UC-NRLF 
 
GIFT 
 
 GIFT OF 
 
 J. Poster Plagg 
 

 
SYSTEM 
 
 USEFUL FORMULAE, 
 
 ADAPTED TO THE PRACTICAL OPERATIONS OF 
 
 LOCATING AND CONSTRUCTING 
 
 RAILROADS: 
 
 A PAPER READ BEFORE THE BOSTON SOCIETY OF CIVIL ENGINEERS, 
 DECEMBER, MDCCCXLIX. 
 
 By SIMEON BORDEX, 
 
 CIVIL E N G I X E E K. 
 
 BOSTON : 
 
 CHARLES C. LITTLE AND JAMES BROWN. 
 MDCCCLL 
 
Entered according to Act of Congress, in the year 1850, by 
 
 CHARLES C. LITTLE AND JAMES BROWN, 
 in the Clerk s Office of the District Court of the District of Massachusetts 
 
 GIFT 
 
 
 
 BOSTON: 
 
 PRINTED BY WILLIAM 
 
PREFACE. 
 
 THE formulae contained in the following pages were written, as their 
 title indicates, for an original paper, which was read before and discussed by 
 the Boston Society of Civil Engineers. 
 
 The Society, having accumulated a number of original papers, appointed 
 a committee to examine and report upon the expediency of having them 
 published. After much consideration, the committee reported that, with 
 their limited resources and present necessities, it would be improper to 
 incur the expense of printing, but recommended an early publication of this 
 paper ; and for that purpose the manuscript has been placed in the hands 
 of the publisher. 
 
 Notwithstanding the obvious importance of constructing the curves of a 
 railroad upon the best practical locations, and giving to their forms or 
 alinement the greatest degree of regularity practicable, the investigation, or, 
 which is more probable, the publication of anything like a system of con 
 venient formulae to aid the young engineer, and such others as have not 
 had the advantage of a good mathematical education, in the proper perform 
 ance of this character of work, it is believed has not yet found a place 
 
 M166992 
 
IV PREFACE. 
 
 upon the shelves of our libraries or book-stores. To supply this deficiency 
 in the library of the civil engineer, particularly the railroad engineer, is 
 the object of the present paper. 
 
 It is not pretended that all the formulae contained in this paper are 
 original. The principles which have governed the investigations for com 
 puting the elements required for tracing curves, where their localities are 
 such as to admit of the most simple and convenient methods, have, it is 
 believed, been published, and are known by most engineers who have been 
 engaged in the construction of railroads, since the commencement of the 
 railroad system. Neither is it pretended that the system of formulae is 
 complete, or that it contains formulae suited to every case that can arise. 
 The writer can only say, that after considerable experience in the construc 
 tion of railroads, he does not recollect a case presenting itself which would 
 not be solved by some one of the formulae ; and it is believed that, with 
 slight modifications, such as any geometer would be able, without difficulty, 
 to make, they may be adapted to all common or ordinary cases. 
 
 Curves in a railroad, unless their radius be very large, are known to 
 be objectionable ; but the contour of the surface, the existence of valuable 
 buildings, of streams, rivers, ponds, oceans, etc., in the line between the 
 points which it is desired to connect, render the adoption of curves neces 
 sary. It is likewise a well-established fact, that the greater the degree of 
 regularity and precision exercised in the construction of curves, the more 
 safely and easily can trains be run over them. 
 
 The main objects of the formulae are twofold; viz., that of enabling the 
 engineer to mark out the curves of a railroad with the greatest degree of 
 
PREFACE. V 
 
 precision and convenience, and to locate them in situations the most desir 
 able. To render this subject clear and perspicuous to every one who may 
 have occasion to locate or mark out curves, upon railroads and other places, 
 the paper is commenced with the investigation of the most simple problems, 
 which are succeeded by the more intricate ; each case being illustrated 
 with diagrams, and accompanied by examples of computation. 
 
 The subject of switches and frogs being blended with the elements of 
 turnout curves, has been considered in connection with them ; and in their 
 arrangement the same objects have been kept in view; and, for this end, 
 each case has been likewise accompanied with a diagram and an example 
 of computation. 
 
 To render the work more useful, there have been added formulae for 
 computing the cubic contents of excavations and embankments, and a formula 
 for computing the difference in height to be observed in laying down the 
 rails upon a railroad curve, based upon its radius and the velocity of the 
 cars. 
 
 BOSTON, December, 1850. 
 
CONTENTS. 
 
 SECTION PAGE 
 
 1. The subject of connecting straight lines of different bearings by 
 
 simple curves considered, with an investigation of formula 
 
 for ascertaining the necessary elements 3 
 
 2. Demonstration of the formula for angles of deflection for locating 
 
 a simple curve 4 
 
 3. Field operations for locating or laying out simple curves de 
 
 scribed 5 
 
 4. Hints to young engineers 7 
 
 5. The subject of dividing curves into a series or system of long 
 
 chords considered and recommended, with practical hints. . . 7 
 
 6. The convenience of making the termination of the long chords 
 
 correspond with even stations considered and recommended 8 
 
 7. A method of correcting the location of tangent points from a re- 
 
 survey considered and explained 9 
 
 8. Formulae for ascertaining the necessary elements for locating 
 
 simple curves, with examples of computation 10 
 
 9. A method of laying out simple curves by a system or series of 
 
 long chords described, with practical examples of computing 
 
 the necessary elements 13 
 
 10. Field operations of laying out simple curves by means of long 
 
 chords described 15 
 
 11. The office and field labors connected with laying out simple curves 
 
 by the aid of long chords, further considered 17 
 
 12.- Further field operations of laying out simple curves described. . 18 
 
Vlll CONTENTS. 
 
 SECTION PAGE 
 
 13. Practical operations and the necessary computations for obtaining 
 
 the elements of simple curves, and for correcting the posi 
 tion of the tangent points, etc., based upon a traverse, con 
 sidered and illustrated 19 
 
 14. The same subject continued, with modifications 25 
 
 15. A method of laying down primitive or instrument points in a 
 
 curve by ordinates measured from the tangent lines continued 
 
 to apex, or the point of their intersection 34 
 
 16. A continuation of the same subject, with investigations of form 
 
 ulas, specimens of computation, and a description of the field 
 work 35 
 
 17. Hints concerning the advantages of the above named method 38 
 
 18. Reasons for not pursuing the subject of simple curves further ... 38 
 
 19. Commencement of the subject of reversed curves simplicity of 
 
 the problem for connecting two parallel lines by curves of 
 equal radii 39 
 
 20. Investigation of formulae 39 
 
 21. Specimen of computing elements 42 
 
 22. The same subject continued, with the given elements varied also 
 
 an investigation of formulae, with specimens of computation 43 
 
 23 and 24 . The propriety of reversed curves with and without a piece 
 
 of straight line between them considered and discussed. . 44 to 48 
 
 25. The same subject continued, and elucidated by diagrams and com 
 
 putations 48 
 
 26. The subject of reversed curves for uniting two lines of different 
 
 bearings briefly discussed 51 
 
 27. Investigation of formulae for obtaining the necessary elements. . . 51 
 
 28. Examples of computation 53 
 
 29. A form of reversed curves for uniting lines of different bearings 
 
 with one radius given also containing an investigation of 
 formulae, with specimens of computation 55 
 
CONTENTS. IX 
 
 SECTION PAGE 
 
 30. A form of reversed curves for uniting lines of different bearings, 
 
 having one tangent point, and a centre angle to one of the 
 curves, fixed also an investigation of formulas, with 
 specimens of computation 59 
 
 31. A form of reversed curves with or without a piece of straight 
 
 line between them, for uniting tracks of different or like 
 bearings, having the tangent points fixed, and the length of 
 the radii governed by the contour of the surface, or by other 
 considerations containing also an investigation of formulae, 
 with specimens of computation G5 
 
 32. The subject of reversed curves and turnout curves briefly noticed 69 
 
 33. Principles connected with the switch-bar briefly discussed 69 
 
 34. 35, and 36. Remarks respecting turnouts from a straight line to 
 
 a side track with investigation of formulae and specimens 
 of computing the necessary elements, including the angle of 
 the frog 71 to 77 
 
 37. Directions for performing the field work for laying out turnout 
 
 curves 77 
 
 38. Investigation of formulae for ascertaining the radius of a turnout 
 
 from a straight track, suited to a given frog ; with specimens 
 
 of computation 77 
 
 39. 40, and 41. Turnout from the outside of a curve in the main 
 
 track investigation of formulae, and specimens of compu 
 tation, including the angle of the frog ." 81 to 87 
 
 42 and 43. Investigation of formulae for ascertaining the radius of a 
 turnout upon the outside of a curve to suit a given frog 
 including examples of computation 87 to 93 
 
 44. Turnout upon the outside of a curve 1 Investigation of formulae, 
 with specimens of computing the necessary elements, in 
 cluding the angle of the frog 93 
 
 45. Investigation of formulae for determining a radius of a turnout 
 upon the outside of a curve to suit a given frog, including 
 
 examples of computation 99 
 
 A 
 
X CONTENTS. 
 
 SECTION PAGE 
 
 46. Discussion of the general principles which combine in the loca 
 
 tion of a turntable by the side of the main track, near a 
 water station, with an investigation of formulae for determin 
 ing the elements, including specimens of computation 103 
 
 47. The field work of locating the turnout to the turntable described 104 
 
 47. Another form of the problem for locating a turnout to a turntable, 
 viz., by commencing at existing joints in the rails of the 
 main track with an investigation of formulae, specimens 
 of computation, and a description of the field work 106 
 
 49 and 50. The subject of locating railroads with curves (or straight 
 lines) in places inaccessible to instruments and persons, ex 
 cept in boats or by some extraordinary device, considered 
 investigation of formulae, followed by specimens of com 
 putation field operations described 109 to 135 
 
 51 . Remarks on the objects of this paper 135 
 
 52. Remarks on the subject of vertical curves formula for deter 
 
 mining their elements examples of computation 137 
 
 53. Hints respecting the management of the elements obtained 146 
 
 54. Subject of vertical curves, continued formula deduced, and 
 
 practical operations described 146 
 
 55. Same subject, continued 150 
 
 56. Remarks upon the subject of bending rails for curves of short 
 
 radii formula for obtaining the elements necessary to mark 
 
 out a pattern board examples of computation 153 
 
 57. Several formulae for determining the radii of curvature in railroad 
 
 tracks, with examples of computation 158 
 
 58. Remarks upon the elevation of the outside rail of a railroad 
 
 curve, with formulae and specimens of computation 161 
 
 59 to 62. Demonstrations of formulae for determining the cubic con 
 tents of pyramids, wedges, parallelepipeds, etc 166 to 174 
 
 63 to 66. Formulae for determining the cubic contents of excavations 
 
 and embankments, with examples of computation. . . . 174 to 188 
 
E R B A T A. 
 
 On page 77, ninth and tenth lines from top, for " = 90 Sw 
 + \ (180 C) " read " (90 + Sw) i (180 C)." 
 
 On page 87, seventh line from bottom, for "= r -f~ \ h -f- d" 
 read " r % h + d," 
 
 On page 113, second line from top, for "S 75 08 35" 07 E" 
 "875 08 35" -37 E." 
 
 On page 119, fourteenth line from bottom, for "We then dis 
 cover " read " We then determine," etc. 
 
AN INVESTIGATION 
 
 F 
 
 USEFUL FORMULAE, 
 
 ADAPTED TO 
 
 THE PRACTICAL OPERATIONS OF LOCATING AND LAYING OUT 
 RAILROAD CURVES. 
 
In 3mie0tigatioE of 
 
 ETC. ETC. ETC. 
 
 ( 1 ) COMMENCING with the most simple operations and forms 
 of computations, we shall assume as the most simple of railroad 
 curves, such as will unite two direct or straight lines of different 
 hearings in such a manner that each of these lines will "become 
 tangents to said curve. Having run these straight or tangent 
 lines to their intersections at A, (which intersection we shall 
 hereafter term their apex, see Fig. 1,) and determined the magnitude 
 of their angle, we then proceed to determine the most favorahle 
 location for the track. In order to do this, we examine the contour 
 of the surface, and select the position we think the most favorahle 
 opposite the angle at A ; then we run and measure from A to this 
 most favorahle place for the track hefore selected, in a direction 
 that will hisect the angle A ; that is, from A to Ss, and this 
 distance we represent "by b. We then proceed to ascertain a radius 
 for the curve, which shall pass through the point 83 and so run into 
 the straight lines that they shall be tangents to said curve. 
 
4 USEFUL FORMULAE. 
 
 Putting r for the radius in feet, or the unit of measure ; 
 
 " C for the angle at the centre of the curve corresponding 
 
 with the subtense of its arc = (180 A ;) 
 " t for the distance from apex to the points of commence 
 ments of the curve, or where the direct line becomes a 
 tangent to said curve, viz., from A to T, (See Fig. ;) 
 We have sin. J C : b : : sin. (J C + i A) : t = * sin ^ + * A) 
 and cos. J A :t :: sin. \ A : r == ^AA^^^+A^ == 
 tan. \ K.I cot. J C. (1) 
 
 Having thus determined the radius of the desired curve for 
 uniting the aforesaid tangent lines, and the distance from their 
 apex to their tangent points, or points of commencement of said 
 curve, we will now proceed to investigate some of the most simple 
 and practical methods of locating or laying out the same. 
 
 ( 2 ) We consider first what we shall call the method of de 
 flections. To explain this operation, let us suppose the arc or 
 curve to be divided into such equal parts as a chord of the length 
 of the chain contemplated to be used will span. We have in the 
 course of our practice generally used for this purpose a chain fifty 
 feet in length ; as, by using a short chain, the chords and the arcs 
 (if the radius be of much magnitude) are nearly of the same length, 
 which affords a great convenience in determining the deflecting 
 angle corresponding to such short chords (consisting of fractions 
 of the chain) as it will frequently be found desirable to use at the 
 commencement and termination of curves, that we may be enabled 
 to keep up a continuous notation of equi-distant stations. 
 
 Many engineers, 1 am aware, use a chain of 100 feet, which, 
 
DEFLECTING ANGLES. 5 
 
 if the radius be not of considerable magnitude, will give a 
 perceptible difference between the length of the chords and the 
 corresponding arcs they span. 
 
 To determine the angle of deflection corresponding to the length 
 of the chain to be used, we represent the length of the chain by cJi ; 
 the angle of deflection by D ; and the angle at the centre of the 
 curve, corresponding to its subtense, by C . 
 
 Supposing T, Si,= the chord ch, we have in the triangle C T Si, 
 
 \ (180 C ) = the angle T = Si ; 
 Now as the angle A T C = a right angle, or 90, we have 
 
 90" (J*=*-) = D; 
 
 Multiplying by 2, 180 (180 C ) = 2D ; 
 Subtracting 180, and changing signs, we have 
 
 2D = C ; 
 
 Consequently, D= i C ; (2) 
 
 Bisecting ch we have r : K : : J cli : sin. \ C = ^~ = sin. D (3) 
 
 It sometimes happens that we wish to know the value of D at 
 
 the commencement of our computations ; expanding the foregoing, 
 
 sin. D = * ch cot - *, A tan " * c (4) 
 
 ( 3 ) Having thus determined the angle of deflection, we now are 
 prepared for locating or marking out the curve. 
 
 Adjusting a good theodolite to the tangent point T, with its 
 principal telescope (or the telescope by which angles are determined) 
 
 * In delineating a curve by the method of deflections, it will be inconvenient to make the 
 stations further apart than the length of the chain used. 
 
6 USEFUL FORMULAE. 
 
 pointing in the direction of A, and with its watch telescope (all 
 theodolites should be provided with watch telescopes) pointing to 
 any convenient well-defined mark, lay off the angle D, and then 
 stretching the chain from T, fix its terminus Si in range with the 
 principal telescope, the point thus marked will he in the curve ; 
 then, laying off another angle of deflection which will read upon the 
 instrument = 2D, stretch the chain from Si to 82, fixing the ter 
 minus at 82 in the range indicated by the main telescope ; the 
 points thus formed will also be in the curve. And in like manner 
 we proceed to fix the points Ss, 84, 85, etc., until the curve is com 
 pleted ; or, as it more frequently happens, as long as the contour 
 of the surface will permit us to sec distinctly. 
 
 Let us now suppose an obstacle which will prevent seeing beyond 
 83. We then remove our theodolite to that station ; and, after 
 having duly adjusted it, with its main telescope pointing at the 
 station thus left, and the watch telescope to some convenient mark, 
 lay off an angle equal to 180, minus the sum of the deflections 
 made at the first station, plus one deflection. Then, stretching the 
 chain from the station where the theodolite is now adjusted, place 
 the other terminus in range with the principal telescope, as here 
 tofore described. Then, proceed in the manner above described, to 
 lay off deflections and chords, until you connect with the straight or 
 tangent lines, or as long as the contour of the country will admit, 
 when the instrument must be again changed, and the like operations 
 performed until the whole curve is completed. If the curve is not 
 measured by whole chords = c/t, the deflection for the fraction of ch 
 will be to that of ch in the proportion the fraction bears to a whole ; 
 which, although not strictly correct, is sufficiently near for practice. 
 \Ve here remark that we do not recommend that more than 10 or 
 
DIVIDING CURVES, 7 
 
 12 chords of 50 feet each should be laid down with the theodolite 
 at one station, it being conducive to accuracy not to permit a great 
 difference between the direction of the chord (cji) and the direction 
 of the pointing of the telescope of the instrument. 
 
 ( 4 ) We have thus endeavored to describe a practical method of 
 laying out one of the simplest of railroad curves which shall unite 
 two straight lines having different bearings ; but, as there is a 
 great variety of methods to accomplish this end, which may be 
 resorted to, some of which seem to possess peculiar adaptation to 
 certain localities, I have thought it would not be uninteresting to 
 describe some of them, believing a few hints of this kind would lead 
 the new beginner to different modes of reasoning and investigation ; 
 and, if he possesses a tolerable knowledge of the elements of plain 
 geometry, he will be able always to select, if not the method best 
 adapted to the circumstances of the case, at least one well suited 
 and convenient. 
 
 ( 5 ) If the curve be of large radius, (and curves cannot well be 
 constructed with too large a radius,) and its location suits the 
 contour or surface of the country, and the apex angle be large, a 
 very convenient and accurate method of proceeding will be to divide 
 the curve into a series of segments of some 500 or 600 feet each, 
 as may be thought best ; taking care as far as possible to make the 
 terminus of each segment an even station ; it will, however, fre 
 quently happen that the number of the stations marking the tangent 
 
 * In the location and construction of a railroad, it has been found convenient in practice to 
 divide the centre line into equal parts, technically called stations, which contain a given number 
 of the units of measure used in the construction. In the United States, the foot has been taken 
 for the unit of measure, and the railroad stations one hundred feet asunder. 
 
8 USEFUL FORMULAE. 
 
 points will contain a fraction, and of course in this case, the segments 
 should contain, besides a number of whole chords, the fractional 
 chord the case requires. 
 
 Having determined upon these preliminary considerations, we 
 first calculate the relative position of points on the line between 
 the tangent points and apex, w r hich shall correspond to radii of the 
 curve passing through the termini of the several segments ; also 
 the distance from the points upon the radii of the curve, together 
 with the angles the radii made with the tangent lines ; the points 
 thus computed are readily determined and marked with a good 
 degree of accuracy, and become instrument stations, from whence 
 the intermediate stations are readily filled up as before described, 
 by chords and deflections, with less liability to practical errors 
 than the preceding method. 
 
 ( 6 ) If the curve be of large radius and the apex angle small, 
 the distance of the tangent points from the apex and from the 
 tangent lines to the curve, will become too great to be measured 
 conveniently with a proper degree of accuracy ; under this condition 
 of the case, we divide the curve into a convenient number of seg 
 ments, taking care as before to have their termini to correspond 
 with the even stations. 
 
 Having thus determined upon the divisions, we compute the 
 chords and angles corresponding therewith, and then proceed to lay 
 off the angles with great care, and measure the chords with as 
 great a degree of accuracy as is practicable, carefully marking 
 the termini of each chord. If the chords and angles when laid 
 down correspond with the tangent points and tangent lines, we 
 
TESTING PRIMITIVE CHORDS. 9 
 
 proceed to put in the intermediate stations by adjusting the 
 theodolite to the terminus of one of these chords ; then, pointing to 
 the other terminus, we proceed to lay down the intermediate chords 
 and stations by the aid of deflections, in the manner described in 
 the foregoing. It is usual to fill up the intermediate stations 
 formed by two of the primitive chords without changing the instru 
 ment ; having done this, we move our instrument to another 
 terminus common to two chords, which have not been filled up with 
 the intermediate stations, and in like manner we proceed until we 
 complete the curve. 
 
 ( 7 ) If, in running our first chords until we have exhausted our 
 computations, we do not find the work to correspond with a proper 
 degree of accuracy to the tangent points, and tangent lines, we 
 make a connection with the tangent points, and carefully ascertain 
 the length of the chord, and the magnitude of the angle with the 
 tangent line ; and, with the elements thus obtained, we recompute 
 the work, and determine the relative position of the tangent 
 points, and fix them anew. We then proceed to perform the work 
 of laying down the primitive chords a second time ; when, if there 
 have been no mistakes made, the work will prove practically 
 accurate. It is very seldom that a second computation will be 
 needed until the road is graded, when the measurement taken upon 
 the graded surface will not be likely to be identical with the 
 primitive measurement; then, a resurvey, and a re-establishment 
 of the tangent points, become, if not absolutely necessary, desir 
 able ; but, if the first survey was performed with any tolerable 
 degree of care, and the grading well finished, no difficulty will be 
 experienced in laying the curve upon the graduated road-bed, 
 without changing its radius. 
 C 
 
10 USEFUL FORMULA. 
 
 ( 8 ) Having thus briefly described the principles which govern 
 us in laying down simple curves, we would now introduce an 
 example, accompanied by specimens of calculation for each par 
 ticular case. 
 
 PRACTICAL EXAMPLES OF CALCULATIONS SUITED TO THE 
 CASES DESCRIBED IN THE FOREGOING PAGES. 
 
 Assuming A = 160, then will C = 180 A = 20 
 b = 132 feet 
 ch = 50 feet 
 
 To find the deflection, we have (4) sin. D = 
 
 Thus, | A = 80 00 00" cot. = 9-2463188 
 
 i C = 5 00 00" tan. = 8-9419518 
 
 ch = 25 feet log. = 1-3979400 
 
 6 = 132 feet co. ar. log. = 7-8794261 
 
 D = 010 02".64 sin. = 7-4656367 
 
 Iii this case it will be seen that D = 10 02". 64 is an incon 
 venient angle to add or subtract, or even read upon the instrument ; 
 we therefore, to remedy this objection, adopt D = 10 , which 
 will not materially change the length of the radius or the location 
 of the curve. This change requires that we base our calculations 
 for determining the elements needed, upon D = 10 ; therefore, 
 to find the radius we have 
 
 Sin.D:l(*::R:r = -A^ D (5) 
 
 and to find t we have 
 
 Cos. J C : r : : sin. J C : t = tan. J C . r (G) 
 
 (representing by t the distance from the apex to the tangent point, 
 or from A to T on the diagram.) 
 
COMPUTATION OF ELEMENTS. 11 
 
 Thus, ch == 25 feet - - - - log. = 1 -3979400 
 
 D = 10 co. ar. sin. = 2-5362745 
 
 Consequently, r = 8594-38 feet log. = 3-9342145 
 
 Then, \ C = 10 00 00" tan. = 9-2463188 
 
 t =1515-42 feet log. = 3-1805333 
 
 Having thus ascertained the radius r = 8594-38 feet, and the 
 distance from apex to tangent point, (t= 1515 42 feet,) we now 
 proceed to find b. By analogy we have 
 
 Sin. (i C + i A) : t :: sin.i C : b = $* = t tan. J C (7) 
 
 Then, t = as above, log. = 3 - 1805333 
 
 C = 5 00 00" tan. = 8-9419518 
 
 b = 132-58 feet log. = 2-1224851 
 
 At commencement ) 
 
 > b = 132-00 feet 
 we assumed ) 
 
 Difference, = 0-58 feet 
 
 Thus it appears that the change of the radius to cause it to 
 correspond with D = 10 will only change the location of the 
 curve from the position designed for it 58 feet; a quantity too 
 small to be generally accounted anything in choosing the position 
 of a curve. 
 
 We now proceed to find the length of the curve. Putting r" for 
 an arc in seconds = radius, and C" for the number of seconds 
 contained in the centre angle which measures the curve, and a the 
 arc subtending the angle C", we have by analogy 
 
 r" : r : : C" : a = ^ (8) 
 
 Thus, r = 8594.38 feet log. = 3.9342145 
 
 C" = 7200" log. = 4.8573325 
 
 r" = co. ar. log. = 4.6855749 
 
 a 3000.00 feet log. = 3.4771219 
 
COMPUTING ELEMENTS. 13 
 
 Now, that we may show this matter as complicated as it is 
 generally found in practice, let us suppose the pin at T marking 
 the first tangent point, to be numbered 140 38; the integer repre 
 senting the whole number of the stations, and the decimals the 
 fractional part of the space beyond station 140. We have found 
 the length of the arc a equal 30 whole stations ; which, added 
 to 140.38, gives 170*38 for the number of the other tangent 
 station T . 
 
 ( 9 ) Let us now proceed to show the method of laying out the 
 curve by the method of a series of long chords corresponding to 
 about 750 feet of the arc, which we afterwards fill up by simple 
 deflections and 50 feet chords = ch. 
 
 In the first place we have supposed the first tangent pin to bear 
 the number 140*38 ; if we now add 7*12 stations, it will make the 
 number 147.50; therefore our first chord will extend from station 
 140*38 to station 147*50. We now add 7* 50 stations to station 
 147*50, which makes the number 155 ; therefore our second chord 
 extends from station 147*50 to station 155. We then add 7*50 
 stations to station 155, which increases the number to 162*50; 
 therefore our third chord extends from station 155 to 162*50. We 
 now add 7*88 stations to 162*50, which increases the number to 
 170*38, and brings us to the tangent point T . 
 
 Having assumed ch = 50 feet, (which is practically the same 
 length of the arc it spans when the radius is of considerable 
 length,) we must now proceed to determine the length of the 
 several long chords we have divided the arc into. (See the diagram 
 on the opposite page.) 
 
14 USEFUL FORMULAE. 
 
 Our first long chord corresponds to 712 feet of arc. We How 
 propose to find the centre angle which it subtends by the following 
 formula. 
 
 Eepresenting the arc by a and the centre angle in seconds by C", 
 we have 
 
 r : r" : : a : C" = -^ (9) 
 
 and further we have 
 
 Cos. \ C" : r : : sin. C" : Ch?= ~~^ lf (10) 
 
 Thus, r = 8594-38 feet co. ar. log. = 6-0657855 
 
 a = 712-00 feet log. = 2-8524800 
 
 r" = log. = 5-3144251 
 
 C" = 17088" log. = 4-2326906 
 
 C" reduced to degrees and minutes = 4 44 48" sin. = 8-9177692 
 
 C" " " " = 2 22 24" ar. co. cos. = 0-0003719 
 
 r log. = 3-9342145 
 
 First chord = Ch = 711 - 79 feet = log. = 2 - 8523556 
 
 The arc corresponding to the succeeding chord is 750 feet long, 
 and is composed of 15 whole deflections, and each deflection being 
 10 , the centre angle will be equal to twice that number of deflec 
 tions, or 300 , which amounts to 5 00 00". 
 
 Now, by formula (10) we have 
 
 We distinguish the primitive or long chords by Cfi in contradistinction with the 50 feet or 
 deflecting chords, which are represented by ch. 
 
DETERMINATIONS OF ANGLES. 15 
 
 C" in degrees = 5 00 00" sin. = 8-9402960 
 
 AC" in degrees = 2 30 00" co. ar. cos. = 0-0004135 
 
 r= log. = 3-9342145 
 
 Second chord = 749-763 feet log. = 2-8749240 
 
 Third chord, of course, will be of the same length, viz., 749 763. 
 
 The arc corresponding to the fourth chord is 788 feet long, and 
 we find C" and Oh by formula (9) and (10) 
 
 Thus, r co. ar. log. = 6-0657855 
 
 a = 788 feet log- = 2-8965262 
 
 r" log. = 5-3144251 
 
 C" = 18912" log. = 4-2767368 
 
 C" reduced to degrees and minutes = 5 15 12" sin. = 8-9617037 
 
 i C" " " " " " = 2 37 36" co. ar. cos. = 0-0004554 
 r = log. == 3-9342145 
 
 Fourth chord = Ch = 787-723 feet log. = 2-8963736 
 
 We have not attempted to make the foregoing calculations 
 strictly exact ; our angles being always taken to correspond with 
 the nearest second, which in most cases gives a greater degree of 
 accuracy than we can practically execute. 
 
 (10) Having thus computed the centre angles and chords 
 corresponding with the proposed division of the arc, we will 
 endeavor to give the method of laying out the work. 
 
 In the first place we will determine the angles at the tangent 
 points, and the several stations which are to mark the termini of 
 the divisions of chords. 
 
16 USEFUL FOKMUL^. 
 
 Commencing at the tangent point numbered 140*38, our first 
 operation is to determine the angles in the several triangles, viz. ; 
 T C 1, 1 C 2, 2 C 3, 3 C T. (See Fig. 2.) 
 
 The angle C", which we have determined, will correspond to C, 
 and in triangle T C 1, C = 4 44 48", (see section 9,) each of 
 these triangles being isosceles, and having computed their centre 
 angles, we have only to deduct them from 180, and half the 
 remainder will be the value of each of the remaining angles. Now, 
 calling triangle TCI No. 1, and 1 C 2 No. 2, and 2 C 3 No. 3, 
 and 3 C T No. 4, we proceed to find their angles in the order we 
 have named them. 
 
 DETERMINATION OF THE ANGLE AT FIRST TANGENT POINT, OR T. 
 
 The angle at centre of curve for triangle No. 1=4 44 48" 
 and 180 - 4 2 044/48 " = 87 37 36" = angle at T, or at sta 
 tion 140-38 and 147-50. 
 " angle at centre of curve for triangle No. 2 = 5 00 00" 
 
 2 
 
 and 155. 
 
 " angle at centre of curve for triangle No. 3 = 5 00 00" 
 and i8o -50 oo GO- = 87 o 3(y = angle at stations 155 and 
 
 162-50. 
 
 " angle at centre of curve for triangle No. 4 = 5 15 12" 
 and 180 - 5 2 015 12 " = 87 22 24" = angle at stations 
 162-50 and 170-38. 
 
 Having thus prepared the angles for the several stations above 
 named, and for the purpose of rendering our description easier to 
 be understood, we arrange them in their order, as in the foregoing 
 diagram. 
 
MEASURING ANGLES. 
 
 17 
 
 (11) Having thus represented our work as shown in the dia 
 gram, we now proceed to ascertain the angles at the termini of 
 the several chords. Thus, 
 
 At T, or station, ---- 140-38 = 90 -f 87 37 36" = 177 37 36" 
 
 "1, " ---- 147-50 = 87 37 36" -f- 87 30 = 175 07 36" 
 
 "2, " ---- 155-00 = 87 30 + 87 30 = 175 00 00" 
 
 "3, " ---- 162-50 = 87 30 -f- 87 22 24"= 174 52 24" 
 
 " T , " ---- 179-38 = 87 22 24" + 90 = 177 22 24" 
 
 Having ascertained the angles for each station, which, for con 
 venience, we write upon the several radii connecting said stations 
 with the centre of the curve in the diagram, we proceed to compute 
 the length of the several chords which span their respective arcs ; 
 commencing with triangle TCI, which is called No. 1, and pursuing 
 the calculations in the order shown in the diagram. Thus, 
 
 T = 87 37 36" . co. ar. 
 r = 8594-38 feet 
 C = 4 44 48" 
 
 Ch = 711-79 feet 
 
 sin. = 0-0003727 
 log. = 3-9342145 
 sin. = 8-9177692 
 
 
 
 = log. = 2-8523564 
 
 sin. = 0-0004135 
 
 log. = 3-9342145 
 sin. = 8-9402960 
 = log. = 2-8749240 
 
 3 =87 22 24" co. ar. sin. = 0-0004565 
 
 r log. = 3-9342145 
 
 QQ 
 
 co 
 
 1 = 87 30 
 
 00" 
 
 .... co 
 
 3 
 
 T3 
 
 
 
 
 
 
 a 
 
 T = 
 
 
 
 fe 
 
 VJ 
 
 
 
 
 
 
 C = 5 00 
 
 00" 
 
 
 1 
 
 i 
 
 Ch = 
 
 749 
 
 76 feet 
 
 C = 5 15 12" 
 
 Ch = 787 -72 feet 
 
 sin. = 8-9617037 
 
 = log. = 2-8963747 
 
 (12) Having thus prepared our work, we proceed to adjust the 
 
18 USEFUL FORMULAE. 
 
 theodolite to station T, or according to the locating stations, to 
 No. 140*38, with its principal telescope pointing in the direction of 
 the line of the road; which, it is presumed, has been properly 
 marked. Then, laying off an angle of 177 37 36", we measure 
 in the direction indicated by the telescope, 711*79 feet to 1 or to 
 station 147-50; then, moving the instrument to 147*50, and 
 pointing the principal telescope to T, we lay off an angle of 
 175 07 36", and measure in the direction indicated 749*76 feet 
 to 2 or station 155. Then, moving the instrument to station 2 and 
 pointing at 1, we lay off an angle of 175, and measure in the 
 direction indicated 749*76 feet to 3 or to station 162*50. Then, 
 moving the instrument to 3 and pointing at 2, we lay off an angle 
 of 174 52 24", and measure in the direction indicated 787*72 feet 
 to T or to station 170*38 ; which, if our angles and measures bring 
 us direct to T or near to it, we presume the work to be correctly 
 done. We should, however, before pronouncing the work correct, 
 place our instrument at T , and pointing the telescope to 3, lay off 
 the angle with the line of the road, and if this agrees with the 
 computed angle, I think we may then, without hesitation, pronounce 
 the work correct. 
 
 But, if our angles and measures do not bring us direct to T or 
 near by it, we then point our telescope to T , and ascertain the 
 angle indicated by the instrument, and measure the distances as 
 correctly as we can, which we duly note down in our field book. We 
 then move to T with our instrument, and pointing its telescope 
 to 3, we measure the angle with the line of the road, which we also 
 note in our field book. 
 
 With the data thus obtained, we proceed to recompute the 
 
CURVE FROM SUPPOSED FIELD NOTES. 19 
 
 elements of a curve that will unite the two lines without materially 
 varying the location of the track from the points which we have 
 just fixed. 
 
 (13) In order to show practically the performance of these 
 operations, we will make up the following as the field notes of a 
 survey for locating the curve ahove described. 
 
 Commencing at T with instrument pointing in the direction of 
 the road, 
 
 we laid off the angle 177 37 36" and measured 711-79 feet to station 1 
 
 At station 1 we laid off the angle 175 07 36" and measured 749-76 feet to station 2 
 
 " 2 " " 175 00 7 00" " 749-76 " 3 
 
 " 3 " 175 00 00" " 751-51 " T 
 
 " T " 177 14 48" in the direction of road. 
 
 Sum, .... 880 GO 7 00" 
 
 Having obtained the field notes of our traverse, our first operation 
 will be to deduce from them the angle at apex, and at the centre of 
 the curve. We here remark that the sum of the angles at apex and 
 at the centre of the curve, always amount to 180, and of course one 
 must be a supplement to the other. 
 
 To ascertain either of the angles, viz., at the apex, or at the 
 centre, a variety of formulae might be deduced, but it is presumed 
 the following is as convenient as any ; viz., subtract the sum of all 
 the angles from as many times 180 as there are angles, and the 
 remainder will be the angle at the centre of the curve. 
 
 It will bo seen that we have noted in our field book five angles, 
 whose sum amounts to 880 ; now 5 X 180 = 900 ; and 900 880 
 
20 USEFUL FORMULAE. 
 
 = 20 == the angle at the centre of the curve, which compares with 
 the angle we had formerly ascertained. 
 
 Our next step is to ascertain the relative position of the present 
 points T and T with respect to A, and also the position they should 
 occupy to suit the radius we have heretofore deduced. 
 
 We know of no more convenient method of determining this 
 problem, than "by working up the traverse, (as the ship captains 
 call it,) and for that purpose we will assume the line of the road 
 extended from T to apex, whatever may he its direction, as hearing 
 due north, and predicate the bearings of the other lines upon it, as 
 indicated by the angles. Thus, 
 
 Angle at T = 177 37 36" and 180 177 37 36" leaves 20 22 24" N. W. to 1 
 
 " " 1 = 1750 07 36" = 3520 45 12" and (2 X 180 o) 352 45 12" = 7O 14 48" " 2 
 
 " " 2 = 1750 oo 00" = 527 45 12" " (3 X 180 o) 527 o 45 12" = 12 14 48" " 3 
 
 " " 3 = 1750 00 00" = 7020 45> 12" (4 X 180) 702 o 45 12" = 17 14 48" " T 
 
 " T = 1770 14 48" = 8800 00" " (5 X 180) 880 o 00 00" = 20 o 00 00" " 
 
 being the direction of the road. 
 
 Computing the northings and westings of the foregoing traverse, 
 we have 
 
 No. 1. N. W. 2 22 24" sin. = 8-6171119 cos. = 9-9996273 
 
 711 -79 feet log. = 2-8523458 log. = 2-8523458 
 
 29-475 " log. = 1-4694577 711 -17 feet log. = 2-8519731 
 
 No. 2. N. W. 7 14 48" sin. = 9-1008572 cos. = 9-9965171 
 
 749- 76 feet log. = 2-8749223 iog. = 2-8749223 
 
 94-576 " log. = 1-9757795 743- 77 feet log. = 2-8714394 
 
WORKING THE TRAVERSE. 
 
 21 
 
 No. 3. N. W. 12 14 48" sin. = 9-3265833 
 749-76 feet log. = 2 8749223 
 
 cos. = 9-9900028 
 log. = 2-8749223 
 
 159-04 " log. = 2-2015056 732-70 feet log. = 2-8649251 
 
 No. 4. N. W. 17 14 48" sin. = 9-4720042 
 751-51 feet log. = 2-8759348 
 
 COS. = 9-9800203 
 log. = 2-8759348 
 
 222-814 " log. = 2-3479390 717-72 feet log. = 2-8559551 
 
 Then, summing up the computed northings and westings, we 
 have as follows : 
 
 
 BEARING. 
 
 DISTANCE. 
 
 NORTHING. 
 
 WESTING. 
 
 From T to No. 1 == 
 
 N. W. 2 22 24 
 
 711-79 
 
 711-17 
 
 29-475 
 
 i 2 = 
 
 " 7 14 48 
 
 749-76 
 
 743-77 
 
 94-576 
 
 " 2 " 3 = 
 
 " 12 14 48 
 
 749-76 
 
 732-70 
 
 159-040 
 
 " 3 " 4 = 
 
 " 17 14 48 
 
 751-51 
 
 717-72 
 
 222-814 
 
 
 Total 
 
 
 2905-36 
 
 505-905 
 
 
 
 
 
 
 Having summed up the traverse, w now proceed to find the 
 
 bearing and distance from T to T . 
 
 Putting N = the sum of the northings, and calling it the cos. ; 
 W = " " westings, " " sin. ; 
 
 C = the bearing sought ; 
 D = the distance from T to T . 
 
 We then have by analogy the following formulae : 
 
 N : W : : ft : tan. C = 
 
 \v 
 
 \\ 
 
 and sin. C : W : : R : I) = sl f- e 
 or cos. C : N : : R : D = N - c - 
 
 (A) 
 (B) 
 (C) 
 
[Fio. 3.] 
 
TESTING COMPUTATIONS. 23 
 
 Performing the computations indicated, we have 
 
 W= 505-905 log. = 2-7040690 
 
 N =2905-36 co. ar. log. = 6-5368001 
 
 C = 9 52 40" tan. = 9-2408691 
 
 C = 9 52 40" co. ar. sin. = 0-7656168 
 
 W = log. = 2- 7040690 
 
 D = 2949-08 feet log. = 3-4696858 
 
 We now have in the triangle ATT , the side T T and the data for 
 finding the unknown angles.* Then, to find the distances A T and 
 A T we have 
 
 sin. A : D : : sin. T : A T =5 
 and sin. A : D : : sin. T : A T == 
 
 P sin. T 
 sin. A 
 
 P sin. T 
 sin. A 
 
 To prevent confusion in our diagram, or to render our work more 
 plain, we reconstruct the figure of the triangle ATT . (See figure 
 on preceding page.) 
 
 To perform the computations indicated, we have 
 
 A =160 00 00" co. ar. sin. = 0-4659483 
 
 D = 2949-08 feet log. = 3-4696861 
 
 T = 10 07 20" sin. = 9-2448918 
 
 A to T = 1515-396 feet log. = 3-1805262 
 
 Sum of logs, of A and D = log. = 3-9356344 
 
 T =9 52 40" sin. = 9 2343832 
 
 A to T = 1479-7 feet = log. = 3-1700176 
 
 * The data for finding the unknown angles are the relative bearings of the sides of the triangles, as 
 found in the preceding computations. 
 
24 USEFUL FORMULAE. 
 
 Having thus computed the distance from apex to the points T 
 and T , we will now ascertain the length those lines should be to 
 suit the contemplated radius ; thus we have 
 
 Cos. i C : r :: sin. J C : * = tan. J C.r 
 
 r = 8594- 38 feet log. = 3-9342145 
 
 1 C = 10 00> 00" tan. = 9 2463188 
 
 t 1515- 42 feet log. = 3-1805333 
 
 t in the present case, as in our former notation, represents the dis 
 tance required from A to T and also from A to T , to suit the con 
 templated curve, which in the present instance, has a radius of 
 8594-38 feet, and an apex angle A = 1GO. 
 
 It appears from the above computations that T should be 
 moved from the apex 1515-42 -- 1515-396 -024 feet, which 
 amount in practice is so small that we should consider T correctly 
 located, and doubtless our calculations would have given its location 
 exact, had we been careful in the management of the fractions ; 
 but it is not so with T. The computations show that T should be 
 moved from apex 1515-42 - - 1479-17 = 3G-25 feet. After 
 having moved T from apex 3G-25 feet, in the direction of the 
 line of the road, there will be no doubt, if the previous traverse 
 upon which the calculations have been based, has been correctly 
 measured, that the contemplated curve could be accurately located 
 between the tangent points as corrected. 
 
 The above calculations have been based upon a traverse consist 
 ing of chords of an arc correctly laid down, with the exception of 
 the last course ; but, had the traverse been a random one, the results 
 arrived at would have been equally exact, with only this dift<erence, 
 
FIELD NOTES OF A TRAVERSE. 25 
 
 that the points T and T would not probably have been found so 
 near their proper location. 
 
 The practice explained above will be found useful in locating 
 curves by the side of rivers, ponds, oceans, mountains, rough sur 
 faces, etc. ; in short, wherever it is found inconvenient to run the 
 direct lines to their intersections, or to apex, and to measure there 
 from to the points T and T . In running traverses for the purpose 
 of obtaining the elements for the location of curves under the con 
 ditions suggested above, it will be found convenient, if not abso 
 lutely necessary, that some portion of the traverse should be so 
 made as to give the relative position of such points as the contour 
 of the surface or other considerations may render it desirable that 
 the location should pass through. We now proceed to give a prac 
 tical example. 
 
 (14) The following practical example supposes the direct or 
 straight lines to be united by the curve, to have been located and 
 marked by some convenient device, and the angles given below to 
 have been measured by a common theodolite, and the lines by a 
 chain, thus : 
 
 Field notes of a traverse for obtaining the elements of a curve 
 for uniting the lines T and T . 
 
 Commencing at station 0, corresponding in the diagram to T, 
 
 * This remark would seem to be uncalled for, as it can make no difference in the computation how 
 the lines and angles are obtained 5 but it frequently happens that the traverse is obtained by means of 
 a triangulation, which would sometimes present the matter in a different form. The writer has had a 
 number of cases that could not have been well performed in any other manner. 
 E 
 
26 USEFUL FOKMULJE. 
 
 with telescope pointing in the direction of the located line, we 
 measured as follows, viz. : 
 
 C The curve should pass through or 
 Station = 169 29 45 = 1200 feet to station 1 J near this point, which of course 
 
 ( governs the radius. 
 " 1 = 170 00 15 = 900 " " 2 
 
 " 2 = 175 04 30 = 750 " " 3 
 
 3 = 1G4 40 10 = 1525 " " 4 corresponding in diagram to T 
 
 " 4 = 140 45 20 = in the direction of the located line of the road. 
 
 820 00 00 
 180 X 5 = 900 00 00 
 
 80 = angle at centre of curve. 
 
 Assuming the line from T to A as hearing due north, whatever 
 its course may he, we deduce the following relative courses for the 
 several lines of the traverse, and hy formula (A) ascertain the 
 relative hearing from T to T . 
 
 STATION 
 
 BEARINGS. 
 
 DISTANCES. 
 
 NORTHINGS. 
 
 WESTINGS. 
 
 to 1 = 
 
 N. W. 10 30 15" 
 
 1200 
 
 1179-890 
 
 218-7685 
 
 1 " 2 = 
 
 " 20 30 00" 
 
 900 
 
 843-005 
 
 315-1866 
 
 2 " 3 = 
 
 " 25 25 30" 
 
 750 
 
 677-361 
 
 321-9970 
 
 3 " 4 = 
 
 " 40 45 20" 
 
 1525 
 
 1163-197 
 
 995-5700 
 
 3863-453 1851-5221 log. = 3-2675289 
 3863-453 log. = 3-5869756 
 
 Relative bearing from T to T = N. W. 25 36 20" -22 tan. = 9 6805533 
 
 Having thus obtained the hearing from T to T , we now proceed 
 to compute the distance ; hy formula (B) and (C) we have 
 Sin. C : W : : R : D = ~r or 
 Cos. C : N :: R : D = c ^ 
 
DISTANCE TANGENT TO APEX. 
 
 27 
 
 In these analogies, C represents the course from T to T ; W the 
 westing ; N the northing ; D the distance. 
 
 We frequently, as we shall in the present instance, use "both 
 formulae, for the purpose of proof. 
 
 W = 1851-5221 log. = 3-2675289 
 
 C = 25 36 20".22 sin. = 9-6356588 
 D = 4284-2 feet log. = 3-6318701 
 
 N = 3863-453 log. = 3-5869756 
 
 C = 25 36 20".22 cos. = 9-9551055 
 
 D = 4284-2 feet log. = 3-6318701 
 
 Our next step in practice is to ascertain the distances from T and 
 T respectively to the point where the direct lines would intersect ; 
 or, in other words, to the apex. 
 
 We have ascertained the angle of the centre of the curve to he 
 80. Of course the angle at apex will he 100. The hearings 
 which we have ascertained also indicate the angles ; thus, in the 
 imaginary triangles we are about to solve, we have supposed the 
 line from T to A to hear due north. Then, by computation, the 
 line from T to T bears N. W. 25 36 20". 22, which gives the 
 angle at T the same number of degrees as the bearing. From the 
 traverse or the table of angles, in our field notes, we deduce the 
 bearing of the located line from T to be N. W. 80 00 00". 
 
 These bearings indicate tlie following angles, viz., at A = 100 00 00". 00 
 
 As before stated, at T = 25 36 20".22 
 
 T = 54 23 39".78 
 
 Proof, Sum = 180 00 00".00 
 
 With these angles, and the distance from T to T = 1), the dis 
 tances T A and T A are readily found ; thus, 
 
[Fia. 4.] 
 
MEASUREMENT TANGENT TO APEX. 29 
 
 sin. A : D : : sin. T : A T 
 
 (10) 
 and sin. A : D : : sin. T : A T l 
 
 A = 100 00 00" ar. co. sin. = 0-0066485 
 
 D = 4284-2 feet log. = 3-6318701 
 
 T = 54 23 39".78 ski. = 9-9101139 
 
 A T = 3536-98 feet log. = 3-5486325 
 
 Again, A = 100 00 00" co. ar. sin. = 0.0066485 
 
 D = 4284-2 feet log. = 3-6318701 
 
 T = 25 36 20".22 sin. = 9-6356588 
 
 A T = 1880-085 feet log. = 3-2741774 
 
 For the triangle A T\ (see diagram) we have by our measurement 
 and computations the sides A T = 3536-98 feet, and 1 T = 1200 
 feet with their included angle = 10 30 15" to find A 1 and the 
 unknown angles. 
 
 For convenience in the enunciation of the formula, let A T = a, 
 1 T = b, C = the given angle, and A and B the unknown angles ; 
 A representing the unknown angle opposite the side a, and B the 
 unknown angle opposite the side b. We then have a -f- b .* 
 a ^ b : : tan. } (180 C) : tan. J (A ^ B) and \ (180 C) + 
 \ (A co B) = the angle opposite the longest side, and J (180 
 C) i (A oo B) = the angle opposite the shortest side. 
 
 In the calculations which follow we change the symbols from 
 those given in the formula, so as to have them conform to the 
 letters and figures given upon the diagram. 
 
30 
 
 USEFUL FORMULAE. 
 
 A T 
 
 3536-98 Given angle T 
 
 180 00 00" 
 = 100 30 15" 
 
 Irp 
 
 
 2) 1690 29 45 / 
 
 JL 
 
 1200 
 
 84 O 44 52" 5 
 
 JQO- 6 344984 
 
 DifF 
 
 
 ] O(y 3-3686550 
 
 i (180 T) 
 
 
 tan 103657 9 2 
 
 
 
 
 (A co 1) = 
 
 79 26 43". 14 
 
 tan. = 0-7297256 
 
 Z_ at 1 = 
 l_ at A = 
 l_ at T 
 
 164 11 35"- 64 
 5 l& 09"- 36 
 10 30 15". 00 
 180 00 00". 00 
 
 
 Z_ at 1 
 AT 
 T 
 A 1 
 
 164 11 35". 64 co. ar. 
 3536-98 feet 
 10 30 15" 
 2367-21982 feet 
 
 sin. =0-5648026 
 log. = 3-5486326 
 sin. =9-2608034 
 
 log. = 3-3742386 
 
 A 
 
 Tl 
 
 T 
 
 Proof, 
 
 = 5 18 09" -36 co. 
 = 1200 feet 
 = 10 30 15" 
 = 2367-21982 feet 
 
 ... &in. = 1 0342539 
 . .. log. = 3-0791812 
 ... sin. =9-2608034 
 
 ,. .. ] jr =3-3742385 
 
 Having obtained all the elements of the triangle A 1 T we rep 
 resent the side A 1 as found above in feet, or in general in the unit 
 of measure by bm, and as we find it in proportion to the radius of 
 the curve = unity, by b. 
 
 As it becomes convenient generally to use the letters standing 
 against the angles in each triangle, and as some of them are com 
 mon at least to three triangles, it becomes necessary to occasionally 
 accent some of them, that we may understand their different values 
 
DETERMINATION OP RADIUS. 31 
 
 in the investigation. We have throughout our investigations rep 
 resented the apex angle by A. We shall continue to give to A that 
 value in the following investigation, with the exception of angle A 
 in the triangle AIT (which in general will not be a multiple of 
 the apex angle) we shall therefore represent it by A . 
 
 Commencing our investigation with the triangle A T" C, and 
 representing T" C the radius of the curve, by unity, and the radius 
 of the tables by E, we have 
 
 Sin.iA: 1 :: B: AC=- 1B J nr (11) 
 
 In the triangle A C 1 we have 01 = the radius of the curve = 1 ; 
 therefore, representing the angle at 1 by G, we have 
 
 1 . sin. (i A A ) : : AC : sin. G = ""1*7* (12) 
 
 For convenience representing the line A C by 6?, we have 
 
 Sin. G : d :: sin. [(J A A ) + G] : b = rf sin [( ^ n ~ G AO+GL (13) 
 then b : bm :: 1 : r= >- (14) 
 
 where r represents the radius of the curve in the unit of measure. 
 
 Or, probably the following formula would be rather more simple 
 for calculation than the above, (13) viz.: 
 
 Sin. (i A -- A ) : 1 :: sin. [(i A - - A ) + G] : b = , 
 
 sin. f(* A A ) + H] / 1 o \ 
 
 sin. i A- A V iJ ) 
 
 then, as above b : bm : : 1 : r = * (14 ) 
 
 where r represents the radius of the curve in the unit of measure. 
 
 Example of calculation, formula (12) sin. G = J 
 
32 USEFUL FORMULA. 
 
 We found A by the reduction of the foregoing traverse (see pre 
 ceding pages) = 100, also A (noted as A) in the triangle 
 A 1 T = 5 18 09". 3G. Therefore, 
 
 A A = 50 5 18 09".3G = 44 41 50". 64 sin. = 9-8471791 
 A 50 00 00".00 co. ar. sin. = 0-1157460 
 
 G (ambiguous) 66 39 38" .22 sin. = 9-9G29251 
 
 G (corrected) 113 20 21".78 
 
 Having found G, we proceed to find the radius = r. By formula 
 (13 ) and (14 ) we have 
 
 7) - - _5i_( ^ 
 
 A +* 
 
 sin. ( A A) b ~ sin. (J A A -f G) 
 
 A A = 44 41 50"- 64 
 
 G = 113 20 21". 78 
 
 A A -fG= 158 02 12". 42 co. ar. sin. = 0-4271153 
 
 A A = 44 41 50". 64 .... .... sin. = 9-8471791 
 
 = 2367-2198 feet log. = 3-3752386 
 
 r = 4462-035 feet log. = 3-6495330 
 
 We thus find the radius of the curve, or r = 4462 035 feet. 
 The deflection for a chord = 50 feet will be (3) sin. D = - i ^- 
 
 Thus, ^ ch = 25 feet log. = 1-3979400 
 
 r = 4462-035 feet co. ar. log. = 6-3504670 
 D =0 19 15"- 67 sin. = 7-7484070 
 
 But, as 19 15" -07 makes an inconvenient number to add or 
 subtract, we choose for the angle of deflection (D) = 19 15", 
 and adopt a radius which shall agree therewith. This change in 
 the radius will not materially alter the location of the curve. 
 
POSITION OF TANGENT POINTS. 
 
 To find a radius corresponding to a deflection of 19 15", we 
 (5) r= 4^ 
 
 i ch = 25 feet log. = 1-3979400 
 
 D = 19 15" co. ar. sin. = 2-2518454 
 
 r = 4464-63 feet log = 3-6497854 
 
 We have thus ascertained the radius of a curve which will cor 
 respond to the location selected. It now remains to ascertain the 
 tangent points, or points of commencement and end. 
 
 We have (6) representing the whole centre angle hy C, 
 t = tan. J C . r 
 
 Thus, . C = 40 00 00" tan. = 9-9238135 
 
 r = 4464-63 feet log. = 3 6497854 
 
 t = 3746-27 feet log. = 3-5735989 
 
 We found (page 29) A T == 3536-98 feet, and A T= 1880-085 
 feet ; and, t being 374G 27 feet, we have 
 
 t = 3746-27 
 AT = 3536-98 
 
 209-29 feet, the distance T should be moved from A 
 
 Again, t = 3746-27 
 AT = 1880-085 
 
 1866-185 feet, the distance T should be moved from A 
 
 Having moved the points T and T to their positions indicated 
 above, and marked in the diagram T " and T", the curve may be 
 laid out and marked by any method the engineer might think best 
 suited to the locality. 
 
34 USEFUL FORMULAE. 
 
 (15) We will add one more method of locating simple curves, 
 principally applicable to large apex angles, and which may in some 
 instances be practised beneficially with apex angles somewhat acute, 
 provided the radii be not of great length. 
 
 This method supposes that the locality will admit of the lines to 
 be connected by the curve, to be run or extended to their intersec 
 tion, so that their apex angle may be measured directly ; and that 
 the contour of the surface is such that measurements may be taken 
 with a good degree of accuracy, from the apex to the tangent points 
 along these extended lines, and from these extended lines to the 
 location of the curve. 
 
 Having premised thus much, let us suppose these extended lines, 
 which we shall hereafter call the tangent lines, intersect each other 
 at an angle of 170, and it be desirable to connect these lines by a 
 curve corresponding to a deflecting angle of 6 for a chord of 50 
 feet. The elements required to be obtained from computation are, 
 
 First, Kadius of the curve. 
 Second, Length of do. 
 Third, Distance from apex to tangent points. 
 Fourth, The number of primitive points convenience requires to 
 be fixed in the curve. 
 
 To ascertain the radius, we have (5 ) r = -^fjj- 
 
 ch = 25 log. = 1-3979400 
 
 D 6 00" co. ar. sin. = 2-7581229 
 
 Radius = r 14323-95 feet locr- = 4-1560629 
 
PRIMITIVE POINTS IN CURVES. 35 
 
 To find the length of the curve, we have (8) a = r " 
 
 r = log. = 4-1560629 
 
 180 170 = C" = 36,000" log. = 4-5563025 
 
 r co. ar. log. = 4-6855749 
 
 Length of arc = a = 2500-02 feet log. = 3-3979403 
 
 To find the distance from apex to tangent points, (G) t = tan. J C . r 
 
 C = 5 00 00" tan. = 8-9419518 
 
 r log. = 4-1560629 
 
 Apex to tan. = t = 1253-18 feet log. = 3-0980147 
 
 We have found the length of arc = 2,500 feet ; if we now sup 
 pose T to hear an even number in the locating stations, say 540, we 
 may divide the arc into five equal parts of five hundred feet each, 
 which will cause every point of division to fall on an even station in 
 the location. This division, of course, divides the centre angle into 
 five equal parts ; and, as C = 180 170 = 10, the centre angle 
 corresponding to each division will be - = 2 00 00". 
 
 (16) Having determined to divide the curve into five equal 
 parts, we now compute the distances from each of these dividing 
 points in the curve to the tangent lines, in the direction of the radii 
 passing through them. (See Fig. 5.) 
 
 Denoting the tangent points by T and T ; and the divisions of the 
 curve by 1, 2, 3, 4 ; and the corresponding divisions of the tangent 
 lines by a b c d; and representing by Ci the centre angle corre 
 sponding to the arc T 1, and by 2 the centre angle corresponding 
 to the arc T 2. (The arcs T 4 and T 3, being similar to T 1 and 
 T 2, will not need separate expressions.) In the general investiga- 
 
[FiQ. 5.] 
 
PRIMITIVE POINTS IN CURVES. 37 
 
 tions, we however denote the centre angles by C, and we have this 
 analogy, 
 
 Cos. C : r : : sin. C : T a = tan. C . r (15) 
 
 and Cos. C : r : : R : C a = -^^ 
 
 and we have - -^ r = a, 1 = etc. (1C) corresponding with the 
 
 centre angle. Now, substituting Ci and C2 for C, as explained 
 above, we have 
 
 Ci = 2 00 00" tan. = 8 5430838 
 
 r = 14323-95 feet log. = 4-1560629 
 
 T to d and T to a= 500-20 feet log. = 2-6991467 
 
 Ci = 2 00 00" co. ar. cos. = 0-0002646 
 
 r =14323-95 log. = 4-1560629 
 
 C to d and C to a = 14332-69 log. = 4-1563275 
 
 .*. d to 4 and a to 1 = 8-74 feet 
 
 Again, Cz = 4 00 00" tan. = 8-8446437 
 
 r = log. = 4-1560629 
 
 T to c and T to 6 = 1001 65 feet log. = 3 0007066 
 
 C 2 = 4 00 00" co. ar. cos. = 0-0010592 
 
 r = 14323-95 feet log. = 4-1560629 
 
 C to c and C to b = 14358-93 log. = 4-1571221 
 
 . . c to 3 and 6 to 2 = 34-98 feet 
 
 Having thus ascertained all the elements necessary to this 
 peculiar method, we may now measure from T to a, in the direction 
 T A= 500*2 feet; and the same distance from T to cZ, in the 
 direction T A ; and then, with the theodolite at a, and pointing to 
 T, lay off the complement angle of Ci, and measure 8 74 feet to 1, 
 fora point in the curve corresponding to station 540 00 -f- 5 00 
 = 545 of the location. Then, remove the theodolite to d, and 
 
38 USEFUL FORMULAE. 
 
 pointing at T, lay off the complement angle of Ci, and measure in 
 the direction indicated 8 74 feet to 4, for a point in the curve cor 
 responding to station (540-00 + 20 00) = 560 of the location. 
 We now measure from d (1001-63 500-20) = 501 -43 to c, and 
 the same distance from a to b ; then, with the theodolite at 6, and 
 pointing at T, lay off the complement angle of 2, and measure in 
 the direction indicated 24-98 feet to 2, for a point in the curve cor 
 responding to station (540-00+ 10*00) =550 of the location. 
 Then, moving with the theodolite to c, and pointing to T , lay off the 
 complement angle of Ca, and measure in the direction indicated 24-98 
 to 3, for a point in the curve corresponding to station (540*00 -f- 
 15-00) = 555 of the location. 
 
 (17) We now take the theodolite successively to each of the 
 points we have established in the curve ; and, by deflections and 
 corresponding chords, complete the work. 
 
 This method of laying out curves is found exceedingly convenient 
 in woodlands, as not being so liable to mistakes which might lead 
 the location astray as other methods, and will frequently save much 
 trouble in chopping timber. 
 
 We are aware that the example we have given in the foregoing 
 is one of the most convenient that the problem admits ; but we think 
 the principle will be sufficiently comprehended to apply it readily 
 and without difficulty in its most complicated form, without further 
 explanation. 
 
 (18) We have now completed all we contemplated respecting 
 the investigations of the practical operations of laying out simple 
 
KEVERSE CURVES. 39 
 
 railroad curves. It was not our purpose to pursue these investiga 
 tions until the subject was exhausted. That would have taken a 
 long time, and might have occupied much room, and it is more than 
 probable that it would have exceeded our ingenuity and ability. 
 But we hope we have said enough to give the proper direction to the 
 inquiries of the young or inexperienced engineer, and to convince 
 him of the necessity of making the study of the elements of 
 geometry and trigonometry (if it be proper to make the distinction, 
 or to class them under different heads) a matter of the first 
 importance. 
 
 (19) We now proceed to the consideration of reverse curves ; 
 the most simple form of which is to unite two parallel lines, which, 
 continued, will not intersect or run into each other, by curves of 
 equal radii. 
 
 This problem is of such simplicity as to admit of many forms of 
 construction, and great variety of formulae ; and, had we not come to 
 the conclusion to give a formula for laying out curves to unite rail 
 road lines under every condition which has occurred in our practice, 
 we think we should have passed by this problem without considering 
 its properties. We therefore content ourselves with giving the fol 
 lowing rules and formulae. 
 
 (20) Let T A and T" B represent two lines of railroad having 
 the same bearing, but so located that they will not intersect by ex 
 tension. 
 
 Let T C and T C = the radii which we suppose to be equal in 
 length, and which in the investigation we shall represent by r. 
 
[Fia. 6.] 
 
REVERSE CURVES. 41 
 
 Draw the line T C and T C at right angles with T A and T B, 
 and make each = r ; then, drawing a line from C to C will inter 
 sect the curves at their reversing point X, (see figure,) so also a line 
 drawn from T to T will intersect the curves at the same point, 
 and will also bisect the line C C . 
 
 We now extend the line B T until it intersects the line or radius 
 T C at E, and as T C is drawn at right angles with T A, and T A 
 and B T having the same bearing, it is obvious that the angle at 
 the intersection will be a right angle. Then, having measured T II 
 and T R, or ascertained their length by computation, we have by 
 considering the line = T T as radius, and the line T R as a 
 cosine, and the line R T as a sine of the angle T T R, this analogy 
 to find the angle T . Representing the sine by , the cosine by 
 c, the radius by R, then will 
 
 c : s :: R : tan. T = -f- (17) 
 
 By letting fall a perpendicular upon the line T X from C, we 
 divide the triangle T C X into two equal right-angled triangles, 
 viz., C M T and C M X ; now, as the triangles T T R and T C M 
 have the angle T common to both, it is obvious they are similar ; 
 the angles T and C must therefore be equal, and in the triangle 
 T C X the angle C will be equal to twice T ; then, resolving the 
 triangle T R T, we find the length T T, half of which is equal to 
 T X ; then solving the triangle C X T, we find the side T C = 
 radius = r. To execute these computations by the aid of the trig 
 onometrical functions, we have in the triangle T T R to find 
 J T T which we represent by a. 
 
 Sin. T : * :: R : T T ; wherefore - T 2 T 1 = a = = -3^-^- 
 or, Cos. T : e :: 11 : T T ; */ - = a =- c ~<, 7 . r 
 
 G 
 
42 USEFUL FOKMUL.E. 
 
 Then, in the triangle T C X we have 
 
 Sin. C : a :: cos. T r = ^^ = ^-^ (18) 
 
 Having thus found the radius, and the centre angle C of the one 
 half the curve, it is evident that the other half must contain iden 
 tical elements. 
 
 (21) The curve may be laid out in accordance with such of the 
 formulae described in the foregoing pages, relating to simple curves, 
 as the engineer may think best suited to the condition of things 
 and the contour of the surface. 
 
 For the purpose of presenting an example of computation, we 
 will suppose 
 
 T E = c = 12GO feet 
 
 E T = s = 150 " 
 Then, by (17) Tan. T = ~ 
 
 Wherefores = 150 feet log. = 2-1760913 
 
 c =1260 " co. ar. log. = 6-8996295 
 
 T = 6 47 20". 31 tan. = 9-0757208 
 
 C being equal to 2 T, we have (18) r = sin c c 2 
 Wherefore T = 6 47 20". 31 
 
 C = 13 34 40"- 63 CO. ar. sin. = 0-6293609 
 
 2 " " log. = 9 6989700 
 
 c = 1260 log. = 3 -1003705 
 
 r =2683-47 log. = 3-4287014 
 
REVERSE CURVES. 43 
 
 (22) We will now vary the given elements, by supposing the 
 radius = r = 2683 47 feet, and the distance between the centres 
 of the tracks = s = 150, to find the relative positions of T and 
 T. This case supposes a beginning point to have been selected. 
 Suppose that point to be at T. Then, from T we draw the radius 
 T C = r at right angles to the line T A, and from T we extend 
 the line T C or r in the opposite direction, a distance = r (or 
 minus) the distance between the tracks (which we have supposed to 
 be 150 feet) to D ; and from D draw the line D C at right angles 
 with the line D C ; then, with the line C C , equal in length to 2 r, 
 intersect the line D C at C ; then will D C = T E, representing 
 T E as in our former proposition by c ; and then, solving the tri 
 angle C C D, we have by trigonometry, 
 
 2 r : E : : (2 r s) : sin. C = ^-f^ (19) 
 
 and Sin. C : (2 r s) : : cos. C : c = cot. C (2 r s) (20) 
 
 The complement of the angle C found by the above formula = 
 the angles T C C == T C C. 
 
 EXAMPLE OF COMPUTATION. 
 
 By formula (19) 
 
 2r = 5367 co. ar. log. = 6-2702685 
 
 (2 r s) = 5217 log. = 3-7174298 
 
 C = 76 25 19".50 sin. = 9-9876893 
 
 By formula (20) 
 
 C == 76 25 19".50 cot. = 9-3829486 
 
 (2 r s) = log. = 3-7174208 
 
 c = 1260 nearly log. = 3-1003694 
 
 * These formula are generally based on trigonometry, as being more convenient or better suited 
 to logarithms. 
 
44 USEFUL FOKMUL^E. 
 
 The complements to C as found above = 90 76 25 19". 5 = 
 13 34 40". 5. 
 
 (23) There being a difference of opinion among practical engi 
 neers respecting the propriety of reversing curves without the 
 intervention of a piece of straight track between them, we have 
 thought it would not be improper, before we proceed further with 
 our investigations of formulae for reverse curves, to add the follow 
 ing discussion of the causes of the lateral shocks experienced in 
 railroad cars when entering upon a curve ; and also the necessity 
 for a piece of straight track between reversing curves. 
 
 From any investigations that we have been enabled to make, we 
 cannot discover any reason why a car should receive a lateral shock 
 when running off a straight track on to a curve, provided the dis 
 tance between the flanges of the car wheels and the distance 
 between the insides of the rails are the same ; and the outside rail 
 of the curve be properly elevated to suit the velocity. We would 
 however remark, if the curve be very short and the velocity rapid, 
 the friction of the flanges of the wheels upon the outside rails of 
 the curve will become so great as to occasionally raise the face or 
 bearing of the wheel from the rail, which falls again suddenly upon 
 the track, producing a perpendicular jar or concussion, which may 
 sometimes occasion lateral motion ; but, if the curves are in good 
 order, and the rails properly curved, this motion will be seldom 
 felt, unless, I repeat, the radius be very short, and the velocity 
 great. In this connection I would add, it is believed to be some 
 what dangerous to run a train over a curve in proper repair or con 
 dition with a velocity so great as to occasion this phenomenon. We 
 have, however, felt this motion when riding on curves not in proper 
 
LATERAL SHOCKS ON A CURVE. 45 
 
 repair, or where the rails have not been curved in a uniform, regu 
 lar manner ; whereas, if the track had been in good condition, 
 nothing of the kind would have been felt. 
 
 Neither can we, from any investigation we have been enabled to 
 make, discover any reason for a lateral shock to a car when running 
 off of a direct on to a reverse curve (if these are the proper terms 
 of expression) provided the condition of the track is good, and the 
 wheel flanges and the rails are the same distance apart. But the 
 question may be asked : Why in practice lateral shocks are so fre 
 quently felt, when running from a straight line on to a curve ; and 
 likewise when running from a direct on to a reverse curve ? We 
 answer, that the practice is almost universal to lay the rails of a 
 track from one half to three fourths of an inch further apart than 
 the flanges of the car wheels. Now if we conceive the flanges of 
 the car wheels to be in contact with the line of rail which forms the 
 inside line of the curve, when the car is about to enter upon that 
 curve, and the track three fourths of an inch wider than the 
 flanges, it will be obvious that the motion of the car, if it meets 
 with no extraordinary obstruction, will continue straight, or in a 
 direct line, after it enters the curve, until the flange of the wheel 
 meets with the outside rail of the curve ; the distance which the 
 car will have then advanced into the curve, before the phenomenon 
 of contact takes place, will be so great that the curve will have 
 obtained a considerable degree of deflection, which of course pro 
 duces a shock, and the shock will be somewhat proportioned to the 
 difference between the width of the flanges of the wheels and the 
 width of the track, the velocity of the cars, and the length of the 
 radius of the curve. The following diagram (Fig. 7) indicates the 
 practice in cases of reversed curves. 
 
[FiG. 7.] 
 
LATERAL SHOCKS ON A CURVE. 47 
 
 (24) Let us now examine the condition of a car running off of 
 a direct curve on to a reverse. The centrifugal force, as well as 
 the position of the wheel axles, direct a car when running on a 
 curve, to the outside rail, which must, of course, become its guide ; 
 and when the car arrives at the reversing point, or rather one 
 fourth of its length between its wheels beyond that point, it will 
 continue its direction in a straight line until its wheel flanges meet 
 with the outside rail of the reverse curve, and if the difference of 
 width between the flanges of its wheels and the rails be consider 
 able, the car will have advanced so far into the curve before the 
 phenomena of contact take place, as to admit the rail taking a 
 considerable degree of deflection ; and, as before stated respecting 
 simple curves, the meeting of the wheel flanges with the deflected 
 rail causes a lateral shock which is sensibly felt, and is proportioned 
 to the velocity of the cars, the difference in width between the 
 flanges of the wheels and the rails, and the length of the radius of 
 the curve. 
 
 The motion of the cars from the direct to the reverse curve will 
 always be of the same character under similar circumstances ; the 
 cars being constantly influenced by the position of their axles, and 
 the centrifugal force of motion. But it is not so with the cars upon 
 a straight track ; there is nothing to uniformly guide them to the 
 side of the track which forms the inner rail of the curve when run 
 ning off of a straight line on to a curve ; and, if the flanges of the 
 wheels are in contact with the side of the track forming the outer 
 rail of the curve, the car will enter upon the curve without lateral 
 shock. 
 
 If our reasoning has been just, it would appear that if the tracks 
 
48 USEFUL FORMULA. 
 
 were laid down to correspond with the flanges of the wheels, a car 
 will meet with no greater lateral shock when running through 
 reversing points of a reverse curve, than when running off of a 
 straight track on to a simple curve ; hence, under restricted circum 
 stances, where reversed curves are required, of short radius, it 
 hecomes an ohject of importance for the curves to occupy the whole 
 line, that is, there should he no straight track between them ; as a 
 straight line of any considerable length will tend to diminish the 
 length of the radius. 
 
 (25) Let us now endeavor to explain this matter by diagrams. 
 We will suppose a car to be running on a straight track from A 
 towards F, with the flanges of the wheels in contact with the rail 
 A B, the flanges of the wheels on the opposite side will not be in 
 contact with the bar a 6, but will describe the dotted line parallel 
 with it ; the car arrives at the tangent point B b, its natural 
 motion will be from b B to c C, in the direction of the dotted lines, 
 where the flange of the wheel running upon the rail a f meets it at 
 the point c, some distance advanced upon the curve, the rail at this 
 point making an angle with the direction of the car, causes a shock 
 and a sudden lateral motion ; the car then proceeds onwards, with 
 the flanges in contact with the rail a/, until it arrives at the point 
 D d; from B to D the flanges have not been in contact with the 
 rail A F, but have described the dotted line parallel with it ; from 
 D d its motion is onward in the direction of the tangent until it 
 arrives at E c, when the flange meeting with the rail at E, which it 
 will be seen forms an angle with the direction of the car, causes the 
 lateral shock felt at this point ; the car then moves onwards, with 
 the flanges in contact with the rail A F, until it passes through or 
 over the curve, and no further shock is felt. In the mean time the 
 
LATERAL SHOCKS ON A CURVE. 49 
 
 flanges upon the opposite side describe the dotted line parallel with 
 ef. If, therefore, our reasoning be correct, it will be obvious that 
 if we would have cars run smoothly over a railroad, the track 
 should, near the tangent points of curves, be laid down to corre 
 spond with the width between the flanges of the wheels ; and we 
 add that the same thing should be observed near the turn-out 
 frogs, as it is important that the scores in the frogs through which 
 the flanges of the wheels traverse, should be just sufficient to per 
 mit them to pass. To render this practice complete calls for a 
 greater degree of care in the adjustment of wheels upon their 
 axles, than is at present practised in many constructing and repair 
 shops ; but, in the present careless condition of adjustment, the 
 management of tracks can be much improved. Near frogs, and the 
 commencement of curves, the rails of the track should be no wider 
 than the widest wheels of the train. A further improvement, 
 adapted to the passage through the score in the frog smoothly, is 
 to have the width between the back side of the wheel flanges as 
 near alike as they well can be, which will much improve the benefits 
 of the guard rails. 
 
 In this connection it may not be amiss to compute the deflections 
 of the rail at the point where it is met by the flanges of the wheels 
 when running into the curve ; assuming the rails of the track from 
 O Ol of a foot wider than the flanges, up to G 06 of a foot; and 
 supposing the car to be tracing the inner rail on its approach to the 
 tangent point, and the curve to be of a radius of 1000 feet. 
 
 The following investigation, and its examples of computation, 
 by referring to Fig. 6, may enable the student to master the whole 
 merits of this subject. 
 H 
 
50 USEFUL FORMULAE. 
 
 INVESTIGATION OF FORMULA, AND EXAMPLES OF COMPUTATION. 
 
 Let r = radius of the curve ; Ji the width between the rails of 
 
 the track ; 
 Aii Ao, As? etc., the differences of distance between the tracks 
 
 and the flanges of the wheels ; 
 CD c-i, c 3 ? etc., the corresponding angles of deflection. 
 
 Then, putting r = 1000 feet ; A = 4-7 feet; Ai> A 2 , A 3 > 
 
 etc. = 0-01, 0-02, 0-03 feet, etc., we have 
 r + J A : B :: r + i A An etc - : cos - = " + ^-^ etc - 
 
 Thus, r -f A = 1002-35 co. ar. log. = 6-9989806 
 
 r -|- 7i Ai = 1002-34 " log. = 3-0010150 
 
 ci = 15 26" cos. = 9-9999956 
 
 r -J- A = 1002-35 co. ar. log. = 6-9989806 
 
 r -f /t A2 = 1002-33 log. = 3-0010107 
 
 c 2 = 2L/ 44" cos. = 9-9999913 
 
 r -f- A = 1002-35 co. ar. log. = 6-9989806 
 
 r + A A 3 = 1002-32 log. = 3 .0010064 
 
 co = 26 34" cos. = 9-9999870 
 
 r -\- ^ h = 1002-35 co. ar. log. = 6-9989806 
 
 r + h A 4 = 1002-31 log. = 3-0010020 
 
 c 4 = 30 45" cos. = 9-9999826 
 
 r -f- -L A = 1002-35 co. ar. log. = 6-9989806 
 
 r + M As = 1002-30 log. = 3-0009977 
 
 c s = 34 20" cos. = 9-9999783 
 
 r+$h = 1002-35 co. ar. log. = 6-9989806 
 
 r + h Ae = 1002-29 log. = 3-0009934 
 
 ce = 37 35" COS. = 9-9999740 
 
ELEMENTS OF REVERSED CURVES. 51 
 
 The foregoing computed deflections of the track, at the point met 
 by the wheel flanges, under the assumed circumstances, show the 
 necessity of narrowing the guage near the commencement of 
 curves, and near the reversing points in reverse curves. There 
 can, however, be no doubt but the cars will run steadier and safer 
 over a narrow track, just suiting the wheel flanges, than over one 
 of greater width. The only argument in favor of the guage of a 
 track being wider than the flanges of the wheels is, that a greater 
 surface of the wheel is exposed to wear upon the rails by the zig 
 zag course which the wide guage allows the cars to take, than 
 would be if that motion was prevented. 
 
 (20) The next form of reverse curves which we shall consider, 
 is that which shall unite two lines having different bearings, which 
 of course would intersect each other were they continued ; but, on 
 account of avoiding some obstacles, or the desire of a near approach 
 to some particular locality, it becomes necessary to connect these 
 lines by reverse curves ; it is, therefore, a matter of great import 
 ance to lay down these curves in the best form possible, particularly 
 if they require short radii, which will be best accomplished, if 
 there be no obstacle in the way, by making the curves of equal 
 curvature, and occupying the whole distance between the tangent 
 points ; (these tangent points are supposed to be fixed by the con 
 tour of the surface, or some other consideration or governing prin 
 ciple, which cannot well be avoided.) 
 
 (27) To proceed with the investigation of the proper formula 
 for determining the elements of these curves, we would first 
 remark, that this problem requires the angles A T T and T T B, 
 and also the length of the line T T to be measured. (See Fig. 8.) 
 
ELEMENTS OF REVERSED CURVES. 53 
 
 Then, by putting T X = c , T X = c, C X = a, C X = b, and 
 T T as measured = m, and the radii C T and C T in proportion 
 = unity ; (these radii by the problem being equal,) and the radius 
 in measure = r. 
 
 Then, by the problem we have the line C C = twice the length 
 of radius, and it will be apparent by a glance at the diagram that 
 the angles X and X must be equal. 
 
 Now, commencing with the radius = unity, we have 
 Sin. X : 1 :: sin. (T 00) : a= i -,^^ 90 ) 
 Sin. X : i :: sin. (T - 90) : I = sl -^-, 90 ) 
 
 Substituting for a -\- b their value, viz., twice radius, and as we 
 have taken radius = unity, we have this equation, 
 
 Sin. (T 90 Q) . sin. (T 9QQ) _ _ 9 
 Si^X BinTX 
 
 Multiplying by sin. X, we have 
 
 Sin. (T 90) + sin. (T 90) = 2 sin. X ; hence 
 
 Sin. X = sin -( T - 9Q ) + sin.(T -9QQ) /g-jx 
 
 Having found the angle X, we next have 
 
 180 (T 90) X = C ; and 180 (T 90) X = C 
 
 Then, Sin. X : 1 : : sin. C : c = -^ (22) 
 
 Sin. X : 1 :: sin. C : c = -^ (23) 
 
 And, c + c : m : : 1 : r = -^- (24) 
 
 Having thus obtained the radius, and the angles required, the 
 remaining elements necessary for making or laying out the curve 
 may, of course, be computed by such of the foregoing formulae as 
 the condition of the locality requires. 
 
 (28) NOTE. Because of the scarcity of extensive tables of nat 
 ural sines, and for the purpose of showing how readily they can be 
 
54 USEFUL FORMULA. 
 
 obtained from logarithmic sines, we have, in the specimens of com 
 putation given below, obtained the natural sine from the loga 
 rithmic sine, and after having found the natural sine of X, we have 
 deduced its logarithm, and then ascertained the corresponding 
 angle from the tables of logarithmic sines. 
 
 We have deemed it proper to give the above hints, for the inform 
 ation of such young engineers as may not be familiar with the 
 principles of trigonometrical tables. 
 
 To proceed with the examples of computation, we have 
 
 T 90 = 87 52 58" log. sin. = 9-9997034 nat. sin. = 0-9993172 
 T _ goo = 740 06 17" loff. sin. = 9-9830685 nat. sin. = 0-9617640 
 
 Nat. sin. (T 90) -f- nat. sin. (T 90) = 2)1 -9610812 
 
 X = 78 40 42" log. sin. = 9 9944656 nat. sin. = 0-9805406 
 
 We have found X = 78 40 42" Again, X = 78 40 42" 
 
 and (T 90) = 87 52 58" and (T 90) = 74 06 17" 
 
 Wherefore C must = 13 26 20" /. C must be = 27 13 01" 
 
 Proof 180 00 00" Proof 180 00 00" 
 
 By (22) we now have X = 78 40 42" co. ar. sin. = 0-0085344 
 
 C = 13 26 20" sin. = 9-3662513 
 
 c = 0-2370204 log. = 9-3747857 
 
 By (23) = again X = 78 40 42" co. ar. sin. = 0-0085344 
 
 C = 27 13 01" sin. = 9-6602591 
 
 c =0-4664375 log. = 9-6687935 
 
 By (24) c -|- c = 0-7034579 co. ar. log. = 0-1527619 
 
 ra = 225-35 feet log. = 2-3528576 
 
 r =320-346 " log. = 2-5056195 
 
PROPORTIONAL RADII TO REVERSED CURVES. 55 
 
 Having ascertained the angles and the radius, we do not think it 
 necessary to extend further the examples of calculation for this 
 particular case. 
 
 (29) We sometimes have another form of the reverse curve, 
 which we will endeavor to investigate. It sometimes happens that 
 in a condition of the tangent lines not very unlike the last de 
 scribed, we have some particular points which we are desirous 
 should govern the location of the track. This state of things 
 necessarily fixes the length of one of the radii, and it is our object 
 in the investigation, to deduce a formula for ascertaining the length 
 of the other, with the centre angles which measure the arcs, etc. ; 
 it being apparent that fixing the length of the radius of one of the 
 curves, governs the radius of the other. 
 
 To proceed with the investigation. Let A T and T B represent 
 the tangent lines, (see figure 9 ;) T and T the measured angles, 
 viz., A T T and B T T ; and T T the line measured, which we 
 represent by m; then, 
 
 Putting a for the line C E == R T or rather = D T + K D 
 
 b " " E T = C K 
 " r = the given radius 
 " x = the radius required 
 
 c f = the line R D 
 we have a 2 + (b + x) 2 = (r~+x)* 
 
 Expanding the equation, a 2 + b 2 + 2 bx + x* = r 2 + 2 rx + x 2 
 Subtracting x 2 leaves a 2 -f- b 2 + 2 bx = r -\- 1 rx 
 
 Trans, and changing signs a 2 + b 2 r = "1 r.r 2 bx 
 
 Dividing by 2 J*!^zli. = ( r b) x 
 
 and by (r - 6) "i - (25) 
 
[Fia. 9.] 
 
COMPUTATION OF ELEMENTS. 
 
 57 
 
 Having deduced the formula, we proceed to give a specimen of 
 calculation. We will suppose m = 630 feet 
 
 r = 555 " 
 T = 169 30 
 T =157 00 
 
 To ascertain a we have first to find T D ; in the triangle T T D 
 we have the 
 
 Angle T = (180 157) = 23 00 00" 
 T = (169-30 900) 790 so/ 00" 
 D = Supplement = 77 SO 7 00" 
 
 D = 77 30 co. ar. sin. = 0-0104185 
 m = 630 log. = 2-7993405 
 
 T =790 30 
 
 sin. = 9-9926661 
 
 Proof = 180 00 / 00" 
 
 TD = 634-4905 
 
 feet 
 
 log. = 2-8024251 
 
 m 
 
 
 log. 
 
 Scno TKnA 
 
 Sin. D 
 
 ouy /oyu 
 
 T 
 
 23 00 
 
 sin. 
 
 = 9-5918780 
 
 T D 
 
 252-1373 
 
 
 = 2-4016370 
 
 T == 
 
 555-0000 
 
 
 
 In triangle CDR, r T D=CD = 
 
 302-8627 
 
 log. 
 
 = 2-4812458 
 
 D 
 
 77-30 
 
 sin. 
 
 = 9-9895815 
 
 b = 
 
 295-6837 
 
 log. 
 
 = 2-4708273 
 
 D 
 
 77 30 
 
 cot. 
 
 = 9-3457552 
 
 cf = 
 
 65-5515 
 
 log. 
 
 = 1-8165825 
 
 From above T D = 
 
 634-4905 
 
 
 
 (T D -f c ) = a 
 
 700-0420 
 
 log- 
 
 = 2-8451241 
 
 
 
 
 2 
 
 a2 = 
 
 490058-8 feet 
 
 log. 
 
 = 5-6902482 
 
 b = 
 
 295-6837 
 
 log. 
 
 = 2-4708273 
 
 
 
 
 2 
 
 6* = 
 
 87428-82 feet 
 
 log. 
 
 = 4-9416546 
 
 r = 
 
 555-00 
 
 log. 
 
 = 2-7442930 
 
 
 
 
 2 
 
 = 308025-00 feet log. = 5-4885860 
 
[FiG. 10.] 
 
VARIETIES OF REVERSED CURVES. 59 
 
 a z _}_ J2 _ r 2 _ 269462-62 log. = 5-4304985 
 
 2 ( r > b) = 5 18- 6326 co. ar. log. = 7-2851402 
 
 x = 519-5636 feet log. = 2-7156387 
 
 b = 295-6837 
 
 x -|- b= E C = 8 15- 24 73* co. ar. log. = 7-0887106 
 a = 700-0420 log. = 2-8451241 
 
 C = 40 39 08" -09 tan. = 9-9338347 
 
 T/ oo T = 12 30 00". 00 
 
 C = 28 09 08" -10 
 
 Having found the centre angle C, we can readily, by several 
 methods, ascertain the value of the other centre angle, C; we how 
 ever shall only give the following method, viz. : 
 
 When the angle D in the triangle T D T, is greater than a right 
 angle, the difference between the angles A.T T and B T T must 
 he added to C, and the sum will be equal to C ; and when D is 
 smaller than a right angle, it must be subtracted. 
 
 The remainder of the elements, which may be needed to facilitate 
 the operations of location, may without difficulty be found, by such 
 of the foregoing formulae as shall be found applicable. 
 
 (30) Another form of the reverse curve is wherein we may 
 have one tangent point fixed, and one centre angle given, the tan 
 gent lines being located in position and direction, but one of them 
 may be shortened or lengthened to adapt it to the unknown or 
 required angles. This case occurs when the point where the curves 
 reverse becomes a governing point in the location, as at the point G 
 in the figure. 
 
 * Having in the triangle C C E found the sides, C E = a, and C E = x + 6, we have to 
 
 find the angle C ; " -- = tan. C. 
 x -f- o 
 
60 USEFUL FORMULA. 
 
 INVESTIGATION OF FORMULAE. 
 
 In the diagram the tangent lines are represented by A T and 
 B T ; and the angles A T T and B T T by T m and Tm ; the line 
 measured, viz., T T by m. 
 
 Then, putting t for the line D T 
 
 And, t " " " D T 
 
 r " " given radius C T 
 
 r " " radius sought G C 
 
 We have in the solution of this problem the following triangles, 
 viz., T D T, which for convenience we denominate No. 1 ; C D E, 
 No. 2 ; E F G, No. 3 ; F G C, No. 4 ; which require to be succes 
 sively solved. 
 
 Commencing with triangle No. 1, we have 
 The angle at T = 180 Tm 
 T = T m - 90 
 " " D = the supplement of the above. 
 
 Then, by analogy, Sin. D : m : : sin. T : t = -^^ D (2G) 
 
 And, Sin. D : m : : sin. T : t = -^j- (27) 
 
 In the solution of triangle No. 2, we have the side C D = r t, 
 and the angle D = the supplement of D in triangle No. 1 ; the 
 angle C being given, the angle E = the supplement of C -f~ D- 
 Denoting the side C E by d, and the side D E by c , we have 
 
 Sin. E : r t :: sin. D : d (28) 
 
 and Sin. E : r t :: sin. C : c (29) 
 
 In triangle No. 3 we have r d = E G ; the angle E the same 
 as E in No. 2 ; the angle G a right angle ; and of course the angle 
 F becomes the complement of E. Denoting the side G F by c, we 
 
COMPUTING ELEMENTS. 61 
 
 have Cos. E : r d :: sin. E : e = tan. E (r d) 
 Then, in the quadrilateral F G C T, we have the angle at F = 
 (180 F), F being the same as in No. 3; bisecting F as thus 
 found by a line from F to C, we have in triangle No. 4, 
 
 Cos. \ F : e : : sin. i F : r = tan. | F . e = 
 
 tan. I F . tan. E . (r d) (30) 
 
 The last expression being equal to the side G C. 
 
 In practice it will be convenient to ascertain the distance of the 
 tangent point from the point T, and for that purpose we have in the 
 triangle C T E, which we call No. 5, 
 
 Sin. E : / : : cos. E : a = cot. E . / (31) 
 
 where c\ represents the line E T. 
 
 Then, will (c + ci) ^ t = the distance of the tangent from the 
 point T, and the direction of course will be known from the relative 
 magnitude of the numbers represented by (ci + c) and t , viz., if t 
 represent the larger number, the tangent point will be in the direc 
 tion of D ; if the smaller, in the direction of B. 
 
 EXAMPLE OF COMPUTATION. 
 
 Suppose m = 030 feet ; r = 555 feet ; T m = 109 30 ; T m = 
 157 00 ; C = 27 49 27" -35. 
 
 Then, T = (180 Tm ) = 180 - 157 = 23 00 
 
 T = (T m - - 90) = 169 30 - 90 = 79 30 
 D = supplement 77 30 
 
 By formula (26) D = 77 30 co. ar. sin. = 0-0104185 
 
 m = 630 feet log. = 2-7993405 
 
 T = 23 sin. = 9-5918780 
 
 t = 252-1373 log. = 2-4016370 
 
62 
 
 USEFUL FORMULAE. 
 
 (27) D 
 
 M 
 
 T/ 
 t> 
 
 = 77 30> 
 
 = 630 feet 
 
 = 79 30 
 
 = 634-4905 feet 
 
 co. ar. sin. =0-0104185 
 log. = 2 7993405 
 sin. = 9-9926661 
 
 loT. = 2-8024251 
 
 (28) E 
 
 555 252-1373 = r t = 302-8627 feet 
 D =102 30 
 
 = 49 40 32"- 65 co. ar. sin. = 0-1178203 
 log. = 2-4812458 
 sin. = 9-9895815 
 
 
 d 
 
 = 387- 
 
 836 
 
 feet 
 
 log. 
 
 = 2- 
 
 5886476 
 
 (29) 
 
 E 
 
 = 49 
 
 40 
 
 32". 65 co. ar. 
 
 sin. 
 
 = 0- 
 
 1178203 
 
 
 r t 
 
 = 302- 
 
 8627 feet 
 
 log. 
 
 = 2-4812458 
 
 
 C 
 
 = 27 
 
 49/ 
 
 27"-35 
 
 sin. 
 
 = 9- 
 
 6690948 
 
 
 c 
 
 = 185- 
 
 4219 feet 
 
 log. 
 
 = 2- 
 
 2681609 
 
 (30) 
 
 
 
 = 69 
 
 50 
 
 16". 32 
 
 tan. 
 
 = o. 
 
 4351232 
 
 
 E 
 
 = 49 
 
 40 
 
 32". 65 
 
 tan. 
 
 = 0- 
 
 0711997 
 
 
 r <f 
 
 = 167- 
 
 164 
 
 feet 
 
 log. 
 
 = 2- 
 
 2231428 
 
 
 r 
 
 = 536- 
 
 3715 feet 
 
 log. 
 
 = 2- 
 
 7294657 
 
 (31) 
 
 E 
 
 = 49 
 
 40 
 
 32". 65 
 
 cot. 
 
 = 9- 
 
 9288003 
 
 
 r ; 
 
 = 
 
 
 
 log. 
 
 = 2- 
 
 7294657 
 
 
 c 
 
 = 455 
 
 267 feet 
 
 log. 
 
 = 2- 
 
 6582660 
 
 
 Cl 
 
 = 185 
 
 -422 
 
 
 
 
 
 c +ci = 640-689 
 
 t = 634-490 
 
 (ci +c ) c^t = 
 
 6- 199 feet = 
 the distance the tangent point must be fixed from T towards B. 
 
 Again, suppose m = 630 feet; r = 555 feet; T m = 169 30 ; 
 Tm= 157; C = 28 09 08" -1. 
 
 Then, T = 160 - 157 = 23 00 
 
 T = 169 30 - 90 = 79 30 
 D = supplement =77 30 
 
 180 00 
 
TESTING COMPUTATIONS. 63 
 
 (26) 
 
 D 
 
 = 77 
 
 30 
 
 co. ar. sin. 
 
 = 0- 
 
 0104185 
 
 
 in 
 
 = 630 
 
 feet 
 
 log. 
 
 = 2- 
 
 7993405 
 
 
 T 
 
 = 23 
 
 00 
 
 sin. 
 
 = 9- 
 
 5918780 
 
 
 t 
 
 = 252 
 
 1373 feet 
 
 log. 
 
 2- 
 
 4016370 
 
 (27) 
 
 D 
 
 = 77 
 
 30 
 
 co. ar. sin. 
 
 = 0- 
 
 0104185 
 
 
 m 
 
 = 630 feet 
 
 log. 
 
 2. 
 
 7993405 
 
 
 T/ 
 
 = 79 
 
 30 
 
 sin. 
 
 = 9- 
 
 9926661 
 
 
 t 
 
 = 634- 
 
 4905 feet 
 
 log. 
 
 = 2- 
 
 8024251 
 
 (28) E = 49 20 51". 9 co. ar. sin. = 0-1199428 
 r t = 302-8627 feet log. = 2-4812458 
 
 D = 102 30 sin. = 9-9895815 
 
 d = 389-7356 feet log. = 2-5907701 
 
 (29) E = 49 29 51". 9 co. ar. sin. = 0-1199428 
 
 r t = 302-8627 feet log. = 2-4812458 
 
 C = 28 09 08"- 1 sin. = 9-6737728 
 
 c = 188- 3482 feet log. = 2-2749614 
 
 (30) |F = 69 40 25"- 95 tan. = 0-4312942 
 E = 49 20 51"- 90 tan. = 0-0661653 
 
 r = 555 ; d = 389-7356, r d = 165 26 44" log. = 2-2181792 
 
 r = 519-5636 feet log. = 2-7156387 
 
 (31) E = 49 20 51". 9 cot. = 9-9338347 
 
 c = 446-1423 feet log. = 2-6494734 
 
 c = 188-3482 
 
 <* +c= 634-4905 
 t = 634-4905 
 
 In this example it appears that the tangent point sought is at T. 
 
 NOTE. The reader will perceive that this example is taken from 
 the results of the example next hut one preceding, and is intended 
 as a test to hoth. 
 
[FIG. 11.] 
 
INSTRUCTIVE EXAMPLE OF REVERSED CURVES. 65 
 
 (31) Another form of reversed curve (if the expression be a 
 proper one) for uniting tracks having different, or like bearings, is 
 where you have the relative position of the tangent points from 
 whence the curves commence in the given tracks, with the bearings 
 of said tracks, or (which is the same thing) the tangent lines, and 
 the radii of the curves given or required by the contour of the sur 
 face, or other considerations which may govern the location. 
 
 In order to give this problem a practical character, we copy from 
 a case which actually occurred in the practice of the writer. We 
 shall not give all the preliminary surveying which was deemed 
 necessary to guide us in the location, (which had been in amount 
 considerable,) but will only state that many lines were run and 
 measured in various directions, to such points as we were desirous 
 of knowing the relative situations of, and the traverses were worked 
 up ; or, in other words, the relative situations of these points were 
 computed in northings and southings, eastings and westings. We 
 copy from those tables such data as we shall find necessary to 
 enable us to explain and solve the problem, and render our com 
 putations intelligible. 
 
 The position of first tangent point, 174 -306 feet northing, 159*617 
 
 feet easting. 
 The bearing of tangent line from first tangent point, N. E. 
 
 85 44 35". 
 The bearing of radius from first tangent point, S. E. 4 15 25", 
 
 and its length = 503 -118 feet, log. = 2. 7016699 
 
 The position of the centre of the curve formed by the above 
 
 radius = 327424 feet southing, 196-963 feet easting. 
 The position of second tangent point, 64-735 feet northing, 509-235 
 
 feet westing. 
 
66 USEFUL FORMULA. 
 
 The bearing of tangent line from second tangent point, X. W. 
 
 61 11 53". 
 The bearing of radius from second tangent point, X. E. 28 48 07" 
 
 and its length, 306 705 feet, log. = 2 5984678 
 
 The position of the centre of the curve formed by the above 
 
 radius = 412-364 feet northing; 318-109 feet westing. 
 
 Representing the first tangent point by T! , and the second tan 
 gent point by T 2 ? and the interior tangent points by T and T , 
 (see Fig. 11,) the centre of the curve with the radius of 503-118 
 feet by C, and the centre of the curve with the radius of 396 705 
 feet by C ; then will C T = 503-118 feet, and C T = 396-705 
 feet. 
 
 Having constructed our diagram in conformity with the data 
 given, we commence by finding the distance C C . We have given 
 
 the 
 
 Position of C = 327-424 feet southing, and 196-963 feet easting, 
 
 c/ = 412-364 feet northing, " 318-109 feet westing, 
 
 Diff. of northing, 739-788 feet 515-072 feet diff. of westing. 
 
 Having obtained the difference of northings and westings between 
 C and C", we have this analogy, to find the bearing from one to the 
 other, viz. Assuming the distance C C = radius, and the differ 
 ence in the northings as a cosine, and the difference of westings as 
 a sine; then, representing these functions, viz., radius by E, the 
 sine by , the cosine by c, the bearing by B, 
 
 We have c . s . . R I tan. B = -| 
 
 or, more practically to express the same thing, we have, tan. B = 
 difference of westings, divided by the difference of northings, and 
 
 Sin. B : westing : : R : C C = -^jr 
 or, Cos. B : northin : : R : C C = 
 
INSTRUCTIVE EXAMPLE OF REVERSED CURVES. 67 
 
 Having thus found C C , we compare it with the sum of the 
 radii, viz., C T + C T"; and if C T + C T" be found greater 
 than C C , the assumed radii will be too great for the relative situ 
 ation of things ; but if C T + C T" be found less than C C the 
 two curves will not run into each other, and must be connected by 
 a piece of straight line. The determination of this straight line is 
 the object of the present investigation. 
 
 For an example of computation we have given the 
 
 Position of C = 327-424 feet southing, and 196 963 feet easting, 
 
 and of C = 412-364 " northing, " 318-109 " westing, 
 
 Diff. of northings, 739-788 " " 515-072 " diff. of westings. 
 
 Difference of westings =515-072 log. = 2-7118680 
 
 " " northings =739-788 log. = 2-8691072 
 
 Bearing from C to C = B = N. W. 34 50 50". 24 tan. = 9-8427608 
 
 B = 34 50 59" -24 co. ar. sin. = 0-2430665 CO. ar. cos. = 0-0858273 
 
 Westing =515-072 log. = 2-2118680 northing 739 - 788 log. = 2-8691072 
 
 From C to C = 901-435 log. = 2-9549345 Proof, log. = 2-9549345 
 
 The given radii are 
 
 From Ti =503-118 feet 
 
 " T2 = 396-705 
 
 C T + C T" = r -f- r = 899-823 
 C C = 901-435 
 
 Difference 1-6 12 feet 
 
 C C being longer than C T -(- C T", it is evident it will require 
 several feet of straight line to connect them. 
 
 * It is obvious that if C C and (C T + C T") are found the same length, the two will run into 
 each other, and form perfect reverse curves. 
 
68 USEFUL FORMULA. 
 
 Eepresenting C T + C T" by r + r , and C C by H, we have 
 
 H : E : : r + / : sin. C or sin. C = ^-^ 
 
 Sin. C : r+r :: cos. C DC = D C == T T" = cot. C (r 
 
 .-. r -f r = 899-823 log. = 2-9541571 
 
 H = 901-435 CO. ar. log. = 7-0450655 
 
 C = 86 34 22" sin. = 9-9992226 
 
 And C = 86 34 22" cot. = 8-7772380 
 
 r -|- r = 899-823 - log. = 2-9541571 
 
 T/ T" = 53-888 feet log. = 1-7313951 
 
 Having thus found the length of the straight line connecting the 
 two curves, it becomes a matter of considerable interest to know the 
 magnitude of the centre angle belonging to each curve respectively. 
 
 We found the bearing from C to C to be N. W. 34 50 50" -24 ; 
 and the angle C in the triangle C C D = 8G 34 22", the comple 
 ment to which will be 3 25 38". 
 
 Then, the bearing from C to C = N. W. 34 50 50 /-24 
 
 It is obvious that if we subtract the complement D C C = 3 25 38" 
 
 Will leave the bearing C T and C T" = N. W. 31 25 12" -24 
 
 The bearing C Ti being = N. W. 4 15 25" 
 
 Gives the centre angle Ti C T = 27 09 47"-24 
 
 Again, C T" bearing (as above) = N. W. 31 25 12"-24 
 
 And the radius T2 C bears = N. E. 28 48 07" 
 
 Gives for the centre angle Ta C T" = 60 13 19". 24 
 
 Such further elements as may be deemed useful in the location 
 might be readily obtained by such of the preceding formula as may 
 be found applicable. 
 
SWITCH-BARS. 69 
 
 (32) There will doubtless arise in practice a great variety of 
 cases, or conditions requiring reverse curves, many of them requir 
 ing formula entirely different from those we have been investi 
 gating, while there are many others which will require merely some 
 slight modifications. But, to repeat what has been more than once 
 stated, it is not our purpose to pursue these investigations until the 
 subject is exhausted, but only to present those cases which we have 
 presumed would most frequently occur in practice. We, however, 
 have another class of curves, the greater portion of them reversing 
 curves, viz., turnouts and side tracks, which may be worthy of 
 consideration. We will therefore proceed to the investigation of 
 formulae for obtaining the necessary elements for locating them, 
 and in the same connection will endeavor to ascertain the magnitude 
 of the angles the rails make with each other at the points of crossing, 
 or, in other words, the dimensions and form of the frogs necessary 
 to be used to best suit each particular case. 
 
 (33) Before we proceed with the investigations, I would make 
 a few remarks upon the switch-bar. The switching of the bar, 
 preparatory to turning a train upon a side track, becomes an 
 important element in our investigation. We have no doubt this 
 element would be considered by persons who have not fully investi 
 gated the subject, as unnecessarily complicating our formulae, and 
 of course our computations. 
 
 The first consideration in preparing the switch, is to ascertain 
 the smallest amount of sliding motion, that will answer to pass the 
 wheels, and, at the same time, give firmness and security to the 
 ends of the rails. The pattern of rails generally used in Massa 
 chusetts requires a movement of about five inches, and the pattern 
 
70 USEFUL FOBMULJE. 
 
 for the switch castings used to secure the ends of the rails, and to 
 give firmness and stability to the structure, are nearly uniform in 
 their dimensions ; hence, whether the switch rail be long or short, 
 whether the turnout be of large or small radius, the switching, or 
 movement of the end of the bars, remains the same. 
 
 It may happen, however, that when the turnout curve is required 
 to be exceedingly severe, and we desire to make the most of the 
 room we have at command, that we determine by calculation the 
 length the switch-bar (switching five inches, or the amount required 
 by the castings) must be to make it exactly correspond to a portion of 
 the intended curve, and the switch rails are accordingly cut to that 
 length. But, if there is nothing to prevent the radius of the turn 
 out from taking such length as may be deemed most desirable, it 
 becomes the better policy to have the switch-bars as long as the bars 
 with which the track is laid, or is being laid. 
 
 The longer the switch-bar is, the smaller will be the angle of 
 deflection occasioned by switching ; and the smaller the deflecting 
 angle, the less the impediment to the passage of the engine and 
 cars, and less springing of the bar than when the bar is shortened ; 
 and of course less liability to accident. 
 
 In general, the deflection of the switch-bar should not be greater 
 than the deflection of the curve for the same length of arc. Cases 
 will, however, occur, when the deflection of the switch-bars of the 
 greatest length in use, will exceed the deflection of the like 
 quantity of arc. These cases occur frequently at the connection of 
 branches ; and, in general, we may say (if this discussion be cor 
 rectly based) that the switch-bar, when switched, should in all cases 
 
SWITCH-BAKS. 71 
 
 be considered the tangent line from whence the curve is to spring, 
 or commence. It may, however, be neglected, when the switching 
 exactly corresponds with the deflection of the same length of curve ; 
 but it will not in that case interfere with the accuracy of the calcu 
 lations to then consider it as the tangent line. Cases may, however, 
 occur, requiring the tangent lines to be continued beyond the end 
 of the bar before the curve commences ; but these cases will not 
 often be met with. 
 
 (34) Having said thus much respecting switches, we commence 
 our investigation by considering the most simple form of the turn 
 out, viz., from a straight track, with curves of equal radii. 
 
 NOTE. I would here state, for the information of the young 
 engineer, that the side-track curves, when there is nothing to inter 
 fere, should be laid to a radius of, say from five to six hundred feet ; 
 but when the nature of things demand it, they may be laid to a 
 radius as short as two hundred and fifty feet. If a radius still 
 shorter is demanded, it becomes necessary to lay extra rails upon 
 the inside of the curve, and as near the rails of the main track as 
 they can be well secured, to assist in supporting the centre driving 
 wheels of the engine, which would otherwise be sometimes unsup 
 ported, and would then cause the engine to run off the track. I 
 hardly need to remark that the double rail will be useless when the 
 engines have only one pair of driving wheels. 
 
 To proceed with the investigation. We first ascertain the relative 
 position of the switch-bar, or the angle it makes with the main 
 track. 
 
[Fia. 12.] 
 
FROG ANGLES. 73 
 
 Let S represent the length of the switch-rail, and d the distance 
 it slides ; Sw the switch angle, or the angle the switch-rail makes 
 with the main track when it is switched. We then have 
 
 8: R : : d . sin. Sw = j- or tan. /Sto (32) 
 
 Having thus obtained the switch angle, we will now put r = 
 radius of the turnout ; a = C Q ; 6 = Q T; q = C A ; d = the 
 distance between the track centres ; e = AT;</ = TB. 
 
 We have in the triangle A C T, to find q and e. 
 
 E : r : : cos. Sw : a = r . cos. Sw ) 
 
 (33) 
 E : r : : sin. Sw : e = r . sin. Sw ) 
 
 Then will b q & + d, and 
 
 2 r : E :: (r + b) ; cos. C = -^ (34) 
 
 Cos. C : (r + 5) : : sin. C : a = tan. C (r -f b) (35) 
 
 We also have a e = g; and the angle C = C $#. 
 
 We have now found the principal elements necessary for locating 
 and marking the centre line of the turnout ; whatever practice 
 requires to fill up the details may readily be supplied from formulae 
 given in the preceding pages. 
 
 (35) The frog angle next claims our attention. Eepresenting 
 the distance between the rails, or in other words, the guage of the 
 track, by h, we have C F = r + J h; and C = r J h + d; 
 
 * To be strictly exact, we make use of the following analogy : 
 
 s : R : : j d : sin. i sw = -~^- 
 
 o 
 
 but this formula is rather a refinement than otherwise, as either of the two first expressions are 
 sufficiently exact for practice, and more convenient. 
 L 
 
74 USEFUL FORMULA. 
 
 and the angle C F = 90 + $w. We now have, in the triangle 
 C F, r + i h : sin. (90 + 8w) : : r J h + d : sin. F = 
 
 sin. (900 -|- sw) (r } 
 
 (r + i A) 
 
 (36) 
 
 Then, drawing at F the tangent line F M, which of course must 
 he at right angles to C F, it will be apparent that the frog angle 
 M F will he a complement angle to F as found above ; where 
 fore, we have 90 F = M F ; and the angle at the centre C, 
 will be equal to 180 (0 + F ;) or, which amounts to the same, 
 C = M F Sw; and the chord, which we represent by ch, from 
 the mouth of the switch upon the outside rail of the turnout track, 
 to the point where the frog angle should be placed in the main 
 track, may be ascertained by the following analogy : 
 
 Sin. i (180 - C) : r + J A : : sin. C : ch = ci- (37) 
 
 The chord just found will be of great convenience to the track 
 layers, as it will show them the proper place for the frog, which 
 should be put into the main track when they are laying it down. 
 
 Having thus obtained our formula}, we now proceed with an 
 example of computation. 
 
 We will assume r = 499 725 feet, which gives a deflection of 
 1 2G for a 25 ft. chord ; h = 4-7 feet ; d = 5 inches; S= 21 
 feet ; and 5=11 feet. 
 
 By formula (32) we have 
 
 d = 5 inches ............ log. = 0-6989700 
 
 S = 252 inches co. ar. ............ loff. = 7 5985995 
 
 1 08 12" tan. = 8-2975695 
 
FROG ANGLES. 
 
 Sw= 1 08 12" cos. = 9-9999145 
 
 r =499-725 log. = 2-6987307 
 
 q = 499- 6262 feet log. = 2-6986452 
 
 Again, Sw= 1 08 12" sin. == 8-2974820 
 
 r = log. = 2-6987307 
 
 e =9-9131 log. = 0-9962127 
 
 2 r = 999 -449 co. ar. log. = 7-0002393 
 
 r -f- b = 988-768 log. = 2-9950944 
 
 (33) 
 
 C = 8 23 02"-7 COS. = 9-9953337 (34) 
 
 C = 8 23 02" -7 tan. = 9-1684391 
 
 r -f6 = log. = 2-9950944 
 
 = 145-7248feet log. = 2-1635335 (35) 
 
 = 9-9131 
 
 = 135-8117 
 
 the distance on the main track from the mouth of the switch to a 
 point opposite T , T being off at right angles from the point. 
 
 C = 8 23 02". 7 
 
 Sw = 1 08 12" 
 
 C Sw = C = 7 14 50"- 7 
 
 r + i h = 502-0746 feet co. ar. log. = 7-2992318 
 
 90 + Sto = 91 08 12" sin. = 9-9999145 
 
 ,- _j_ d i h = 497-7912 feet log. = 2-6970473 
 
 F = 82 25 31" sin. = 9-9961936 
 
 (36) 
 
 9QQ _ F = frog angle = 7 34 29" 
 Sw = 1 08 12" 
 Frog angle Sw = C" = 6 26 17"; C" = 
 
 the angle at C in the triangle T C F. 
 
76 USEFUL FORMULA. 
 
 (180 C) = 86 46 51". 5 co. ar. sin. = 0-0006858 
 r + h = 502-0746 feet log. = 2-7007682 
 
 C" = 6 26 17" sin. = 9-0497178 
 
 ch = 56-386 feet log. = 1-7511718 = 
 
 chord distance from mouth of switch to the angle of the frog upon 
 the outside rail of the turnout. 
 
 Recapitulation of the elements ohtained, viz., 
 
 Centre angles C = ........................................ 8 23 02". 7 
 
 C = ........................................ 7 14 50"- 7 
 
 Frog angle M F = ........................................ 7 34 29"-0 
 
 Chord distance from the mouth of switch to mouth of frog, (outside rail,) 56-386 feet. 
 
 (36) We find wanting the relative position of the point where 
 the curves reverse. The formula will be 
 
 Sin. } (180 - C) : r : : sin. C : e = 
 
 wherein c = the chord distance from the centre point between the 
 mouth of the switch-bars when switched, and the point where the 
 curve reverses. 
 
 Then, to find the chord of the reverse curve = c , we have 
 
 Sin. i (180 C ) :r i: sin. C : c =- ^^ (3D) 
 representing by c in the foregoing analogy, the chord from the 
 reversing point to the tangent point T. 
 
 EXAMPLE OF COMPUTATION. 
 
 ^ (180 C) = 86 22 34"-G5 co. ar. sin. = 0-0008692 
 r = 499- 725 feet log. = 2-6987,307 
 
 C = 7 14 50" sin. = 9-1008914 
 
 c =63-167 feet log. = 1-8004913 (38) 
 
KELATION BETWEEN TURNOUT AND FROG. 77 
 
 (180 C ) = 85 48 28" -65 co. ar. sin. = 0-0011635 
 r = 499- 725 feet log. = 2-6987307 
 
 C = 8 23 02"*7 sin. = 9-1631819 
 
 c =72- 958 feet log. = 1-8630761 (39) 
 
 (37) To lay down the chords c and <? , we commence by placing 
 the theodolite at the point in the centre of the mouth of the switch, 
 when switched ; then, pointing the telescope in the direction towards 
 B, in a range parallel with the main track, lay off an angle towards 
 the side of the road to which the switch-bar switches = (90 Sw) 
 -f- 5 (180 C,) then measure from the instrument the distance c 
 for the reversing point. Then, moving the theodolite to the reversing 
 point, and directing the telescope to the point just left, (to wit, in 
 the centre of the mouth of the switch,) lay off an angle towards 
 the main track = 180 \ (180 C) + J (180 C ,) and 
 measure the distance c for the tangent point of the turnout. The 
 remainder of the laying out may be performed by deflections, or 
 other methods, as explained in the foregoing pages. 
 
 We have now, I think, obtained every element necessary for 
 locating and marking out a turnout from a straight track, and for 
 making a frog pattern to suit. 
 
 N. B. If the tangent points, the reversing point, and the place 
 for the frog be distinctly and properly marked, and the rails 
 properly curved, a skilful tracklayer would put in a turnout 
 without further laying out. 
 
 (38) We now proceed to the investigation of formulae for 
 determining the radius of a turnout from a straight track suited 
 to a given frog. 
 
[Fio. 13.] 
 
RELATION BETWEEN TURNOUT AND FROG. 79 
 
 Let Sw represent the switch angle. 
 
 Fr " " frog " 
 
 C " " centre " 
 
 A " one of the equal angles in isosceles triangle AFC. 
 
 h " the distance between the rails. 
 
 d " " " the switch slides. 
 
 r " " radius. 
 
 We now have in the triangle B C F, (See Fig. 13,) the angle 
 at B = 90 + Suo; the angle at F = 90 Fr ; and the angle at 
 C = 180 (B + F.) Then, in the triangle A B F, representing 
 F by F 2 , and B by B 2 , we have the angle A = i (180 C ;) and 
 the angle F 2 = (A F,) F being = (90 Fr) as above, and 
 the angle BS = 90 Sw. 
 
 Having thus determined the angles, we have in the triangle 
 FAB, A B = ^-; that is, sin. B 2 : h d I : B : A B ; then, 
 representing F A by ch, and A B by w, we have 
 
 Sin. F 2 : w ll sin. B 2 : ch = ~^- J F -- = sin ~^ * F = Sin 7r 3 
 
 Sin. C : ch : : sin. A : r + J h = -^5^- = ( ^7 y f s f c - 
 then, by subtracting J h we have r. 
 
 EXAMPLE OF COMPUTATION. 
 
 Let Fr = 7 34 29"; Sw = 1 08 12"; h = 4-7 feet; d = 
 0-41GG; then, 
 
 900 90 o I 80 o 
 
 JFV= 70 34 29" Sw= 10 08 12" C = 6 o 26 17" 
 
 900_Fr = 82025 31" = F 9QO + Sw = 91 08 12" = B 2)l73O33 43" 
 
 910 08 12" = B 90 o A = 86O46 51"-5 
 
 180 (B + F) = 6 26 17" = C Sw = 1 08 12" F = 82 25 31" 
 
 1800 oo 00" 900 Sw = 88 51 48" = B 2 F., = 40 21 20" -5 
 
TURNOUTS OUTSIDE THE CURVE. 81 
 
 F 2 = 4 2V 20"-5co. ar. sin. = 1-1194836 
 
 C =6 26 17" co. ar. sin. = 0-950282? 
 
 A = 86 46 51"-5 sin. = 9-9993142 
 
 h d =4-2834 log. = 0-6317886 
 
 r 4. 1 h = 502-19 feet ........ log. = 2-7008686 
 
 k = 2-35 
 
 r = 499-84 
 
 We have thus found the radius == 499*84 ; it was intended as 
 a reverse of the previous problem, which gave 499*725 feet. The 
 difference, it will be perceived, is a mere trifle, and is owing to the 
 loss of small fractions in the frog angle, and by using tables of 
 limited extent. 
 
 (39) The next form of a turnout which we shall consider, is 
 one which shall turn out upon the outside of a curve. 
 
 Retaining the same length of switch we had in our preceding cal 
 culation, of course the switch angle will remain the same. Then, 
 representing the switch angle by Sw ; the slide motion by d ; the 
 radius of the main track by r; the radius of the turnout by /; 
 the radius of the side track by / ; and the line C C by a, we have 
 in the small triangle C A CT, two sides, and an included angle, viz., 
 the angle at A = the supplement of the switch angle Sw, and 
 C A = r; C A = r, to find the remaining side a, and the angles 
 C and C . 
 
 This problem has been so often investigated, and is so well under 
 stood, that I have deemed it unnecessary to give an investigation. 
 We however give a formula in connection with the investigation 
 of the turnouts, so that the computer is enabled, without being 
 
USEFUL FORMULA. 
 
 obliged to look up at the time, elementary works to supply the 
 deficiency of such papers, or to refresh his memory, where, per 
 chance, he may he somewhat in doubt. 
 
 In the following formula we shall use the symbols by which we 
 represent the triangle under consideration. 
 
 Tan. X = ta^H^o-AH(r+ d) c.. rr . and x + i (180 o _ A ) = CT ; 
 
 and X | (180 A) == C. 
 
 And sin. C : r + d : : sin. A : a = ^^^- (40) 
 
 or, by way of proof, we have 
 
 Sin. a:/ :: sin. A : a = -=^1 
 
 which, if our previous computations have been correctly performed, 
 the results of this and the preceding analogy will be alike. 
 
 (40) Having solved the small triangle, we next endeavor to 
 find the magnitude of the frog angle, and its relative position in 
 the main track. For the accomplishment of which object, we have 
 in the triangle C F C the three sides, to obtain their angles. 
 
 Having obtained their angles, we then, by the solution of the 
 triangle C A F, obtain the chord, which we shall represent by ch, and 
 which will give us the distance of the frog angle in the main track 
 from the mouth of the switch upon the outside rail of the turnout, 
 or from A to F. We also give the following formula, without going 
 into a general investigation, using the symbols by which we repre 
 sent the lines of the present triangle under consideration. Repre 
 senting the guage of the track by h, we have, in the triangle 
 CFG , the line C C = a, obtained by (40 ;) the line C F = r + 
 J h, which we represent by b; the line C F = / + \ h, which we 
 represent by c. 
 
TURNOUTS OUTSIDE THE CURVE. 83 
 
 If wo now put p = J (a -f- b -f- c) we have 
 Tan. t F = (Jt.J^)* ; and tan. J C - (^"^1^-)*; 
 
 and tan. I C^;;^^)*; (41) 
 
 and 180 F = frog angle == F. 
 
 Substituting C 2 for C as obtained by formula (40,) we have 
 Sin. i [180 (C C O] : c :: sin. (C C 2 ) : ch = 
 
 c . sin. (C C a) 
 
 sin. i (1800 C C a) 
 
 or, probably, in practice, the following may be substituted with con 
 venience, viz., 
 
 E : 2c : : sin. i (C C 2 ) : ch = sin. J (C C 2 ) 2c (42) 
 
 (41) The next step in our investigation will be to ascertain the 
 reversing point, M ; and the terminus of the reverse curve, T. For 
 this purpose, in the triangle C C C" we have three sides, viz., the 
 side C C = a, from (40 ;) side C C" = 2 r ; and the side C C" == 
 C T /; and C T = r + 8; therefore, C C" = r + 8 /. 
 
 Ecpresenting by S the distance between centre lines of the main 
 and side tracks, and substituting b for C C", and c for C C", we 
 obtain the angles by (41 ;) and then, to find A M, which we repre 
 sent by ch , we have 
 
 E : 2r : : sin. i (C C ) : cli = 2r . sin. \ (C " C 2 ) (43) 
 
 Then, to find the chord M T, which we represent by ch" , we have 
 B : 2r : : sin. i (180 C") : ch"=2r. sin. i (180 C") (43) 
 
 To lay off these chords, we place the instrument at the centre of 
 the mouth of the switch, when switched, pointing in the direction 
 
84 USEFUL FORMULA. 
 
 parallel to the tangent of the curve of the main track, (the tangent 
 to the curve of the turnout will be found to vary from the tangent 
 of the main track in amount equal to the switch angle ;) lay off an 
 angle towards the turnout side of the road = 90 Snv + 
 2 (180 C ;) then, measure the chord ch to M, the reversing 
 point ; then, moving the instrument to M, and pointing it to the 
 station between the mouth end of the switch-bars just left, lay off 
 an angle on the side towards the main track equal to 180 
 i (180 C ) + i (180 C";) then measuring the chord ch" to 
 the tangent point T. 
 
 We think that further details of the method of locating the 
 curves need not be here given, the principles having been fully 
 explained in the foregoing pages. 
 
 To proceed with an example of computation. We put r = 
 5729*597 feet; / = 499 -72.5 feet; 7* = 4-7 feet; <2 = 0-416 
 feet ; 5 = 11 feet ; Sw = 1 08 12" ; and of course A = 
 
 1780 si/ 48" 
 
 2) 1 08 12" 
 
 00 34 06" (40) 
 
 r + d = 5730-013 
 
 r = 499-725 
 
 (r + d) + r = 6229-738 co. ar. log. 6-2055302 
 
 (r + d) ~ r = 5230-288 log. 3-7185256 
 
 (1800 A) = 00 34 06" tan. 7-9964947 
 
 X = 00 28 37" -772 tan. 7-9205505 
 
 C =1 02 43" -772 
 
 C 05 28" -228 
 
 A = 178 5i/ 48" 
 
 Proof 1800 QO 7 00" -000 
 
COMPUTATION OF ELEMENTS. 
 
 85 
 
 C =1 02 43" -772 co. ar. sin. = 1 7388259 
 
 r -f- d = 5730 013 feet log. = 3 7581556 
 
 A =178 51 48" sin. = 8 2974820 
 
 a = 6229 647 feet log. = 3 7944635 
 
 C = 00 05 28" -228 co. ar. sin. =2-7982497 
 
 499-725 feet log. = 2 6987311 
 
 8-2974820 
 
 3-7944628 
 
 NOTE. By a more strict computation, the second analogy gave 
 the same results as the first. 
 
 = 5729-597 
 = 2-350 
 A = 5731-947 = b 
 
 r> = 499-725 
 
 1 h = 2-350 
 
 r + J h 502-075 = c 
 
 (41) 
 
 a 
 
 = 6229-641 
 
 
 
 b 
 
 = 5731-947 
 
 
 
 c 
 
 = 502-075 
 
 
 
 
 2^12463-663 
 
 
 
 P 
 
 = 6231-831 
 
 log. 
 
 = 3-7946157 
 
 p>- a 
 
 2-190 
 
 log. 
 
 = 0-3404441 
 
 p - ft 
 
 = 499-884 
 
 log. 
 
 = 2-6988693 
 
 p c = 5729-756 log. = 3-7581362 
 
 p co. ar. log. = 6-2053843 
 
 p a co. ar. log. = 9 6595559 
 
 p b log. = 2-6988693 
 
 p c log. = 3-7581362 
 
 2)22-3219457 
 
 F = 86 03 04" -44 tan. = 11-1609728 
 2 
 
 F =172 06 08" 
 
 p co. ar. log. = 6-3053843 
 
 p -- b co. ar. log. = 7-3011307 
 
 p c log. = 3-7581362 
 
 p a log. = 0-3404441 
 
 6050953 
 
 JC =* 3 37 53" -46 tan. = 8-8025476 
 2 
 
 70 is/ 46" -92 
 
 7 53 51" -12 = the frog angle. 
 
 p co. ar. log. = 6-2053843 
 
 p c co. ar. log. = 6-2418638 
 
 p a log. = 0-3404441 
 
 p b log. = 2-6988693 
 
 2) 15 -48656 15 
 
 C = 00 19 02" -09 tan. = 7-7432807 
 
 F,=s 1720 06 08" -88 
 C = 7 15 46" -93 
 C = 00 38 04"- 19 
 
 1800 00 00" -00 
 
 Proof by 
 
 adding angles. 
 
 C =00 
 
 04" -18 
 
USEFUL FORMULA. 
 
 C =70 15 46" -93 
 
 C 2 =1 02 43" -77 
 
 2^6 13 03" -16 
 i (C C a ) = 30 06 31" -58 
 T -\- \ h = 502-075 feet 
 
 2- 
 ch = 54-457 feet 
 
 sin. = 8-7342531 
 log. = 2-7007686 
 log. = 0-3010300 
 
 log. = 1-7360517 
 
 (42) 
 
 a = 6229-641 
 
 r + <5 r = b = 5240-872 
 
 2 r = c = 999-450 
 
 2^12469-963 
 
 p = 6234-981 log. = 3-7948352 
 
 p a = 5-340 log. = 0-7275413 
 
 p b = 994-109 log. =st 2-9974341 
 
 p c = 5235-531 log. = 3-7189608 
 
 p CO. ar. log. = 6-2051648 
 
 p a co. ar. log. = 9 2724587 
 
 p b log. = 2-9974341 
 
 p c log. = 3-7189608 
 
 2 j 22 -1940184 
 
 85 25 37" -62 tan. = 11-0970092 
 2 
 
 C" = 170 51 15" -24 
 
 p co. ar. log. = 6-2051648 
 
 p b co. ar. log. = 7-0025659 
 
 p c log. = 3-7189608 
 
 p a log. = 0-7275413 
 
 3 50 32" -15 tan. 
 
 C = 70 41 04" -30 
 
 6542328 
 
 8271164 
 
 p co. ar. log. = 6-2051648 
 
 p c co. ar. log. = 6-2810392 
 p a = 7275413 
 
 P 
 
 = 2-9974341 
 2^16-2111794 
 
 43 50" -21 tan. = 8-1055897 
 2 
 
 10 27 40" -42 
 
 C" = 1700 5 1/ 15". 25 
 C = 7 41 04" -31 
 C = 1 27 40" 43 
 
 1800 00* 00" -00 
 
 $ (1800 _ c") = 40 34 22" -37 sin. B- 9016660 
 
 r = 499 -725 feet log. 2 698731 1 
 
 2-000 log. 0-3010300 
 
 ch" = 79 -683 feet log. 1-9013671 
 
 (43) 
 
TURNOUT TO GIVEN FROG. 87 
 
 We have thus completed the computations necessary to find all 
 the leading or principal elements of a turnout upon the outside of 
 a curve in the main track; and whatever of unexplained detail 
 may he required can he computed in the field, as the computations 
 will he hoth short and simple. 
 
 Let us now reverse the prohlem hy supposing that we possess a 
 frog of given dimensions, and are desirous to make it serve us in a 
 turnout from the outside of a curve in the main track, whose radius 
 of course we know. It will then "become necessary to ascertain a 
 radius for the turnout which will he suited to, or compare with, the 
 angle of the frog. 
 
 (42) Without further remark we will proceed to the investiga 
 tion of a formula. To render this investigation plain to the under 
 standing, it may he necessary to "become rather more particular in 
 describing the figure or diagram upon which it is hased, than it has 
 heretofore heen our custom. Making use of the same notation of 
 the preceding prohlem, as far as applicahle, we will commence the 
 construction of the figure at the point where the angle of the frog 
 is to be placed in the outside rail of the main track, viz., at F ; 
 from thence we draw a line to C = r -f~ | h, for the radius of the 
 outside rail of the curve in the main track, and describe a portion 
 of the arc ; then, from the same centre, with a radius = r + | h -f- 
 d, describe the arc S s; and with a radius r J h, describe the 
 inner rail of the main curve. At F lay off an angle from F C = 
 the frog angle + the switch angle, and in accordance therewith 
 draw the line F C" = r J h + d; and then, from the point C" 
 as a centre, with a radius = r -f- \ h, describe the arc A S ; the 
 intersection of the arc with the little arc S s at S will be the place 
 
TURNOUT TO GIVEN FROG. 89 
 
 of the mouth end of the switch-rails, when switched; then, draw 
 the radius from C to S, and continue the same indefinitely on the 
 opposite side ; from S draw the line S C , making an angle with the 
 continued radius = the switch angle ; then, continue the line C" F 
 indefinitely, and draw the line F C , making an angle with the con 
 tinued line F C" = the switch angle ; the intersection of the lines 
 S C with F C at C will determine the length of the radius of the 
 turnout. If we now unite S C" and C C", we shall have a sym 
 metrical figure containing two triangles, S C C" and F C" C, which 
 are similar and equal. We shall have also the triangles S C F and 
 F C" S, which are similar and equal. 
 
 (43) Having thus completed our figure, we commence our 
 investigation hy endeavoring first to find the angle S C F, 
 
 We have the angle C S C" = the angle C F C", and the angle 
 C F C" as "before stated = the frog angle + the switch angle. In 
 the triangle CSC" we have the angle S = the frog -f- the switch 
 angle ; and the side S C = r \h^rd; the side S C" = r + i 7a. 
 As we have before stated, the angles S C C" and F C" C are equal ; 
 hence, it is obvious that the angle sought, viz., S C F, is = to the 
 difference between the angles S C" C and S C C"; to find which, we 
 have (r + \ li) + (r \li + d) : (r + \ Ji) - (r \ h + d) 
 I : tan. J [180 (Fr + Sw)~] : tan. X, and 2 X = C, the angle 
 sought. 
 
 Having found the angle C, our direct course would be to find the 
 
 angles S and F, and the side S F in the triangle C S F, which could 
 
 readily be done by formulae similar to the above, viz., (44 ;) but, 
 
 believing the following to be more convenient, we pass that by. 
 
 N 
 
90 USEFUL FOEMUL2E. 
 
 We therefore have, in the quadrilateral figure C S C F, the angle 
 at C = 2 X ; the angle at S = 180 + the switch angle; the 
 angle at F = (180- frog angle,) and the angle at C = the 
 explementary angle, or which shall make the sum of all the 
 angles = 360. It is now apparent that it will he convenient to 
 represent hy the letters C S C F, the angles belonging to three 
 distinct figures, viz., the angles of the quadrilateral just named ; 
 the angles of the triangle C S F, and of the triangle C S F. For 
 the purpose of preventing confusion, when we use the letters to 
 denote an angle of the quadrilateral, they will not be accompanied 
 by any distinguishing mark. When to denote an angle in the 
 triangle C S F, they will be marked thus, Ci Si Fi ; and when to 
 denote an angle in the triangle C S F, they will be marked thus, 
 2 82 F2. We shall also denote the frog angle by Fr, and the 
 switch angle by Sw. 
 
 To proceed with the investigation, we have 
 
 i (180 C 2 ) = S 2 = F 2 ; and F -- F 2 = Ft ; 
 
 and S S 3 =Si. 
 
 Having thus obtained all the angles of both triangles, we have 
 
 Sin. Si : r + |- h : : sin. Ci : S F = -^^;.^L. (45) 
 
 Substituting c for S F, we have an equal expression, which we fre 
 quently use by way of proof to our work, viz., 
 
 Sin. Fi : r H + d::sin.C, : 0=^-^^^- (45) 
 
 We next have sin. C 2 : c :: sin. Fa or 82 : r = the radius of 
 the turnout sought. (46) 
 
TURNOUT TO GIVEN FROG. 91 
 
 EXAMPLE OF COMPUTATION. 
 
 We will suppose r = 5729 -597 feet; Fr= T 53 51" -12; Sw = 
 1 08 12", to find the radius of turnout /. 
 
 Fr 
 
 Sw 
 Fr -f Sw 
 
 = 7 53 51" 12 
 = 10 08 12" 
 
 = 90 02 03"- 12 
 
 2)l70 57 56" 88 
 i (180 [pr + SW]) ^850 28 58" 44 
 
 r + 4 h 5731-947 
 
 r _ h + rf 5727-663 
 
 ( r _|_ i fc) _|_ ( r _ j A + d) = 11459-610 co. ar. log. = 5-9408302 
 
 (r + J- A) co (r i A + d) = 4-284 log. = 0*6318495 
 
 (1800 [Fr + Su>])= 850 28 58" -44 tan. = 1-1023618 
 
 X = 00 16 16" -03 tan. = 7-6750415 
 
 (44) 
 
 C = 2X = 00 32 32"-06 
 
 (1800 _ pr) = F = 1720 06 08" -88 
 
 18QO 4. Sw = S s= 1810 08 12"-00 
 
 C 60 13 07" -06 
 
 3600 0(X 00" -00 
 
 1800 00 00" -00 
 C a =60 13 07" -06 
 
 2 j 1730 46 52" -94 
 
 S 2 = F 2 = i (1800 c a ) = 860 53 26"-47 S 2 = 86 53 26"-47 
 F = 1720 06 08" 88 S =181 08 12" 
 
 F! =850 12 42" -41 S x = 94O 14 45" -53 
 
 Si =940 14 45" -53 
 
 2X = C, = 00 32 32"-06 
 
 180 00 00" -00 
 
 S, = 940 14 45"-53 co. ar. sin. = 0-0011936 
 
 r + i h = 5731-947 feet log. = 3-7583021 
 
 C, = QO 32 32" -06 sin. = 7-9760615 
 
 c log. == 1-7355572 
 
[Fia. 16.] 
 
TURNOUT FROM INSIDE OF CURVE. 93 
 
 Fi =850 12 42" -41 co. ar. sin. = 0-0015183 
 
 r h + d = 5729-663 feet log. = 3-7579775 
 
 G! = 00 32 32" -06 sin. = 7-9760615 
 
 c = log. = 1-7355573 
 
 C a =60 13 07" -06 co. ar. sin. = 0-9652810 
 
 F 2 =86 53 26" -47 sin. = 9-9993601 
 
 r = 501-41 feet log. = 2-7001984 (44) 
 
 We have thus found / = 501 41 feet. It was intended as a reverse 
 of the previous case; there we assumed r = 499 725 ; the differ 
 ence is a trifle, being only 1-685 feet, which is not astonishing 
 when we consider the acuteness of the angles we have to use in 
 some of the triangles. 
 
 I ought not to close my remarks without an acknowledgement 
 of my indebtedness to Mr. Percival, of Sandwich, for the man 
 ner of constructing the figure which has led us to the foregoing 
 investigation. 
 
 (44) We next examine a turnout from the inside of a curve 
 in the main track. 
 
 Eetaining our former notation as far as practicable, we have in 
 the triangle S C C , the line S C = r d; S C = /; angle S == 
 switch angle to find the side C C (which we denote by ,) and the 
 angles C and C ; wherefore, (r d) + / I (r d) ^ r : : tan. J 
 
 (180 S) : tail. X = t"-H180^S).|r- d ) ~ rO and j ^QQO _ g) 
 
 + X = C ; and \ (180 S) X = C. 
 
 Then, sin. C : r d :: sin. S : s = r ~^ s g; s ; or, we have 
 
 sin. C : / : : sin. S : s = ^f- (47) 
 
 Having found s, we have in the triangle C C C" the side C C 
 
94 USEFUL FORMULA. 
 
 = s, as found above; the side C C" = 2/; and the side C C" = 
 (r S) + / to find the angles, which we do by formula (41) 
 
 Having thus found the angles required, we will denote the 
 triangle SCO No. 1, and represent the angles in said triangle 
 by Si, Ci, C i ; and the triangle C C C" No. 2, and represent 
 the angles by Cs, C 2 , C" 2 ; and the triangle S C M No. 3, and 
 represent the angles in said triangle by Sa, C s, M 3 , and the 
 triangle C" M T No. 4, and shall accompany the letters denoting 
 the angles by 4 ; and so on of such other triangles as may enter 
 into our investigation in the order they are presented. 
 
 We will now proceed to find the chord S M, which we shall 
 denote by ch 3. In triangle No. 3, we have C a = C i C 2 ; and J 
 (180 C 3 ) = S 3 = M 3 ; then will 
 
 Sin. S 3 : r : : sin. C 3 : ch z = r/sin -^_ (48) 
 
 sin. Si 
 
 In triangle C" M T = No. 4, we have C" 4 = C" 2 ; and \ 
 (180 C" 4 ) = M 4 = T 4 ; then, 
 
 Sin. M 4 : r :: sin. C" 4 : ch* = ^ sin - c// * (49) 
 
 sin. M 4 
 
 Then, putting CFG for No. 5, we have C F = r + } h; C F 
 = r J Ji; and C C as found in No. 1 , (which we called s,) to find 
 the angles. See formula (41) 
 
 Having found the angles, the angle C F C, or Fs,will = the 
 frog angle. Then, to ascertain the chord S F, we have in the triangle 
 C F S = No. G,C z C 5 = C 6 ; andi(180 C e) S 6 = F c ,and 
 
 sin. S 6 : / + i h : : sin. CT . : ch, = 0" + * ) . "in. c 6 (50) 
 
 sin. S 6 
 
 which represents the distance from the mouth of the switch of the 
 outside rail of the turnout to the frog angle. 
 
TURNOUT FROM INSIDE OF CURVE. 
 
 95 
 
 EXAMPLE OF CALCULATION. 
 
 Let r== 5729 -597 feet;/ 499 -725 feet; A ==4 -7 feet; 
 feet; 5=11 feet; Si , or switch angle, = 1 08 12". 
 
 r d 
 r 
 
 = 5729-181 
 = 499-725 
 
 (r _ d) -f r 6228-906 
 (r d) c~ r = 5229-456 
 (18QO _ Si) = 890 25 54" 
 X =890 19 23" 
 
 180 00 00" 
 Si = 1 08 12" 
 
 2^1780 51 48" 
 J (1800 i) = 890 25 54" 
 co. ar. log. = 6-2055882 
 log. = 3-7184566 
 tan. = 2-0035053 
 
 tan. = 1-9275501 
 
 C\ = 178 45 17" 
 
 C a =00 06 31" 
 
 S 1 =10 08 12" 
 
 1800 00 00" 
 
 Ci =1780 45/ 17" C o. ar . sin. = 1-6628906 
 
 r d = 5729-181 feet log. = 3-7580926 
 
 Si =10 08 12" sin. = 8-2974820 
 
 s, =5229-532 log. = 3-7184617 
 
 (47) 
 
 C, = 00 06 31" co. ar. sin. = 2-7222486 
 
 r = 499-725 feet log. = 2-6987311 
 
 sin. = 8-2974820 
 
 log. = 3-7184617 
 
 To find the elements of triangle No. 2, we have 
 
 r 5 + r = 6218-322 
 
 
 c 2r 
 
 = 5229-532 
 = 999-450 
 
 
 
 
 
 
 2)l2447-304 
 
 
 
 
 
 P 
 
 = 6223-652 
 
 log. = 
 
 3-7940453 
 
 
 
 p a 
 
 = 5-330 
 
 log. = 
 
 0-7267272 
 
 
 
 p - b 
 
 = 994-120 
 
 log. = 
 
 2-9974388 
 
 
 
 p c 
 
 = 5224-202 
 
 log. = 
 
 3-7180200 
 
 
 P 
 
 co. ar. log. = 
 
 6-2059547 
 
 P 
 
 co. ar. log. 
 
 = 6-2059547 
 
 p a 
 
 co. ar. log. = 
 
 9-2732728 
 
 p - b 
 
 co. ar. log. 
 
 = 7-0025612 
 
 p - b 
 
 log. = 
 
 2-9974388 
 
 p c 
 
 log. 
 
 = 3-7180200 
 
 p c 
 
 log. = 
 
 3-7180200 
 
 p a 
 
 log. 
 
 = 0-7267272 
 
 2)22-1946863 
 85 25 50" -22 tan. = 11-0973431 
 
 2 
 
 2) 17 -6532631 
 30 50 16" -79 tan. = 8-8266315 
 2 
 
 C a = 1700 5 i> 40" -44 
 
 C" 2 = 
 
 33" -58 
 
96 USEFUL FORMULA. 
 
 p co. ar. log. i= 6-2059547 
 
 p c co. ar. log. = 6-2819800 
 p a log. = 0-7267272 
 
 p b log. = 2-9974388 
 
 2J16-2121007 
 00 43 53" 
 2 
 
 C 2 = 10 27 46" 
 
 C a 1700 5i/ 40" -45 
 
 C" 2 = 7 40 33" -58 
 
 C 2 =10 27 46" -00 
 
 Proof 1800 otX 00" -00 
 
 In triangle No. 3, we have 
 
 C , = 1780 45/ i7 / 
 
 C 3 =70 53 36" -56 
 
 2 ) 1720 06 23" -44 
 
 i (180 C ,) = M 3 = 860 03 11" -72 co. ar. sin. = 0-0010312 
 
 r = 499-725 feet log. = 2-6987311 
 
 C , = 70 53 36" -56 sin. = 9-1377717 
 
 ch 3 = 68-791 feet log. = 1-8375340 
 
 In triangle No. 4, we have 
 
 C" 2 = C" 4 = 70 4(X 33" -58 
 
 J (1800 C" 4 ) = M 4 =860 09 43". 21 co. ar. sin. = 0-0009751 
 
 r =499-725 feet log. = 2-6987311 
 
 C 4 =7 40 33" -58 sin. = 9-1257121 
 
 ch 4 = 66-899 feet log. = 1-8254183 
 
 In triangle No. 5, we have C F = r 5 h; C C" = s 
 5229-565 ; C F = / + \ h, to find the angles. Let 
 
 a = r h = 5727-247 
 b = r + i ; A = 502-075 
 
 log. = 3-7581123 
 log. = 0-3416323 
 log. = 3-7182831 
 
 log. = 2-i 
 
 
 2)ll458-887 
 
 P 
 
 = 5729-443 
 
 p a 
 
 = 2-196 
 
 p -b 
 
 = 5227-368 
 
 p c 
 
 = 499-878 
 
TUENOUT FROM INSIDE OF CURVE. 
 
 97 
 
 p co. ar. log. = 6-2418877 
 
 p a co. ar. log. = 9-6583677 
 p-b 
 p c 
 
 86 oi 50" -13 
 
 C 5 = 172 03 40" -26 
 
 p co. ar. log. = 6-2418877 
 
 p c co. ar. log. = 7-3011359 
 
 p a log. = 0-3416323 
 
 p b log. = 3-7182831 
 
 2^17-6029390 
 
 30 37 21" -13 tan. = 8-8014695 
 2 
 
 F, =70 14 42" -26 Frog angle. 
 
 In triangle No. 6, we have 
 
 C s 
 
 p co. ar. log. = 6-2418877 
 
 p b co. ar. log. = 6-2817169 
 
 log. = 3-7182831 
 log. = 2 6988641 
 
 p c 
 p a 
 
 00 20 48" -73 
 2 
 
 log. = 2-6988641 
 log. = 0-3416323 
 
 2)22-3174026 
 
 2^15 -5641010 
 
 tan. = 11-1587013 
 
 tan. = 7.7820505 
 
 C. = 00 41 37" -46 
 
 C s 1720 Q3/ 40" -26 
 F K 70 14 42" -27 
 
 1800 00 00" -00 
 
 = 1780 45 17" 
 = 1720 05 24" -65 
 
 = 60 39 52" -35 
 
 211730 20 07" -65 
 
 = J (1800 C 6 ) = 860 40 03"- 
 
 S 6 =860 40 03" -82 
 
 r 1 + J h = 502 075 feet 
 
 C 6 =60 39 52" -35 
 
 ch 6 = 58-233 feet 
 
 co. ar. sin. = 0-0007349 
 log. = 2-7007686 
 sin. = 9-0646679 
 
 log. = 1-7661714 
 
 the distance from the mouth of the switch on the outside of the 
 turnout to the frog angle in the main track. 
 
TURNOUT FROM INSIDE OF CURVE. 99 
 
 (45) Having thus completed our investigation of the problem 
 direct, we will now examine it reversed, by supposing the radius of 
 the main track given, as before, viz., r = 5729 597; h = 47 feet ; 
 d = -416 feet ; and the frog angle Fr = 7 14 42". 27 ; the switch 
 angle, Sw = 1 08 12"; to find the radius of the turnout = /, and 
 the position of the frog. 
 
 Draw the lines ee, cc representing the outer and inner rail of the 
 main track, and the dotted line //, corresponding to the switching 
 of the outer rail ; then, draw the radius F C ; then, from F draw the 
 line F C indefinitely, making an angle with F C = the frog angle ; 
 then, from F draw the line F C", making an angle with F C = the 
 switch angle -f- the frog angle, and equal in length to the radius 
 of the dotted line = C G; then, with a radius = C F and with C" 
 as a centre, draw the angular dotted line at S, and this dotted line 
 will intersect //at the mouth of the switch. From this intersection 
 draw the radius S C ; then, draw the line S C , making an angle with 
 S C = the switch angle, and the lines S C and F C will intersect 
 each other at the centre of the curve of the turnout, viz., at C ; 
 then, with a dotted line, join S C" and C" C, which will complete our 
 diagram. 
 
 If we now examine our diagram, we shall find it to contain two 
 equal and similar triangles, viz., FCC" and S C" C with the 
 angles F C" C = S C C"; then it will be apparent that the angle 
 S C F will be equal to the difference between the angles F C C" and 
 F C" C. Having thus shown the relative magnitude of the angles 
 last named, we will noAv proceed to find the angle S C F. In the 
 triangle F C C" we have the angle at F = the frog angle -f- the* 
 switch angle; the side F C = r } li ; the side F C" =r+ (i h 
 
100 USEFUL FOKMUL^E. 
 
 - d;) then, (r + \ li d) + (r J h) : (r + \ h d) ^ (r 
 4 h) : : tan. 4 [180 (Fr + Sw)] : tan. X ; that is, 
 
 ~ * h > tan - * t 180 (fr + SuQ] /*1\ 
 
 d) r-iA) 
 
 And 2 X = F C S, (51,) which we shall hereafter represent by 2. 
 
 In the above notation, we have represented the frog angle by Fr, 
 and the switch angle by Sw. 
 
 We now have the triangle F C S, which we shall hereafter 
 denominate No. 2 ; the angle C = C 2 , as found above ; the line C F 
 = r 4 h> the line CS = r + iA d, to find the angles at F 
 and S, which we shall denote thus, by Fa and 82; then, (r -\~ 5 h 
 - d) + (r 4 h) : (r + %h d)c~ (r 4 h) : I tan. 4 (180 
 C 2 ) : tan. X 2 and 4 (180 C 2 ) + X 2 =F 2 , and J (180 - 
 
 n \ V 2 __ g (52^ 
 
 Then, sin. F 2 : r ~\- \ li d : I sin. C 2 : S F, which we shall 
 denote by ch, or sin. 82 : r J 7i . : sin. 2 : ch. (53) 
 
 We then have in the triangle S F C , (which we denominate No. 3, 
 and mark the letters denoting the angles accordingly,) the line S F 
 = ch, found above; the angle 83 = (82+ Sw;) the angle F 3 = 
 (F 2 Fr) = (82+ Sw,) and the angle C 3 = 180 [(F 2 Fr) 
 + (82 + Sw;)] and C 3 : ch : l Fs : / + 4 h (54) 
 
 Then, by subtracting J h, we have /. 
 
 Having thus obtained our formula, we now give an example of 
 calculation. 
 
TURNOUT FROM INSIDE OF CURVE. 
 
 101 
 
 To find / from the frog angle, 
 
 Fr 
 
 Sw 
 
 Fr + Sw 
 
 * [1800 (Fr + Sw;)] 
 
 d = 5731-531 
 
 = 5727-247 
 
 = 70 14 
 = 1 08 
 
 42" -27 
 12"- 
 
 = 8 22 54"- 27 
 
 2)171 37 05"- 73 
 
 85 48 32" 86 
 
 Sum = 11458-778 
 
 Difference = 4-284 
 
 i [1800 (Fr -f SuOl =850 48 32"- 46 
 X = 00 17 32"- 38 
 
 2 
 
 log. = 5-9408619 
 log. = 0-6318495 
 tan. = 1-1350427 
 
 tan. = 7-7077541 
 
 (1800 c a ) 
 
 = 00 35 04" -76 
 2) 1790 24 55"- 24 
 = [890 42 27"- 62 
 
 i (1800 C 2 ) 
 
 = 5731-531 
 
 = 5727-247 
 
 11458 778 
 
 4-284 
 
 = 890 42 27"- 62 
 = 40 11 27"- 14 
 
 log. = 5-9408619 
 log. = 0-6318495 
 tan. = 2-2922459 
 
 tan. = 8-8649573 
 
 F s =930 53 54"- 76 
 
 S a = 850 31 00"- 48 
 
 C 2 =00 35 04"- 76 
 
 1800 00 00"- 00 
 
 = 930 53 54"-76 co. ar. sin. = 0-0010062 S 2 = 
 
 i d = 5731-531 feet log. = 3-7582707 r fc = 
 
 C 2 = 35 04"-76 sin. = 8-0087699 
 
 cA a = 58-620 feet log. = 1-7680468 
 
 850 31 oo"-48 co. ar. sin. = 0-00133C9 
 
 5727-247 feet log. = 3-7579458 
 
 sin. = 8-0087699 
 
 log. = 1-7680468 
 
 In triangle No. 3, we have 
 
 Fr 
 F 3 
 
 = 930 53 54"- 76 
 
 _ 70 i4/ 42"- 27 
 
 = 860 39 12"- 49 
 
 = 860 sg/ 12"- 48 
 
 5 2 =850 31 00"- 48 
 Siv = 1 08 12" 
 
 5 3 =860 sg/ 12"- 48 
 
 1730 is 24" -97 
 18QO _ (F, 4- S 3 ) = 60 41 35"- 03 
 
[Fia. 18.] 
 
TURNOUT FROM INSIDE OF CURVE. 103 
 
 By the problem, F 3 and S should be equal. Then, 
 
 1800 - (F, + S 3 ) = 6 41 35"-03 co. ar. sin. = 0-9334859 
 
 ch t log. = 1-7680467 
 
 F 3 =860 39 12". 48 sin. = 9-9992587 
 
 r + A = 502-101 
 
 } & = 2-350 
 
 r 1 = 499-751 
 
 It was expected the radius would be found to be 499*725. The 
 error only amounted to 026 feet, or a little more than one fourth 
 of an inch. 
 
 (46) The following problem has frequently presented itself in 
 the practice of the writer, viz., the situation of a turning table 
 with respect to the main track, (the radius of the turnout curve 
 being given to find the relative situation of the switch,) and such 
 additional elements as will be required to locate the turnout. 
 
 To explain : we have in several instances found it necessary so to 
 place a turntable by the side of the railroad track, that a building 
 erected over it might answer the purpose of shielding the table 
 from the weather, and the engine during the night, occasionally, if 
 not constantly ; and also, to afford a convenient situation for a water 
 tank to distribute water to the engines when upon the main track, 
 and when sheltered upon the turntable. 
 
 The ruling principles which govern in this matter may be stated 
 thus: 
 
 First, The proper distance between the centre of the table and the 
 
 main track. 
 Second, A suitable amount of straight track to guide the engine 
 
104 USEFUL FORMULA. 
 
 steadily upon the table. (A turning table should never be directly 
 connected with a curve, as the engine will have a tendency to force 
 it out of place.) 
 
 Third, The remainder of the track to be united to a curve of fixed 
 radius, which shall just connect the straight track adjoining the 
 table with the main track. 
 
 INVESTIGATION. Kepresenting the mouth of the switch by S ; the 
 length of the switch by s, and the switch angle by $iv; the centre 
 of the table by O ; the point where the curve unites with the straight 
 line adjoining the table by T ; the centre of the curve by C ; the 
 radius by / ; the point on the main track where, a line being drawn 
 therefrom to the centre of the table shall form a right angle with the 
 centre line of said track, by Q ; we shall then have in the triangle, 
 Sw C S, for finding the line Siv C, which line we represent by q, 
 
 R : / -|- d cos - Sw : q = (r -f- d) . cos. Sw (55) 
 
 And in the triangle COT, (by assuming C O as radius,) we have 
 this analogy : 
 
 / : T : : (C O = radius,) : tan. C 2 = - r ^- (56) 
 And therefore, cos. C 2 : r : : R : C = ^ (57) 
 
 Representing the distance of the centre of the turning table 
 from the main track (viz., Q O) by n, and the line C O by^>, we 
 then draw the line O A, parallel with the centre line of main track, 
 and in the triangle A O C \vc have 
 
 p:*R::q n: cos. C = -^ (58) 
 
 If we now deduct from the angle C, the angles Sw and 2, we 
 shall have left the angle C of the triangle S C T, which we repre 
 sent by Cs, then will J (180 Cs) = S = T ; and 
 
 Sin. T : r : : sin. C 3 : S T = J ^~^ 2 - (59) 
 
TURNING TABLE AND MAIN TRACK. 105 
 
 To give an example of calculation, we will assume ft = 3G 4 
 
 feet; / = 499 -725 feet; s= 21 feet; T O = 60 feet ; d = 0" 
 feet; Sw = 1 08 12". Then, 
 
 r 1 -j- d = 500-141 log. =r 2-6990925 
 
 Sw == 1 08 12" cos. = 9-9999145 
 
 q = 500-043 log. = 2 -6990070 
 
 TO =60 feet log. = 1-7781513 
 
 r = 499-725 co. ar. log. = 7-3012689 
 
 C 4 =6 50 47" -41 tan. = 9-0794202 
 
 C 2 =60 50 7 47" -41 co. ar. cos. = 0-0031080 
 
 r 
 
 P 
 
 Sw = 10 08 12" q n 
 C 2 = 60 50 47" -41 C 
 
 = 499-725 
 = 503-314 
 = 463-643 
 = 220 54/ 02" -78 
 
 log. = 2-6987311 
 
 log. = 2-7018391 
 log. = 2-6661837 
 
 cos. = 9-9643446 
 
 7 58 59"-41 (Sw + C,) = 70 58 59" -41 
 C 3 =140 55/ 03" -37 
 
 2^1650 04 56" -63 
 
 T = (180 C 3 ) = 82 32 28" -31 co. ar. sin. = 0-0036904 
 
 r> = 499-725 log. = 2-6987311 
 
 C, =140 55/ 03" -37 sin. = 9-4106586 
 
 S T = 129-742 feet log. = 2-1130801 
 
 Having thus ascertained the elements deemed necessary, before 
 we commence the business of location, we will now proceed to 
 describe the operations necessary to execute the work. 
 
 (47) An examination of the figure will render it apparent that 
 taking from the complement of Cu the complement of C will leave 
 the angle A O T. Having thus obtained A O T, we place our 
 instrument at O and lay off said angle from A, and measure the 
 
106 USEFUL FORMULAE. 
 
 distance O T to T ; then, moving the instrument to T, and point 
 ing it to O, we lay off the angle O T S == (90 + S) or (90 + T) 
 and measure the distance T S to S, the place of the mouth of the 
 switch ; and if the work has heen correctly prepared, we shall ho 
 the distance d from the centre line of the main track, upon the 
 side towards the turnout curve. The curve may now he further 
 marked by deflections, agreeably to directions given in the foregoing 
 pages. 
 
 (48) Having thus obtained the formula for computing the 
 elements of a turnout to a turntable, with a given amount of 
 straight line to guide the engine, it will frequently be found con 
 venient to have a formula to lay out a track to a turntable from an 
 existing joint in the rails of the main track, with a fixed radius, 
 and a fixed position for the table. This method of proceeding will 
 save the trouble of cutting rails, and making unnecessary joints in 
 the main track ; and another consideration will be that of affording 
 side track room for cars to stand upon. 
 
 In the following investigation we shall preserve the notation of 
 the preceding formula, as far as applicable. 
 
 Making 5 = the distance S Q, (as measured,) and a == S O, we 
 have, by taking a as radius, the following analogy : 
 
 5 : n d : I (rad.) : tan. S = -^-7 (CO) 
 
 wherein S will be equal to the angle at S in the triangle Q S O ; 
 then, cos. S : * :: B : a= C08 3 8 (61) 
 
 then, representing the angle S in the triangle C S O by 82, we 
 have 1)0 + Sw S = Sa, and 
 r + alr f <~a : : tan. \ (180 S 2 ) : tan. X = ^L^A 
 
TURNING TABLE AND MAIN TRACK. 107 
 
 and X + i (180 82) = the angle ; and X i (180 Sa) 
 = the angle C (62) 
 
 We then have sin. O r : : sin. 82 : C O ; or, sin. C : a 1 1 sin. 
 S 2 : C O ; then, putting p = C O, we have 
 
 p : E : : / : cos. C 2 = ^- (63) 
 
 And cos. C 2 : / : : sin. C 2 : T O = tan. C 2 / (64) 
 
 Then, deducting Ca from C leaves C 3 = the angle C in the triangle 
 S C T ; and \ (180 C 3 ) = the angle S == the angle T ; and 
 
 Sin. T : / : : sin. Ca : S T = ^^- (65) 
 
 To give an example of calculation, we assume n = 36*4 ; / = 
 499-725; d = 0-416; Sw = 1 08 12"; 5 = 200 feet. 
 
 & = 
 
 200 feet co. ar. log. = 7-6989700 
 
 900 00 00" -00 
 
 n d = 
 
 35-984 log. = 1-5561094 
 
 10 08 12" 
 
 S = 
 
 100 n 58" -32 tan. 9-2550794 
 
 90 + Su>= 91 OS 12" 
 
 
 
 S = 100 n 58" -32 
 
 S = 
 
 co. ar. cos. = 0-0069179 
 
 S 2 =800 56 13" -68 
 
 S = 
 
 200 feet log. = 2-3010300 
 
 2(990 03 46" -32 
 
 a = 
 
 203-211 feet log. = 2-3079479 i 
 
 (180 S 2 )= 490 31 53" -16 
 
 
 
 (GO) 
 
 r 
 
 499 -725 feet 
 
 
 a = 
 
 203-211 
 
 
 Sum = 
 
 702-936 co. ar. log. = 7-1530842 
 
 
 Difference = 
 
 296-514 log. = 2-4720452 
 
 
 i(18(P-S a ) = 
 
 49 31 53" -16 tan. = 0-0689836 
 
 
 X = 
 
 26 18 34"-57 tan. = 9-6941130 
 
 
 O = 
 
 750 50 27" -73 
 
 
 C 
 
 230 13 18" -59 
 
 
 S, 
 
 80 56 13" -68 
 
 
 180 00 00" -00 
 
108 USEFUL FORMULAE. 
 
 =75 50 27" -73 co. ar. sin. = 0-0133981 C = 23 13 18" -59 co. ar. sin. = 0-4041819 
 
 r = 499 -725 feet 
 
 log. = 2-6987311 a =203 -211 feet 
 
 log. = 2 3079472 
 
 S., =800 5 6 13". 68 
 
 sin. = 9-9945442 
 
 sin. = 9-9945442 
 
 P 
 
 log. = 2-7066734 
 
 log. = 2-7066733 
 
 r = 
 
 log. = 2-6987311 
 
 
 = 10 55 27" -50 cos. = 9-9920577 C =23 13 18" -59 (63) 
 
 C 2 = 100 55/ 2 7//. 50 
 
 = 100 55/ 27"-50 tan. = 9-2855789 C 3 = 12 17 51" -09 
 
 r = log. = 2-6987311 2J167O 42 08"-91 
 
 TQ = 96-452 feet log. = 1-9843100 i (180 C 3 ) = 83O 51 04"-45 (64) 
 
 T = 830 51 04" -45 co. ar. sin. = 0-0025057 
 
 r 1 =499 -725 feet log. = 2-6987311 
 
 C 3 = 120 i7> si /- 09 sin. = 9-3283555 
 
 S T = 107-051 feet log. = 2-0295923 
 
 Having thus ascertained the elements of the turnout, it remains 
 to describe the method of locating or marking the same upon the field. 
 
 We have found, formula (63,) Ca = 10 55 27" 5, the comple 
 ment to which = 79 04 32" -5 = the angle COT; we have also 
 found, formula (62,) O = 75 50 27" -73 = the angle COS. 
 
 We now place our instrument at O, and lay off from S the dif 
 ference between 79 04 32" -5 and 75 50 27" -73 = 3 14 04" -77, 
 and measure from O to T 96 -452 feet ; we then move the instru 
 ment to T, and lay off the angle S T O = J (180 C) + 90 
 that is = 90 + T, as found above (65) = 90 + 83 51 04" -45 
 = 173 51 04"-45, and measure 107 -051 feet to S; and if the 
 field work and computations have been correctly performed, the 
 point s will be found directly between the joints in the rails in the 
 main track, and 41(> feet from the centre line on the side of the 
 turnout. The curve may then be further marked by deflections, as 
 heretofore explained. 
 
TRACK OVER WATER. 109 
 
 (49) It is not an unfrequent occurrence for an engineer to be 
 required to locate railroad and other curves in situations inaccessible 
 to the making of measurements in the common or ordinary methods. 
 These cases occur where railroads are located across bays and 
 inlets of the ocean or lakes, and across rivers or estuaries, etc., etc. 
 We know of no better method of managing this matter than by 
 projecting a system of triangles from a well-selected base to points 
 desirable to mark or permanently fix. The calculations necessary 
 for the arrangement of such a series of triangles, when connected 
 with the choice of the location of the curve, and the determination 
 of the necessary elements to carry forward the whole work with 
 accuracy and convenience, may, in some instances, be too compli 
 cated for the invention of the young and inexperienced engineer. 
 To aid such in the performance of their task is the object of the 
 present article. 
 
 We have chosen as an example, an imaginary river of some 250 
 feet wide ; but, before we proceed with the investigations, let us 
 suppose the straight or tangent lines upon both sides of the river to 
 have been located, and sufficiently marked to show their relative 
 bearings. Our first operation, then, is to select such a situation as 
 may be thought, upon a thorough examination of the whole subject, 
 the most desirable for the location of the curve. The point we 
 have selected will be seen in Fig. 19, marked 1 ; and in Fig. 20, 
 marked as station 10 of the railroad location. 
 
 Having determined on this point, we commence the survey by 
 running a line from it to a point in the tangent line upon the same 
 
 * Care should be taken to so select the termini of the base, that the lines projected therefrom 
 should intersect with each other at the points to be located as near at right angles as they well 
 can. 
 
[Fio. 19.] 
 
 A 
 
TRACK OVER WATER. Ill 
 
 side of the river, marked on the diagram. Then, by considering 
 the tangent line from to A to hear due east, (whatever may he its 
 direction,) we measure the angle A 1, hy which we determine the 
 line from to 1 to hear S.E. 75 08 35" -37, and hy measurement 
 we find the distance 152*5 feet. We then place a signal at station 
 2, and hy a triangulation we ascertain the hearing from station 1 
 to 2 to he S.E. 85 02 37", and distance = 350 feet ; then, crossing 
 the river to station 2, we run and measure a line therefrom to 
 station 3, situated in the other tangent line ; the hearing of this 
 line we determined to he S. 72 59 3G"-41 E., and its length = 
 1604-264 feet; then, removing to station 3, we ascertain the hear 
 ing of the tangent line to he S. 60 E., or (which is the same,) 
 K. 60 W. 
 
 Having thus connected the straight or tangent lines, hy a traverse 
 running through the point selected as the most suitable for the 
 location of the curve, our next step will he to prepare our data to 
 ascertain its radius. 
 
 Upon examination of the foregoing, we find the courses and dis 
 tances noted in the following table, viz., 
 
 BEARINGS. DISTANCES. 
 
 O / // 
 
 Station to Station 1 = S. 75 08 35-37 E. = 152-5 feet 
 "1 " 2 = S. 85 02 37-00 E. = 350-0 " 
 
 " 2 " 3 = S. 72 59 36-41 E. = l04-264 " 
 
 and the bearing of the tangent lines from to A due East, and from A to 3 S. 60 E. 
 
 Having thus collected and arranged our courses and distances, 
 we then compute their northings and southings, eastings and west 
 ings, according to the requirements of the case. 
 
USEFUL FORMULA. 
 
 Eepresenting the northing by N, southing by S, easting by E, 
 and westing by W ; and, for the convenience of a general expression 
 in our formula, we call the northings or southings the latitudes, 
 which we represent by L ; and the eastings and westings the depar 
 tures, which we represent by D. Then, putting E for the radius 
 of the tables, B for the bearings, and d for the distance, we have 
 
 II : d : : sin. B : D = sin. B . d ; 
 and E : d : : cos. B : L =d. cos. B (M) 
 
 COMPUTATIONS OF LATITUDES AND DEPARTURES. 
 
 No. 1. B = S 750 08 35" -37 E sin. = 9*9852331 cos. = 9-4089262 
 
 d = 152-5 feet log. = 2 1832698 log. = 2 1832698 
 
 D = 147-401 feet log. = 2-1685029 L = 39-1017 log. = 1-5921960 
 
 No. 2. B = S 85 02 37" E sin. = 9-9983730 cos. 8-9365008 
 
 d = 350 feet log. = 2-5440680 log. = 2-5440680 
 
 D = 348-691 feet log. = 2-5424410 L = 30-239 log. = 1-4805688 
 
 No. 3. B = S 720 59 / 36 //. 41 E sin. = 9-9805811 cos. = 9-4660977 
 
 d 1604-264 feet log. = 3-2052779 log. = 3-2052779 
 
 D = 1534-119 feet log. = 3-1858590 L = 469-219 log. = 2-6713756 
 
 Having computed the latitudes and departures, or, in other 
 words, the southings and eastings indicated by the tables of courses 
 and distances ; we then, to render these operations as perspicuous 
 as we well can, re-arrange in a tabular form, our courses and dis 
 tances, with the southings and eastings belonging to each ; and 
 having summed them up, we proceed to compute the bearing and 
 distance from to station 3. Thus, by making use of the symbols 
 of the preceding formula, with the addition of 5, by which we repre 
 sent the distance from station to 3, we have 
 
 Tan. B = - ; and sin. B : D : : 11 : $ = -J^-; 
 or, Cos. B : L : : K : a = J,; , (N) 
 
BEARINGS OF STATIONS. 
 
 
 BEAEINGS. 
 
 DISTANCES. 
 
 SOUTHINGS. 
 
 EASTINGS. 
 
 Station to 1 
 
 S 75 08 35"- 07 E 
 
 152-5 feet 
 
 39-102 
 
 147-401 
 
 " 1 to 2 
 
 S 85 02 37"-OOE 
 
 350-0 
 
 30-239 
 
 348-691 
 
 " 2 to 3 
 
 S 72 59 36 /- 41 E 
 
 1604-264 " 
 
 469-219 
 
 1534-119 
 
 
 
 
 L = 538- 560 
 
 D = 2030- 211 
 
 D = sum of eastings = 2030-211 
 
 L = sum of southings = 538-560 
 
 log. = 3-3075411 
 log. = 2-7312341 
 
 Bearing from station to 3 = B = S 75 08 35"-37E tan. = 0-5763070 
 
 B =75 08 35" -37 sin. = 9-9852331 cos. = 9-4089261 
 
 D = 2030-211 log. = 3-3075411 L = 538-560 log. = 2-7312341 
 
 <5 = 2100-429 feet log. = 3-3223080 Proof 
 
 log. = 3-3223080 
 
 Having thus obtained the "bearing from station to 3, viz., 
 S 75 08 35" -37 E, and distance = 2100-429 feet, our next step 
 will he to ascertain the distances A and A 3. In the triangle 
 A 3 0, we have to find the several angles. The bearing from 
 
 o to A 
 o to 3 
 
 A to o 
 A to 3 
 
 3 to A 
 3 toO 
 
 = due East 
 
 = S 75 08 35"- 37 E 
 
 > which gives angle at = 14 51 24". 
 
 = due West ) 
 
 = S 60 00 00"-OOE ) 
 
 = N 60 00 00"- 00 W 
 
 = N 75 03 35". 37 W 
 
 150 00 "-00 
 
 15 08 35". 37 
 
 Having found the angles, we have 
 
 Sin. A : 5 : : sin. : A 3 
 and Sin. A : 8 : : sin. 3 : A 
 
 S . sin. . 
 sin. A 
 . sin. 3 
 
 sin. A 
 
 Thus, 
 
114 
 
 USEFUL FORMULA. 
 
 A = 1500 00 00" -00 co. ar. sin. = 0- 
 
 = 14 51 24? -63 sin. = 9-4089261 
 
 5 = log. = 3 3223080 
 
 A 3 = 1077-12 feet log. = 3-0322641 
 
 A = co. ar. sin. = 0-3010300 
 
 6 = log. = 3/3223080 
 
 3 = 150 08 35" -37 sin . = 9 -4170259 
 AO = 1097-4 feet 
 
 log. = 3-0403639 
 
 For the purpose of ascertaining the angle G in the triangle 
 A G C, we assume a radius = unity, which we represent in our 
 formula by 1, retaining r as the radius in the unit of measure. 
 Then, putting A for the angle at apex ; T and T for the tangent 
 points ; C for the angle at the centre of the curve ; G for the angle 
 at station 1 in the traverse ; A for the angle T A G ; C for the 
 angle A C G ; we have in the triangle A G, the angle and the 
 sides A and 1, to find the side A 1 = A G, and the angles A 
 and G. Putting a for the side A, and b for the side 1 = G, 
 we have 
 
 a + b : a c~ b :: tan. J (180 0) : tan. J (A co G) = 
 
 (a co ft) . tan. 4- (180 0) 
 
 and J (180 0) + J (A ^ G) = G ; 
 and \ (180 0) i (A co G) = A 
 
 (0) 
 
 Having found the angles A and G, we find the side A G, which 
 we represent by m, by either of the analogies following : 
 
 Sin. A : b I : sin. : m = jj^r- ; 
 or, Sin. G : a 1 1 sin. . m = ^ n ^ ~ (P) 
 
 Having obtained m, and putting G for the angle G in the 
 triangle A C G, we have 
 
 Sin. | A: 1 ::E:AC=- sta 1 jX -; 
 then, representing A C by n, we have 
 
 1 : sin. (| A A ) : : n : sin. G = n . sin. (J A A ) (Q) 
 
 And sin. (J A A -f- G ) : m : : sin. (J A 
 
 _ m sin - (* A A/ ) 
 sin. (i A A + G ) 
 
 A ):r = 
 
 (R) 
 
BEARINGS OF STATIONS. 115 
 
 We now introduce an example of computation [formulae (0) 
 and (P.)] 
 
 a = 1097 -4 feet 180 00* 00" -00 
 
 b = 152-5 " = 1451 24"-63 
 
 a + b = 1249-9 co. ar. log. = 6-9031247 2)165 OS 35" -37 
 
 a ~> b = 944-9 log. = 2-9753858 i (180 0) = 82 34 17" -68 
 
 i (180 0) = 82 34 17" -68 tan. = 0-8847831 
 
 J (A 7 co G) = 80 12 52" -64 tan. = 0-7632936 
 
 G =162 47 10" -32 G =162 47 10" -32 co. ar. sin. = 0-5287991 
 
 A =2 21 25" -04 a = 1097-4 feet log. = 3-0403650 
 
 = 14 51 24" -63 = 14 51 24" -63 sin. = 9-4089261 
 
 Proof = 180 00 00" -00 m = 950-8 feet log. = 2*9780902 
 
 By formulae (Q) and (R) we have 
 
 i A = 75 00 00" -00 co. ar. sin. = 0-0150562 = n 
 
 J A A =72 38 34" -96 sin. = 9-9797599 
 
 Ambiguous G = 81 09 53" -75 * sin. = 9-9948161 
 
 True G =98 50 06" -25 
 
 i A A = 72 38 34". 96 
 
 i A A + G = 171 28 41" -21 co. ar. sin. = 0-8291893 
 
 m =950-8 feet log. = 2-9780892 
 
 k A A = 72 38 34" -96 sin. = 9-9797599 
 
 r =6124-05 feet log. = 3-7870384 
 
 Having ascertained the radius which the problem requires, we 
 proceed to ascertain the deflection for a chord of 50 feet. 
 
 By formula (3) we have sin. D = _**. hence 
 
 r = 6124-05 feet co. ar. log. = 6-2129616 
 
 4 ch = 25 " log. = 1-3979400 
 
 I) = 14 02" sin. = 7-6109016 
 
 * As the problem under all its forms requires G to be larger than a right angle, it is evident that 
 the true G must be the supplemental angle, inasmuch as the sine of an angle is the same as the sine 
 of its supplement. 
 
116 USEFUL FORMULAE. 
 
 The deflection thus found being an awkward sum to add or sub 
 tract in the field, we may assume one more convenient without 
 materially changing the location of the curve ; we therefore assume 
 14 as the measure of a deflection ; then, by formula (5) we 
 have r = \^ ; and, for the purpose of ascertaining by what 
 amount this change in the length of the radius will affect the loca 
 tion of the curve, we will endeavor to find the distance from the 
 apex to the middle of the curve for each radius. By formula (6) 
 we have t = tan. \ C . r ; and by (7) we have 
 b == t. tan. J C == tan. J C . tan. J C . r = * c .J 
 
 THE VALUE OP *, COMPUTED FROM THE RADIUS 
 
 ALREADY OBTAINED. 
 
 \ C = 15 00 00" tan. = 9-4280525 
 
 i C = 7 30 00" tan. = 9-1194291 
 
 r =6124-05 log. = 3-7870384 
 
 b = 216-033 feet log. == 2-3345200 
 
 THE VALUE OF b, COMPUTED TO CORRESPOND TO 
 
 A RADIUS BASED UPON A DEFLECTION OF 14 . 
 
 D = 00 14 00" co. ar. sin. = 2-3901470 
 
 J C = 15 00 7 00" tan. = 9-4280525 
 
 i C = 70 30 00" tan. = 9-1194291 
 
 4- ch = 25 feet log. = 1-3979400 
 
 = 216-555 feet log. = 2-3355686 
 
 216-033 " 
 
 0-522 " 
 
 Thus we see that the proposed change in the deflection will affect 
 the location of the curve only 0*522 feet, an amount in most cases 
 too small to produce any practical inconvenience. 
 
 Having shown that whenever convenience requires a change of a 
 few seconds in the angle of deflection, the change may be made 
 without materially affecting the location of the curve, we now pro 
 ceed to determine a radius which shall correspond with the desired 
 deflection, viz., of 14 , as explained in the foregoing. By 
 formula (5) we have r = -f^- Thus, 
 
 D = 14 co. ar. sin. = 2-3901470 
 
 i ch. = 25 feet log. = 1-3979400 
 
 r = 6138-853 feet log. = 3-7880870 
 
BEARINGS OF STATIONS. 117 
 
 To find the position of the tangent points at stations T and T 
 we compute their distance from apex, and compare them with the 
 distances of and 3, which have been already determined. By 
 formula (G) we have t = tan. J C . r. Thus, 
 
 r = 6138-853 feet log. = 3-7880870 
 
 J C = 15 00 00" tan. = 9-4280525 
 
 t = 1644-90 log. = 3-2161395 
 
 A to 3, heretofore computed 1077-12 
 567-78 
 
 We thus find the point T 567 78 feet further from A than the 
 point 3. Again, 
 
 t = 1644-90 feet 
 
 A to 0, heretofore computed = 1097-40 " 
 
 547-50 " 
 
 We thus find the point T 547-50 feet further from A than the 
 point 0. 
 
 To find the length of arc from point T to point G, corresponding 
 to point 1 in the traverse, we have in the triangle A C G, the angle 
 at C = 180 (the angle G + angle A.) Thus, we find 
 
 A = (A A ) = 75 2 21 25" 04 = 72 38 34" 96 
 G =980 50 06" -25 
 
 C = the supplement = 8 31 18" -79 
 
 Proof 180 00 00" -00 
 
 Then, we have half the centre angle = \ C = 15 00 00" minus 
 the supplementary angle C found above = the angle C in the 
 triangle GCT. Thus, 
 
 * C =15 00 00" -00 
 
 Supplementary C = 8 31 18" -79 
 
 Hence the C sought 6 28 41" 21 
 
BEARINGS OF STATIONS. 119 
 
 Then, representing the angle C thus found and reduced to seconds 
 by C", and the arc sought by a, we have, by formula (8,) a = - 
 
 C" = 23321-21 log. = 4-3677511 
 
 r = log. = 3-7880870 
 
 r" = co. ar. log. = 4 6855749 
 
 a = 694-09 feet log. = 2 8414130 
 
 If we now consider T as numbered 3*06, in the stations of loca 
 tion, and the numbers in the location to be increasing as we enter 
 the curve, we shall find the point at G, or rather near G, (as we 
 have slightly changed the radius,) to be equal to 3 -06 -f- 6 94, 
 which increases the number of the locating stations to 10. For the 
 purpose of avoiding fractions, we ascertain the point T by measure 
 ment from 0, and then locate the curve to station 10 ; from this 
 point we ascertain the direction of the radius, and select the point 
 A, which should be so situated as to command a distinct view of 
 the locality where the work is to be laid out. We then discover the 
 relative direction and length of the line 10 A, by ascertaining the 
 angle which it makes with the radius of the curve, and measuring 
 the distance between the termini. Let us now suppose the angle 
 to measure 20 00 , and the length of the line to be 200 feet. We 
 next cross the river or bay, and select the point B, which should 
 likewise command a distinct view of the locality where the work is 
 to be laid out ; we then measure the angles which form the triangle 
 A B 10, and compute their relative positions. Now, supposing the 
 angles at 
 
 830 00 
 32 0(X 
 
 650 (X 
 1800 00 
 
120 USEFUL FORMULA. 
 
 Then, putting 5 for the distance between 10 and A, we have 
 Sin. B : 6 : : sin. 10 : A B ; and sin. B : 6 : : sin. A : 10 B (S) 
 Thus, 
 
 B =32 00 co. ar. sin. = 0-2757903 
 
 S = 200 feet log. = 2-3010300 
 
 10 =65 0(X sin. = 9-9572757 
 
 A B = 342-05 feet log. = 2-5340960 
 
 j 
 
 log. = 2-5768203 
 
 Sin. B 
 
 A = 830 00 sin. = 9-9967507 
 
 10 B = 374-61 feet 
 
 (50) We have, from station 10 to 10 50, a distance of 50 feet 
 of arc; then, from 10 50 to 11, a like distance; and so on, from 
 station to station, to station 13 50. (See Fig. 20.) To compute in the 
 readiest manner the relative position of these several points, or 
 rather the relative positions of the points a b c and c?, (which repre 
 sent the corners of the piers situated about these stations,) we assume 
 the radius of the curve from station 10 to bear due south, and C 
 as a zero point. Then, by ascertaining the distances and relative 
 bearings to each of the points, we compute what we shall (for the 
 want of more appropriate terms) call their northings or southings, 
 eastings or westings, without consideration of their astronomical or 
 geodetical bearings. To ascertain the angles of the radii from C to 
 these several points, we put a for the arc connecting them, r for the 
 radius in the unit of measure, and r" for an arc in seconds equal in 
 length to radius. We then have, by formula (9,) C" = ^- 
 Thus the angle at C, between stations 10 and 10-50, gives a = 50 
 feet. Then, 
 
 a =50 feet log. = 1-6989700 
 
 r" = log. = 5-3144251 
 
 r = co. ar. log. = 6-2119130 
 
 C" = 1680" = nearly log. = 3-2253081 
 
BEARINGS OF STATIONS. 121 
 
 and 1680", reduced to degrees and minutes, will give = 28 00". 
 We have given this computation for the purpose of explaining a 
 general rule which will apply in all cases. 
 
 In the present case, the arc a = the chord of one of our 
 deflections ; and, as the difference between the chord of fifty feet 
 in length and the arc it spans (based upon a radius of 6138*853 
 feet) is so small that the one may be taken for the other, in the 
 practical operations of locating a railroad, we therefore may, without 
 sensible error, take the angle at C for 50 feet of arc = two 
 deflections = 28 , or the same as above. 
 
 But we shall, notwithstanding, when we come to consider .the 
 dimensions of the piers, find a necessity for the formulae. Let us 
 assume the foundations of the piers to be 8 feet broad and 18 feet 
 in length. Now, as the stations named above correspond to the 
 centre of these piers, we find it necessary to determine half the 
 angular width of them from C. Thus, 
 
 a =4 feet log. = 0*6020600 
 
 r" = log. = 5-3144251 
 
 r co. ar. log. = 6-2119130 
 
 C" = 134"- 4 log. = 2-1283981 
 
 Reducing C" to degrees and minutes, gives us the angle = 
 2 14" 4: ; but, for the purpose of avoiding (in the computations) 
 the fractions of a second, we may, without varying the dimensions 
 of the piers perceptibly, assume the angular width of the half pier 
 to be 2 15". For like reasons, with a radius of the length we 
 have adopted, we assume both ends of the pier to be of the same 
 angular width. 
 
122 USEFUL FORMULA. 
 
 Having made these explanations, we proceed to construct a table 
 of the angular positions of the corners of the piers represented by 
 abed. We have stated above that the centres of the piers are two 
 deflections, or 28 , apart from C. Then, taking station 10 for a 
 starting point, and the radius from point C through this point, as 
 bearing due north, we have the angle to station 10 -50 = 28 ; and 
 the angle to station 11, twice 28 ; and so on. Having determined 
 the position of the primitive stations, we may, by additions and 
 subtractions of the angular half widths and .widths of the piers, 
 determine the angular positions of the points abed; and upon 
 these principles we construct the following table, viz., 
 
 RELATIVE RELATIVE RELATIVE 
 
 BEARINGS OF BEARINGS OP BEARINGS OF 
 
 PRIMITIVE STATION b AND d a AND c 
 
 N.E. N.E. N.E. 
 
 Station 10 to 10-50 angle = 28 -f 2 15" = 30 15" 
 
 " " 11-00 = 56 + " = 58 15 / 4 30" = 53 45" 
 
 " 11-50 " = 84 + " = 126 15" " = 121 45" 
 
 " " 12-00 " = 112 + " = 1 54 15" " = 1 49 45" 
 
 " " 12-50 " =140 + " = 2 22 15" " = 2 17 45" 
 
 13-00 " =168 + " = " *=245 45" 
 
 13-50 " = 196 = N.E. 3 16 
 
 Having thus arranged a table of bearings from the centre of the 
 curve, or C, of a b c d, with the primitive station to which they are 
 connected, we next prepare a table containing both bearings and 
 distances, leaving a space for the northings and eastings to be added 
 after computation. 
 
BEARINGS OF STATIONS. 
 
 123 
 
 A TABLE OF BEARINGS, 
 
 DISTANCES, NORTHINGS AND EASTINGS, OF STATIONS, 
 FROM C, OR THE CENTRE OF THE CURVE. 
 
 PRIMITIVE STATIONS. 
 
 BEARINGS. 
 
 DISTANCES. 
 
 NORTHINGS 
 
 IN FEET. 
 
 EASTINGS 
 
 IN FEET. 
 
 10 -00 from C to 10 
 
 Due 
 
 North 
 
 r + feet 
 
 = 6138-85 feet 
 
 6138-85 
 
 00-000 
 
 10-50 " " " d 
 
 N.E. 
 
 = 00 30 15" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6129-61 
 
 53-938 
 
 " " " b 
 
 a 
 
 
 
 r + 9 
 
 = 6147-85 " 
 
 6147-61 
 
 54-096 
 
 11-00 " " " c 
 
 N.E. 
 
 = QO 53 45" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6129-10 
 
 95-838 
 
 " " a 
 
 u 
 
 (C it 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6147-10 
 
 96-119 
 
 U i( t( (C d 
 
 N.E. 
 
 = QO 58 45" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6128-97 
 
 103-860 
 
 " " " " b 
 
 a 
 
 u u 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6146-97 
 
 104-165 
 
 11-50 " " " c 
 
 N.E. 
 
 = 10 21 45" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6128-12 
 
 145-755 
 
 d 
 
 " 
 
 it U 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6146-11 
 
 146-183 
 
 " " " " d 
 
 N.E. 
 
 = 10 26 45" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6127-92 
 
 153-176 
 
 " " " " b 
 
 u 
 
 u a 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6145-92 
 
 154-228 
 
 12-00 " " " c 
 
 N.E. 
 
 = 10 49 45" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6126-73 
 
 195-662 
 
 " " " " a 
 
 a 
 
 a u 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6144-72 
 
 196-237 
 
 " " " " d 
 
 N.E. 
 
 = 10 54 15" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6126-46 
 
 203-682 
 
 " " " " b 
 
 a 
 
 u u 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6144-45 
 
 204-280 
 
 12-50 " " " c 
 
 N.E. 
 
 = 2 17 45" r 9 " 
 
 = 6129-85 " 
 
 6124-93 
 
 245-557 
 
 " " " " a 
 
 ic 
 
 a a 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6142-92 
 
 246-177 
 
 " " " " d 
 
 N.E. 
 
 20 22 15" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6124-60 
 
 253-574 
 
 " " " " b 
 
 u 
 
 u u 
 
 r + 9 " 
 
 = 6147-85 " 
 
 6142-59 
 
 254-318 
 
 13-00 " " " c 
 
 N.E. 
 
 = 2 045 45" 
 
 r 9 " 
 
 = 6129-85 " 
 
 6122-81 
 
 295-435 
 
 " " " " a 
 
 u 
 
 11 U 
 
 r + 9 
 
 = 6147-85 " 
 
 6140-79 
 
 296-302 
 
 13-50 " " " 
 
 N.E. 
 
 = 30 16 00" 
 
 r " 
 
 = 6138-85 " 
 
 6128-87 
 
 349-811 
 
 Point A from C 
 
 
 
 
 
 5950-91 
 
 68.404 
 
 " B " " 
 
 
 
 
 
 
 5873-97 
 
 269-884 
 
 Having prepared our table of bearings and distances from the 
 centre C to the corners of the piers, we introduce examples of 
 computation of northings and eastings ; all the bearings the table 
 contains being north-east. Substituting <5 for d, we have, by 
 formula (M,) 
 
 D = sin. B <5 ; and L = cos. B 5. 
 
124 USEFUL FORMULAE. 
 
 10-50 d = N.E. 30 15" sin. = 7 9444459 cos. = 9 9999832 
 
 (5 = 6129-85 log. = 3-7874499 log. = 3-7874499 
 
 D = 53-938 log. = 1-7318958 L = 6129-61 log. = 3-7874331 
 
 10-50 b = N.E. sin. = 7-9444459 cos, = 9 9999832 
 
 d = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 54-096 log. = 1-7331692 L = 6147-61 log. = 3-7887065 
 
 11-00 c N.E. 00 53 45" sin. = 8-1940869 cos. = 9 9999469 
 
 (5 = 6129-85 log = 3-7874499 log. = 3-7874499 
 
 D = 95-838 log. = 1-9815368 L = 6129-10 log. = 3 "7873968 
 
 11-00 a = N.E. 53 45" sin. = 8-1940869 cos. = 9-9999469 
 
 6 = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 96-119 log. = 1-9828102 L = 6147 10 log. = 3-7886702 
 
 11-00 d = N.E. Q t 58f 15" sin. = 8-2290013 cos. = 9"9999377 
 
 <J = 6129-85 log. = 3-7874499 log. = 3-7874499 
 
 D = 103-86 log. = 2-0164512 L = 6128-97 log. = 3 7873876 
 
 11-00 b = N.E. 00 58 15" sin. = 8-2290013 cos. = 9-9999377 
 
 (5 = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 104-165 log. = 2-0177246 L = 6146-97 log. = 3-7886610 
 
 11-50 c = N.E. 10 21 45" sin. = 8-3761729 cos. = 9.9998772 
 
 d = 6129-85 log. = 3-7874499 log. = 3-7874499 
 
 D = 145-755 [log. = 2-1636228 L = 6128-12 log. = 3 7873271 
 
 11-50 a = N.E. 10 21 45" sin. = 8-3761729 cos. = 9*9998772 
 
 d = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 146-183 log. = 2-1648962 L = 6146-11 log. = 3-7886005 
 
 11-50 d = N.E. 1 26 15" sin. = 8-3994397 cos. = 9-9998633 
 
 5 = 6129-85 log. = 3-7874499 log. = 3-7874499 
 
 D= 153-776 log. = 2-1868896 L = 6127*92 log. = 3-7873132 
 
BEARINGS OF STATIONS. 125 
 
 11-50 b = N.E. 10 26 15" sin. = 8 3994397 cos. = 
 
 J = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D= 154-288 log. = 2-1881630 L = 6145-92 log. = 3-7885866 
 
 12-00 c = N.E. 1 49 45" sin. = 8*5040569 cos. = 9-9997786 
 
 d = 6129-85 log. = 3-7874499 log. = 3-7874499 
 
 D = 195-662 log. = 2-2915068 L = 6126"73 log. = 3 7872285 
 
 12-00 a = N.E. 1 41 45" sin. = 8-5040569 cos. = 9-9997786 
 
 r5 = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 196-237 log. = 2-2927802 L = 6144-72 log. = 3-7885019 
 
 12-00 d = N.E. 10 54 15" sin. = 8 5215024 cos. = 9 9997601 
 
 (J = 6129-85 log. = 3-7874499 log. 3-7874499 
 
 D = 203-682 log. = 2 "3089523 L = 6126-46 log. = 3 "7872100 
 
 12-00 b = N.E. 1 54 15" sin. = 8-5215024 cos. = 9-9997601 
 
 (5 = 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 204-280 log. = 2-3102257 L = 6144*45 log. = 3- 
 
 12-50 c = N.E. 20 17 45" sin. = 8-6027015 cos. = 9-9996513 
 
 (5 6129-85 log. = 3-7874499 log. = 3 -7874499 
 
 D = 245-557 log. = 2 3901514 L = 6124-93 log. = 3-7871012 
 
 12-50 a = N.E. 2 17 45" sin. = 8-6027015 cos. = 9-9996513 
 
 d =s 6147-85 log. = 3-7887233 log. = 3-7887233 
 
 D = 246-177 log. = 2-39142-18 L =r 6142 92 log. = 3-7883746 
 
 12-50 d N.E. 20 22 15" sin. S 6166545 cos. = 9*9996281 
 
 d = 6129-85 log. = 3-7874499 log. = 3-7874499 
 
 D = 253-574 log. = 2-4041044 L = 6124 60 log. = 3-7870780 
 
 12-50 b = N.E. 20 22 15" sin, = 8 6166545 cos. = 9-9996281 
 
 f 5 = 6147-85 log. = 3-7887233 log. 3 7887233 
 
 D= 254-318 log. = 2-4053778 L = 6142-59 log. = 3-7883514 
 
126 
 
 USEFUL FORMULA. 
 
 13-00 c == N.E. 20 45 45" sin. =r 8 6830114 
 d = 6129-85 log. = 3-7874499 
 D = 295-435 log. = 2*4704613 L = 6122-81 log. = 3-7869510 
 
 cos. = 9-9995011 
 log. = 3-7874499 
 
 cos. = 9-9995011 
 log. = 3-7887233 
 
 13-00 a = N.E. 2 45 45" sin. = 8-6830114 
 d = 6147-85 log. = 3-7887233 
 
 ^ 
 
 D= 296-302 log. = 2-4717347 L = 6140-79 log. = 3-7882244 
 
 13-50 station N.E. 3 16 00" sin. = 8-7557469 cos. = 9-9992938 
 
 d = 6138-85 log. = 3-7880870 log. = 3-7880870 
 
 D = 349-811 log. =; 2-5438339 L 6128-87 log. = 3-7873808 
 
 Having computed all the points connected with the centre C, and 
 carried the results into the preceding table of bearings, distances, 
 etc., we next compute the relative situation of the points A and B 
 from station 10. We have before stated that the line 10 A made 
 an angle with the radius of the curve of 20 ; the radius being taken 
 to bear due south from 10, gives the bearing of 10 A = S. 20 W. 
 The angle at A, in the triangle 10 A B, being 83, gives the bearing 
 of A B = S. 77 E. And the angle at 10, being 65, gives the 
 bearing 10 to B = S. 45 E. Having thus ascertained the bearings, 
 and the distances being already computed, we now make up a table 
 of bearings and distances, leaving room to put in the latitudes and 
 departures when obtained. 
 
 BEARINGS AND DISTANCES, 
 FROM STATION 10 TO A AND B, AND FROM A TO B. 
 
 BEARINGS AND DISTANCES. 
 
 SOUTHINGS. 
 
 EASTINGS. 
 
 WESTING. 
 
 From 10 to B = S.E. 45 = 374-61 feet 
 
 264-884 
 
 264-884 
 
 
 " 10 to A = S.W. 20 = 200-00 feet 
 
 187-939 
 
 
 68-404 
 
 " A to B = S.E. 770 = 333-288 feet 
 
 76-945 
 
 338-288 
 
 
 We now proceed to compute the latitudes and departures. 
 
BEARINGS OF STATIONS. 127 
 
 COMPUTATIONS OF THE LATITUDES AND DEPARTURES. 
 
 S.E. 450 00 00" sin. = 9-8494850 cos. = 9-8494850 
 
 d = 374-61 log. = 2-5735710 log. = 2-5735710 
 
 D = 264-884 log. s= 2-4230560 L = 264-884 log. = 2-4230560 
 
 S.W. 20 00 00" sin. = 9-5340517 cos. 9-9729858 
 
 d = 200 feet log. = 2-3010300 log. = 2-3010300 
 
 D = 68-404 log. =1-8350817 L 187-939 log. =2-2740158 
 
 S.E. 77 00 00" sin. = 9 9887239 cos. = 9-3520880 
 
 d = 342-05 log. = 2-5340960 log. = 2-5340960 
 
 D = 333-288 log. = 2-5228199 L = 76 945 log. = 1-8861840 
 
 Having computed our latitudes and departures, and written them 
 in the above table, it now becomes necessary to ascertain their 
 relative position to the point C. We find by our table, computed 
 from the point C, that station 10 has a northing of 6138 853 feet, 
 and easting it has nothing. We see by the above table that A is 
 west of 10 = G8 404 feet. For the purpose of preventing the 
 necessity of an additional column, we shall note this in our table 
 of northings and eastings from point C, as minus eastings, distin 
 guishing it by the negative sign, thus, ( 68*404,) in the easting 
 column. Now A, being south of station 10 187*939 feet, we 
 subtract this sum from the northing of 10, which gives the northing 
 of A. Thus, 
 
 Station 10 northing = 6138-853 
 
 A from 10 " = 187-939 
 
 Leaves the northing of A = 5950-914 feet 
 
 Then, managing in a like manner with station B, we have B east 
 of 10 = 264-884 feet ; and, as 10 has neither easting nor westing, 
 we put this into our table as the proper easting. 
 
128 USEFUL FORMULA. 
 
 For the northing we have 
 
 Station 10 = 6138-853 
 
 and " B = 264-884 
 
 Leaves the northing of B = 5873-969 
 
 Again: for the purpose of proving a portion of our work, we 
 ascertain the relative position of B from A. We have previously 
 found the situation of A to be 
 
 A northing = 5950-914 and easting 68-404 
 
 B (from the preceding table) " = 76-945 " 338-288 
 
 B = 5873-969 " 269 884 
 
 Adding, as indicated by the algebraic signs, gives the northings 
 and eastings of B as above. We now carry these results into the 
 preceding table of northings and eastings, etc., from C ; and it will 
 then contain all the points which are needed. 
 
 Having completed our table of bearings, distances, northings, and 
 eastings, etc., from the point or centre C, we next compute the 
 bearings from A and B to the several corners of each pier and 
 abutment noted in the aforesaid table, and to station 13 -50 of the 
 general location ; the bearing to station 10 being already known. 
 Beginning with corner b, in the abutment at station 10 50, the 
 formula may be thus enunciated, applying the affix N to the 
 expression representing the station, for northing, and E for casting, 
 we then have 
 
 AN co N : 1 : : AE co JE : tan. B = -fj^-ff- 
 wherein B expresses the bearing sought. 
 
BEARINGS OF STATIONS. 
 
 129 
 
 EXAMPLE OF COMPUTATIONS. 
 
 Station 10-50 AN ~ &N = 5950-914 co 6147-61 = 196*696 log. 
 
 AE co bE = 68-404 co 54-096 = 122 500 log. 
 
 B or bearing from A to 10 -50 b = N.E. 31 54 51" * tan. =: 9-7943405 
 
 2-2937956 
 2-0881361 
 
 Station 10-50 Ax <~ <L\ = 5950 914 co 6129-61 = 178-696 log. 
 
 AE co dE = 68-404 co 53-938 = 122-312 log. 
 
 B or bearing from A to 10 50 d = N.E. 31 23 49" tan. 
 
 Station 11-00 AN cs> ON = 5950-914 co 6147-10 = 196-186 
 AE co aR 68-404 co 96-119 = 164-523 
 Bearing from A to 11-00 a = N.E. 39 59 00" 
 
 Station 11-00 AN co &N = 5950-914 ^ 6146-97 = 196-056 
 AE co bE = 68-404 co 104-165 = 172-569 
 Bearing from A to 11 -00 
 
 : 2- 2521148 
 ; 2-0875756 
 9-8354608 
 
 log. = 2-2S2G680 
 log. = 2-21C22G6 
 tan. 9 -235586 
 
 log. = 2-2923301 
 log. = 2-2369628 
 = N.E. 41 21 15" tan. = 9-9445827 
 
 Station ll OO AN co C N = 5950-914 co 6129-100 = 178-186 
 AE co CE = 68-404 co 95 838 = 164 242 
 Bearing from A to 11-00 c = N.E. 42 40 29" 
 
 Station 11-00 AN co dx = 5950 914 co 6128-97 = 178-056 
 AE co dE = 68-404 co 103-860 = 172*264 
 
 log. 
 log. 
 
 2-2508736 
 2-2155842 
 tan. = 9-90471G6 
 
 log. 
 log. 
 
 2-2505566 
 2-2361945 
 
 Bearing from A to 11-00 d = N.E. 44 03 10" tan. = 9-9856379 
 
 Station 11-50 AN co x == 5950-914 co 6 146 -11 = 195-196 log. = 2*2904709 
 
 AE co fl E = 68-404 co 146-183 = 214-587 log. = 2-3342064 
 
 Bearing from A to 11-50 a = N.E. 47 52 48" tan. = 0-0437355 
 
 Station 11-50 AN co &N = 5950-914 co 6145-920 = 195-006 log. = 2-29C0480 
 
 AE co bs = 68-404 co 154-228 = 222 632 log. = 2-3475876 
 
 Bearing from A to 11-50 b = N.E. 48 47 04" tan. = 0-0575396 
 
 * The northing of A being less than the northing of 6, and the easting of A being less than 
 the bearing must of course be northeasterly. 
 
130 USEFUL FORMULAE. 
 
 Station 11-50 Ax <*> ex = 5950 914 co 6123 120 = 177-206 log. = 2-2184784 
 
 AE co CE = G8 404 co 145-755 = 214-159 log. = 2-33G7364 
 
 Bearing from A to 11-50 c = N.E. 50 23 38" tan. = 0-ce22580 
 
 Station 11-50 Ax co dx = 5950-914 co G127 920 = 177-006 log. = 2-2179880 
 
 AE co dE = 68-404 co 153-176 = 221-580 log. = 2 -3455306 
 
 Bearing from A to 11-50 d = N.E. 51 22 56" tan. = 0-0975426 
 
 Station 12-00 Ax co a x = 5950-914 co 6144-720 = 193-806 log. = 2-2873672 
 
 AE co CE 68-404 co 196-237 = 264-641 log. = 2-4226571 
 
 Bearing from A to 12-00 a N.E. 53 47 00" tan. = 0-1352899 
 
 Station 12-00 Ax co fcx = 5950-914 co 6144-450 = 193-536 log. = 2-2867617 
 
 AE co IE = 68-404 co 204 2SO = 272 684 log. = 2 4356597 
 
 Bearing from A to 12-00 b = N.E. 54 38 06" tan. = 1488980 
 
 Station 12-00 Ax co ex = 5950-914 co 6126-730 = 175-816 log. = 2-2450584 
 
 AE co CE = 68-404 co 195-662 = 264-066 log. = 2-4217125 
 
 Bearing from A to 12-00 c = N.E. 56 2(X 39" tan. = 0-1766541 
 
 Station 12-00 Ax co ds = 5950-914 co 6125-460 = 175-546 log. = 2-2447125 
 
 AE co dE = 68-404 ^ 203-682 = 272-086 log. = 2-4347062 
 
 Bearing from A to 12-00 d = N.E. 57 09 04" tan. = 0-1899937 
 
 Station 12-50 Ax co fl x = 5950-914 co 6142-920 = 192-006 log. = 2-2833148 
 
 AE co OE = 68-404 co 246-177 = 314-581 log. = 2-4977325 
 
 Bearing from A to 12 50 a = N.E. 58 35 37" tan. = 0-2144177 
 
 Station 12-50 Ax co b* = 5950-914 co 6142-590 = 191-676 log. = 2-2825677 
 
 AE co &E = 68-404 co 254-318 = 322 722 log. = 2-5088286 
 
 Bearing from A to 12-50 ft = N.E. 59 17 33" tan. = 0-22G2609 
 
 Station 12-50 Ax co C x = 5950-914 co 6124 930 ra 174-016 log. = 2-2105892 
 
 AE co CE = 68-404 co 215-577 =s 313-961 log. = 2-4968757 
 
 Bearing from A to 12-50 c = N.E. 61 00 08" tan. = 0-2562865 
 
BEARINGS OF STATIONS. 
 
 131 
 
 Station 12-50 Ax 
 
 AE 
 
 dx = 5950-914 ^ 6124-600 = 173-686 
 dE = 68-404 <^> 253-574 = 325 978 
 
 log. = 2-2397648 
 log. = 2-5078262 
 
 Bearing from A to 12-50 d = N.E. 61 39 22" tan. = 0-2680614 
 
 Station 13-00 Ax <*> ax = 5950-914 
 AE co aE = 68-404 
 
 6140-790 = 189-876 
 296-302 = 364-706 
 
 log. = 2-2784701 
 log. = 2-5619430 
 
 Bearing from A to 13-00 a = N.E. 62 29 50" tan. = 0--834729 
 
 Station 13-00 Ax co ex = 5950-914 co 6122-810 = 171-896 log. = 2-2352658 
 
 AE co CE = 68-404 co 295-435 = 363-639 log. = 2-5609065 
 
 Bearing from A to 13-00 c = N.E. 64 42 41" tan. = 3256407 
 
 Station 13-50 Ax co 13 50x = 5950-914 co 6128-870 177-956 log. = 2-2502980 
 
 AE co 13-50K = 68-404 co 349-821 = 418-215 log. = 2-6213996 
 
 Bearing from A to station 13-50 = N.E. 66 57 01" tan. = 0-3711016 
 
 Having computed the bearings from A to the corners of each pier 
 and abutment shown in the diagram, including stations 10 and 13-50 
 in the alinement of the road, we now proceed to compute the bearing 
 of the corners of each pier, and the stations 10 and 13 -50 from 
 station B. 
 
 Station 10-50 Bx co bs = 5873-969 co 6147-610 = 273-641 
 BE co frE = 264-881 co . 54-096 = 210-783 
 
 log. 2-4371812 
 log. = 2-3233459 
 
 Bearing from B to 10-50 B = N.W. 37 36 27" tan. = 9-8836647 
 
 Station 10-50 Bx co dx = 5873-969 co 6!21 610 = 255-641 log. = 2-4076305 
 
 BE co dE = 264-881 co 53-933 210 916 log. = 2-C241713 
 
 Bearing from B to 10-50 d = N.W. 39 31 41" tnn. = 9-9KJ5408 
 
 Station ll OO Bx co ax = 5873-969 co 6117-100 = 273-131 
 BE co aE = 264-884 co 96-119 = 168 765 
 
 log. = 2-43G3710 
 log. = 2-2272824 
 
 Bearing from B to 11-00 a = N.W. 31 42 41" tan. = 9-7909114 
 
 Station 11-00 Bx co fcx = 5873-969 co 6146-970 = 273-001 log. = 2-4361642 
 
 BE co 6 E = 264-884 co 104-165 = 160 719 - leg. = 2-2060672 
 
 Bearing from B to 11-00 b = N.W. 30 29 09" tan. = 9 7699030 
 
132 USEFUL FORMULAE. 
 
 Station 11-00 Bx co ex = 5873-969 co 6129 IOG = 255-131 log. = 2-4067632 
 
 BE co CE = 264-884 co 95 "838 = 169-046 Jog. = 2-2280049 
 
 Bearing from B to 11 -00 b = N.W. 33= 31 40" tan. = 9-8212417 
 
 Station 11-00 Bx co ds 5873-969 co 6128-970 = 255-001 log. = 2-4065419 
 
 BE co dE = 264-881 co 103-860 = 161-024 log. = 2-2C68906 
 
 Bearing from B to 11-CO d = N.W. 32 16 15" tan. = 9-8003487 
 
 Station 11-50 Bx co x = 5873-969 co 6146-110 = 272-141 log. = 2-4347940 
 
 BE co OE = 264-884 co 146-183 = 118-701 log. = 2-0744544 
 
 Bearing from B to 11-50 a N.W. 23O 3 56" tan. = 9-C3S6604 
 
 Station 11-50 Bx ^ ^ = 5873-969 <*= 6145-920 = 271-951 log. = 2-4344907 
 
 BE co ^ E = 264-884 co 154-223 = 110-656 log. = 2-0439751 
 
 Bearing from B to 11-50 b = N.W. 22 08 29" tan. 9-6194844 
 
 Station 11-50 Bx ^ ex = 5373-969 co 6123-120 254-151 log. =r 2-4050918 
 
 BE co CE = 264-884 co 145-755 = 119-129 log. = 2-0760175 
 
 Bearing from B to 11-50 e = N.W. 23 06 51" tan. = 9 6709257 
 
 Station 11-50 Bx co rfx = 5873-969 co 6127-920 = 253-951 log. = 2*4047499 
 
 BE co dE = 264-884 co 153-176 = 111-708 log. = 2-0480883 
 
 Bearing from B to 11-50 d = N.W. 23^ 44 38" tan. = 9-6433384 
 
 Station 12 00 Bx co ax = 5783-969 co 6144-720 = 270-751 log. = 2-4325701 
 
 BE co OE = 264-884 co 196-237 = 68-647 log. = 1-8366216 
 
 Bearing from B to 12-00 a = N.W. 14 13 38 tan. = 9-4040515 
 
 Station 12-00 Bx ^ b* = 5783-969 co 6144-450 = 270*481 log. = 2-4321368 
 
 BE co br. = 264-884 co 204-280 = 60-604 log. = 1-7825013 
 
 Bearing from B to 12 -00 b = N.W. 12 37 45" tan. = 9-35C3645 
 
 Station 12-00 Bx ex = 5873-969 ~ 6126-730 = 252-761 log. = 2 4C27100 
 
 BE co CE 264-884 co 195-662 = 69 222 log. = 1-8402441 
 
 Bearing from B to 12-00 c = N.W. 15 18 5C" tan. = 9 -43753-11 
 
BEARINGS OF STATIONS. 
 
 Station 12 00 Bx <*> dx = 5873-969 <*> 6126-460 = 252 491 log. = 2-4022459 
 
 BE co dE = 264-884 <*> 203-682 = 61*202 log. = 1-7867656 
 
 Bearing from B to 12-00 d = N.W. 13 37 31" tan. = 9-3845197 
 
 Station 12-50 Bx v= a* = 5873-969 <*> 6142-920 =s 268-951 log. = 2 4296731 
 
 BE co aE = 264-884 co 246-177 = 18-707 log. = 1-2720041 
 
 Bearing from B to 12-50 a = N.W. 3 58 44" tan. = 8-8423310 
 
 Station 12 50 BN <~ by = 5873-969 co 6142 590 = 268-621 log. = 2-4291399 
 
 BE co &E = 264-884 ^ 254-318 = 10-566 log. = 1-0239106 
 
 Bearing from B to 12-50 b = N.W. 2 15 09" tan. = 8 5947707 
 
 133 
 
 Station 12-50 Bx co ex = 5873-969 co 6124-930 = 250-961 
 BE co CE = 264-884 co 245-557 = 19-327 
 
 log. r= 2-3996062 
 log. = 1-2861644 
 
 Bearing from B to 12-50 c = N.W. 4 24 13" tan. = 8-8865582 
 
 Station 12-50 Bx <*> dx = 5873-969 co 6124 600 = 250*631 
 BE co dE = 264-884 co 253-574 = 11-310 
 Bearing from B to 12-50 d = N.W. 2 35 02" tan. = 8-6544278 
 
 log. = 2-3990348 
 log. = 1-0534626 
 
 Station 13-00 BN co ax = 5873-969 co 6140*790 = 266 821 
 BE co aE = 264-884 co 296-302 = 31-418 
 
 log. = 2-4262200 
 log. = 1-4971785 
 
 Bearing from B to 13 -00 a = N.E. 6 42 56" tan. = 9-0709585 
 
 Station 13 00 BN co C N = 5873*969 co 6122-810 = 248-841 
 BE co CE = 264-884 co 295-435 = 30-551 
 
 log. = 2-3959220 
 log. = 1-4844564 
 
 Bearing from B to 13*00 c = N.E. 6 59 25" tan. = 9-0885344 
 
 Station 13-50 Bx co 13 50x = 5873-969 co 6128 870 = 254*901 Jog. = 2-4063715 
 
 BE co 13-50E =: 264 "884 co 349*811 = 84*927 log. = 1* 9290458 
 
 Bearing from B to station 13*50 = N.E. 18 25 37" tan. = 9*5226743 
 
 We have thus completed our computations of bearings, from the 
 points A and B to the corners of the piers, etc. We now arrange 
 them in the following table, and from the bearings between A and 
 
134 
 
 USEFUL FOKMUL.E. 
 
 B, and between A and the corner of the piers, etc., we readily deduce 
 the angles required to be measured at A, and in like manner those 
 required to be measured at B. 
 
 A TABLE OF BEARINGS, 
 
 FROM STATIONS A AND B TO THE CORNERS OF THE SEVERAL PIERS AND 
 ABUTMENTS, AND TO STATIONS 10 AND 13-50 ; 
 
 WITH THE ANGLES TO MEASURE FROM A AND B TO EACH CORNER AND STATION INDICATED. 
 
 PIERS 
 
 AND 
 
 CORNERS. 
 
 BEARINGS FROM 
 STATION A TO THE 
 
 POINTS INDICATED 
 IN COLUMN 1. 
 
 BEARINGS FROM 
 STATION B TO THE 
 
 POINTS INDICATED 
 
 IN COLUMN 1. 
 
 ANGLES 
 AT STATION A 
 
 WITH B. AND 
 THE POINTS 
 INDICATED IN 
 
 COLUMN 1. 
 
 ANGLES 
 AT STATION B 
 
 WITH A, AND 
 THE POINTS 
 
 INDICATED IN 
 COLUMN 1. 
 
 10-00 station 
 
 / / 
 
 N.E. 20 00 00 
 
 / // 
 
 N.W. 45 00 00 
 
 / / 
 
 83 00 00 
 
 / " 
 
 32 00 00 
 
 10-50 b 
 
 " 31 54 51 
 
 " 37 36 27 
 
 71 05 09 
 
 39 23 23 
 
 d 
 11-00 a 
 
 " 34 23 49 
 " 39 59 00 
 
 39 31 41 
 
 " 31 42 41 
 
 68 36 11 
 63 01 00 
 
 37 23 19 
 45 18 19 
 
 " b 
 
 " 41 21 15 
 
 " 30 29 09 
 
 61 38 45 
 
 46 30 51 
 
 " c 
 
 " 42 40 29 
 
 " 33 31 40 
 
 60 19 31 
 
 43 28 20 
 
 " d 
 
 " 44 03 10 
 
 " 32 16 15 
 
 58 56 50 
 
 44 43 45 
 
 11-50 a 
 
 " 47 52 48 
 
 " 23 33 56 
 
 55 07 12 
 
 53 26 04 
 
 " 6 
 
 " 48 47 04 
 
 " 22 08 29 
 
 54 12 56 
 
 54 51 31 
 
 " c 
 
 " 50 23 38 
 
 " 25 06 51 
 
 52 6 22 
 
 51 53 09 
 
 " c/ 
 
 " 51 22 56 
 
 " 23 44 38 
 
 51 37 04 
 
 53 15 22 
 
 32-00 a 
 
 " 53 47 00 
 
 " 14 13 38 
 
 49 13 00 
 
 62 46 22 
 
 " 6 
 
 " 54 38 06 
 
 " 12 37 45 
 
 48 21 54 
 
 C4 22 15 
 
 " c 
 
 " 56 20 39 
 
 " 15 18 56 
 
 46 39 21 
 
 Cl 41 04 
 
 " d 
 
 " 57 09 04 
 
 " 13 37 31 
 
 45 50 56 
 
 63 22 29 
 
 12-50 a 
 
 " 58 35 37 
 
 " 3 58 44 
 
 44 24 23 
 
 73 01 16 
 
 // 
 
 " 59 17 33 
 
 " 2 15 09 
 
 43 42 27 
 
 74 44 51 
 
 " c 
 
 " 61 00 08 
 
 " 4 24 13 
 
 41 59 52 
 
 72 35 47 
 
 d 
 
 " 61 39 22 
 
 " 2 35 02 
 
 41 20 38 
 
 74 24 58 
 
 13-00 a 
 
 " 62 29 50 
 
 N.E. 6 42 56 
 
 40 30 10 
 
 83 42 56 
 
 c 
 
 " 64 42 41 
 
 " 6 59 25 
 
 38 17 19 
 
 83 59 25 
 
 13-50 station 
 
 " 66 57 01 
 
 " 18 25 37 
 
 36 02 59 
 
 95 25 37 
 
 A 
 
 
 N.W. 77 00 00 
 
 
 
 B 
 
 S.E. 77 00 00 
 
 
 
 
DIFFERING GRADES AND VERTICAL CURVES. 135 
 
 We have thus completed our table of angles which are to be used 
 in the location of the points indicated as follows. Having two 
 observers with instruments, one at station A and the other at B ? 
 each having a suitable instrument, they proceed to lay off upon 
 their respective instruments the angles indicated in the table to 
 any one of the corners or stations desired. Having done this, an 
 assistant repairs with a boat to the place of intersection of the lines 
 corresponding with their instruments ; and, if the water be not too 
 deep, fixes a stake by driving it into the mud or sand which forms 
 the bottom of the river. Or, if piles are being driven from a scow, 
 the position of the pile may be brought to the intersection indicated 
 by the instruments, and driven. Or, the point may be otherwise 
 marked by mooring a buoy or float by the aid of two or three lines ; 
 and doubtless, by many other devices, marks may be fixed which 
 will be found equally simple and exact, the whole of the mechanical 
 operations being so simple in their character as not to need further 
 description. I will only add that we have made use of the method 
 which we have here endeavored to develope in several instances, and 
 have found it very convenient and accurate. 
 
 (51) We have thus, in the foregoing pages, completed our 
 contemplated essays upon railroad curves connected with the aline- 
 ments of the main tracks, side tracks, and turnouts; we now 
 propose to add a formula for uniting the different gradients of 
 railroad tracks with vertical curves. 
 
 Before we proceed to the investigation of formula, we would 
 remark, it is not our purpose to give anything like a full description 
 of the operations for laying down the gradients of a railroad 
 track, (the operations being so simple in character as to be readily 
 
[FiG. 21.] 
 
VERTICAL CURVES. 137 
 
 comprehended by every one,) but merely to develope a formula 
 which has been found convenient and useful in our practice of 
 rounding off the salient angles and hollowing the re-entering angles 
 (if they may be so termed) formed by the intersections of the 
 gradients of a railroad track. 
 
 (52) In our practice we have never laid down a vertical curve 
 of a less radius than forty thousand feet, but in general our curves 
 have embraced two hundred feet upon either side of the intersecting 
 point of the gradients ; that is to say, the vertical arc has usually 
 been about 400 feet in length ; but, when the inclinations of the 
 gradients have been such as to make the angles to be rounded 
 or hollowed, comparatively acute, we have sometimes used a 
 shorter arc. 
 
 Presuming that the inclinations of the gradients and the relative 
 positions of the angles have been determined, we commence with 
 the investigation of formula for determining the value of these 
 angles in degrees. 
 
 The problem presents four different cases, viz., the salient angle 
 formed by an ascending and descending grade. The re-entering 
 angles formed by two descending or ascending grades, one of which 
 being much more inclined than the other. The re-entering angle 
 formed by a level line, and one ascending or descending grade. The 
 re-entering angle formed by a descending and an ascending grade. 
 
 To explain the foregoing angles, and the method of ascertaining 
 their value in degrees and minutes, (See Fig. 21,) let a a represent 
 
 a level line, A S an ascending grade, and S D a descending grade, 
 T 
 
138 USEFUL FORMULA. 
 
 then will the angle at S be a salient angle. To determine the 
 measure of this angle, viz., A S D, we will suppose A S to ascend 
 at the rate of forty feet to the mile, and S D to descend at the rate 
 of twenty-five feet to the mile. We will now suppose S a to equal 
 one mile, or 5280 feet; S D and S A are also taken as a mile each; 
 inasmuch as there will be no practical difference between the 
 length of a line inclining 40 or even 80 feet to the mile, and the 
 same line reduced to a level, (and this remark will apply almost 
 universally, or to the gradients of railroads in general :) therefore 
 the lines S A, S a, S D , are taken each as one mile. 
 
 From an inspection of the figure, it will be obvious that the angle 
 A S D will equal 180 (A S a + D S a ) and the angle A S a = 
 A S a ; wherefore, A S a -}~ ^ S a = D S A ; an( j ^ e f n ow i n g 
 method of determining which, though not strictly accurate, will be 
 found sufficiently exact for every practical purpose. Taking S D 
 and S A = one mile each, then will a A = 40 feet, and D a = 
 25 feet, and the angles S D A and S A D being each so near a 
 right angle that we may take either of them as such. Taking the 
 angle D as a right angle, S D will be a cosine, and D A a sine, 
 and S A will be a radius. 
 
 Then, taking radius = unity, and representing it by E, we have 
 this analogy, cos. : sin. :: R : tan. = ~- ; which, in practice, will 
 stand thus, 
 
 S D : (I) a + a A ) : : R : tan. A S D = (D a/ s V A 5 
 and 180 A SD = ASD; 
 
 and consequently, the angle A S D = C in the quadrilateral 
 C T S T. (63) 
 
VERTICAL CURVES. 139 
 
 Having obtained the angles S and C in the quadrilateral, we will 
 now proceed to give an example of computation. As before stated, 
 (D a + a A) = D A = 25 + 40 = 65 feet, and S D = 5280 
 feet = 1 mile; then, we have 
 
 D A = 65 feet log. = 1-8129134 
 S D = 5280 " log. = 3-7226339 
 
 C = 42 19" tan. = 8-0902795 
 
 We take the angle C to the nearest second ; it is not necessary 
 that we should be more exact. 
 
 Then, taking the distances S T and S T = 260 feet each, we 
 ascertain the radius, (in our computations it is not necessary to 
 know the radius, and we merely ascertain it this time as a matter 
 of curiosity rather than use ;) to find which, we have 
 
 Sin. i C : 260 feet : : cos. |- C : radius = cot, | C 260 (64) 
 
 Thus, 260 log. = 2-4149733 
 
 C = 00 21 10" cot. = 2-2106159 
 
 Radius = 42226 feet log. = 4-6255892 
 
 We have thus found our radius = 42226 feet, the angle S being 
 very acute, (speaking comparatively,) it becomes necessary to take 
 the distances S T and S T somewhat greater than it is our habit. 
 We would remark here, that whatever distance we assume for S T, 
 it will be found convenient that it should divide even by the 
 number 20, because in setting the grade-pins for laying down the 
 rails, it is usual to place them 20 feet apart, which is as long as 
 we can conveniently have the straight-edged board used as a guide 
 in placing the sleepers or ties to the proper height or grade. 
 
 We have taken S T = 260 feet, which, divided by 20 feet, will 
 
[Fia. 22.] 
 
VERTICAL CURVES. 141 
 
 give thirteen divisions upon each side of S; consequently, the portion 
 of the arc spanned by one of these divisions will be equal to--. 
 
 We have demonstrated in section (2) that the angle of deflection 
 is equal to half the centre angle, spanned by the chord governing 
 said deflection. We therefore have in the large triangle TCI, the 
 angle at T = 90, consequently the angle at 1 is equal to the 
 complement of the angle C ; and the angle C = ^-~p- = -* ^ 1C 
 = Or 37" -7; hence the angle at 1 = 89 58 22" -30, which 
 is common to the triangles TIC and T I a; then, in the triangle 
 T 1 a, we have T= 0!/ 2 37 "- 70 -=- Q 00 48" 85; hence the angle 
 at a will be equal to 180 (89 58 22" -30 + 00 48" -85) = 
 90 00 48" -85. That is, the angle at a is equal to 90 + the 
 angle at T ; and the angle b, in the triangle T b 2, will also be 
 equal to 90 + the angle at T; and the same remark will apply to 
 the triangles T c 3, T d 4, etc., to T m S. 
 
 Having thus explained the method of deducing the angles, we 
 find the ordinates a 1, 62, c 3, etc., to m S, by the following formula. 
 Taking the triangle T a 1 for an example, and making use of the 
 symbols belonging to the same, we have sin. 90 -f- T I T 1 : : sin. 
 T : a I ; but, since the sin. of 90 + T is equal to the cos. T, 
 the formula may be rendered thus, cos. T : T 1 : : sin. T : a 1 = 
 tan. T X T 1 (65) 
 
 We wish to remark, that in the investigation above, we have 
 considered the arc T m T and the tangent lines T S and S T 
 of equal length ; consequently, the computed ordinates will be 
 practically the same as if they were perpendiculars to the tangent 
 line T S. 
 
1-t- USEFUL FORMUUB. 
 
 We will now show, by actual computations, that the formula, 
 though not in the strictest sense exact, still it presents by far a 
 greater degree of accuracy than would be possible to practise in 
 laying down railroad tracks. 
 
 As a test to our formula, we will now determine the length of 
 the curve T m T, and compare the same with the tangent lines 
 T S, S T . In the triangle S T C, we have 
 
 Sin. C : S T : : cos. C : rad. = cot. C X S T. 
 
 Then, representing S T by , the radius in seconds by r" , and the 
 angle C in seconds by C", and the radius in the unit of measure 
 (found above) by 7% and the length of the curve by c, we shall have 
 
 r . C" cot. C . t . C" 
 
 Thus, C = 21 00" cot. = 2-2106159240 
 
 t = 260 feet log. = 2 4149733480 
 
 C" = 2540 " log. = 3-4048337166 
 
 r" = co. ar. log. = 4-6855748668 
 
 c = 519-9934 feet log. = 2-7159978554 
 2t = 520 
 
 Difference = 0066 
 
 We find thus, that the arc and the tangent lines agree within 
 10060" f a ft whi^ is a little larger than ^ of an inch, a 
 quantity quite too small to be considered an error in laying down 
 a railroaj track, especially in the ordinates where the error in the 
 greatest will be reduced in the proportion the length of radius 
 bears to the ordinate, which can never amount to any thing -worth 
 noticing, especially when we consider that the case we are examining 
 is of that class which produces errors greater in amount than the 
 most of cases which come under consideration ; so that I think we 
 may be warranted in pronouncing our formula practically exact. 
 
VERTICAL CURVES. 
 
 143 
 
 Having investigated the necessary formula, we will now proceed 
 to give a specimen of calculation. 
 
 We have found in the foregoing, the angle T in the triangle T a 1 
 = 00 48" -85, and in the triangle T b 2 it will be twice that amount, 
 and in the triangle T c 3 it will be three times that amount; and so 
 on to the centre ordinate m S, which will be thirteen times the 
 amount. 
 
 Commencing with T a 1 we have T 
 
 T 1 
 a I 
 
 00 48" -85 tan. = 6 "3744395 
 
 20 feet log. = 1-3010300 
 
 0-00473663 feet log. = 7-6754695 
 
 In the triangle T ft 2 we have T 
 
 T 2 
 ft 2 
 
 QO oi 37" -07 tan. = 6-6754695 
 
 40 feet log. = 1-6020600 
 
 0-0189465 feet log. = 8-2775295 
 
 In the triangle T c 3 we have T 
 
 T 3 
 c 3 
 
 02 26" -55 tan. = 6-8515608 
 
 60 feet log. = 1-7781513 
 
 0-0426297 feet log. = 8-6297121 
 
 In the triangle T d 4 we have T 
 T 4 
 d 4 
 
 03 15"-40 tan. = 6-9764996 
 
 80 feet log. = 1-9030900 
 
 0-0757861 feet log. = 8-8795896 
 
 In the triangle T e 
 
 we have T 
 T 5 
 e 5 
 
 00 04 04" -25 
 100 feet 
 0-118415 feet 
 
 tan. = 7-0734097 
 log. = 2-0000000 
 
 log. = 9-0734097 
 
 In the triangle T/6 we have T 04 53"-01 tan. = 7-1525910 
 
 T 6 =120 feet log. = 2-0791812 
 
 / 6 = 0-170518 feet log. = 9-2317722 
 
 In the triangle T g 7 we have T 05 41"- 95 tan. = 7-2195379 
 
 T 7 =140 feet log. = 2-1461280 
 
 gl = 0-232095 feet log. = 9-3656659 
 
144 
 
 USEFUL FORMULAE. 
 
 In the triangle T h 8 we have T 
 
 T 8 
 h 8 
 
 In the triangle T i 9 we have T 
 T 9 
 i 9 
 
 : 06 30" -08 tan. = 7-2775300 
 
 : 160 feet log. = 2-2041200 
 
 0-303144 feet log. = 9-4816500 
 
 07 19"- 65 tan. = 7-3286827 
 
 180 feet log. = 2-2552725 
 
 0-383668 feet log. = 9-5839552 
 
 In the triangle T j 10 we have T =0 08 08" 50 
 
 T 10 = 200 feet 
 
 tan. = 7-3744403 
 log. = 2-3010300 
 
 j 10 = 0-473664 feet log. = 9-0754703 
 
 In the triangle T k 11 we have T =0 08 57"-35 
 
 T 11 = 220 feet 
 
 tan. = 7-41583S2 
 log. 2 3!2 227 
 
 k 11 = 0-573134 feet log. = 9-7562559 
 
 In the triangle T / 12 we have T = 09 46" -20 tan. = 7 4536218 
 
 T 12 = 240 feet log. = 2-38(2112 
 
 12 = 0-682076 feet log. = 9-8338330 
 
 In the triangle T m S we have T = 10 35" -05 tan. = 7-4883842 
 
 T S =260 feet log. = 2-4149733 
 
 m S = 0-800493 feet log. = 9- 
 
 We have thus computed the thirteen ordinates according to the 
 formula. It will be seen that the first angle in the triangle T a 1 
 is taken a very small amount too large, which will make all the 
 angles used something large, hut not sufficiently so as to practically 
 affect our results. 
 
 The reason why we did not correct the angles in the course of 
 our operations was, that we may he enabled to compare the results 
 obtained by another method, which will much abridge the work; 
 and, although not strictly accurate, still we may state, as we have 
 before, respecting the preceding formula, that it is practically 
 exact. 
 
VERTICAL CURVES. 
 
 145 
 
 The other method may be explained thus. Having divided the 
 line T S, in a suitable number of parts, which, for the purpose of 
 comparing with our previous computations, we will suppose to be 
 thirteen, of 20 feet each ; we must then compute the first ordinate, 
 viz., a 1, in the triangle T a 1, which we call?/; which, however, 
 in the present case will be needless, as we have it already computed. 
 We therefore take the value of y from our previous computations, 
 which of course will need no comparison ; we then find the re 
 mainder of the ordinates, b 2, c 3, d 4, etc., to m S, according to 
 the expressions given in the following table. 
 
 COLUMN 
 
 OF 
 
 ORDINATES. 
 
 EXPRESSION 
 OP COMPUTED 
 FORMULA. RESULTS. 
 
 COMPUTED RESULTS 
 BY PREVIOUS FORMULA, 
 FOR COMPARISON. 
 
 
 No. 1. 
 
 No. 2. No. 3. 
 
 No. 4. 
 
 
 No. 
 
 1 or a 1 
 
 = y = 0-00473663 feet 
 
 0-00473663 
 
 feet 
 
 2 
 
 "62 
 
 = 2 2 y 0-01894652 " 
 
 0-01894650 
 
 u 
 
 3 
 
 " c 3 
 
 = 3 2 y = 0-04262967 
 
 0-0426297 
 
 (( 
 
 4 
 
 " d 4 
 
 = 4 2 y = 0-07578608 
 
 0-0757861 
 
 ({ 
 
 5 
 
 " e 5 
 
 = 5 2 y = 0-11841575 " 
 
 0-1184150 
 
 
 
 6 
 
 " / 6 
 
 = 6 2 y 0-17051868 " 
 
 0-1705180 
 
 u 
 
 7 
 
 " <7 7 
 
 = 72 y 0-23209487 " 
 
 0-2320950 
 
 
 
 8 
 
 " h 8 
 
 = 8 2 y = 0-30314432 " 
 
 0-303145 
 
 u 
 
 9 
 
 " i 9 
 
 = 9 2 y = 0-38366703 " 
 
 0-383668 
 
 u 
 
 10 
 
 " j 10 
 
 = 10 2 y = 0-47366300 " 
 
 0-473664 
 
 u 
 
 11 
 
 " k 11 
 
 = U 2 y = 0-57313223 " 
 
 0-573134 
 
 
 
 12 
 
 " Z 12 
 
 = 12 2 y = 0-68207472 " 
 
 0-682076 
 
 1C 
 
 13 
 
 " m S 
 
 = 13 2 y = 0-80049047 " 
 
 0-800493 
 
 K7 
 
 EXPLANATION OF THE TABLES. The first column contains the 
 number of the ordinates, arranged in numerical order, from one to 
 thirteen. The second column contains tjbe notation expressing the 
 
146 USEFUL FORMULAE. 
 
 method of computation. Third column contains the resulting com 
 putations. Fourth column contains the same elements, computed 
 "by the preceding formula, which is placed here for the purpose of 
 conveniently comparing the results of the two methods. 
 
 The comparison shows, that the method by squares, is sufficiently 
 accurate for the most exact work, when we consider we can only 
 make use of the three first decimals, in practice, while the two 
 methods do not differ at all in the fifth decimal. I repeat, we may 
 without hesitation, pronounce the rule practically exact. 
 
 (53) In applying the foregoing results to practice, doubtless 
 different engineers will pursue different methods; but, a convenient 
 method is to ascertain the total heights (as they are generally 
 called) of stations in the inclined lines, which shall correspond to 
 stations of the same number, belonging to the vertical curves ; 
 then adding or subtracting the computed elements of the curve, or 
 length of ordinates, corresponding to the station set out in the 
 inclined lines, accordingly as the nature of the case requires. 
 
 (54) It sometimes happens that the locating stations are ob 
 literated, and in that case, the position of the angle at S, (which is 
 the main starting point,) cannot be readily found. In such cases the 
 engineer must measure a portion of the road anew, extending the 
 measurement sufficiently far from the apex or point S, upon both 
 sides, so as to ascertain the grades correctly. Having completed the 
 measurements, and marked and numbered the stations, (which are 
 usually fixed one hundred feet apart,) and determined their relative 
 levels, and the inclination of the grades which govern our operations, 
 we proceed to ascertain the points of intersection. 
 
VERTICAL CURVES. 147 
 
 First, draw the lino k k to represent a level, (See Fig. 23 (a);) 
 then, draw the line n I , at an angle with k k , equal to the inclination 
 of the grade S n. Again, through the point S, draw the line n I at 
 an angle with k k , equal to the inclination of the grade S n . From 
 the point n let fall the perpendicular n m; and also, from the point 
 n t let fall the perpendicular n I , until it intersects the line n I . 
 Then, from the intersection of n I with n I , draw the line I m 
 parallel with k k ; and, from the intersection of n I with n I, draw 
 the line I m parallel with k k . 
 
 It will now be obvious, from an inspection of the diagram, that 
 the angle n S k is equal to the grade or inclination of n S ; and the 
 angle k S I is equal to the grade or inclination of S n . And it will 
 also be obvious that the angle n S k will be equal to the grade or 
 inclination S n ; and the angle k S I will be equal to the grade 
 or inclination S n. Then, representing by 
 d, the distance between tho stations, (usually 100 feet;) 
 </, the difference in heights between the stations in grade n S ; 
 g , " " " " " n S; 
 
 n and n , the numbers of the stations at the points they represent ; 
 h, the height of station n; 
 h , the " " n ; we have 
 
 d g .: n co n . n m ; 
 and d . g \\ n <=o n : n m 
 
 Then, by substituting p for n m, and p for n m ; 5 for n S, 
 and <$ for n S ; and, subtracting p from h, (which gives the height 
 of the line m I above the datum line, and which we will represent 
 by ;) then, subtracting p from Ji , (which gives the height of m I 
 above the datum line, and which wo will represent by 0,) we have 
 
2.992 7J1 
 
 71 12.452 
 
 4.452m 
 
 [Fia. 23(6).] 
 
 1=44:52 
 
VEETICAL CURVES. 
 
 149 
 
 : a :: (K-o) : y =<*=, 
 
 We will suppose, for an example of calculation, station n to be 
 numbered 10, and station n to be numbered 20; h, to be 10 567 
 feet above the datum line, and li to be 7 385 feet above the datum 
 line ; g to be 0*7575 feet, and g to be = 0*3787 ; and both grades 
 to be ascending from S ; with d = 100 feet. We then have 
 
 Again, 
 
 n m = p 
 h, 
 
 p h = o 
 d 
 
 n oo n 
 
 = 100 feet co. 
 
 = 0-7575 feet 
 
 = 10 stations, or 1000 feet 
 
 = 7-575 feet 
 = 10-567 
 
 log. = 8-0000000 
 log. = 9-8793826 
 log. = 3-0000000 
 
 2-992 
 
 = 100 feet co 
 
 = 0-3787 feet 
 
 = 10 stations, or 1000 feet 
 
 log. = 0-8793826 
 
 log. = 8-0000000 
 log. = 9-5782953 
 log. = 3-0000000 
 
 
 
 m = p 
 
 = 3-787 feet 
 
 V 
 
 
 = 7-385 
 
 It: 
 
 p = o 
 
 = 3-59H 
 
 g* 
 
 + S 
 
 = 1-1362 feet 
 
 d 
 
 
 = 100 feet 
 
 ft 
 
 
 
 = 6-969 
 
 log. = 0-5782953 
 
 log. = 9-9445452 
 log. = 2-0000000 
 log. = 0-8431705 
 
 <r + <r 
 a 
 
 h o 
 
 = 6-1336 
 
 = 1-1362 
 = 100 feet 
 = 4-395 
 
 log. = 2-7877157 
 
 log. = 2-9445452 
 log. = 2-0000000 
 log. = 0-6427612 
 
 log. = 2-5873064 
 
 Having thus ascertained the distances n S, and n S = <5 and <T ; 
 if we now take n + 5 = the number of the station, represented 
 by S, (the point of the intersection of the grades ;) then will n <5 
 = the same number S, if the computations be correctly prepared. 
 
150 USEFUL FORMULAE. 
 
 STATIONS. STATIONS. 
 
 Thus, n 10 n = 20 
 
 f 5 = 6-1336 f5 = 3-8664 
 
 16-1336 n 6 = 16-1336 
 
 We may now further prove our work by ascertaining the height 
 of the point S, from the datum line, by computing the descent 
 from n to S, and from ri to S. If our computations are correct, the 
 results should be alike. Thus, we have h 8.g = the height of 
 S ; and h -- S .g = height of S. 
 
 6 
 
 = 6-1336 
 
 log. 
 
 = 0-7877155 
 
 8 
 
 = 0-7575 
 
 log. 
 
 = 9-8793826 
 
 6>g 
 
 = 4-6462 
 
 
 0-6670981 
 
 h 
 
 = 10-567 
 
 
 
 S 
 
 = 5-9208 above datum line. 
 
 
 
 <y 
 
 = 3-8664 
 
 log. 
 
 = 0-5873068 
 
 s 1 
 
 = 0-3787 
 
 log. 
 
 = 9-5782953 
 
 Height of 
 
 fj .ff 7 = 1-4642 0-1656021 
 
 h 1 7-385 
 
 Height of S = 5 9208 above datum line. 
 
 Having thus found the station corresponding to the intersecting 
 point of the grades, and its relative height, the necessary stations 
 for laying down the vertical curves can be readily prepared, and 
 the work can be proceeded with in the manner set forth in the 
 foregoing. 
 
 To make the formula just enunciated applicable to every case 
 would require several modifications ; we shall, however, only give 
 one, believing that the ingenuity of the reader will readily supply 
 whatever may be deficient. 
 
 (55) The case we propose, is, when we have one grade de- 
 
VERTICAL CURVES. 151 
 
 scending say O SOO feet per 100 feet, which will intersect another 
 descending grade of 0*400 feet per 100 feet. 
 
 To describe the construction of a figure applicable to this case we 
 have only to cop} r verbatim the description of Fig. 23 (a); we there 
 fore refer to that description as a substitute. 
 
 After having constructed the figure, it will be obvious that we 
 have but few modifications to make in the formula already given ; 
 but, lest we should not be fully understood, we repeat our former 
 formula, with the necessary modifications. Thus we have 
 
 d : g : : n ^ ri : n m; and d : g : : n ^ ri : ri m (70) 
 
 Then, as before, substituting p for n m, and p for ri m ; d for 
 n S, and <$ for n S ; and then, subtracting p from h, we get the 
 height of m I above the datum line, (which height we represent by 
 o ;) and then, by adding p to li we obtain the height of m I above 
 the datum line, (which height we represent by o.) We then have 
 
 g <^> g : d : : h o : 5 
 and g ^ g : d : : //, o : X (71) 
 
 EXAMPLE OF COMPUTATION. 
 
 Let d = 100 feet ; g = 0-800 feet ; g = 0-400 feet ; and n = 
 10; ri = 20; h =12-452; li = 8 -234. We then have 
 
 Firstly, d: g :: n <** ri - n m ; 
 Secondly, d . g . . n <^> n *. ri m . 
 
 Thus, d = 100 feet co. ar. log. = 8-0000000 
 
 S =0-800 feet log. = 9-9030900 
 
 n w n 1 =10 stations, or 1000 feet log. = 3-0000000 
 
 n m p = 8-00 feet log. = 0-9030900 
 
[Fia. 24.] 
 
 2 3 4 5 B 7 8 9 ID 
 
 6789 /lO 
 
CURVING BOARDS. 153 
 
 Secondly, d = 100 feet co. ar. log. = 8-0000000 
 
 ^ 0-400 feet log. = 9-6020600 
 
 n *> ri 10 stations, or 1000 feet log. = 3-0000000 
 
 n m = p = 4-00 feet log. = 0-6020600 
 
 h = 12-452 h = 8-234 
 
 p = 8-000 p = 4-000 
 
 (^ _ k) O i 4-452 (p 1 + A ) = o = 12-234 
 
 h, = 8-234 A 12-452 
 
 (k o 1 ) = 3-782 (h 0} = 0-218 
 
 Again, we have 
 
 g csz g d I , h o : 8 ; 
 
 and g ts*g . d II ft d . S ( 7 1 ) 
 
 oo g- 0-400 co. ar. log. =z 0-3979400 
 
 d = 100 feet log. = 2-0000000 
 
 h = 0-218 feet log. =: 9-3384565 
 
 ,5 = 0-545 stations, or 54-5 feet log, = 1-7363965 
 
 cx3 5- = 0-400 co. ar. log. =; 0-3979400 
 
 d 100 feet log. = 2 0000000 
 
 h o 3-782 feet log. = 0-5777215 
 
 <Y = 945-5 feet, or 9-455 stations log. 5= 2-9756615 
 
 (56) We conclude our remarks upon tracklaying with a for 
 mula for the computation of elements convenient for setting out 
 curving boards or patterns for bending rails to suit the horizontal 
 curves of short radii. 
 
 The principles of our. present formula are based upon a chord 
 
 of the arc equal in length to the longest rails, being divided into 
 
 equal spaces or abscissa ; and then ascertaining the length of 
 
 the corresponding ordinates which shall extend therefrom to the 
 
 V 
 
154 USEFUL FORMULAS. 
 
 periphery or curve. Calculations based upon strict formula, being 
 somewhat lengthy, will require considerable time and labor to 
 perform them. The engineer being frequently called upon to give, 
 in great haste, the elements for making a pattern to guide the 
 tracklayer in curving his rails, it becomes desirable to obtain a 
 formula as short and convenient as practicable. 
 
 These considerations have led to the adoption of the following 
 formula, which, though not strictly correct, is nevertheless as 
 accurate as mechanical skill requires. 
 
 By way of explanation, suppose it desirable to form a pattern for 
 bending rails of twenty feet in length, it will be found convenient 
 to divide the chord into equal parts of one foot each. 
 
 From an examination of the sketch, it will be obvious that one 
 of these divisions will bisect both the chord and the arc, and that 
 the parts thus bisected will be similar and equal ; therefore, the 
 computations made for the one part will apply to the other. 
 
 To proceed to the investigation, we first ascertain the angle at 
 the centre of the curve spanned by an absciss at the periphery of 
 one foot. 
 
 Representing this angle by C ; the absciss by a ; the ordinatc 
 corresponding to No. 1, by?/; and the radius of the curve by r; 
 then, by considering r a cosine ; and the absciss a, which spans the 
 arc, a sine ; we have the following analogy, 
 
 Cos. : sin. : : R : tan. C ; 
 
 which, by substituting for the cosine its value = r; and, for the 
 sine, its value = a = unity ; we then have 
 
CURVING BOARDS. 155 
 
 Tan. C = y- and cos. \ C : a \\ sin. \ C : y = tan. J C . a = 
 tan. \ C . 1 ; it is now obvious that ---- = 2JL= - ; wherefore, 
 
 2 r 
 
 tan. i C . 1 =#= 7 !a~ ( 72 ) 
 
 Performing the computations indicated, by logarithms, we have 
 log. y = (ar. co. log. r + ar. co. log. 2.) 
 
 It will be seen by the above expression that we have considered 
 the arc and the tangent of the same length, which will be found 
 sufficiently exact for every practical purpose, and that the ordinate 
 represented by y = 1 a, as represented in the figure. To find the 
 remainder of the ordinates we have, for ordinate 
 
 No. 2 = b 2 = 2 2 y 
 
 3 = c 3 = 3 2 y 
 
 4 = d 4 = 4 2 y 
 
 5 = e 5 = 5 2 y 
 
 etc., to the number of ordinates contained in half the chord or 
 tangent line. 
 
 For the purpose of testing the degree of accuracy of the formula 
 enunciated above, it may be necessary to obtain from strict com 
 putations the ordinate D d = (10) (10.) This ordinate will 
 correspond to 10 2 ?/, or the greatest ordinate of the computation, 
 and will contain a greater error than any one of the others. 
 
 INVESTIGATION OF EXACT FORMULA. Let r represent the radius of 
 
 the curve ; II the radius of the tables ; \ ch the half of the chord AB ; 
 
 C the angle D C B ; then will r : K : : \ ch : sin. C = ^- ; and 
 
 Cos. \ C : J ch : : sin. } C : D d = tan. \ C . J ch (74) 
 
156 
 
 USEFUL F R M U L JE . 
 
 which expression corresponds to the greatest ordinate of the 
 calculations ; and which, in the practical examples we shall give, 
 will be represented by 10 2 y. So that the difference between the 
 ordinate found by the exact formula and 1C) 2 y will constitute the 
 error. 
 
 Example of computation, according to the approximate formula. 
 Assuming r = oOO feet ; \ eh = 10 feet ; and a = 1 foot ; we shall 
 then have 
 
 r = 300 i eet 
 2 
 
 0-0016666 feet 
 
 co. ar. log. 1= 7-5228787 
 
 co. ar. log. = 9-6989700 
 
 log. = 7-2218487 
 
 Having found !No. 1 
 
 = a 
 
 1 
 
 = 
 
 y 
 
 = 0- 
 
 0016666 
 
 we have 2 
 
 = b 
 
 2 
 
 __ 2 2 
 
 y 
 
 = 0- 
 
 0066666 
 
 3 
 
 == c 
 
 3 
 
 = 3- 
 
 y 
 
 = 0- 
 
 0149999 
 
 4 
 
 = d 
 
 4 
 
 =i & 
 
 y 
 
 = 0- 
 
 0266666 
 
 5 
 
 = e 
 
 5 
 
 = 5- 
 
 y 
 
 = 0- 
 
 0416666 
 
 6 
 
 = f 
 
 6 
 
 = 6 2 
 
 y 
 
 = 0- 
 
 0599999 
 
 7 
 
 = 9 
 
 7 
 
 = 7- 
 
 y 
 
 = 0- 
 
 0816666 
 
 8 
 
 = h 
 
 8 
 
 = 8- 
 
 y 
 
 = 0- 
 
 1066666 
 
 9 
 
 = i 
 
 9 
 
 = 9- 
 
 y 
 
 = 0- 
 
 1349999 
 
 10 
 
 J 
 
 10 
 
 = 10- 
 
 y 
 
 = 0- 
 
 1666666 
 
 Example of computation of the greatest ordinate, according to 
 the exact formula, as a test to the above, viz., by formulae (73) and 
 (74.) We have 
 
 Sin. C = 
 
 , ch 
 
 and tan. I C . i ch = 1) d = 10 . 10 
 
 r = 300 feet 
 
 =V ch = 10 " 
 
 C = 10 5V 36 -767 
 
 co. ar. log. = 7-5288787 
 log. = 1-0000000 
 
 sin. = 8-5228787 
 
CURVING BOARDS. 157 
 
 C = 00 57 18" -383 tan. = 8-2219693 
 
 ch = 10 feet 1- 0000000 
 
 D d = (10) . (10) = 0-16671 feet 9-2219693 
 
 10 a y = 0-16666 
 
 Difference = 0-00005 
 
 It will be seen, by comparing D d, with 10 2 y, that the errors of 
 the approximate formula are too small to be noted in the practical 
 operations of curving rails. 
 
 It will be noticed that the above ordinates extend from the 
 tangent line D 10, D 10", to the curve A B D. It frequently 
 happens that it will be more convenient to set out the curve 
 from the chord line A d B, which line may be readily represented 
 upon the pattern board by straining a small string or wire from 
 A to B. To prepare ordinates to be thus used, we subtract the 
 ordinates found successively from the greatest ordinate, (which, in 
 our example, is == 10 2 y.) Thus, 
 
 10 2 y 
 
 
 
 
 
 = D d 
 
 = 0-1666666 
 
 10 2 y 
 
 
 
 y 
 
 = a 
 
 V 
 
 = 0- 
 
 1650000 
 
 I0 2 y 
 
 2 2 
 
 y 
 
 = b 
 
 2 
 
 = 0- 
 
 1500000 
 
 10 2 y 
 
 3 2 
 
 y 
 
 = c 
 
 3 
 
 0-1516666 
 
 10 2 y 
 
 4 2 
 
 y 
 
 = d 
 
 4 
 
 = 0- 
 
 1400000 
 
 10 2 y 
 
 5 2 
 
 y 
 
 ^^ e 
 
 5 
 
 = 0- 
 
 1250000 
 
 10 2 y 
 
 G 2 
 
 y 
 
 = f 
 
 6 
 
 = 0- 
 
 1066666 
 
 10 2 # 
 
 7 2 
 
 y 
 
 = 9 
 
 7 
 
 = 0- 
 
 0851111 
 
 10 2 y 
 
 8 2 
 
 y 
 
 = h 
 
 8 
 
 = 0- 
 
 0600000 
 
 10 2 y 
 
 9 2 
 
 y 
 
 = i 
 
 9 
 
 = 0- 
 
 0316666 
 
 10 2 ?/ 
 
 10 2 
 
 y 
 
 = J 
 
 10 
 
 = 0- 
 
 0000000 
 
 Before we leave this subject, we would remark, that all curves 
 of a less radius than 3000 feet, provided they are laid with rails 
 
158 USEFUL FORMULAE. 
 
 twenty-one feet in length, should be curved ; and a board prepared 
 as a pattern will facilitate, and add both to the convenience and 
 accuracy of the operations. 
 
 (57) It frequently becomes necessary for the engineer to 
 ascertain the radius of a small part of a curve in a railroad track ; 
 as, for example, when called upon to lay down a side track, a 
 turnout, or to connect a branch road with a curve, the radius of 
 which is unknown. Many formula may be deduced for the solution 
 of this problem, each possessing nearly equal convenience; we shall, 
 however, limit our investigations to some two or three of those in 
 common use. 
 
 Let T A T represent a portion of a curve, the radius of which it 
 is desirable to ascertain. We measure the chords A T and A T , 
 each of the same length, which we represent by c; then ascertain 
 the point B in the middle of the chord T T; and measure the 
 ordinate A B, which we represent by b. Then, representing the 
 radius of the curve by r; and the radius of the tables by 11 ; we 
 have, in the triangle A B T , c . R : : b *. cos. A =~ ( r 5 and 
 
 For an example of computation, we will suppose c = 100 feet; 
 and b = 8 feet. Then, 
 
 c 2 = 100 2 log. = 4-0000000 
 
 b = 8 co. ar. log. = 9-0969100 
 
 2 co. ar. log. = 9-6989700 
 
 r = 625 feet log. = 2-7958800 
 
 V 
 
 Again, let T A T represent, as before, a portion of a curve, the 
 radius of which we desire to ascertain ; let the chord T T be 
 
DETERMINATION OF RADIUS. 159 
 
 represented by a; and let a = 199*359 feet; and 6=8 feet; 
 as above. We then, by dividing a by 2, have in the triangle 
 A B T, (taking b as a cosine, and J a as a sine,) the following 
 analogy, 
 
 Cos. .* sin. : : R : tan. A = -=p- (7 6) 
 
 Then, Sin. A : _ 
 
 (77) 
 
 XX 
 
 A 
 
 !/ T? <* ^ c 
 
 - = 2 
 
 * 
 
 . 2 " -LI / . 
 
 sin. A cos. A 
 
 
 EXAMPLE OF C ALC 
 
 U L ATIO N. 
 
 *0 
 
 = 99-6795 
 
 log. 
 
 = 1-9986059 
 
 b 
 
 = 8 co. ar. 
 
 log. 
 
 = 9-0969100 
 
 A 
 
 = 850 24 45" -17 
 
 tan. 
 
 = 1-0955159 
 
 A 
 
 = 850 24 45" -17 co. ar. 
 
 sin. 
 
 = 00 13942 
 
 A 
 
 = co. ar. 
 
 cos. 
 
 =; 1 0969102 
 
 
 2 co. ar. 
 
 log. 
 
 == 9-6989700 
 
 ka 
 
 = 99-6795 
 
 log. 
 
 =r 1-9986059 
 
 r 
 
 = 625 feet radius 
 
 log. 
 
 = 2-7958803 
 
 Again, let T A T represent the segment of the curve whose 
 radius is desired, and the half chord T B === a; the ordinate A B 
 = b; and the chord T A = c; and the diameter A D = d; the 
 radius = r. We here remark, that the angle A is common to 
 the triangle T A B, and the triangle T A D ; and the angle B in 
 the triangle T B A, and the angle T in the triangle D T A are 
 each a right angle ; consequently, the two triangles are similar. 
 We now have 
 
 a 2 + b 2 = c z 
 and b : c : : c : d = -- = " j^ = -- + b = 2 r (78) 
 
 EXAMPLE OF COMPUTATION. Let a = 99-6795 feet; b =^ 8 feet. 
 Then, 
 
ELEVATING OUTSIDE RAIL FOR CURVE. 1G1 
 
 a* 5= 99-G795 2 feet 
 b = 8 feet 
 
 1242 
 
 log. = 3-9972118 
 co. ar. log. 9-0969100 
 
 log. = 3-0941218 
 
 2)1250 
 
 = 625 
 
 We have here thus endeavored to apply the same elements in 
 each of our examples of computations, by way of testing the 
 different methods, and we find each of them to give the same 
 results. 
 
 (58) Before we take our final leave of railroad tracks, we will 
 add a formula for elevating the outside rail of curves. We extract 
 what we shall say upon the subject from Article XV. of the 22d 
 vol. of the American Journal of Science, 1832. 
 
 The article was written by J. Thompson, (Engineer, and late Pro 
 fessor of Mathematics in the University of Nashville, Tenn.,) and 
 commences with a discussion of a formula given in Colonel Long s 
 work on railroads, which he shows to be erroneous. With this 
 criticism it is not our intention to meddle ; but, as Mr. Thompson 
 has, in the course of his remarks, developed a convenient and an 
 accurate formula, we shall endeavor to extract only so much as 
 may seem to be connected with its explanation. 
 
 [w] 
 
162 USEFUL FORMULA. 
 
 " Let CAB represent a horizontal surface on which a railway is 
 situated ; A and B the rails placed in a circular curve around C as 
 a centre. A car moving over the rails A and B, around the centre 
 C, will be acted upon by two forces, one horizontal and centrifugal, 
 arising from the motion of the car in a curved line, and acting in 
 a direction from the centre C ; the other, the force of gravity, acting 
 in a vertical direction. I omit here, as not necessary in the present 
 investigation, the moving force derived from animal or other power 
 acting in a direction of a tangent to the curve. Let the horizontal 
 line A K represent the centrifugal force above mentioned, and the 
 line E A the force of gravity. It is evident that the resultant 
 of these two forces will be E K, which will represent both the 
 intensity and the direction of the pressure of the loaded car upon 
 the rails. The line E K, therefore, representing the direction of 
 pressure, the rails should be so placed that this line may be perpen 
 dicular to the plane passing through them. Draw the vertical line 
 B D, and through A draw A D, perpendicular to E K ; B D will be 
 the elevation of the exterior rail above the interior, and the angle 
 DAB will be the inclination of the plane of the rails to the 
 horizon. The centrifugal force A K, compared with the force of 
 gravity A E, is easily found, when the radius of curvature of the 
 track and the velocity of the car are given. The distance between 
 the centre C, and the middle of the track, may be considered as 
 the radius of curvature. 
 
 " We may obtain a very simple algebraic expression for the 
 elevation of the exterior rail. Let g = force of gravity; c = 
 centrifugal force ; d = distance between the rails ; and E = required 
 elevation ; R, and V representing radius and velocity. Then, by the 
 similar triangles E A K and A B D, we have E= cd ; but, by 
 
ELEVATING OUTSIDE KAIL FOR CURVE. 163 
 
 central forces, c = -^-; hence, E = ~~- in this expression; g is 
 always a constant quantity, and equal to 32 -2 feet. 
 
 " If the velocity of a car on a railway were always the same, we 
 should have no difficulty in assigning the proper elevation of the 
 exterior rail. But, as there must be necessarily a great variety in 
 rates of travelling, an elevation for a rate of twenty miles per hour 
 would he much too great for a rate of eight, twelve, or fifteen miles 
 per hour. Perhaps the elevation required by the mean velocity 
 would be most eligible. There is one view of the subject, however, 
 which ought to be taken into consideration in the location of the 
 exterior rail. When a car moves with great velocity on a curved 
 road, and the planes of the rails are horizontal, the flange of the 
 fore wheel on the exterior rail is exposed to very great friction, 
 which operates as a retarding force, and injures both the car and 
 the railway ; this friction is diminished, though not altogether 
 removed, by giving the exterior rail the elevation which the velocity 
 and radius require. In order to reduce the friction still further, or 
 remove it altogether, it would perhaps be advisable to increase by 
 a small quantity the elevation obtained as above. f It is evident 
 that a car moying on the inclined plane A D, will tend by its own 
 weight to approach A, and recede from D ; this will oppose the 
 centrifugal force by which the flange is pressed against the rail D, 
 and thus the friction will be in whole or in part removed. I know 
 it has been maintained that the flange of the hind wheel on the 
 interior rail produces as much friction as the flange of the exterior 
 
 * It has been the practice of the author of the foregoing papers to elevate the exterior rail to suit 
 the highest velocity with which the regular trains are supposed to run over the curve. At the present 
 time we should not think thirty-five miles the hour too great. Quick trains produce greater friction 
 Upon the exterior rail, and are more liable to accident than slow trains. 
 
 t It has been the practice of the author to add one fourth of an inch to the computed elevation. 
 
164 USEFUL FORMULAE. 
 
 fore wheel. It may, however, be shown, from various considerations, 
 that if either of the hind wheels produces friction, it is rather the 
 exterior one ; indeed, we may suppose that motion is communicated 
 to the hind wheels by a force which acts precisely in the same 
 direction as if they were moved by animal power, the direction 
 being nearly a tangent to the curve. This being admitted, the 
 flanges of the two exterior wheels sustain all the friction occasioned 
 by curvature. It may be observed, however, that when the distance 
 between the fore and the hind wheels is comparatively very great, 
 the direction of the force moving the hind wheels will vary con 
 siderably from the tangent, and consequently the friction will be 
 diminished." 
 
 * Although we agree with Mr. Thompson in the main, we do not fully agree with his concluding 
 remarks. Mr. Thompson says, " It may, however, be shown, from various considerations, that if 
 either of the hind wheels produces friction, it is rather the exterior one ; indeed, we may suppose that 
 motion is communicated to the hind wheels by a force which acts precisely in the same direction as 
 if they were moved by animal power, the direction being nearly a tangent to the curve. This being 
 admitted, the flanges of the two exterior wheels sustain all the friction occasioned by the curvature. 
 It may be further observed, however, that when the distance between the fore and hind wheels is 
 comparatively very great, the direction of the force moving the hind wheels will vary considerably 
 from the tangent, and consequently the friction will be diminished." 
 
 The reasoning of Mr. Thompson, doubtless was applicable to cars sustained npon two axles and 
 four wheels only, one axle being situated near the forward end of the car, and the other near the back 
 or hind end. Now, if the car be short, the axles must of course be near each other , in this condition, 
 the flanges of both the forward and hind wheels may grind the exterior rail, the forward wheels of 
 course grinding much the hardest. A distance, however, between the axles can be readily ascertained 
 which will relieve the hind wheels from the friction of the flanges against either the interior or 
 exterior rail , then, expanding the distance between the axles, the flanges of the hind wheel will begin 
 to grind against the interior rail, and the greater the distances between the axles the greater will be 
 the friction. I would observe, however, that the forward wheel flange will, under all distances 
 between the axles, grind upon the exterior rail, and will grind more and more severely in proportion 
 as the distance between them increases. These notions are based upon the condition that the axles 
 are firmly and permanently secured to the car, and at right angles with its frame. We might easily 
 demonstrate the position we have here taken by diagrams if it were thought necessary, but the change 
 produced by the adaptation of what we term four-wheel trucks to our long car bodies, which permits 
 
ELEVATING OUTSIDE RAIL FOR CURVE. 165 
 
 We have thus copied Mr. Thompson s article, with his remarks, 
 omitting only the portion relating to Mr. Long s formula. 
 
 EXAMPLE OF COMPUTATION. Assuming a radius, K = 4000 feet ; 
 and a velocity of 35 miles per hour ; 35 miles per hour = V = 
 51 *33 feet per second of time; the width between the rails being s 
 d = 4 7 feet. 
 
 Formula E = -^ 
 
 R = 4000 feet co. ar. log. = 6-3979400 
 
 g = 32-2 co. ar. log. = 8 4921441 
 
 V 2 = 51-33 log. = 3-4207990 
 
 d = 4-7 log. = 0-6720979 
 
 E =s 0-09615 log. = 8-9829810 
 
 Again, for the purpose of showing the changes of E, consequent 
 upon the changes of E, we assume E, =* 2000 feet; the other 
 expressions remaining the same. Thus, 
 
 R = 2000 feet co. ar. log. = 6-6989700 
 
 g = 32-2 co. ar. log. = 8-4921441 
 
 V s = 51-33 S log. = 3-4207990 
 
 d = 4-7 log. = 0-6720979 
 
 E = 0-1923 feet log. = 9-2=40110 
 
 We believe we have in the foregoing pages examined every 
 distinct species of curve that enters into the construction of a 
 railroad. We were aware, as we proceeded in our investigations, 
 
 the arrangement of the axles to a very near approximation with the radii of the curve, by the force 
 of the wheel flanges against the exterior rail, seems to render such an undertaking unnecessary, 
 particularly as the railroad companies appear to be universally adopting them. Butj it is not our 
 intention to discuss generally the principles which should govern the construction of cdrs. Having 
 concluded to adopt the formula of Mr. Thompson, we thought, in justice to him, we Were bound to 
 copy his remarks. We would only further mention, respecting Mr. Thompson s remarks, that we 
 cannot discover any material difference in the action of the flanges of the wheels upon the curved 
 rails, whether the car receives its motion from a force pulling in front or pushing behind. 
 
166 USEFUL FORMULA. 
 
 that many modifications of the formula we have deduced would 
 frequently be called for ; but, as we have before stated, it has not 
 been our intention to exhaust the subject, but merely to give a 
 formula for the most prominent of each class, or rather for that 
 class which most frequently present themselves to the engineer 
 while engaged in construction. 
 
 We contemplated, when we commenced our work, closing our 
 paper here; but it has occurred to us, that the inexperienced 
 engineer might feel the want of some convenient plan or system 
 of computing the cubic contents of excavations and embankments. 
 For the purpose of supplying those wants, we add the following. 
 
 An investigation of formula for the computation of the cubic 
 contents of earth, excavations, embankments, masonry, etc., in 
 constructing railroads. 
 
 (59) An article quoted from Sillimaris Journal, by Professor 
 Eaton, states, in effect, as follows ; that whereas the sections into 
 which the engineer would divide the excavations upon a railroad, 
 readily admit of being subdivided into pyramids, wedges, and paral 
 lelepipeds; therefore, if you add the area of both ends of the sec 
 tion, to four times its middle area, divide the sum by six, and 
 multiply the quotient by the length of the section, the product 
 will give its solid contents. 
 
 This problem can be readily demonstrated to be strictly correct, 
 provided the sides of the section are perfect planes. It is the 
 constant endeavor of every skilful engineer so to arrange the 
 sections that, were the irregularities of the earth to be pared down, 
 
MEASURING SOLIDS AND SUPERFICIES. 167 
 
 so as to produce regular planes between the points which he has 
 chosen to take his levels at, the solid contents of the earth con 
 tained in the section would he just sufficient to fill the hollows, 
 and make the surfaces planes between them. 
 
 (60) In Professor Eaton s enunciation of the rule above 
 quoted, I did not discover any method of ascertaining the middle 
 area of the section ; it being evident that an arithmetical mean of 
 the areas of the ends of the section would not uniformly produce 
 the desired result. To supply this deficiency in the formula is the 
 main object of the present paper ; but, as the original formula may 
 not be within the reach of every individual who may feel interested 
 in seeing an investigation and demonstration of it, we have thought 
 a brief investigation might not be out of place ; besides, it will 
 aid much in rendering the subject more plain and intelligible. I 
 shall, however, take it for granted that the interested reader will 
 know enough of geometry to be familiar with the common formulae for 
 measuring the solid contents of pyramids, parallelepipeds, wedges, 
 and surfaces of the cross sections of the excavations of a railroad ; 
 we shall, therefore, only allude to the most common formulae for 
 measuring solids and superficies, as we may have occasion to 
 compare them with the formula to be deduced. 
 
 Commencing with the pyramid. 
 
 (61) According to the rule, we have to add the area of the 
 base of the pyramid to four times the area of its middle section, 
 (taken parallel with said base,) divide the sum by six, and multiply 
 the quotient by the height of the pyramid ; the product will give 
 its solid contents. 
 
[Fia. 26(i) 
 
 THE PYRAMID. 
 
 G. 26( 2 ).] 
 
 THE WEDGE. 
 
PYRAMID AND WEDGE. 169 
 
 For the purpose of proof, or demonstration, let us suppose a 
 pyramid of four sides, Fig. 1, the angles of its base being right 
 angles, and the lines circumscribing said base of equal length, 
 which length we represent by a; it is evident that the lines which 
 circumscribe the middle section of said pyramid (taken parallel with 
 said base) will be equal to J a; if then we represent the height by 
 h, we have for the area of the base a X a = a 2 ; for the area of 
 the middle section |aXi^==i 2 ; then, by the rule, ( fl3 + * 34 j h 
 = A? a JL = ^ a 2 h = the solid contents ; which corresponds exactly 
 
 with the formula in common use (1.) 
 
 This equation shows the area of the middle section of a pyramid 
 to be one fourth of the area of the base ; and this proposition is 
 universal and equally correct in every species of pyramid, whether 
 it be three, four, or many-sided, regular or irregular. 
 
 Secondly, the wedge. 
 
 (62) We next apply the rule to the measurement of the wedge. 
 Let us now suppose a regular or symmetrical wedge, with a base 
 circumscribed by lines of equal length, which we represent by a; it 
 is obvious that of the lines which circumscribe the middle section of 
 the wedge, (taken parallel with the base,) two of them will equal 
 a; and the other two will equal J a. Then, making, as "before, h 
 equal the height, or length ; we have, by the rule, 
 
 For the area of the base a X a = a z ; 
 For middle section a X \ a = J <- 
 
 Then, ( a2 +/ fl2 4 )h= *^ h - = i a 2 h =* the solid contents ; 
 which corresponds with the formula in common use for determining 
 
 the solid contents of the wedge (2.) 
 x 
 
170 USEFUL FORMULA. 
 
 From this equation we learn, that the area of the middle section 
 is equal to one half the area of the base. By cutting the wedge 
 into pyramids we may compute its solid contents in a way some 
 what different, with the same formula. 
 
 Let A B C D, in Figure 2, represent the base end of the wedge; 
 and a b the edge or sharp end ; cut the wedge diagonally through 
 the plane 6 B C. The wedge is thus divided into two pyramids; 
 A B C D b, and a b B C ; the pyramid A B C D b being a four- 
 sided one ; and the pyramid a b B C being a triangular or three- 
 sided one. We may further cut the four-sided pyramid in the 
 plane b B D, which divides that pyramid into two three-sided pyra 
 mids. The wedge will then consist of three triangular pyramids ; 
 but, as the same rule for determining the cubes applies to three, 
 four, and many-sided pyramids, we shall, in our further investi 
 gations of the mensuration of the wedge, only use the four-sided 
 pyramid, in connection with the blind pyramid a b B C ; (this 
 pyramid is so named because it presents no area in either of the 
 surfaces of the cross section of a cut in a railroad excavation.) 
 Pyramids of this character enter into the calculations of nearly 
 every cross section, and my principal object in introducing it in the 
 wedge, is, for the purpose of testing the method of computing its 
 solid contents* 
 
 If we now compare the solid contents of the four-sided pyramid, 
 it will be observed that we make use of the whole area of the base 
 of the wedge, and add thereto four times the quarter area of the 
 base ; which quarter is equal to one half the area of the middle 
 section of the wedge ; this sunl, divided by six, and the quotient 
 multiplied by the length of the wedge, gives the solid contents of 
 
PYRAMID AND WEDGE. 171 
 
 the pyramid. Then, to complete the measurement of the wedge, 
 according to the rule ; having divided the area of the base, plus 
 four times one fourth of the area of the base, etc. ; as explained 
 before, the remainder of the area of the middle section will of course 
 be equal to one fourth of the area of the base ; which, multiplied by 
 four, its product divided by six, and the quotient multiplied by the 
 length of the wedge, the operation will be complete ; (and have been 
 performed in a different method,) the result being the same as in 
 equation (2.) 
 
 I would however mention, that we found by equation (2) that 
 the middle area of the wedge was equal to half the area of the 
 base ; and, in measuring the four-sided pyramid, the middle area of 
 which is equal to one quarter of the area of the base, of course it 
 is equal to one half of the middle area of the wedge, leaving the 
 other half for the middle area of the blind pyramid. 
 
 To elucidate this, let us introduce the calculations. 
 
 Using the former notations, we have, in the mensuration of the 
 four-sided pyramid, for the base a X a = a 2 , the middle area J a 
 X i a = k a* ; then, (J*+*) . h = - 2 ^ A (3.) 
 
 In the mensuration of the blind pyramid a b B C, we have for 
 the middle area, (a b being equal to a, and B C being also equal 
 to a) \ a X i a = J a z ; and (_* <A.) . h = - .*- (4.) 
 
 If we now add the results of the above equations, (3 and 4,) 
 their sum will be equal to the contents of the wedge, as found in 
 equation (2 ;) thus proving that the mensuration of the blind pyra 
 mid is exact, according to the rule. As a further proof, we may 
 
[FiG. 27.] 
 
 IN WHICH THE TWO SEPARATE DIAGRAMS DEFERRED TO BY THE TEXT MAY BE TRACED. 
 
EXCAVATIONS AND EMBANKMENTS. 178 
 
 DOW measure the blind pyramid in this manner. Making a B C 
 the base, and a b the height, which we shall denote by h; then, 
 according to rule, and using the same notation, we have 
 
 / i a h -a L a_xjr _*)_ii ) . a = a \ h (5;) 
 
 the same as in equation (4.) 
 
 We might produce a great variety of proofs to show the accuracy 
 of the rule. Thus, if we should endeavor to measure a four-sided 
 square parallelepiped by the rule, the simplest manner of proceed 
 ing will be to divide it into two wedges, and then apply the rule ; 
 or, we may cut off four pyramids, leaving a large blind pyramid, 
 which, in order to determine its contents, will require that we should 
 determine the length of the diagonals, (this method will be exact 
 only when the diagonals are at right angles), and then multiplied 
 into each other, will give four times the middle area required ; this, 
 divided by six, and multiplied by the length, will give the solid 
 contents of the blind pyramid. 
 
 But, this last method is somewhat complicated, inasmuch as we 
 should be obliged to find the diagonals by extracting their square 
 roots from the sum of the squares of the other sides ; or, we may 
 find the diagonal by trigonometry. I shall, therefore, only give an 
 example of determining the cubic contents of the parallelopiped, by 
 dividing, first, into two wedges. This example we give merely as 
 an illustration of the method, 
 
 Retaining the former notation, we have for the measurement of 
 the apparent pyramids, the bases and four times the middle area of 
 each pyramid, equal to twice the area of the bases. See equation 
 (1,) which, for both pyramids, is equal to 4 2 , and four times the 
 middle area of one of the blind pyramids, will be a X ct = a* each 
 
174 USEFUL FORMULAE. 
 
 there being two of them, we have, therefore, 4 a 2 -f- 2 a z = 6 a z , 
 and 6 a \ h = a 2 h ; the same result as by the ordinary method of 
 determining the cubic contents of a parallelepiped. 
 
 Of course the above is not the most convenient method, but will 
 serve to show the application of the rule to the measurement of the 
 frustrum of a pyramid. 
 
 The rule is also peculiarly applicable to the measurement of 
 wedges in which one end is wider than the other, and to almost 
 every figure imaginable which is bounded by right lines and plane 
 surfaces. 
 
 We will now give the calculations of a few imaginary figures of 
 different forms, after the manner of our practice, to determine the 
 cubic contents of excavations, embankments, masonry, etc. 
 
 (63) Let A B C I L K represent one end of the supposed section, 
 which we will denominate No. 1, (in Fig. 27,) and D E F G M H, the 
 other end, which we denominate No. 2. After preparing diagrams 
 of the ends of the sections, and marking the heights of the points 
 A B C of diagram No. 1, and of D E F in No. 2, we then divide the 
 diagrams into figures which we shall now proceed to describe. 
 
 Firstly, we divide No. 1 by the perpendicular line L B, the point 
 L representing the centre of the road bed. 
 
 Secondly, draw the line A c parallel to the base, or the road bed, 
 K I ; which is always level. 
 
 Thirdly, draw the line B a parallel to the line K I ; then the 
 diagram will be divided into the triangles A B c and B C a, and the 
 
EXCAVATIONS AND EMBANKMENTS. 175 
 
 trapczoids A K L c and B L I a. We now propose for the dimensions 
 that K L and L I shall each be 9 feet long, and that the height of 
 A he 4 feet, the height of B 8 feet, the height of C 12 feet. In 
 order to determine the areas of the above mentioned trapezoids and 
 triangles, we first determine the length of the lines A c and B a; 
 the length of A c being equal to the height of A, plus one half the 
 height of A -j- K L, upon the supposition that the slopes of the 
 cuttings are as three to two. Thus, 
 
 The height of A = 4 feet 
 
 & " " =2 
 
 K L =9 
 
 Length of A c =15 feet 
 
 which we mark upon diagram No. 1. 
 
 To find the length of B a, we have the height of B L + } the 
 height of B L + L I = B a. Thus, 
 
 The height of B L =8 feet 
 
 * " " = 4 
 
 Length of L I =9 
 
 Length of B a . =21 feet 
 
 which we mark upon the diagram also. 
 
 Having prepared the diagram, we proceed to determine the area 
 of the trapezoid A K L c. We have found 
 
 Ac = 15 feet 
 
 K L =9 
 
 AC + KL 2)24 
 
 Mean, or } (A c + K L) =12 
 
 Height of A above K = L c =4 
 
 Area = 48 feet 
 
USEFUL FORMULA. 
 
 To find the area of the trapezoid a B L I, we have found 
 
 a B = 21 feet 
 
 LI = 9 
 
 a B + L I) = 2)30 
 
 k (a B + L 1) = 15 
 
 Height of B 8 
 
 Area = 120 feet 
 
 To find the area of the triangle A B c, we have found 
 
 Ac = 15 feet 
 
 i difference of heights B and A = 2 
 
 Area . =30 feet 
 
 To find the area of the triangle a B C, we have found 
 
 B a = 21 feet 
 
 difference of heights of C and B . . . . = 2 
 
 Area =42 feet 
 
 Having determined the areas of each figure composing diagram 
 No. 1, and marked the same upon it, we then proceed to divide 
 diagram No. 2 into triangles and trapezoids, and compute their 
 areas in a manner similar in principle to that adopted in No. 1. 
 
 Assuming the height of F == 8 feet; the height of E = 13 -6 
 feet ; the height of D == 10 feet ; and G M and M H == 9 feet 
 in length each. Firstly, we have, in the triangle F g E, 
 
 The length of the line F #, equal the height of F . =8 feet 
 
 + half the height of F =4 
 
 + GM =9 
 
 Wherefore, F 5- = 21 feet 
 
 We then have half the difference in heights of g and E = half 
 the difference of F and E. Thus, 
 
EXCAVATIONS AND EMBANKMENTS. 177 
 
 The height of F = 8 feet 
 
 E = 13.6 
 
 Difference 2^5.6 
 
 Half difference of g E = 2-8 feet 
 
 We now find the area of the triangle F#E = 21X2 8 = 
 58-8 feet. 
 
 Secondly. In the triangle D E /we have 
 
 The length of the line D/ equal the height of D . =10 feet 
 
 + " " } D . =5 
 
 + M H . . . =9 
 
 Wherefore / D = 24 feet 
 
 Then we have half the difference in height of / and E equal to 
 half the difference of D and E ; thus, 
 
 The height of D = 10-00 feet 
 
 " " E = 13-60 
 
 Difference 2) 3-60 
 
 Half difference 
 
 We now find the area of the triangle DE/=24Xl 8 = 
 43-2 feet. 
 
 Thirdly. To find the area of the trapezoid F Gr M g ; we found 
 in the foregoing, 
 
 F g- = 21-00 feet 
 
 G M = 9-00 
 
 FS- + GM 2)30-00 
 
 Mean length = 15-00 
 
 Then, M G = height of F = 8-00 
 
 Area = 120*00 feet 
 
 Y 
 
[FiG. 28.] 
 
EXCAVATIONS AND EMBANKMENTS. 179 
 
 Fourthly. Then, to find .the area of the trapezoid D H M/; we 
 find in the foregoing, 
 
 D/ = 24-00 feet 
 
 M II = 9-00 
 
 D/+ M H . . . " . . . . 2W 
 
 00 
 
 Mean length = 16-5 
 
 Then, M /= height of D = lO OO 
 
 Area = 165-00 feet 
 
 Having ascertained the several areas of the divisions of the cross 
 sections, and marked the same upon the diagram, our next operation 
 will be to ascertain the cuhic contents of that portion of the half 
 section lying above the lines F g, g c, c A, and A F. 
 
 (64) It will be evident, from an inspection of the drawing, that 
 the upper portion of the half section may be divided into two appar 
 ent, and one blind pyramid. This solid admits of two distinct methods, 
 or plans of divisions, so that, if the surface is not twisting, it matters 
 not which of the plans is adopted. For the purpose of a test, we 
 will consider the divisions under both aspects ; firstly, as represented 
 in the drawing of the section, (viz., Fig. 27 ;) and secondly, as in 
 Fig. 28. Whatever difference there may be, if any, will be seen in the 
 different dimensions the blind pyramid will assume ; hence we will 
 confine ourselves to the comparison of the contents of this solid. 
 
 Firstly, we have A c X g E = 4 times the centre area of the 
 blind pyramid ; and, in the second form, we have F g X c B = 4 
 times the centre area, as above. 
 
 FIRST FOEM. SECOND FORM. 
 
 Ac. . . = 15- feet 
 
 g E . . . =5-6 
 
 9-0 
 75- 
 4 times middle area . =84-00 
 
 F g . . . . = 21-00 feet 
 c B . = ~ 4 00 
 
 4 times middle area = A 84-00 
 
. 29.] 
 
EXCAVATIONS AND EMBANKMENTS. 181 
 
 It thus appears that the top surface is a perfect plane : had it 
 been warped, or twisted, the middle areas of the blind pyramid 
 under both aspects would not have been alike. 
 
 We have now, in the remainder of the first half section, the solid 
 A K L eg F G M. The lines F g, G M, K L, A c, being parallel, 
 the solid admits of a division into two wedges ; and these wedges, 
 as we have already shown, may be divided into one apparent and 
 one blind pyramid each ; the area F g M G being the base of one of 
 the apparent pyramids, and the area A c L K being the base of the 
 other. The surfaces of this solid being perfect planes, it matters 
 not in what manner we form the wedges, and cut from them the 
 blind pyramids, as their combined measurements will be the same. 
 For example, if the wedges be formed by cutting the solid through 
 the plane A c M G, four times the middle area of one of the blind 
 pyramids will be equal to M g X A c, and four times the middle 
 area of the other will be equal to L c X G M. Or, we may form 
 the wedges by cutting the solid through the plane K L g F ; then, 
 four times the middle area of one of the blind pyramids will be 
 equal to L c X F g, and the other will be equal to g M X K L. 
 Before we proceed further with our investigation, we will compare 
 the contents of the blind pyramids as ascertained by both methods 
 of division. 
 
 Firstly. We have A c X M g = 4 times the middle area of one 
 of the blind pyramids. 
 
 Ac =15 feet 
 
 MS- =8 
 
 4 times the middle area, =120 feet 
 
182 USEFUL FORMULA. 
 
 Again, we have L c X G M = 4 times the middle area of the 
 other blind pyramid. Thus, 
 
 Lc = 4 feet 
 
 G M =9 
 
 4 times the middle area =36 
 
 Then add the area found above = 120 
 
 Gives the sum of the middle areas of both . . = 156 
 
 Secondly. We have L c X F g = 4 times the middle area of one 
 of the blind pyramids. Thus, 
 
 Lc =4 feet 
 
 F g =21 
 
 4 times the area =84 feet 
 
 Again, we have M g X K L = 4 times the middle area of the 
 other blind pyramid. Thus, 
 
 M g = 8 feet 
 
 KL =9 
 
 4 times the area = 72 
 
 Add the area found above =84 
 
 Sum of middle areas of both =156 feet 
 
 Hence we see that the computations prove that the final result 
 will be the same under both methods of computation. 
 
 We will now sum up the measurement of the supposed divisions 
 of the first half section. 
 
EXCAVATIONS AND EMBANKMENTS. 183 
 
 1st. We have the area of the base of the triangle F g E A = 58 -8 feet, and + 4 times 
 
 its middle area = 58-8 ;* (See Fig. 27) = 117 -6 feet 
 
 2d. We have the area of the base of the pyramid A B c E = 30 feet, and + 4 times its 
 
 middle area = 30 = 60*0 
 
 3d. We have 4 times middle area of blind pyramid A c E g- . . . . = 84*0 
 
 4th. We have the area of the base of the pyramid F G M g A = 120 feet, and + 4 
 
 times the middle area = 120 =240-0 
 
 5th. We have the area of the base of the pyramid A c L K g = 48 feet, and -f 4 
 
 times the middle area = 48 = 96-0 
 
 6th. We have the sum of the middle areas of the remaining two blind pyramids = 156*0 
 
 753-6 
 7th. Taking the length of the section = 100 feet 100 
 
 Dividing by 6^75360-0 
 
 Solid contents = 1256.0 
 
 (65) It may not be amiss here to remark, that we always, when 
 the nature of the case will admit, divide each section into two parts 
 by a plane passing through the vertical lines M E and B L, and 
 compute each portion separately, (see Fig. 27.) The reason for this is, 
 that a level is always taken over the centre of the road bed in every 
 cross section, and, in a majority of the cases which occur, there are 
 only two other levels taken ; viz., one at the right, and one at the 
 left hand slope stakes ; and whenever it becomes necessary to take 
 other levels at the right or left of the centre, it will still be con 
 venient to preserve the centre division. And it will be apparent 
 when we have completed our computations, that our system of 
 dividing the cross sections into triangles and trapezoids, is pecu- 
 
 * In the early part of this discussion, we proved that the middle area of a pyramid, taken parallel 
 to its base, was equal to one fourth of the area of its base ; hence, 4 times the middle area will be 
 equal to that of the base. 
 
184 USEFUL FORMULA. 
 
 liarly adapted to this method of computing cubic contents, and 
 affords a very convenient, as well as an accurate method of ascer 
 taining the area of said cross section. 
 
 Before we proceed to the computation of the cubic contents of the 
 remaining, or second half section, we would remark that the portion 
 of the second half section which lies above the plane passing 
 through B/D a, having its upper surface much twisted or warped, 
 admits of two forms of division, which will be found by computation 
 to give different results ; one of which will be applicable to one 
 form of surface, and the other to another. And that the portion of 
 the half section lying below the plane B / D a having all of its sur 
 faces perfect planes, the computation will give correct, and of course 
 like results, from whichever of the forms the divisions may take. 
 
 We now proceed to the examination of the upper portion of said 
 half section. 
 
 Firstly. If we suppose the upper surface to have this form, viz., 
 that of a plane through DEB, and intersecting in the line D B, a 
 plane passing through DBG, the solid will then contain only two 
 apparent pyramids, with no blind pyramid; viz., the apparent 
 pyramids D f E B and B a C D ; the measurement of which is as 
 folio AYS : 
 
 For the area of base of D E / B = 43-2 feet, and -f 4 times middle area = 43-2 = 86-4 feet 
 a B C D = 42 _}- 4 = 42 = 84 
 
 Dividing by e)l70 4 
 
 28-4 
 Multiplying by the length of section 100 
 
 Solid contents 2840-0 
 
EXCAVATIONS AND EMBANKMENTS. 185 
 
 Secondly, if we suppose the upper surface to have the following 
 form ; viz., that of a plane passing through the points B C E, and 
 intersecting in the line C E, a plane passing through the points 
 C D E ; the solid will then contain two apparent pyramids and two 
 blind pyramids, viz., the apparent pyramids B a C E and D E /C, 
 and the blind pyramids C a f D and E / a B ; to measure which 
 we have 
 
 For the area of base of D E/ C = 43 -2 feet, and + 4 times middle area = 43-2 = 86 -4 feet 
 
 " " " aBCE = 42 " +4 " " " = 42 = 84 
 
 " 4 times middle area of blind pyramid C a/D = C a X/D =4 X 24 , . = 96-0 
 
 " " " aBE/=aB X E/=21 X 3-6 , . = 75-6 
 
 Dividing by 6^342-0 
 
 57-0 
 
 Multiplying; by length of section 100 
 
 5700-0 
 
 We thus find the upper portion of the second half section under 
 consideration computed. 
 
 For the 1st form of surface gives solid contents = 2840 feet 
 
 u u 2ll u u . = 5700 
 
 Difference in cubic feet 27^2860 
 
 " " " yards =106 
 
 The wide difference in the results shows the necessity of noting 
 while in the field the form of the twisted surface, that the proper 
 method of computation may be applied. But it is not supposed 
 that surfaces like those we have been considering will very fre 
 quently occur in practice, as the engineer would be likely to divide 
 z 
 
186 USEFUL FORMULAE. 
 
 the section into two or three, which would have a tendency to lessen 
 the differences much. But, to repeat, if the form of a surface he 
 noticed when the field work is performed, so that the proper form of 
 dividing the cross section may he applied, large sections may 
 frequently he computed with as good a degree of accuracy, or even 
 "better, than smaller sections without such notice. 
 
 To complete the computation of the cuhic contents of the figure, 
 we have 
 
 The double area of the base of pyramid H M/ D a = 330 feet 
 
 Also " " " " " a B L I H =240 
 
 Then, 4 times the middle area of theblind pyramid =/M X B a = 21 X 10 . . =210 
 
 And 4 " " " " " " " = B L X M II = 9 X 8 . . =72 
 
 Dividing by 6)852 
 
 142 
 Multiplying by 100 
 
 Cubic contents of aBLIHM/D = 14200 
 
 Having thus discussed the operations necessary to ohtain the 
 cuhic contents of every portion of the section, we now add an ex 
 ample of summing up the contents after the manner in common 
 practice. Before entering on our work, we remark that we contem 
 plate two summations ; the first containing the computation of the 
 upper portion of the second half section, noticed in the foregoing as 
 containing no hlind pyramid ; and the second containing the com 
 putation of the upper portion of said second half section, noticed 
 as containing two hlind pyramids. 
 
 (66) Now, as we have the areas computed, and marked upon 
 the diagram of the cross sections, as described in the foregoing, we 
 have as follows : 
 
EXCAVATIONS AND EMBANKMENTS. 187 
 
 FIRST SUMMATION. SECOND SUMMATION. 
 
 No. 1 30-00 feet taken twice = GO OO feet 60 -00 feet 
 
 " 2 48-00 " " " = 96-00 96 00 
 
 " 3 58-8 " " = 117-60 117-60 
 
 " 4 120-00 " " " = 240-00 240 00 ^ 
 
 Blind pyramid " 1 E g X A c = 15 X 5 6 84-00 84 00 
 
 " 2 g-MxAc 15X8 = 120 00 120-00 
 
 "3LcxGM = 4x9 = 36-00 36 00 
 
 " 5 42-00 feet taken twice = 84-00 84-00 
 
 " 6 120-00 " " " = 240-00 240 00 
 
 " 7 43-2 " " " = 86-4 86-4 
 
 " 8 165-00 " " " = 330-00 330 00 
 
 Blind pyramid " 1 = M/ X B a 10 X 21 = 210-00 210-00 
 
 " 2 = B L X M H = 8 X 9 = 72-00 Blind pyramid 72-00 
 
 Dividing by 6^1776-00 acX/D 4x24 = 96-00 
 
 296-00 E/X B A = 3-6 X 21 = 75 6 
 
 Multiplying by length of section ... 100 Dividing by . . 6)1847-6 
 
 27^29600-00 324.6 
 
 Cubic contents, according to 1st summation = 1096-30 yds. Length of section = 100 
 
 1202-22 27 \ 32460 -00 
 
 Difference ^= 105-92 Cubic yards . . 1202 22 
 
 The figures considered in the foregoing pages are those most 
 commonly met with in railroad excavations. We frequently meet 
 with modifications, however, containing points necessary to he 
 noticed, between the slope and centre stakes ; hut it is believed that 
 the ingenious engineer will, with a little practical experience, be 
 enabled so to arrange the division of the sections, whatever may be 
 their form, into pyramids, so as to admit of a ready and satisfactory 
 method of computing their solid contents. 
 
 Before we leave this subject, we would remark, that the general 
 rule we have been endeavoring to demonstrate gives the measure 
 ment of the middle area of the blind pyramid C /D in the second 
 
188 USEFUL FORMULAE. 
 
 half section, a trifle large ; but so near the truth, that it has not 
 been deemed necessary to change or modify the form of division or 
 computation. In the supposed figure, named above, which presents 
 rather an uncommon case, the error is something less than a square 
 yard. If the length of the section had been taken at some thirty, 
 or even fifty feet, the error would have been much diminished. 
 
 We had thought of adding some formula for computing the 
 cubic contents of what we technically term borrowing pits ; but, as a 
 great majority of the figures composing these pits possess forms or 
 solids so simple in their character that they will, at first thought, 
 suggest ready and appropriate methods of computation, we deem 
 it unnecessary to further enlarge upon the subject. 
 
 (T 
 
YC 13171 
 
 M166992 
 
 THE UNIVERSITY OF CALIFORNIA LIBRARY 
 
 SBH.OII.Y&C1 
 
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