. Let I = length of spiral from A to E, S E = the spiral angle for point E and i E = its deflection angle from tangent AK. 29. To find the value of S E . From the circular curve EF of radius R and degree D, *B = 2X I D(EF= 2> a PP rox -)- 5730 2 X 100 R (24) 80 THE SLIDE RULE Equation (24) gives the value of S E in degrees. In length of arc of unit radius (25) From Equation (23) R=; substituting this value of R in Equation (25) 30. To find the relation between deflection angles from the tangent at the point of spiral to points on spiral. Assuming that the part of the spiral from A to E is a circular arc of D curvature (prac- tically true if AE is a very small part of the spiral), the deflection *- or iE=D - - (27) D The curvature of the spiral at point D = -*-> z where D is the curvature of the spiral at E , , n AE I and AD = . Considering the spiral as a circular curve of D -Tj- from A to Z), THE EASEMENT CURVE 81 L 1 22 (28; -- IE 200 4 , 00 v hence = -? - = T ..... (29) to * n o 1 800^ Hence deflections from the tangent at the P.S. to points on the spiral vary as the squares of the distances from the P.S. to the points on the spiral. 31. To find the relation between the offsets from the tangent thru the P.S. to points on the spiral. pi = DB = ^ sin i D , pz = EK = I sin i E , 2 = I sin i E Pi I . . gSiniD Since the sines and arcs for small angles are proportional, = -r-^and substituting the value of -r* from Equation (29), ID - <30) 82 THE SLIDE RULE Hence offsets from the tangent thru the P.S. to points on the spiral vary as the cubes of the distances to the said points from the point of spiral. 32. To find the value of the deflections from the tangent thru the P.S. to points on the spiral. Since the spiral changes in curvature uni- formly with the distance, the amount that it will deflect from the tangent for the distance AD = - is the same that it deflects from the & circular arc for the distance EF = = or FD Zi = DB. Hence F = 2DB = 2 P1 = - 2 From the circular curve FE, 2 or P _3p 2 8R^ 4 ' then 4P ' ' = i sin IE* THE EASEMENT CURVE 83 For small angles sines and arcs of unit radii are equal, then or ** = O ...... (32) j-, _ R c l c T hence ** = P , . SE hence ^E = -$- o Since ^7 may be taken as any point on the spiral, 33. To show P = 24TR> p = FB = ^, but p 2 = /^from Eq. (31), P hence p = ..... (35) 34. To find a value for the deflection angle, i, to any point on the spiral. 84 THE SLIDE RULE Let I = distance from P.S. to the point where the deflection is i. N = number of chords from P.S. to the same point. C = rate of change in curvature of the spiral per chord length. i = INC X O.I 7 * giving i in minutes. From Equation (32) I D 5730 f 1 = OR ' ~~D~ ( a PP roxlmatel y)- * From Kellogg' s Transition Curve by N. B. Kellogg, C. E. NOTE. If a spiral of six chords in length is always used the deflection for the end of the first chord is | X 1 X - X 0.1 in minutes = ^ -| D c or in seconds oo oU o is IN D CJ where IN is the length of a chord, which, ex- pressed as a rule, is: RULE. For a six-chord spiral the deflection for the end of the first chord in seconds is the length of the chord in feet multiplied by the degree of the circular curve. Deflections for the ends of the other chords may be found by the rule that the deflections vary as the square of the distances from the point of spiral; i.e., the deflection for the end of the second chord is four times that for the end of the first chord and for the end of the third, nine times that for the end of the first, etc. THE EASEMENT CURVE 85 Hence ID 6 X 5730' hence INC i = 6 X 5730 in length of arc of unit radius, and multiplying by 57.3 gives i in degrees, INC i = 6 X 100 Multiplying this value of i by 60 gives i in minutes. i = INC X 0.1. ... (36) In Equation (36) I is expressed in feet, N in chord lengths and C in degrees. N may have a fractional value. 35. To find the distance AB in Fig. 34, AB = q = ^ (approximately). & The following gives a value for q more nearly correct. In any right-angled triangle the approximate difference between the hypotenuse and base is equal to the square of the altitude divided 86 THE SLIDE RULE p.r. P. S. by twice the known side, either base or hy- potenuse, hence Then 1 P 2 = 2~n (37) P=^V (Equation (35)). l_ V 2 4) .... (38) THE EASEMENT CURVE 87 36. To find the distance of the P.S. from V, the intersection of the tangents. Ordinarily the same spiral will be used at each end of the circular curve. The distance from P.S. to V will be found under this condi- tion. AV = AB + BH + HV. Let T 8 = AV. T 8 = g + fl c tan + Ptan. . . (39) 37. To find a tangent at any point on the spiral. In Fig. 33, the angle AEH = EHK - EAR. AEH=s E -i E =3i E -i E = 2i E . . (40) Set up the transit at E and with instrument set at 000', sight to A. Transit the tele- scope and turn off an angle of 2 i E , in the same direction as the curve is running, and the line of sight will be in a tangent to the spiral at E. 38. To find the deflection from a tangent at any point on the spiral to any other point on the spiral. 1. In the direction from the P.S. toward the circular curve. Let KFL be an arc of a circle of the same radius as that of the spiral at F. 88 THE SLIDE RULE Since the spiral changes in curvature uni- formly with the distance, the angular amount that the spiral deflects from the curve FL is the same that it deflects from the tangent AB in a distance qual to FL, or the angle EFL equals the deflection from the tangent thru the point of spiral for a distance of FL. The angle LFM is the deflection angle for a chord of FL for the circular curve of radius R. The angle EFM = EFL + LFM. Then the deflection angle from the tangent at any point on the spiral to any other point on the spiral is equal to the sum of the deflec- tions for the spiral and for the circular arc for the length of chord between the two points on the spiral, the circular arc having the same radius as the spiral at the point thru which the tangent runs, and the spiral being run in toward the circular curve which it connects with main tangent. THE EASEMENT CURVE 89 2. In the direction from the circular curve toward the tangent. By a proof similar to that just given, it may be shown that the deflection from the tangent at any point on the spiral to any other point on the spiral is equal to the difference of the deflections for the circular arc and the spiral for the length of chord between the points on the spiral, the circular arc having the same radius as the spiral at the point thru which the tangent runs, when the spiral is run in from the circular curve toward the main tangent. 39. There are two common methods of lay- ing out easement curves, viz.: first, by deflec- tion angles, and second, by offsets from the tangent thru the P.S. The following steps give the deflection angle method by the use of tables derived from the formulas and by the use of the slide rule. 1. From the assumed speed of the train, the degree of the curve and the rate of gaining the elevation of the outer rail, find the length of the easement curve by Equation (22) or from a table or diagram made from this equation. The slide rule readily solves the equation. 2. From tables, or by slide rule, find values of p and q. By formula or table find the tan- 90 THE SLIDE RULE gent distance, T c , for the circular curve for a central angle equal to the deflection angle between the tangents. Find value of T 3 in Equation (39). From station of V, intersection of tangents, subtract T 8 expressed in stations and result is the station P.S. 3. By the deflections and their correspond- ing distances from the P.S. run in the spiral as far as the P.S.C., where it joins the circular arc (Fig. 34). From the deflection angle between the tan- gents subtract twice the spiral angle, S c , for the spiral used, and the result is the central angle of the circular arc, from which its length may be determined. Set the transit up at P.S.C. and run in the circular arc by deflections from the tangent thru the P.S.C. , to the P.C.S., where it joins the ending spiral. Set up the transit at P.C.S. and run in the ending spiral to the P.T. by deflections from the tangent thru the P.C.S. Set up the transit at the P.T. and check work by sighting on the P.C.S. and turning off one-half of the deflection angle for the P.T., THE EASEMENT CURVE 91 and if line found runs thru V the work is correct. 40. The method of laying out a circular curve with spirals at each end, the spirals to be located by offsets from tangents, is as follows: See Fig. 34. 1. The circular curve is to be shifted from the tangents toward the center, until the shifted P.O. is a distance of "p" from the tangent. As the curve is moved toward the center every point on the curve will move along a line parallel to the line joining the center of the curve with its vertex. The shifted P.O. will Fig. 36 be directly opposite a point on the tangent at a distance of p tan - from the original position 2^ of the P.O., measured backward along the tangent. 92 THE SLIDE RULE Set the instrument up at E, the shifted P.O., and run in the circular curve by deflections from the tangent EF parallel to AV. 2. From the length of the spiral found as described in paragraph 39, find the value of q and p. From station of original P.O. for the circular curve subtract q + p tan = to find the Zi station of A, the point of spiral. The offset from the tangent A V to the middle fQ point of the spiral is . By paragraph 31, the m offsets to the spiral from any point on the tan- fY\ gent A V is to ^ as the cube of the distance of 'g said point from A is to the cube of ~. 40 Let AB = -: and the offset to the spiral at B 3 I Let AC = -T- and c = offset at (7, 27 then C== P- AD = I and d = offset at D, then d = 4 p. THE EASEMENT CURVE 93 Find the location of A, the P.S., as described, then lay off on the tangent AB, AK, AC and AD. At B lay off 6; at K, | ; at C, c; at D, d. Zt The ends of the offsets are in the spiral. This method is not theoretically correct, as it assumes that the distance from the P.S. to any perpendicular to the tangent is the same measured along the tangent as along the spiral. However, the resulting curve is prac- tically the same as the one found by a precise method. 41. To connect the two branches of a com- pound curve by a spiral, find the difference in degrees of curvature between the two branches. Then from the adopted speed, rate of elevation of the outer rail and the difference in degree of curvature, find the length of the spiral. Use the spiral given in the tables, that is practically of this length, or figure the spiral deflections, etc., by the slide rule. The number of chords in this spiral multi- plied by the change in curvature per chord must be equal to the difference in the degrees of curvature of the two parts of the compound curve. 94 THE SLIDE RULE The "p" of the spiral will be the distance GH in Fig. 37. AG = BH is equal to "q" of the spiral used. Fig. 37 Assume that the curve is to be run in from F to E. Let H be the original P.C.C. Find B on the curve FH, at a distance of "q" back from H. Set the transit at B and run in the spiral by deflections from the tangent at B y as given in paragraph 38, to A. Set the transit at A and run in the circular curve AE by deflections from the tangent at A. 42. In the tables on pages 100 and 102 are given the deflections from the tangent through the point of spiral, for different lengths of spiral and different changes in curvature per chord length for ten, twenty, thirty, forty and fifty foot chords, and also the values of p and q for spirals of different lengths and changes in THE EASEMENT CURVE 95 curvature per chord. From the tables, data for running in the center line spirals for street railway tracks may be found, using chords either 10 or 20 feet in length. 43. The table on page 96 shows the spirals given in the tables on pages 100 and 102 that may be used for different degrees of the cir- cular curves. A similar table can be made for the spirals to be used with any circular curve up to 20. After selecting the spiral for any particular curve the deflections for the spiral may be taken directly from the tables, or be figured by the slide rule. 44. Where a spiral of the proper length for a given circular curve cannot be taken directly from the table, the following example shows a method that may be used : It is desired to use a spiral 165 feet long for a 5 30' circular curve. Take from the tables the deflections for the spiral 150 feet long having a change of curva- ture of 1 for each 30' chord. This leaves only the deflection for the end of the spiral to be determined. This may be found by formula (36), i = INC X 0.1. In this case I = 165', N = 5J and C = 1. i = 1 30.75'. For ease in computation, the last chord of the 96 THE SLIDE RULE Degree of circular curve. Length of spiral. Change of curvature Per 30' chord. Per 40' chord. Per 50' chord. 3 00' tt ft tt it u u n u tt 4 00' it tt it n tt tt tt n tt n 5 00' tt tt tt tt tt tt ti tt 60' 80' 90' 100' 120' 160' 180' 200' 240' 300' 60' 80' 100' 120' 160' 180' 200' 240' 300' 320' 400' 90' 120' 150' 160' 200' 250' 300' 400' 500' 130' 130' 100' 130' 045' "6 30'"' 100' 045' 045' "6 30' 030' 2 00' 2 00' 2 00' 100' 100' 040' 100' "6 40'"' "030' 030' 040' 030' 140' 115' 100' 140' 140' 115' 100' 115' 100' 030' 030' 030' THE EASEMENT CURVE 97 spiral should be taken some even fractional part of a full chord length as J, J, \ or J of it. The length of the spiral expressed in chord lengths multiplied by the change in curvature per chord length must always equal the degree of the circular curve. 45. The following problem shows the method of obtaining by use of the tables the necessary quantities to locate a spiral by deflections. Problem 32. Given two tangents intersecting at station 187 + 40, with a deflection angle of 33 40', to find the stations of the P.S. and the P.T., and the deflections for a 4 curve, with equal spirals at each end to connect the given tangents; the speed of train to be 40 miles per hour and the rate of superelevation to be 1.6" per second. The length of the spiral, from S 3 D the formula l c = innn , is 160 feet. 1UUU T The curve selected from the tables may be either a 160-foot spiral consisting of four 40' chords with a change of curvature of 1 per chord or a 180-foot spiral consisting of six 30' 2 chords with a change of curvature of -^ per o chord. Assuming that the 180' spiral is used, the following is the solution for the various quantities: 98 THE SLIDE RULE Tc = + o.08 = . . . . 433.5 p tan i / = 0.94 tan 16 50' log 0.94 = 9.973128 - 10 log 16 50' = 9.480801 - 10 9.453929 - 10 log of 0.28 . 90.0 523.8* feet V at station 187 + 40 5 + 23.8 P.S. at station 182 + 16.2 1 + 80 P.S.C. at station 183 + 96.2 Deflections from P.S. to P.S.C., for end of each chord 30 feet long, may be taken from the table. For other points deflections may be found from the formula i = INC X 0.1'. Deflection for station 183 on the spiral is found as follows: Distance from P.S. to 183 = 83.8'. Distance in chords of 30' = 2.79. i = 83.8 X 2.79 X f X 0.1 = 15.6'. I c the amount of curvature in the circular arc =I-2S C . I c = 33 40' - 7 12' = 26 28'. * Results need be found only to nearest j^ of a foot. THE EASEMENT CURVE 99 X 100 = 661.7 feet. P.S.C. is at 183 + 96.2 6 + 61.7 P.C.S. is at 190 + 57.9 1 + 80.0 P.T. is at 192 + 37.9 Stations. Deflections. Description of curve. 182+16 2 P S D = 4R -j-46 2 0-02' 7=33 40' +76 2 0-08 T c =433.5' 183 0-15.6 T=523.8' +06 2 0-18 Spiral is +36 2 0-32 180' long +66 2 0-50 p = 0.94' +96 2 P.S.C 1-12 g=90.0' 184 0-04.6 c =336' 185 2-04.6 7 C =2628' 186 4-04.6 L c =661.7' 187 6-04.6 188 8-04.6 189 . . 10-04.6 190 12-04.6 190+57. 9 P.C.S 13-14 190x87 9 0-34 191 0-46.6 +17 9 1-04 +47 9 1-30 +77.9. . ... 1-52 192 2-05.6 +07.9 2-10 37. 9 P.T. 2.24 100 THE SLIDE RULE CHORD LENGTH Change of curvature 100' 2 00' 3 00' 5 00' 8 00' 10 00' 100' I Deflections. 10 0-01 0-02 0-03 0-05 0-08 0-10 P Q 20 0-04 0-08 0-12 0-20 0-32 0-40 0.01 10 30 0-09 0-18 0-27 0-45 1-12 1 30 0.02 15 40 0-16 0-32 0-18 1-20 2-08 2-40 0.05 20 50 0-25 0-50 1-15 2-05 3-20 4-10 0.09 25 60 0-36 1-12 1-48 3-00 4-48 6-00 0.16 30 CHORD LENGTH Change of curvature 100' 2 00' 3 00' 5 00' 8 00' 10 00' 100' I Deflections. 20 0-02 0-04 0-06 0-10 0-16 0-20 P Q 40 0-08 0-16 0-24 0-40 1-04 1-20 0.02 20 60 0-18 0-36 0-54 1-30 2-24 3-00 0.08 30 80 0-32 1-04 1-36 2-40 4-16 5-20 0.19 40 100 0-50 1-40 2-30 4-10 6^0 8-20 0.36 50 120 1-12 2-24 3-36 6-00 9-36 12-00 0.63 60 140 1-38 3-16 4-54 8-10 13-04 16-20 1.00 70 CHORD LENGTH Change of curvature 30' 40' 45' 100' 115' 130' 140' 2 00' 30' I Deflection angles. 30 001.5 0-02 0-02.25 0-03' 003.75 004.5 005' 006' P q 60 0-06 0-08 0-09 0-12 0-15 0-18 0-20 0-24 0.03 30 90 0-13.5 0-18 0-20.25 0-27 33.75 0-40.5 0-45 0-54 0.09 45 120 0-24 0-32 0-36 0-48 1-00 1-12 1-20 1-36 0.21 60 150 0-37.5 0-50 0-56.25 1-15 1-33.75 1-52.5 2-05 2-30 0.41 75 180 0-54 1-12 1-21 1-48 2-15 2-42 3-00 3-36 0.71 90 210 1-13.5 1-38 1-50.25 2-27 3-03.75 3-40.5 4-05 4-54 1.12 105 240 1-36 2-08 2-24 3-12 4-00 4-48 5-20 6-24 1.68 120 270 2-01.5 2-42 3-02.25 4-03 5-03.75 6-04.5 6-45 8-06 2.40 135 300 2-30 3-20 3-45 5-00 6-15 7-30 8-20 10-00 3.27 150 THE EASEMEN*T,.CUHVE Oiifcj 10 FEET, per chord. 2 00' 3 00' 5 00' 8 00' 10 00' p and q. P P a P q P q P q 0.01 10 0.02 10 0.03 10 0.05 10 0.06 10 0.04 15 0.06 15 0.10 15 0.16 15 0.20 15 0.09 20 0.14 20 0.23 20 0.37 20 0.46 20 0.18 25 0.27 25 0.45 25 0.73 25 0.91 25 0.32 30 0.48 30 0.80 30 1.27 30 1.59 30 20 FEET, per chord. 2 00' 3 00' 5 00' 8 00' 10 00' p and q. P Q P q P q P q P q 0.05 20 0.07 20 0.12 20 0.19 20 0.23 20 0.16 30 0.24 30 0.39 30 0.63 30 0.78 30 0.37 40 0.56 40 0.93 40 1.49 40 1.86 40 0.73 50 1.09 50 1.82 50 2.91 50 3.64 50 1.25 60 1.88 60 3.13 60 5.02 59.9 6.27 59.9 2.00 70 3.00 70 5.00 70 8.00 69.9 10.00 69.8 30 FEET, per chord. 40' 45' 100' 115' 130' 140' 2 00' p and q. P q P q P q P q P q P q P q 0.03 30 0.04 30 0.05 30 0.07 30 0.08 30 0.08 30 0.10 30 0.12 45 0.13 45 0.18 45 0.22 45 0.27 45 0.29 45 0.35 45 0.28 60 0.31 60 0.42 60 0.52 60 0.63 60 0.70 60 0.85 60 0.55 75 0.61 75 0.82 75 1.02 75 1.22 75 1.36 75 1.63 75 0.94 90 1.06 90 1.41 90 1.76 90 2.12 90 2.35 90 2.82 90 1.50 105 1.68 105 2.28 105 2.65 105 3.36 104.9 3.74 104.9 4.48 104.8 2.24 120 2.51 120 3.35 120 4.16 119.9 5.00 119.9 5.53 119.9 6.59 119.8 3.19 135 3.60 134.9 4.80 134.9 5.95 134.9 7.18 134.8 7.95 134.8 9.52 134.7 4.36 149.9 4.89 149.9 6.51 149.9 8.13 149.8 9.77 149.7 10.4 149.6 13.0 149.4 &LIDE RULE CHORD LENGTH Change of curvature 30' 40' 45' 100' 115' 130' 140' 2 00' 30' 40' I Deflection angles. 40 0-02 0-02.7 0-03 0-04 0-05 0-06 0-06.7 0-08 P Q P 80 0-08 0-10.7 0-12 0-16 0-20 0-24 0-26.7 0-32 0.05 40 0.06 120 0-18 0-24 0-27 0-36 0-45 0-54 1-00 1-12 0.16 60 0.21 160 0-32 0-42.7 0-48 1-04 1-20 1-36 1-46.7 2-08 0.37 80 0.50 200 0-50 1-06.7 1-15 1-40 2-05 2-30 2-46.7 3-20 0.73 100 0.97 210 1-12 1-36 1-48 2-24 3-00 3-36 4-00 4-48 1.26 120 1.68 280 1-38 2-10.7 2-27 3-16 4-05 4-54 5-26.7 6-32 2.00 140 2.64 3202-08 36012-42 2-50.7 3-36 3-12 4-03 4-16 5-24 5-20 6-45 6-24 8-06 7-06.7 9-00 8-32 10-48 2.98 4.25 160 180 3.97 5.65 400J3-20 4-26.7 5-00 6-40 8-20 10-00 11-06.7 13-20 5.80 199.9 7.70 CHORD LENGTH Change of curvature 1 30' 40' 45' 100' 115' 130' 140' 2 00' 30' Deflection angles. 50 0-02'.5 0-03.5 0-03.75 0-05 0-06.25 0.07.5 0-08.5 0-10 P () 2-02.5 2-43.5 3-03.75 4-05 5-06.25 6-07.5 6-48.5 8-10 3.12 175 400 2-40 3-33.5 4-00 5-20 6-40 8-00 8-53.5 10-40 4.67 200 450 3-22.5 4-30 5-03.75 6-45 8-26.25 10-07.5 11-15 13-30 6.65 225 ,500 4-10 5-33.5 6.15 8-20 10-25 12-30 13-53.5 16-40 9.04 250 THE EASEMENT CURVE 103 40 FEET, per chord. 40' 45' TOO' 115' 130' 140' 2 00' p andg. q P q P q P Q. P q P q P Q. 40 0.07 40 0.09 40 0.12 40 0.14 40 0.16 40 0.19 40 60 0.24 60 0.31 60 0.39 60 0.47 60 0.52 60 0.63 60 80 0.56 80 0.75 80 0.93 80 1.12 80 1.24 80 1.49 80 100 1.08 100 1.45 100 1.81 100 2.18,100 2.42 100 2.90 100 120 1.89 120 2.51 120 3.15 120 3.78J119.9 4.19 119.9 5.02 119.9 140 2.98 140 3.98 139.9 4.97 139.9 5.95(139.9 6.61 139.8 7.95 139.8 160 4.46 159.9 5.95 159.9 7.42 159.8 8.90 159.8 9.90 159.7 11.90 159.6 179.9 6.38 179.9 8.50 179.8 10.60 179.7 12.70 179.6 14.10 179.4 16.90 179.2 199.8 8.70 199.8 11.60 199.7 14.50 199.5 17.40 199.3 19.30 199.0 23.10 198.7 50 FEET, per chord. 40' 45' 100' 115' 130' 140' 2 00' p and q. P q P q P q P q P q P q P q 0.10 50 0.11 s' 044' 002.5' +98.6 117.5' 128' 10.4' 417 l19' 129.5' 10.5' +38.6 151.4' 2 14' 022.6' +78.6 2 15. 5' 2 56' 040.5' 418 2 27' 3 19' 052' +18.6 2 37' 3 40' 103' +48.6 2 48' 4 12' 124' This solution shows the flexibility of this method which may be used for any length of spiral with any unit chord length. With a little experience with the slide rule solutions can be made more rapidly than by the use of tables. 47. The following gives in a concise form the equation for finding the deflections for a spiral and for which the use of the slide rule is particularly adapted. By Equation (36), Let Then 0.1 L = the chord length. NL = I and 10 (41) THE EASEMENT CURVE 113 The next table gives the deflections, for a change of curvature of one degree per chord, at the end of chords of 10 feet each. Deflections for chords of 10 feet and for a change of curvature of one degree per chord. 1 2 3 4 5 6 7 8 9 10 Gor 004' 009' 016' 025' 036' 049' 104' 121' 140' For any other chord length and change of curvature per chord, multiply the tabular amount by the given change of curvature, in degrees, per chord, and then multiply this product by the given chord length divided by 10, the result being the required deflection. For fractional chords multiply the deflection, found for the first chord, by the squares of the fractions which represent the distances of the points from the beginning of the spiral, expressed in chords. 48. The following diagram is for finding e in inches from the speed of the train and the degree of the curve. In using it find the intersection of the curve and the line of the adopted speed, then go parallel to the x axis to the intersection with the line of the degree of the curve, then go 114 THE SLIDE RULE THE EASEMENT CURVE 115 parallel to the y axis and read the result on the bottom line of the diagram, e.g., to find the value of e for a speed of 40 miles per hour and for a 4 curve, the result would be 4.6 inches. 49. The next diagram is for finding l c in feet from the adopted speed in miles per hour, the degree of the curve and the adopted rate of gaining the elevation of the outer rail. In using it, find the intersection of the line of the adopted speed and the curve, then go parallel to the x axis to the intersection with the line of the degree of the curve, then go parallel to the y axis to the intersection with the line of the adopted rate of gain of the elevation of the outer rail, then go parallel to the x axis to the left line of the diagram and read the length of the spiral, e.g., to find the length of the spiral for a 4 curve, speed 40 miles per hour and a rate of gaining the eleva- tion of the outer rail of 1.6 inches per second, find the intersection of the 40 miles per hour line with the curve, then go parallel to the x axis to the 4 line, then go parallel to the y axis to the 1.6 line, then go parallel to the x axis to the left line of the diagram and read 160 feet. 116 THE SLIDE RULE -o Q. y axis i j ? Qo co O c^o bfl c^ Q O O O IO O to cr> ^- ^ CHAPTER VII EARTHWORK 50. For finding the volumes of earthwork in railroad grading, it is necessary to take cross- sections in the field and from the notes taken the volumes may be figured. The method of taking cross-sections where an ordinary level is used is as follows: 1st. From the grade line established on the profile made from the levels run over the center line of the survey, find the elevation of grade at the station where the cross-section is to be taken. 2nd. From the readings taken on some bench mark and on turning points, find the H.I. of the level. 3rd. Subtract the elevation of grade from the H.I. of the level and the result is r g , the rod to grade. 4th. Read the rod held at the center line stake and subtract this reading, r 8 , from r g and the result is the center cut or fill, cut if + and fill if -. 5th. Assuming the section to be a level 117 118 THE SLIDE RULE section ; figure, from the given width of roau- bed and ratio of the side slope, the distance from the center line to the edge of the side slope (equal to one-half of the width of the road- bed plus the side slope times the center cut or fill). Going out this distance estimate the difference of elevation between this point and the center line point. Apply this difference to the cut or fill at the center and from the resulting amount of cut or fill figure anew the distance to the edge of the side slope. Go out this distance from the center stake and on a line at right angles to the center line. At the point thus found take a rod reading. Sub- tract this reading from the r g and the result is the cut or fill. From this cut or fill figure anew the distance out from the center and if this exceeds the actual distance out, go out further and if less than the actual distance go nearer the center line, until a point is found where the reading of the rod gives a cut or fill that corresponds with the distance from the center line. Where the section is not level across the top, the three-level form of section is usually found. 51. The following is a form for the notes of cross-sections: EARTHWORK 119 Station. Eleva- tion. Grade. L c R 85 105.3 105.2 0.0/10.0 +0.1 +0.4/10.6 84 107.4 104.6 +2.0/13.0 +2.8 +3.2/14.8 83 110.2 104.0 +4.2/16.3 +6.2 +7.0/20.5 82 109.4 103.4 +4.2/16.3 +6.0 +7.2/20.8 81 106.6 102.8 +3.0/14.5 +3.8 +4.0/16.0 80 107.8 102.2 +4.6/16.9 +5.6 +5.8/18.7 52. Where the grade contour does not cross the center line at right angles at a point where there is a change from cut to fill or vice versa, Fig. 39 shows the points that must be taken in cross-section work. 312 s Center D\ SIS &?2~~nr~L*ne ~ 814 p O "Tfl 7 oft*- - ^i^ " I. " g I Fig. 39 D is at station 312 + 60, F is opposite station 312 + 78, K is at station 313 + 06, H is at station 313 + 38, and E is opposite station 313 + 62. The lower diagrams show the forms of the cross-sections at D and H respectively. 120 THE SLIDE RULE From station 312 to D is in cut and may be figured by the average end area method. From D to E is in cut and may be figured as a pyramid of the base ABCD and of the height equal to the distance from D to E measured along the center line. From F to H is in fill and may be figured as a pyramid of the base GHJI and of the height equal to the distance from F to H measured along the center line. From H to station 314 is in fill and may be figured by the average area method. The following gives the notes for this part of the cross-section work: Station. Eleva- tion. Grade. L c R 314 313+62 164.0 166.1 -1.8/9.7 0.0/10.0 -2.1 -2.4/10.6 313+38 313+06 165.7 167.0 166.7 167.0 0.0/7.0 -1.0 0.0 -1.2/8.8 313 167.4 167 1 +0 3 312+78 0/7.0 312+60 312 170.2 172.4 167.5 168.1 +3.4/15.1 +5.2/17.8 +2.7 +4.3 0.0/10.0 +3.2/14.8 53. A level section is one at which the sur- face of the ground at right angles to the center line is horizontal. The area of a level section is found as follows: EARTHWORK 121 Let 6 = the width of the roadbed in feet, s = the ratio of the side slope, horizontal to vertical. c = the center cut or fill in feet. Fig. 40 From Fig. 40, it is seen that the area of the section is equal to be + sc X c, or A = (b + sc)c. . . . (42) 54. A three-level section is shown in Fig. 41. The amounts of the cut or fill at the center and the edges of the side slopes must be found. The area may be expressed in two ways: 1st. By the use of the grade triangle. In Fig. 41, ABFDE is the three-level section and BHF is its grade triangle. EG b The figure AHDE may be divided into two triangles with the common base EH = c+^- and of altitudes di and d r respectively. 122 THE SLIDE RULE Area^FD = (c + A)^_^ X Let Then \ \ N \j Fig. 41 G / H D = d r + di. B G Fig. 42 2nd. Without the use of the grade triangle. In Fig. 42, the area ABFDE is divided into four triangles, AEG and GFD having equal EARTHWORK 123 bases, each equal to ~, and AGE and GED hav- Zi ing the same altitude c. Area ABFDE = %(d r + di) + \ ^y^ - (44) 55. The two methods most commonly used for figuring volumes of earthwork are: 1st. The average end area, which is ob- tained by simply dividing the sum of the end areas by two and multiplying the result by the distance between the end sections. The last result divided by 27 gives the volume in cubic yards, if the measurements given in the notes are in feet. 2nd. The use of the prismoidal formula for precise results. If A\ and A 2 are the end areas, A m the middle area and h the length of the earthwork section, then, by the prismoidal formula, V = ^(A l + A m + A 2 ). . . (45) A m may be found by taking a mean of the dimensions of the end sections and figuring its area from these as its dimensions. 56. By Equation (46) the prismoidal correc- tion may be found. This correction sub- 124 THE SLIDE RULE tracted from the volume given by the average end area method gives the precise result. Cp = ~ Da) - (46) Cp is in cubic yards if ci, Co, DI and Z) are in feet. In Fig. 43, the volume A\GiFiF^G^ is the same, whether figured by the average end area method or by the prismoidal formula, because GiFi = GoF . Likewise, the volume FiDiEiEoFoDQ needs no correction if figured by the average end area method. If the entire volume of earthwork between sections zero and 1 is figured by the average end area method, then the correction to be applied, EARTHWORK 125 to get precise results, comes from the parts o and B^^^^D^. For the sum of these parts the area at section 1 = - , JU at section zero= -^, and at the middle sec- 4H Co + Ci DO + DI , tion = ^ , where CQ and a are Zi A /\ the respective center heights and D and DI are the respective total widths of the sections. Substituting these values in the equations for the volumes, VB = 2 ' 2 . _ I 2 ~ L ~\ I /CjDi Cp + Ci DQ + DI CoD \ +; -- + - = (2 Cl Z)! + CoDi + ciA> + 2 c D ). Subtract Vp from F E for Cp. Cp = 126 THE SLIDE RULE This expression shows that if the similar dimensions of the end section have about the same values, the average end area method gives close results. If I = 100 feet and Cp is expressed in cubic yards = 3 (ci - Co) (Di - A>.) 57. For finding the amounts of earthwork from cross-section notes, the slide rule does not give as rapid solutions as diagrams or tables. Where these are not available its use is advisable. For this class of problems the following equations are in forms readily solved by the slide rule. 1st. For volumes, in cubic yards, of level sections for lengths of 50 feet, 50 , . 50 F50 = 27 c6 + 27 sc ' Where c is the center cut or fill, b is the width of roadbed and s is the ratio of the side slope, horizontal to vertical. For b = 20 feet and s = 1J, Fso = 37.04 c + 2.78 c 2 ; EARTHWORK 127 For b = 14 feet and s = 1|, Foo = 25.95 c + 2.78 c 2 ; For b = 33 feet and s = 1^, F 50 = 61.11 c + 2.78c 2 ; For b = 27 feet and s = 1^, 7 5 o = 50 c + 2.78 c 2 . 2nd. For volumes, in cubic yards, of three level sections for lengths of 50 feet, T7 50^ , 50 b /rk ,, V*>=U D * + &2- 8 ( D -V- Where A is the area of the section in square feet and D is the entire width of the section, from left side height to right side height, and the other letters represent the same quantities as in the first case. For 6 = 20 feet and s = 1J, ion For 6 = 14 feet ands = 1J, 100 128 THE SLIDE RULE For 6 = 33 feet and s = 1 1, 100 For 6 = 27 feet and s= li, inn = Dc + 8 * 33 (- The following problems show the use of the slide rule for solving earthwork problems. Problem 34. Find the volume, in cubic yards, of earthwork from station 116 to sta- tion 118, assuming that the sections are level sections. 6 = 14 feet. s = 1J. F 50 = 25.95 c + 2.78 c 2 . Stations. Elevations. Grade. c 316 317 318 315.1 313.6 316.8 318.2 319.0 319.8 - 3.1 - 5.4 - 3.0 316 317 318 80.5 + 26.6 (140.1 + 81.0)X2 77.9 + 25.0 = 107.1 = 442.2 = 102.9 yds. 652.2 cu. Problem 35. Find the volume, in cubic yards, of the earthwork from station 212 to EARTHWORK 129 station 214, assuming that the sections are level sections. b = 33 feet. s = If. 750 = 61.11c + 2.78c 2 . Stations. Elevations. Grade. c 212 213 214 153.9 150.4 146.9 147.2 146.6 146.0 + 6.7 + 3.8 + 0.9 212 213 214 409.0 +124.5 =533.5 2(232.0 + 40.0) =544.0 54.9 + 2.2 = 57.1 1134. 6 cu. yds. Problem 36. Find the volume, in cubic yards, of the earthwork from station 272 to station 274. i b = 20 feet. s = 1 J. Sta- tions. Eleva- tions. Grade. L C R 272 273 274 153.9 150.4 146.9 147.2 146.6 146.0 + 7.8 +6.7 +3.8 +0.9 + 6.0 21.7 + 5.4 19.0 + 3.2 18.1 + 2.2 14.8 + 0.6 13.3 10.9 272 273 274 252.5+127.5 2(115.5+ 79.5) 20.1+ 25.9 = 380.0 = 390.0 = 46.0 816.0 cu. yds. 130 THE SLIDE RULE Problem 37. Find the volume, in cubic yards, of earthwork from station 376 to sta- tion 378. b = 27 feet. s = 1J. 100 F M = c + 8.33 (D - 6). Sta- tions. Eleva- tions. Grade. L C R 7fi Q1 K 1 ,1<3 9 - 1.2 Q 1 -4.6 077 01 q (\ 01 Q n 15.3 -2.8 C A 20.4 -7.8 070 01 f* Q 01Q 17.7 -2.2 * n 25.2 - 3.4 0/0 376 OIO .0 102 5 4- 72 5 16.8 = 175 18.6 377 2(214 5 +132.5) = 694.0 378 98.3 + 70.0 = 168.3 1037. 3 cu. yds. Problem 38. Find the prismoidal correction for the volume found in Problem 36. Cp = ~ Co) (Dl ~ From station 272 to station 273 2.9 X 7.8 3.24 = 6.98. EARTHWORK 131 From station 273 to station 274 2.9 X 8.7 7.78 C P = 3.24 " 14.76 cu. yds. Problem 39. Find the prismoidal correction for the volume found in Problem 37. From station 376 to station 377 2.3 X 7.2 Cp - -24- From station 377 to station 378 = 2.4 X 7.5 = 5.55 3.24 10.66 cu. yds. There are other problems in curves and earthwork in which the slide rule may be used to advantage. However, a sufficient number of problems has been given to guide the stu- dent in deciding when to use it. 132 THE SLIDE RULE 05 1 -o 03 8 II II I B I-O U IrO U I ^ ^ ^ ^ ^ > s > I "** 3 rss S n 0) 5 S ^?s? oco SS o u ^.^-J^^^f cococ : : : : : : :?H^ : : : S : : : : '^ ' o :.2 : : : H3 . . . io : : : ; :||||| H || j j ; _ S ' ; ; S IS|| n | ; ; 3 : : eJ G 3 el c3 e OS ^ g^3 : C3 c3 iS Jf ' .S.S-S'S III S S S .PH "^ "^ I 18 1 ffc fa cj ^JS a! s.a 2 ^ 3 g S J2 g "o FORMULAS 133 ^ I !5 ! ^ i S ^ TTJ o S ^ a d oq II -H S5si: ^^ H- I 3 II ' s ' S ? ? a 5 ^ + i + i 4 22 2 3 8 i 5 I 05 ' I l^ ' + 'S + ^ i OQ ! ^J O oq - -' -C 11 J,' + o "a o O i-H ^ 134 THE SLIDE RULE 'a s a Q o !i x 10 s 8 03 03 ^ <=> d II 3 X * P tf 5 S S3 05 cq - T + ^> 05 O i-t O &3 i 1 ^. * I j M II g ^* tf 1 ^ >-H e 2 o a g ft 8 + ^-1 II i p a ' ^ -9 ii D OQ II 1 IT^ * ^"T e e. | ft > S a 1 ^ a* S II ft* ft> %|^ J, Q >!* 3 ^ 1 ii ii ii ii II 136 THE SLIDE RULE . I I I S i +3 2- I | i ft) c C)l^ + +" g s a S " + + " II S" ,. % 7 M ^ i, IcOOOrHC^ c ^' ! f OOOi-H-t>-'-'i-l ! ; x" !>. 00 OJ OO OO OO THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. 6 196948 A/o \/ t REC'D LD MV 15 "69 -9 A SEMTONJLL IMAY o ? tase C. BERKELEY / / 288606 UNIVERSITY OF CALIFORNIA LIBRARY