/.
/S,/ X^^/^>~~~~~~








A
SYSTEM
OF
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 BORDEN
CIVIL  ENGINEER.
BOSTON:
CHARLES C. LITTLE AND JAMES BROWN,
M DCCC L.




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.
BOSTON:
PRINTED BY WILLIAM CHADWICK,
EXCHANGE STREET.




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 convenient formulae to aid the young engineer, and such others as have not
had the advantage of a good mathematical education, in the proper performance of this character of work, it is believed has not yet found a place




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 computing 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 construction 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 necessary. 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 desirable.  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 described...........................................     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 resurvey 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




viii                         CONTENTS.
SECTION                                                            PAGE
13. -Practical operations and the necessary computations for obtaining
the elements of simple curves, and for correcting the position of the tangent points, etc., based upon a traverse, considered  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 formulae, specimens of computation, and a description of the field
w ork..............................................    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 formula................................    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 computations...........................................   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 formulae, 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......................   65
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 computation, 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 - Investigation of formulae,
with specimens of computing the necessary elements, including 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.
SECT1ON                                                            PAGE
46. - Discussion of the general principles which combine in the location of a turntable by the side of the main track, near a
water station, with an investigation of formulae for determining 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, except in boats or by some extraordinary device, consideredinvestigation of formulae, followed by specimens of computation - field operations described............. 109 to 135
51.- Remarks on the objects of this paper...................  135
92.- Remarks on the subject of vertical curves - formula for determining 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 contents of pyramids, wedges, parallelopipeds, 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  RR ATA.
On page 77, ninth and tenth lines from top, for " - 90~ -  Sw
+ i (180-    C)" read " (90~ + Sw)-   i (180~- C)."
On page 87, seventh line from bottom, for "- r + I h + d"
read " r -    h + d."
On page 113, second line from top, for "S 75~ 08' 35" 07 E"
read "S 75~ 08' 35"-37 E."
On page 119, fourteenth line from bottom, for "We then discover " read " We then determine," etc.








AN INVESTIGATION
F
USEFUL FOR M U LE,
ADAPTED TO
THE PRACTICAL OPERATIONS OF LOCATING AND LAYING OUT
RAILROAD CURVES.




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In        ^rfestigaflfn nf OspfiAI Dnun Cl,
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
bearings 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 favorable
location for the track. In order to do this, we examine the contour
of the surface, and select the position we think the most favorable
opposite the angle at A; then we run and measure from A to this
most favorable place for the track before selected, in a direction
that will bisect the angle A; that is, from A to S3, and this
distance we represent by b. We then proceed to ascertain a radius
for the curve, which shall pass through the point S3 and so run into
the straight lines that they shall be tangents to said curve.




4                  USEFUL FORMULA.
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 commencements of the curve, or where the direct line becomes a
tangent to said curve, viz., from A to T, (See Fig.;)
We have sin. I C  b   sin. ( C + I A):t      i. ( +  A)
and cos.  A: t:: sin. I A: r = tan. A.bsin(    i) 
tan. I A.b cot. ~ 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 deflections. 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, I 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 ch;
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,
1 (180~- C')= the angle T     S1;
Now as the angle A T C -a a right angle, or 90~, we have
90~- ( 180~-) =D;
Multiplying by 2, 180~-(180~-C')= 2D;
Subtracting 180~, and changing signs, we have
2D    C';
Consequently,        D. -= C';                     (2)
Bisecting ch we have r::: R  ch: sin. I C'-=  h- =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    FORMULA.
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 S1 in range with the
principal telescope, the point thus marked will be in the curve;
then, laying off another angle of deflection which will read upon the
instrument =2D, stretch the chain from S1 to S2, fixing the terminus at S2 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 S3, S4, S5, etc., until the curve is completed; or, ag it more frequently happens, as long as the contour
of the surface will permit us to see distinctly.
Let us now suppose an obstacle which will prevent seeing beyond
S3. 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 heretofore 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 = ch, 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.
We 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 (ch) 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, frequently 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    FORMULA.
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, which 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.
( ) 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 segments, 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 instrument; 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, desirable; 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 particular case.
PRACTICAL    BXAMPLES 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
g ch cot.. A tan. ~ C
To find the deflection, we have (4) sin. D =   h cotA tan 
Thus,   i A = 80~ 00 001r...... cot. = 9.2463188
~ C =  5~ 00' 00"...... tan. = 8.9419518
ch =  25 feet    *...*.  log. = 1.3979400
b   = 132 feet co. ar...... log. = 7.8794261
D      00 101 02t.64...... sin. = 7.4656367
In this case it will be seen that D = 0~ 10' 02".64 is an inconvenient angle to add or subtract, or even read upon the instrument;
we therefore, to remedy this objection, adopt D = 0~ 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 = 0~ 10'; therefore,
to find the radius we have
Sin. D:   h::: R  = r                     (5)
and to find t we have
Cos. I C: r:: sin. I C  t =tan. I C. r        (6)
(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,........ i ch = 25 feet    *.*.  log. = 1.3979400
D   =       0o 10 co. ar. *.. sin. = 2.5362745
Consequently, r  = 8594-38 feet........ log. = 3.9342145
Then,........ l   C = 10~ 001 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 6. By analogy we have
Sin. (  C +IA)      t:sin.~  C: b6     t in.ct A)=t tan. IC (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     0~ 10' will only change the location of the
curve from the position designed for it 0'58 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      r/-t                 (8)
Thus,   r = 8594.38 feet      log. = 3.9342145
CO =- 720011           log. = 4.8573325
r/l =          co. ar.  log. = 4.6855749
a = 3000.00 feet       log. = 3.4771219




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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 representing 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     FORMULE..
Our first long chord corresponds to 712 feet of arc. We now
propose to find the centre angle which it subtends by the following
formula.
Representing 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": C%'I-:    os n            (10)
Thus,  r = 8594.38 feet  co. ar.  log. = 6.0657855
a =   712-00 feet        log. = 2.8524800
rtt =                     log. = 5.3144251
CIf=  17088!f            log. - 4.2326906
C'/ reduced to degrees and minutes = 4~ 444 4851  sin. = 8.9177692
i C/"         "          "c  = 2~ 22t 244t  ar. co. cos. = 0-0003719
r........................................  log.  -  39342145
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 deflections, or 300', which amounts to 5~ 00' 00".
Now, by formula (10) we have
We distinguish the primitive or long chords by Ch in contradistinction with the 50 feet or
deflecting chords, which are represented by ch.




DETERMINATIONS           OF   ANGLES.               15
C/I in degrees = 5,~ 001 001,      sin. = 8.9402960
4 C/ in degrees = 20 301 001 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 Ch by formula (9) and (10)
Thus,.... r............   co. ar. log. = 6.0657855
a = 788 feet          log. = 2.8965262
r"f..........        log. = 5.3144251
C" = 189121           lo1g. = 4.2767368
Ct" reduced to degrees and minutes = 5~ 151 12"t  sin. = 8.9617037' C/l  "   "   "    "    "   =- 2~ 371 3611 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 FORMULA.
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, 2C 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  z-; 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 T C 1 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 18o04i44 48   87~ 37' 36" = angle at T, or at station 140'38 and 147-50.
"  angle at centre of curve for triangle No. 2 -- 5 00' 00"
and 1800 -5 000" = 87~ 30' =  angle at stations 147-50
2
and 155.
angle at centre of curve for triangle No. 3  5~ 00' 00"
and 1800-5o 00' 00  870 30' = angle at stations 155 and
162-50.
" angle at centre of curve for triangle No. 4 = 50 15' 12"
and 1805 15'2   =  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 diagram, we now proceed to ascertain the angles at the termini of
the several chords. Thus,
At T, or station,.... 140.38 - 90~ + 87~ 371 3611  - 177~ 371 36t1
c 1,   ".... 147-50 = 87~ 371 361! + 87~ 301= 175~ 071 3611
" 2,   "     *... 155.00 = 87~ 301 + 87~ 301  = 175~ 001 00"1
"3,    "'... 162.50 = 87~ 30' + 87~ 221 24-1= 174~ 521 2411
" TI,  "... 170.38 = 87~ 22' 24"1 + 90  = 177~ 221 241t
Having ascertained the angles for each station, which, for convenience, 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 T C 1, which is called No. 1, and pursuing
the calculations in the order shown in the diagram. Thus,
T = 870 371 3611.*.. co. ar..**  sin. = 0.0003727. r =- 8594.38 feet                log. - 3.9342145
~    I  C =   4~ 441 4811             sin. = 8.9177692
l,            Ch = 711.79 feet.... = log. = 2.8523564: 1 = 87~ 301 001.... co. ar.... sin. = 0.0004135
c   m    r =                           log. = 3.9342145
<     i C =   5~ 001 001l             sin. = 8.9402960
Ch = 749.76 feet.... = log. = 2.8749240
3 = 87~ 221 2411.... co. ar.... sin. = 0.0004565
c  v    r  =                          log. = 3.9342145
C = 50 151 12"1                  sin. = 8.9617037
i             Ch = 787.72 feet...  = log. = 2.8963747
(12) Having thus prepared our work, we proceed to adjust the
D




18                 USEFUL    FORMULA.
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
1750 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 above described.
Commencing at T with instrument pointing in the direction of
the road,
We laid off the angle  1770 37' 36i  and measured 711 79 feet to station  1
At station 1 we laid off the angle  1750 07' 36"  and measured 749 76 feet to station  2
"   2     "    "    175 0 00 00"  "  749'76   "    3
"   3     "    "    1750 0  00'  "  "  751-51  "   T
"   T     "     "   1770 14' 48"  in the direction of road.
Sum,... 880~ 00 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 be 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      FORMU LE.
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 be its direction, as bearing
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 1800 - 1770 37' 36" leaves  20 22' 24" N. W. to 1
" 1 = 1750 07' 36" = 3520 45' 12" and (2 X 1800) - 3520 45' 12" - 70 14' 48"  "  2
"  " 2 - 1750 00' 00" = 5270 45' 12" " (3 X 1800) - 5270 45' 12" = 120 14 48"  "  3
"  " 3 = 1750 00' 00" = 7020'45' 12" " (4 X 1800) - 7020 45' 12" = 170 14' 48"  "  T'
" T' = 1770 14' 48" = 8800 00' 00" " (5 X 1800) - 8800 00' 00" = 20o 00' 00"  "
being the direction of the road.
Computing the northings and westings of the foregoing traverse,
we have
No. 1. N. W. 2~ 221 241/ sin. = 8.6171119        cog. = 9.9996273
711-79 feet log. = 2.8523458        log. = 2.8523458
29.475 " log.= 1.4694577 711.17feet log. = 2.8519731
No. 2. N. W. 7~ 14! 481! sin. = 9.1008572        cos. = 9.9965171
749.76 feet log.= 2.8749223         log. = 2.8749223
94.576 " log.= 1.9757795 743.77feet log. = 2.8714394




WORKING     THE    TRAVERSE.               21
No. 3. N.W. 12~ 141 4811 sin. = 9.3265833   cos. = 9.9900028
749.76 feet log. = 2.8749223   log. = 2.8749223
159.04 " log.= 2.2015056 732.70feet log. = 2.8649251
No. 4. N. W. 17~ 141 48'1 sin. = 9.4720042  cos. = 9.9800203
751.51 feet log. = 2.8759348   log. = 2.8759348
222.814 " log.= 2.3479390 717.72feet log. = 2.8559551
Then, summing up the computed northings and westings, we
have as follows:
BEARING.   DISTANCE. NORTHING. WESTING.
0   /  it
From T to No. 1    N. W.2 22 224  711.79  711.17  29.475
"   1  "  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
Totals =..........  2905.36  505.905
Having summed up the traverse, we 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:: R: tan. C w     -                  (A)
and sin. C: W:: R: D        -in. C                  ()
or cos.  C: N:: R: D      = —                     (C)
cos.C                 C




PI t   ~ —.7  / 133J 80-6 6a It.-j -.-'   g S  L   s0
AZS          i 
[.  OL{




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! 40I!........ tan. = 9.2408691
C  = 9~ 52t 401t co. ar. -..... sin. = 0.7656168
W........ log. = 2-7040690
D  = 2949.08 feet....... log. = 3.4696858
We now have in the triangle A T T', 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'               Dsin T
sin. A
and sin. A: D:  sin. T'   A T =         Din. T
To prevent confusion in our diagram, or to render our work more
plain,, we reconstruct the figure of the triangle A T T'. (See figure
on preceding page.)
To perform the computations indicated, we have
A      -  1600 00   I 001t co, ar......... sin. = 0.4659483
D      =  2949.08 feet........ log. = 3.4696861
T      -- 10~ 071 20t!........ 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' 401........ sin. = 9.2343832
A to Tt = 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    FORMULA.
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. C: t = tan. - C.r
r   = 8594.38 feet........ lo. = 3.9342145
i C = 10~ 00! 00o........ tan. = 9.2463188
t   = 1515.42 feet........ log. = 31805333
t in the present case, as in our former notation, represents the distance required from A to T and also from A to T', to suit the contemplated curve, which in the present instance, has a radius of
8594-38 feet, and an apex angle A = 160~.
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 =   36-25 feet.  After
having moved TV from apex 36-25 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 consisting 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 difference,




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 surfaces, etc.; in short, wherever it is found inconvenient to run the
direct lines to their intersections, or to apex, and to measure therefrom to the points T and T'. In running traverses for the purpose
of obtaining the elements for the location of curves under the conditions suggested above, it will be found convenient, if not absolutely 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 practical 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; 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      FORMULA.
with telescope pointing in the direction of the located line, we
measured as follows, viz.:
ANGLES.   DISTANCES.
0  1                          ( The curve should pass through or
Station 0 = 169 29 45 = 1200 feet to station 1   near this point, which of course
(governs the radius.
"   1 = 170 00 15 =   900 "      "    2
"   2 = 175 04 30-    750 "      "    3
"   3 = 164 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 bearing due north, whatever
its course may be, we deduce the following relative courses for the
several lines of the traverse, and by formula (A) ascertain the
relative bearing from T to T'.
STATION        BEARINGS.  DISTANCES. NORTHINGS.  WESTINGS.
0 to I = N.W. 10~ 301 1511  1200   1179.890    218-7685
1 " 2 =   "   20~ 301 0011   900    843.005    315.1866
2 " 3 =   "   25~ 251 30t1   750    677.361    321.9970
3 " 4 =   "   40~ 451 2011  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~ 361 201. 22 tan. = 9.6805533
Having thus obtained the bearing from T to T', we now proceed
to compute the distance; by formula (B) and (C) we have
Sin. C: W:: R: D   =       w   or
Cos. C: N::R: D      -   N
cos. 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  N = 3863.453   log. = 3.5869756
C = 25~ 36r 201I.22 sin. = 9.6356588  C = 25~ 361 20II.22 cos. = 9.9551055
D = 4284.2 feet  log. = 3.6318701  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 be
80~.  Of course the angle at apex will be 100~. The bearings
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 bear 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 the following angles, viz., at A = 1000 001 00f.00
As before stated, at                  T =     25~ 36/ 20l.22
TI =  54~ 23/ 39II.78
Proof,  Sum = 180~ 00 00l^.00
With these angles, and the distance from T to T' - D, the distances T A and T' A are readily found; thus,




*1
tt~ ~ ~~~c
-yo  oo  / ca-.0)
1-4  A-  -^ —--------------------'A~
M./   \   y      ^*~~.4
F3<  ^^  \  ^  /   /^41~ 




MEASUREMENT        TANGENT      TO  APEX.         29
sin.A: D: sin. T': AT )
and sin. A: D: sin. T: A T'             ()
A   = 100~ 001 001 ar. co......... sin. = 0.0066485
D   = 4284.2 feet........ log. = 3.6318701
T   = 54~ 231 3911.78........ sin. = 9.9101139
A T = 3536.98 feet........ log. = 3.5486325
Again, A   = 1000 00t 001' co. ar......... sin. = 0.0066485
D    = 4284.2 feet........ log. = 3.6318701
T    = 25~ 361 20t1.22........ 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 = 6, 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 -        b:
a c b:: tan. i (180~-  C): tan. I (A c- B) and  (180~ - C) +
i (A c- B) =  the angle opposite the longest side, and I (180~C) -  ~ (A c 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      FORMULA3.
1800 00' 00"
A T         =  3536-98  Given angle T =    10o 30' 15"
2) 1690 29' 45"
I T        =   1200                        840 44/ 52".5
Sam         =  4736.98 co. ar.......... log. = 6.3244984
Diff.       =  2336.98........ log. = 3.3686550
i (180~- T) =  84~ 441 52".50........ tan. = 1.0365722
i (A co 1)  =  79~ 26' 4311.14       tan. = 0.7297256
L at 1      - 164~ 11 3511.64
/ at A      =   5~ 18' 091/.36
Given / at T     =   100 30' 151".00
180o 00' 00".00
L at I      = 164~ 11' 35".64 co. ar.... sin. = 0,5648026
AT          = 3536.98 feet.... log. = 3.5486326
T           =  100 30' 15.... sin -= 9.2608034
A I         - 2367.21982 feet.... log. = 3 3742386
A           -   5~ 18' 09/.36 co. ar.... sin. = 1.0342539
T 1         = 1200 feet.... log. = 3.0791812
T           =  10~ 301 1511      *... sin. = 9.2608034
Proof,  = 2367.21982 feet.... log. =3.3742385
Having obtained all the elements of the triangle A 1 T we represent 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 common at least to three triangles, it becomes necessary to occasionally
accent some of them, that we may understand their different values




DETERMINATION        PF  RADIUS.              31
in the investigation. We have throughout our investigations represented 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 A 1 T (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 R, we have
Sin. A: 1:: R: AC          siA              (11)
In the triangle A C 1 we have C 1 = the radius of the curve =  1;
therefore, representing the angle at 1 by G, we have
1   sin. (A-A):: AC: sin. G=         sin. A-A)     (12)
For convenience representing the line A C by d, we have
Sin. G: d:: sin. [(2 A-  A') + G]: b    d in- [(A-)+] (13)
then            b   bm:1    r    --                     (14)
b
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. ( A -A'):1: sin. [( A -              A') +  G]: bsin. [( A-A'-                    (13 )
sin. A - A'(13')
b
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 =  sin. (A A)




32                   USEFUL      FORMULAE.
We found A by the reduction of the foregoing traverse (see preceding pages)-      100~, also A' (noted as A) in the triangle
A 1 T      5~ 18' 09".36.   Therefore,
i A - A' = 50~ - 5~ 181 0911.36 = 44~ 41' 5011.64  sin.  9.8471791
i A       =                      50~ 00, 00,1.00 co. ar. sin. = 0.1157460
G  (ambiguous)                   66~ 391 38I1.22  sin. = 9.9629251
G  (corrected)                  113~ 201 2111.78
Having found G, we proceed to find the radius = r.    By formula
(13') and (14') we have
b-    sin. (j A - A' + G) and r  bm     s. (i A. A').bm
sin. (a A-A)              b      sin. ( A-A' + G)
A- Al      =    44~ 411 5011.64
G           =   1130 201 2111.78
A-Al     G =   158~ 021 121".42.... co. ar..... sin. = 0.4271153
A- A       =    44~ 411 5011.64........ sin. = 9.8471791
bm          = 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 h
Thus,.... i ch = 25 feet             log. = 1.3979400
r   = 4462.035 feet  co. ar.  log. = 6.3504670
D   = 00 191 15". 67          sin. = 7.7484070
But, as 0~ 19' 15"'67 makes an inconvenient number to add or
subtract, we choose for the angle of deflection (D) = 0~ 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.            33
To find a radius corresponding to a deflection of 0~ 19' 15", we
have (5) r -   i. D
i ch = 25 feet                 log. = 1.3979400
D   = 0o 191 15t.... co. ar. *-.. sin. = 2.2518454
r   - 4464.63 feet              log = 3.6497854
We have thus ascertained the radius of a curve which will correspond to the location selected.  It now remains to ascertain the
tangent points, or points of comnmencement and end.
We have (6) representing the whole centre angle by C,
t - tan.' C. r
Thus, *.. i C = 400 00o 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 3746 - 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.
F




34                 USEFUL    FORMULE.
(1 5 ) 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 intersection, 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, Radius 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  -  ch^i
i ch =  25                  log. = 1.3979400
D   =  00 6t 0011t... co. ar.... sin. = 2.7581229
Radius = r  =  14323-95 feet       log. = 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~ = CI = 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, (6) t = tan. 2 C. r
C =   5~ 00 001l   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 suppose T to bear 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 -1-  20 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 e d; and representing by C1 the centre angle corresponding to the arc T 1, and by C2 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



[FIG. 5.]
A
Y




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    Ca      - 
cos. C
and we have --    -   r -  a, 1 =  etc. (16) corresponding with the
centre angle. Now, substituting C1 and C2 for C, as explained
above, we have
Ci - 2~ 00, 00Ot........ tan. = 8.5430838
r   = 14323-95 feet...... log. = 4.1560629
TI to d and T to a =  500-20 feet........ log. = 2.6991467
Ci - 20~ 00 Ow 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,        C2 = 4~ 00o 00t"........ tan. = 8.8446437
r   -........ log. = 4.1560629
T' to c and T to b =  1001.65 feet....... log. = 3.0007066
C2 = 4~ 00' 00t1 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 b 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 d, in the
direction T' A; and then, with the theodolite at a, and pointing to
T, lay off the complement angle of C1, and measure 8'74 feet to 1,
for a point in the curve corresponding to station 540 00 +   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 C1, and measure in
the direction indicated 8*74 feet to 4, for a point in the curve corresponding 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 b, and
pointing at T, lay off the complement angle of C2, and measure in
the direction indicated 24-98 feet to 2, for a point in the curve corresponding 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 C2, and measure in the direction indicated 24 -98
to 3, for a point in the curve corresponding to station (540 00 +
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.
(1 ) We have now completed all we contemplated respecting
the investigations of the practical operations of laying out simple




REVERSE     CURVES.                  39
railroad curves. It was not our purpose to pursue these investigations 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 railroad 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 following 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 extension.
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.




[FIaG. 6.]
r^   -      -..  -...........- - ~..................-..........-...........-......I
II^-  ^                     A_
pi
C-                                                         ID~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~I




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 intersect 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 R, 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' R
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 s, the cosine by
c, the radius by R, then will
c; s: R: tan. T' =-                     (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 trigonometrical functions, we have in the triangle T T' R to find
i T T' which we represent by a.
Sin. T: s::: T T'; wherefore T   =  a =snT
2           sin. T. 2
or, Cos. T'::: R: T T';   "         T' -  a =      -
2           cos. T.
G




42                   USEFUL     FORMUL E.
Then, in the triangle T C X we have
Sin. C: a:: cos. T'     r      s= i C T,  -     2  (18)
sin. C     sin. C. 2
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 identical 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' R =   c    1260 feet
R T      s      150  "
Then, by (17)           Tan. T'      -
Wherefore s =  150 feet..... log. = 2-1760913
c = 1260 " co. ar........ log. = 6.8996295
T' = 6~ 47t 20l.31........ tan. = 9-0757208
C being equal to 2 T, we have (18) r -   in c. 
Wherefore T! =  6~ 47/ 20f.l31
2
C = 13~ 341 40tt.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' R, representing
T' R as in our former proposition by c; and then, solving the triangle C C' D, we have by trigonometry,::
2 r: R:: (2 r -       ): sin. C'  (2 r s)      (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 OP COMPUTATION.
By formula (19)
2 r     = 5367 co. ar......... log. = 6.2702685
(2 r - s) = 5217........ log. = 3.7174298
C       =- 760 25' 1911.50........ sin. = 9.9876893
By formula (20)
C/      = 76~ 251 19i.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     FORMULiE.
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 engineers 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 following 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 distance 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 somewhat dangerous to run a train over a curve in proper repair or condition 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, regular 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 frequently 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 produces 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.]
c ~ ~/ 
9'~~




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 considerable, 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 running 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.




48                  USEFUL    FORMULAE.
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 circumstances, where reversed curves are required, of short radius, it
becomes an object of importance for the curves to occupy the whole
line, that is, there should be 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 b, 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 af 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 f, 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 e, 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 correspond 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 permit 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
0 01 of a foot wider than the flanges, up to 0-06 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      FORMUL E.
INVESTIGATION OF FORMULA, AND EXAMPLES OF COMPUTATION.
Let r =-   radius of the curve; h the width between the rails of
the track;
A/, A2,   As, etc., the differences of distance between the tracks
and the flanges of the wheels;
C1,   2, C3, etc., the corresponding angles of deflection.
Then, putting    r    1000 feet; h =    4-7 feet; Ai,    A2, A,,
etc. =  0 01, 0-02, 0 03 feet, etc., we have
r +    h   R:: r +      h -  A1, etc.: cos. e      r+  h-    etc.
r+~n
Thus,  r + - h        = 1002.35 co. ar........ log. = 6.9989806
r + " h - AI = 1002.34........ log. = 3.0010150
C1             = 0~ 15/ 261........ cos. = 9.9999956
r + I h        = 1002.35 co. ar......... log. = 6.9989806
r +   h A - A2 = 100233........ log. = 3.0010107
C2                00 211 44/........ cos. = 9.9999913
r + i h        = 1002.35 co. ar. *~..... log. = 6.9989806
r +   h - A3 = 1002.32........ log. = 3.0010064
C3             = 00 26/ 341/........ cos. = 9.9999870
r + jA h       = 1002.35 co. ar......... log. = 6.9989806
r + 4 h - A4 = 1002.31........ log. = 3.0010020
c4             = 00 30' 4511........  cos. = 9.9999826
r + 2 h        = 1002.35 co. ar....... log. = 6.9989806
r +   h h- A5 = 1002.30........ log. = 3.0009977
C5             = 00 341 201........ cos. = 9.9999783
r +   hA       = 1002.35 co. ar......... log. = 6.9989806
r + j h - As = 1002.29........ log. = 3.0009934
C6             = 0~ 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 zigzag course which the wide guage allows the cars to take, than
would be if that motion was prevented.
(2 6) 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 importance 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 contour of the surface, or some other consideration or governing principle, 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.)




[FIG. 8.]
I~          T
/i     i: Ad0F)
^ ---- ~~-~I   ---
Jfc^-^^^^^MW^^'^-'^^./,v 
\ e^ 




ELEMENTS      OF  REVERSED     CURVES.          53
Then, by putting T' X'=  c', T X =f 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 - 90~):   a=   i (T- 90o)
sin. X
Sin. X'::: sin. (T'- 90~): b    =-  s- 900)
Substituting for a + b their value, viz., twice radius, and as we
have taken radius = unity, we have this equation,
Sin. (T - 90) +  sin. (T'- 900)  2
sin. X       sin. X 
Multiplying by sin. X, we have
Sin. (T -  90~) + sin. (T'- 90~) = 2 sin. X; hence
Si.       s(T90)in. X  T(T'-900)                  (21)
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  - sin.             (22)
Sin. X: 1:: sin. C': c   sin              (23)
And,         c +':m::     1:r=   c                (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 natural sines, and for the purpose of showing how readily they can be




54                     USEFUL       FORMULAE.
obtained from logarithmic sines, we have, in the specimens of computation given below, obtained the natural sine from the logarithmic 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 information 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' - 900 = 740 06' 17" log. sin. = 9.9830685 nat.sin. =  0.9617640
Nat. sin. (T - 90~) + nat. sin. (TI - 900)     = 2)1.9610812
X         = 78~ 40' 421! log. sin. = 9 9944656 nat. sin. =  0.9805406
We have found X = 78~ 40' 42t!          Again,     X = 78~ 40' 42/1
and (T - 90~)   = 87~ 52' 58"t          and (T — 900) = 74~ 061 1711
Wherefore C must = 13~ 261 20r.'. C must be  = 27~ 13! 01"1
Proof       180~ 001 00"1               Proof     180~ 00, 001
By (22) we now have X = 78~ 401 42't co. ar........ sin. = 0.0085344
C = 13~ 26t 20"I........ sin. = 9.3662513
c =   0.2370204........ log. = 9.3747857
By (23) = again    X = 78~ 40! 4211 co. ar......... sin. = 0.0085344
C' = 27~ 131 011O........ sin. = 9.6602591
c/ = 0.4664375........ log. = 9.6687935
By (24)       c + c' = 0.7034579 co. ar........ log. = 0.1527619
m      = 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 described, 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 + R D
b     "    "  E T    C' R
r - the given radius
x = the radius required
"    -= the line R D
we have a2 + (   x)2   (r + x)2
Expanding the equation,  a2 + b2 + 2 bx + x2 - r2 + 2 rx - x2
Subtracting x2 leaves   a2 + b2 + 2 bx     = r2 + 2 rx
Trans. and changing signs a2  b2 - r2      = 2 rx - 2 bx
Dividing by 2            a2+ b-r2          =  (r- b) x
and by (r - b)         -a2b2r2             =            (25)
2 (r —b)




[FIG. 9.1
A                       E
R'/        0.   ~        2;'....'-.-.. —-. —  ~-... —...-..~-.....-..... — —...




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 = (1800 - 1570) = 23~ 00' 00"  D  - 770 30' co. ar. sin. = 0'0104185
T' = (169-30 - 900) = 790 30 00"  m  = 630      log. = 2-7993405
D = Supplement  =  770 30' 00"  T' = 790 80'    sin. = 9-9926661
Proof         = 180~ 00' 00"  T D = 634 4905 feet log. = 2-8024251
Sin.D                       log. = 2.8097590
T          =    23~ 001      sin. = 9.5918780
T D        -    252.1373        = 2.4016370
r          =    555.0000
In triangle CD R, r - TI D = C D =   302.8627    log. = 2.4812458
D          =     77.30       sin. = 9*9895815
b          =    295.6837     log. = 2.4708273
D          -    77~ 301      cot. = 9.3457552
ct         =     65-5515     log. =  1.8165825
From above          T D         -    634.4905
(T D + cl) =        a          =    700.0420     log. = 2.8451241
2
a2         = 490058.8 feet  log. = 5.6902482
b          =    295.6837    log. = 2.4708273
29
b2         =  87428.82 feet  log. = 4.9416546
r          =    555.00      log. = 2.7442930
2
r2= 308025.00 feet        log. = 5.4885860
I




a:
/'OT'oiD




VARIETIES     OF   REVERSED      CURVES.         59
a2 + b2 - r      269462.62    log. = 5.4304985
2 (r - b)    =   518.6326 co. ar. log. = 7.2851402
x            =   519.5636 feet  log. - 2.7156387
b            =   295.6837
x + b = E C  =   815.2473* co. ar. log. = 7.0887106
a            -   700.0420     log. = 2.8451241
C            - 40~ 391 0811.09  tan. = 9.9338347
TI co- T     = 12~ 30' 0011.00
Cl           = 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 however 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
be 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 tangent 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 + b, we have to
find the angle C - --  = tan. C.
a? 3- 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 successively solved.
Commencing with triangle No. 1, we have
The angle at T = 180~- Tm
"    "     T'   T'm — 900
"    "   " D    the supplement of the above.
Then, by analogy, Sin. D: m:: sin. T: t =      miT   (26)
And,             Sin. D: m     sin. T': t=  msin.T'    (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' + 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 e, 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. 1 F: r'    tan. i F. e
tan.  Ftan. t. (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: r':: cos. E: cl=  cot. E. r'     (31)
where cl represents the line E T.
Then, will (c +- c) - e = 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 (c + c') and t, viz., if t
represent the larger number, the tangent point will be in the direction of D; if the smaller, in the direction of B.
EXAMPLE OF COMPUTATION.
Suppose m = 630 feet; r - 555 feet; T'm = 169~ 30'; Tm =
157~ 00'; C' = 27~ 49' 27"-35.
Then,    T = (180~- Tm)       180     - 157~ 0  23~ 00'
T'   (T'm - 90~)     1690 30' -  90    79~ 30'
D = supplement                         77~ 30'
By formula (26) D  = 77~ 301    co. ar. sin. = 0.0104185
m    = 630 feet         log. = 2.7993405
T    = 23~              sin. = 9.5918780
t    = 252.1373         log. = 2.4016370




62                    USEFUL      FORMULE.
(27) D    = 77~ 30r        co. ar. sin. - 0.0104185
m     = 630 feet             log. = 2.7993405
TI    -- 79~ 30              sin. = 9.9926661
tl    = 634.4905 feet       log. = 2.8024251
(28) E    = 49~ 401 321".65 co. ar. sin. = 0.1178203
555 - 252.1373 = r - t = 302.8627 feet       log. = 2.4812458
D      = 102~ 301            sin. = 9.9895815
d      = 387.836 feet        log. = 2.5886476
(29) E     -49~ 40' 321".65 co. ar. sin. = 0.1178203
r - i = 302.8627 feet        log.   2.4812458
Cl    - 27~ 491 271".35      sin. = 9.6690948
ct    = 185.4219 feet        log. = 2.2681609
(30) j F   - 69~ 501 1611.32      tan. = 0.4351232
E     - 49~ 40' 3211.65      tan. = 0.0711997
r- d = 167.164 feet          log. = 2.2231428
r     =_ 536.3715 feet       log. = 2.7294657
(31) E    - 49~ 40t 3211.65      cot. = 9.9288003
rf     -                     log. = 2.7294657
cl     = 455.267 feet        log. = 2.6582660
C1       185-422
c- + cl = 640.689
t      -= 634,490
(c + c') c t- =  6.199feet
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
By formula (26) D    = 77~ 30'      co.ar. sin. = 0.0104185
m     = 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
T1    = 79~ 30'             sin. = 9.9926661
it    = 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~ 301            sin. -  9.9895815
d     = 389.7356 feet       log. = 2.5907701
(29) E    = 49~ 29' 51t".9 co. ar. sin. = 0.1199428
r - t = 302.8627 feet       log. = 2.4812458
C     = - 28~ 09' 08I.-1   sin. = 9.6737728
c/    = 188.3482 feet       log. = 2.2749614
(30)     ~ F  -   69~ 401 25tt.95 tan. = 0.4312942
E    -   49~ 201 51"i.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"l.9  cot. = 9.9338347
c    = 446.1423 feet   log. = 2.6494734
cl   = 188.3482
c/ + 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 but one preceding, and is intended
as a test to both.




(L
a.\1
[I T'* Ij]




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 surface, 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 computations 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 = 327 424 feet southing, 196'963 feet easting.
The position of second tangent point, 64 735 feet northing, 509 *235
feet westing.
K




66                 USEFUL    FORMULAE.
The bearing of tangent line from second tangent point, N. W.
61~ 11' 53".
The bearing of radius from second tangent point, N. 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 T1, and the second tangent point by T2, 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 difference in the northings as a cosine, and the difference of westings as
a sine; then, representing these functions, viz., radius by R, the
sine by s, the cosine by c, the bearing by B,
We have   c: s:: R: tan. B  -- 
C
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'=   westins
or, Cos. B: northing  R   C C' -northin
~~~~~n ~ ~ ~ -  COS. ——'B ~




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 situation 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 "           " 515072 " 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~ 50o 501t.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 + r = 899.823
C C/                 = 901.435
Difference                1.612 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      FORMULAE.
Representing C T' +    C' T" by r +   r', and C C' by H, we have
H: R:: r+  r: sin. C or sin. C'         =     r+-H
Sin. C: r+   r':: cos. C   D C =    D' C' =  T' T" =   cot. C (r +r').. r + r = 899.823............ log. = 2.9541571
H      = 901.435 co. ar............. log. = 7.0450655
C      = 86~ 34! 2211........... sin. = 9.9992226
And   C      = 86~ 34' 221............ cot. = 8.7772380
r + r = 899.823............ log. = 2.9541571
TI 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 = 86~ 34' 22", the complement to which will be 3~ 25' 38".
Then, the bearing from C to Ct.....................= N. W. 34~ 50' 50/1.24
It is obvious that if we subtract the complement DI C C/ =  3~ 25! 38f"
Will leave the bearing C T' and C/ Tl.......... = N. W. 31~ 25! 12/!.24
The bearing C Ti being........................ = N. W. 4~ 15' 2511
Gives the centre angle Ti C T'...............=    27~ 09' 47/1.24
Again, C/ TIt bearing (as above).................... = N. W. 31~ 25! 121!.24
And the radius T2 C bears.......................... = N. E. 28~ 481 07"/
Gives for the centre angle T2 C' TI.........  =  60~ 13' 1911.24
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
(3 2) There will doubtless arise in practice a great variety of
cases, or conditions requiring reverse curves, many of them requiring formula entirely different from those we have been investigating, 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 investigated the subject, as unnecessarily complicating our formulae, and
of course our computations.
The first consideration in preparing the switch, is to ascertain
the smallest anount 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 Massachusetts requires a movement of about five inches, and the pattern




70                  USEFUL    FORMULAE.
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 turnout 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 correctly based) that the switch-bar, when switched, should in all cases




SWITCH-BARS.                       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 calculations 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 turnout, 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 interfere, 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 unsupported, 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.]
A               13
PT^,~~~~~~T
t." —---' //~~~~~~~~~~~r
/      a       r5IQ




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  -  or tan. Swc        (32)
Having thus obtained the switch angle, we will now put r 
radius of the turnout; a -C Q; b=-QT'; q      CA;    == the
distance between the track centres; e = A T; g = T B.
We have in the triangle A C T, to find q and e.
R: r:: cos. Sw: q = r. cos. Sw 
5 i         ~(33): r:: sin. Sw: e    r.sin. Sw 
Then will b = q- 6 + d, and
2 r: R:: (r + b): cos. C' = r b              (34)
Cos. C':(r + b):: sin. C':a = tan. C' (r + b)  (35)
We also have a - e= g; and the angle     C'- &w.
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 formulhe
given in the preceding pages.
(3 5) The frog angle next claims our attention. Representing
the distance between the rails, or in other words, the guage of the
track, by h, we have C F == r + i h; and C   - r-i h + d;
* To be strictly exact, we make use of the following analogy:
S: R: d: sin. Sw = -
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     FORMULE.
and the angle C e F = 90~ + Sw. We now have, in the triangle
C E F, r + I h: sin. (90~ + Sw):: r -  h + d: sin. F =
sin. (900 + Sw)(  (r-I3 h +d6) 6)
(r +)                        (  ))
Then, drawing at F the tangent line F M, which of course must
be at right angles to C F, it will be apparent that the frog angle
M F ED will be a complement angle to F as found above; wherefore, we have 90~ -  F = M F; and the angle at the centre C,
will be equal to 1800 - (E + F;) or, which amounts to the same,
C - M F d -    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+ I h:: sin. C: h = - sn..( i+h) (37)
sin. J (1800-c)
The chord just found will be of great convenience to the tracklayers, 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 formulae, we now proceed with an
example of computation.
We will assume r = 499 725 feet, which gives a deflection of
1~ 26' for a 25 ft. chord; h - 4-7 feet; d= 5 inches;  -= 21
feet; and 6 =  11 feet.
By formula (32) we have
d = 5 inches........... log. = 0.6989700
S = 252 inches co. ar............ log. = 7.5985995
Sw= 1~ 085 12I............. tan. = 8.2975695




FROG     ANGLES.                          75
Sw — 1~ 081 12..........cos. =c. - 9.9999145
r = 499.725............ log. = 2.6987307
q - 499.6262 feet.....   log. = 2.6986452
(33)
Again,    Sw= 1~ 081 121............ sin. = 8.2974820
r -......   log. - 2.6987307
e =9.9131........... log. = 0.9962127  J
2 r   = 999.449 co. ar......... log. = 7.0002393
r + b - 988.768.......... log. = 2.9950944
C'   = 8~ 23! 0211.7.......... cos. = 9.9953337   (34)
C'     - 8~ 231 021.7......     tan. - 9.1684391
r+     =.......... log. = 2.9950944
a      = 145.7248 feet.......... log. = 2.1635335   (35)
e       -   9.9131
(a -e) - 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! 0211.7
Sw  -   1~ 081 121"
C - Sw = C =     7~ 141 501.7
r + I h      = 502.0746 feet co. ar. log. = 7.2992318
90~ + Sw'    = 91~ 081 12!1        sin. = 9.9999145
r + d - 2 h = 497-7912 feet       log. = 2.6970473
F            = 82~ 251 3111       sin. = 9.9961936
(36)
90~ - F = frog angle = 7~ 341 2911
Sw = 1~ 08' 121/
Frog angle - Sw = C/I   = 6~ 26/ 1711; C! =
the angle at C in the triangle T C F.




76                    USEFUL      FORMULA.
j (180~- C) = 86~ 46I 51 1.5 co. ar. sin. = 0.0006858
r + - h     = 502.0746 feet        log. = 2.7007682
C/r         = 6~ 261 17fr         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 obtained, viz.,
Centre angles  C  =.......................................  8~  231  02ft.7
"      "  C  =........................................  7~  141  50/1.7
Frog angle M F  =........................................ 7~ 341 29"I.0
Chord distance from the Ihouth 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. 1 (180~ -  C): r:: sin. C:'       -.sin. ( c)    (38)
sin.I(18OO0 -C)
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. 2 (180-    C'): r:sin. C':      -  sin. (180 -C')   (39)
representing by a' in the foregoing analogy, the chord from the
reversing point to the tangent point T'.
EXAMPLE     OF  COMPUTATION., (180~- C) = 86~ 22' 3411-65 co. ar. sin. = 0.0008692
r = 499.725 feet         log. = 2.6987307
C = 7~ 141 50/!      sin. = 9.1008914
c = 63.167 feet          log. = 1.8004913   (38)




RELATION     BETWEEN     TURNOUT     AND   FROG.    77
j (180o- C') = 85~ 481 28 6.65 co. ar. sin. = 0.0011635
r = 499.725 feet     log. = 2.6987307
C/ = 8~ 231 021.7    sin. = 9.1631819
cr   72.958 feet     log. = 1.8630761  (39)
( 3 7 ) To lay down the chords c and c, 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)
-+ (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) +  1 (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.




[FIG. 13.]
c/




RELATION     BETWEEN     TURNOUT     AND   FROG.    79
Let Sw represent the switch angle.
Pr   ~"      " frog     "
C     "      " centre   "
A     "     one of the equal angles in isosceles triangle A F C.
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~ + Sw; the angle at F = 90~ - Pr; and the angle at
C = 180~ - (B + F.) Then, in the triangle A B F, representing
F by Fa, and B by B2, we have the angle A  1 (180~- C;) and
the angle Fa = (A - F,) F being =   (90~- IF) as above, and
the angle B2 = 90~ - Sw.
Having thus determined the angles, we have in the triangle
F A B, A B =    -.B-; that is, sin. Be: h - d  R: A B; then,
representing F A by ch, and A B by w, we have
Sin. F2  *w                            h- sin. B2  ch  wsin. B  h - d.sinB,  h - d
sin. F,  sin. B,  sin.  F2  sin. F2
Sin. C: ch:: sin. A: r +-  h   ch sn.   sin.. d).in. A
2     sin. C   sin. F,. sin. C
then, by subtracting 1 h we have r.
EXAMPLE OF COMPUTATION.
Let rr -  7~ 34' 29"; Sw    1~ 08' 12"; h =  4'7 feet; d
0-4166; then,
900                 900               1800
Fr= 7034'29"        Sw= 1 08' 12"   C =   60 26' 17"
900 -Fr =820 25'31"t= F  900+ Sw = 91 08' 12" = B  2)1730 33' 43"
910 08'12" = B      900          A =  860 46'51"'-5
1800- (B + F) = 626' 17" = C  Sw = 1008'12"  F =  82025' 31" 
1800 00' 00/'  90 - Sw = 880 51' 48" = B2  F, =  40 21' 20" 5




Ii I
Il  II          L
I 
II  II
T-A                  I
t~-    ~-C-C(I-1
-r-                                                             L
t.)




TURNOUTS      OUTSIDE    THE   CURVE.           81
F2      =   4~ 211 201.5 co. ar.........  sin. --  11194836
C      =   6~ 261 1711 co. ar. ~....... sin. = 0.9502822
A       -  86~ 461 5111.5........  sin. =  9.9993142
h —d    = 4.2834........ log. = 0.6317886
r+-h    = 502.19 feet....... log. = 2,7008686
h   =     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.
(3 9 ) 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 calculation, 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 r';
the radius of the side track by /'; and the line C C' by a, we have
in the small triangle C A C', 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 understood, 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
M




82                 USEFUL    FORMULA.
obliged to look up at the time, elementary works to supply the
deficiency of such papers, or to refresh his memory, where, perchance, he may be somewhat in doubt.
In the following formula we shall use the symbols by which we
represent the triangle under consideration.
Tan. X -     i t.(l8 —A) [(r+d),r; and X + -  (180- A)  C';
r' + (r + d)            2
and X - - (180~- A)= C.
And sin. C': r + d:: sin. A: a =   -. in. A  (40)
or, by way of proof, we have
Sin. C: ~:: sin. A: a   sin.A. r
sin. C
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 represent the lines of the present triangle under consideration. Representing the guage of the track by h, we have, in the triangle
C F C', the line C C' = a, obtained by (40;) the line C F = r +
2 h, which we represent by b; the line C' F = r' + i h, which we
represent by c.




TURNOUTS     OUTSIDE     THE   CURVE.          83
If we now put p = i (a + b + c) we have
Tan.  F = (         C(p-b). ))  and t(an.  C'  ( (p-a(-) 
PTa tn.              i (p- a)  2  - p.(p-;b)
andtan.   C    ((P-a)' (p-b));           (41)
and 180~- F = frog angle = F'.
Substituting C'2 for C' as obtained by formula (40,) we have
Sin. i [180~-  (C'-  C'2)]:: sin. (C'-  C'2): ch
c. sin. (C' -  C'2)
sin. j (180 - C - C'2)
or, probably, in practice, the following may be substituted with convenience, viz.,
R: 2e:: sin. (C' - C'2): ch = sin. i (C' - C'a) 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 - r'; and C T = r + 6; therefore, C C" = r +  - r'.
Representing by 6 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 represent by ch', we have
R: 2r':: sin. - (C' - C'2): h' = 2r. sin. - (C' - C'2)  (43)
Then, to find the chord M T, which we represent by eh", we have
R: 2r':: sin. i (180~-  C"): eh"= 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~ -  &  +i (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; r'= 499-725 feet; h = 47 feet; d = 0416
feet;  = 11 feet; Sw - 1~ 08' 12"; and of course A
1780 51' 48".
2) 10 08' 12"
QO 34' 06"                  (40)
r + d    = 5730 013
r        = 499 725
(r +  d) +  r' =  6229-738  co. ar.  log. 6-2055302
(r +  d) o   r' =  5230-288  log. 3-7185256
~ (1800- A) =  00 34' 06"  tan. 7-9964947
X        =  00 28' 37"/772  tan. 7-9205505
C'       =  10 02' 43".772
C        =     05/ 28"/ 228
A        = 1780 51' 48"
Proof     1800 00' 00/.000




COMPUTATION          OF    ELEMENTS.                   85
C'   = 1~ 02' 43"' 772 co. ar. sin. = 1 7388259  C = 00 05' 28'" 228 co. ar. sin. = 2 7982497
r - d = 5730 013 feet   log. = 3 7581556   499 725 feet     log. = 2 6987311
A    = 178o 51' 48"     sin. = 8 2974820                         8*2974820
a    = 6229' 647 feet   log. = 3 7944635                          3-7944628
NOTE. By a more strict computation, the second analogy gave
the same results as the first.
r       = 5729-597           r/      = 499-725
h            -= 2-350       & h        2 350
r +- f I- =5731-947=b       r'+ t i = 502-075 =         (41)
a       = 6229-641
b       = 5731-947
c       =  502-075
2)12463-663
p       = 6231-831  log. = 3-7946157
p' - a  =    2190   log. = 0'3404441
p - b   =  499-884  log. = 2'6988693
p - c   = 5729-756  log. = 3'758136S
p      co. ar. log. =  6,2053843      p     co. ar. log. =  6'2053843
p - a co. ar. log. =  9'6595559       p   b co. ar. log. =  7'3011307
p - b       log. =  2 6988693         p - c       log. =  3'7581362
p - c       log. =  3-7581362         p - a       log. =  0-3404441
2)22 3219457                          2)17-6050953
j F =  860 03' 04"'44 tal. = 11-1609728  fC' - 30 37' 53"'46 tan. =  8'8025476
2                                     2
F = 172~ 06' 08"'88                   C'    7~ 15 46/"92
70 53' 51"'12 = the frog angle.
p      co.ar. log. = 6-2053843      F.= 1720 06' 08"'88
p - c Co. ar. log. = 6-2418638      C' =  70 15' 46"'93  Proof by
p - a       log. = 0-3404441        C =   00 38' 04"'19  adding angles.
p - b       log. = 2-6988693             180 00' 00"'00
2)15-4865615
2 C = 0~ 19' 02" 09 tan. = 7-7432807
2
C   = 00 38' 04"-18




86                       USEFUL        FORMULSA.
C'         = 70 15' 46"/93
C'2        = 10 02' 43".77
2)60 13' 03".16
(C'- C'2) = 30 06' 31"/58    sin. = 8-7342531
r +   h    = 502'075 feet     log. = 2'7007686
2'             log. = 0'3010300
~h         =   54-457 feet    log. = 17360517             (42)
a               = 6229 641
r + d - 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           =i 994 109  log. 2 2-9974341
p -c            = 5235'531   log. = 3-7189608
p      co. ar. log. =  6-2051648         p      co. ar. log.  6 6-2051648
p - a Co. ar. log. =  9*2724587          p - b co. ar. log. =  70025659
p - b        log. =  2-9974341           p - c        log. =  3-7189608
p - c        log. =  3-7189608           p - a        log. =  0-7275413
2)22 -1940184                            2)17 6542328
850 25' 37" 62 tan. = 11-0970092         30 50' 32"/ 15 tan. r  8'8271164
2                                        2
C" = 1700 51' 15i 24                       C' = 70 41' 04"-30
p      co. ar. log. =t 62051648        C" 3 1700 51' 15'25
p - c co. ar. log.  6'2810392          C'     70 41' 04''-31
p - a            - 07275413            C  =   10 27' 40/t.43
p - b            = 2'9974341                 1800 00' 00" 00
2)16 2111794
00 43' 50"'21 tan. = 8-1055897
2
C' = 10 7' 40//"42
~ (1800 - C") = 40 34' 22"-37   sin. 8-9016060
r             = 499-725 feet     log. 2-6987311
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 be required can be computed in the field, as the computations
will be both short and simple.
Let us now reverse the problem by 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 be suited to, or compare with, the
angle of the frog.
(42) Without further remark we will proceed to the investigation of a formula. To render this investigation plain to the understanding, it may be necessary to become rather more particular in
describing the figure or diagram upon which it is based, than it has
heretofore been our custom. Making use of the same notation of
the preceding problem, as far as applicable, 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 + I 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 +
d, describe the arc Ss; and with a radius = r - I 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 -    + h + d; and then, from the point C"
as a centre, with a radius = r + i A, describe the arc A S; the
intersection of the arc with the little arc S s at S will be the place




PIa. 15.
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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 continued 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 symmetrical 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 by 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 C S C" we have the angle S = the frog -+ the switch
angle; and the side S C  r -  h+ d; the side S C"  r   2- h.
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 — + h) -- (r —  h -- d): (r +  h), (r -  I h + d):: tan. I [180~- (Fr + Sw)]: tan. X, and 2 X = C, the angle
sought.                                               (44)
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 FORMULAE.
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 be convenient to
represent by 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 S1 F1; and when to
denote an angle in the triangle C' S F, they will be marked thus,
C'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 )       = S2  F; andF-F — Fi;
and S — S2=S1.
Having thus obtained all the angles of both triangles, we have
Sin. S: r + I h:: sin. Ci: S  F =(r +Isn.C     (45)
Substituting c for S F, we have an equal expression, which we frequently use by way of proof to our work, viz.,
Sin.Fi: r-   h+d::sin. C     c e  (r: - + d).sin. C,  (45)
We next have sin. C'2: sin. F2 or S2  r' — the radius of
the turnout sought.                                   (46)




TURNOUT      TO   GIVEN     FROG.                91
EXAMPLE OF COMPUTATION.
We will suppose r=   5729 597 feet; Fr=    7~ 53' 51"-12; Sw
1~ 08' 12", to find the radius of turnout r'.
Fr                = 70 53' 51"12
Sw                = 10 08' 12"
Fr + Sw           = 90 02' 03"'12
2)1700 57' 56"'88
~ (1800 - [Fr + Sw])  = 850 28' 58"'44
r + i h                   5731-947
r --   h - d              5727-663
(r +  h) + (r - ~ h + d) = 11459-610  co. ar. log. = 5-9408302
(r + j h) c (r -  h + d) =  4'284         log. = 0-6318495
~ (1800 - [Fr + Sw]) =  =  850 28' 58"'44  tan. = 1-1023618
X                      =   00 16' 16"'03   tan. = 7-6750415
2                    (44)
C = 2X                 =   00 32' 32"'06
(1800 - Fr) = F        = 1720 06' 08"'88
1800 + Sw = S          = 1810 08' 12"'00
C'                     =   60 13' 07" 06
3600 00' 00" 00
1800 00' 00" 00
C'2                    =   60 13' 07"'06
2)1730 46' 52" 94
S2 = F2 = -  (1800 - C'2)  =  860 53' 26" 47  S2 = 860 53' 26"-47
F                      = 1720 06' 08"'88  S = 1810 08' 12"
F,                     = 850 12' 42"'41  Si = 940 14' 45"*.53,S- 940 14' 45"-53
2X = C,                =   00 32' 32"'06
1800 00' 00" 00
St                     -  940 14' 45"'53 co. ar. sin. = 0-0011936
r +   A h              = 5731-947 feet     log. = 3'7583021
C,                     =   00 32' 32"'06   sin. = 7-9760615
c                                           log. = 1-7355572 (45)




[FIG. 1 6.]
/ l




TURNOUT     FROM    INSIDE    OF  CURVE.         93
F1                  =  850 12' 42"'41 co. ar. sin. = 0'0015183
r - i h + d         =  5729-663 feet  log. = 3 7579775
C,                  -   00 32' 32"'06  sin. = 7'9760615
c                   -                  log. = 1-7355573
C'2                 =   6~ 13 07"'06  co. ar. sin. =  0'9652810
F2                   =  860 53' 26"/47  sin. =  9-9993601
r          -= 501-41 feet             log. = 2'7001984 (44)
We have thus found r'  501 41 feet. It was intended as a reverse
of the previous case; there we assumed r' = 499 725; the difference 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 manner 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.
Retaining our former notation as far as practicable, we have in
the triangle S C C', the line S C  = r - d; S C'  r'; angle S
switch angle to find the side C' C (which we denote by s,) and the
angles C' and C; wherefore, (r - d) + r' (r - d):r': tan. i
(180~ - S): tan. X =   tan.(180-).r- d) r and    (1800 - S)
+ X= C'; and i (180~ - S) - X = C.
Then, sin. C': r —d:: sin. S   s=    r -d;8    or,wehave
sin. C: r':: sin. S: s =   sin.                       (47)
Having found s, we have in the triangle C C' C" the side C C




94                 USEFUL    FORMULAE.
= s, as found above; the side C' C" - 2 r'; and the side C C"=
(r - 5) + r' to find the angles, which we do by formula  (41)
Having thus found the angles required, we will denote the
triangle S C C' No. 1, and represent the angles in said triangle
by Si, C1, C'; and the triangle C C' C" No. 2, and represent
the angles by C2, C 2, C" 2; and the triangle S C' M No. 3, and
represent the angles in said triangle by S3, C'3, M3, 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' 3  C' i - C' 2; and i
(180~-  C3) = S3= M3; then will
Sin. S3  r':: sin. C' 3: C=  r' sin. C's   (48)
sin. S1
In triangle C" M T =   No. 4, we have C" 4 =     2C"; and 2
(180- C"4) =    M4 = T4; then,
Sin. M4: r:: sin. C"4: Ch4 =  r sin. C,4  (49)
sin. M4
Then, putting C F C' for No. 5, we have C' F = r' +  h; C F
r- r    h; 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 F5, will =  the
frog angle. Then, to ascertain the chord S F, we have in the triangle
C' F S =No. 6, C' -- C5= C6; and - (180- C  ) = S6  F6,and
sin. Se:r' -+  h::. C'          (+) sin. C'6       (50)
sin. S6
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.
Letr   - 5729 597 feet; r' 499 725 feet; h-=4 -7 feet; d=0-416
feet; 6 =  11 feet; Sl, or switch angle, =  1~ 08' 12".
1800 00' 00"
S =-    1~ 08' 12"
r'- d     = 5729-181            2)1780 51' 48"
r    = 499-725     ( (1800- Si)= 890 25' 54"
(r - d) + r' = 6228-906  co. ar. log. =- 6 2055882
(r - d) cv r'= 5229'456   log. = 3 7184566
i (1800 - S) = 890 25' 54"  tan. -- 2'0035053
X         = 890 19' 23"   tan. = 1'9275501
C'l       = 1780 45' 17"
C1           00 06' 31"
Sl        -   10 08/ 12"
1800 00' 00"                          (47)
C,   = 1780 45' 17" co. ar. sin. = 1-6628906  C, = 00 06' 31" co. ar. sin. = 2'7222486
r - d = 5729-181 feet  log. = 3-7580926  r' = 499 725 feet  log. = 2-6987311
S1   =  10 08' 12"    sin. = 8'2974820                sin. = 8'2974820
81   = 5229-532      log. = 3'7184617                log, = 3 7184617
To find the elements of triangle No, 2, we have
a = r - 6 + r' = 6218'322
b = s        = 5229'532
c = 2 r'     = 999'450
2)12447.304
p            = 6223'652  log. = 317940453
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                     2)17 6532631
850 25' 50"'22 tan. = 11-0973431   30 50/ 16" 79 tan. = 8-8266315
2                                 2
C'2 = 170~ 51' 40"'44              C"/ = 70 40( 33"1158




96                         USEFUL FORMULA.
p        co. ar.  log. = 6*2059547
p - c    co. ar.  log. = 6-2819800
p - a             log. = 0-7267272
p - b             log.   2'9974388
2)16 2121007
00 43' 53"  tan. = 8-1060503
2
C2      10 27' 46"
C'a               = 1700 51' 40"-45
C"2               =   70 40' 33"'58
Ca                =   10 27' 46"-00
Proof        1800 00' 00"'00
In triangle No. 3, we have
C'1.              = 1780 45' 17"
C'2               = 1700 51' 40"'44
C'3               =   70 53' 36"'56
2 )1720 06' 23"-44
~ (180 - C'a) = Ma =  86~ 03' 11" 72     co. ar. sin. = 0-0010312
7'                = 499 725 feet                log. = 2-6987311
C'3               =   70 53' 36"'56             sin. = 9-1377717
ch,               =   68-791 feet               log. = 1-8375340
In triangle No. 4, we have
C"2 = C"4          =   70 40' 33"-58
2)1720 19' 26"'42
1 (1800- C"4) = M4 = 860 09' 43".21      co. ar. sin. = 0-0009751
r/                 = 499-725 feet               log. = 2-6987311
C4                 =   70 40' 33"-58            sin. = 9 1257121
ch4                -   66-899 feet              log. = 1-8254183
In triangle No. 5, we have C F =                 r-        h; C C"= s
5229 565; C' F =         r' +  I h, to find the angles.       Let
a = r -   h   = 5727-247
b = r' + I h  =  502-075
c = s,        = 5229-565
2)11458.887
p             = 5729-443                 log. = 3-7581123
p - a          =    2-196                 log. = 0-3416323
p - b         = 5227-368                  log. = 3-7182831
p-c            =  499-878                 log. = 2-6988641




TURNOUT FROM                  INSIDE         OF CURVE.                   97
p       co. ar. log. = 6-2418877              p      co. ar. log. = 6 2418877
p - a co. ar. log. = 9'6583677                p - b co. ar. log. = 6'2817169
p - b          log. = 3-7182831               p - c          log. = 2-6988641
p - c          log. = 2-6988641               p - a          log. = 0-3416323
2)22-3174026                                 2)15-5641010
860 01' 50"'13   tan. = 11-1587013             00 20' 48" 73   tan. = 7.7820505
2                                             2
C'5 =  172~ 03' 40"'26                          C, = 0o 41' 37"'46
Cs   1720 03' 40" 26
F,    70 14' 42" 27
1800 00' 00" 00
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,   = 7~ 14' 42"'26 Frog angle.
In triangle No. 6, we have
C'                   -=    1780 45' 17"
C'5                   =    1720 05' 24" 65
C'6                          60 39' 52" 35
2)1730 20' 07"-65
S6 =    (180o- C'6)    =    860 40' 03"'82
S6       = 860 40' 03" 82           co. ar.  sin. = 0'0007349
r' + i h = 502-075 feet                     log. = 2-7007686
C6       =   60 39' 52"-35                   sin. = 9'0646679
ch6      =   58-233 feet                     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.
0




[Pia. 17.1
S ~~ -~ —. —....-   L^^




TURNOUT     FROM   INSIDE    OF CURVE.         99
(4 5) 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   0 416 feet; and the frog angle  r-= 7~ 14' 42".27; the switch
angle, Sw  - 1~ 08' 12"; to find the radius of the turnout =-r', 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 ff, 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 + 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 ff 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., F C C" 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 now proceed to find the angle S C F. In the
triangle F C C" we have the angle at F = the frog angle + the
switch angle; the side F C = r-  ht; the side F C" = r + (- h




100                 USEFUL     FORMULE.
- d;) then, (r +   a h - d) + (r -  I h): (r+ I h - d) c (r -
h):: tan. 2 [180- (Fr + Sw)]: tan. X; that is,
Tan. X     (r + + h h-d) (r - h ). tan. j [180 - (Fr+Sw)]  (5l)
(r + j h - d) + (r - j h)
And 2 X = F C S, (51,) which we shall hereafter represent by C2.
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 = C2, as found above; the line C F
= r -- 1 h; the line C S - r + -  h - d, to find the angles at F
and S, which we shall denote thus, by F2 and S2; then, (r + -  h
-d) + (r-      ): (r +  I  - d)    (r-      ): tan. I (180
-C2): tan. X2 and 1 (180-    C2) +   X2=F2, and      I (180 -
C2) -  X2= S2                                             (52)
Then, sin. F2: r +    h -d:: sin. C2:S F, which we shall
denote by ch, or sin. S2: r -  A h: sin. C2: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 S3 =  (S2+  Sw;) the angle F3 
(F2- - Fr) =  (S2 +  Sw,) and the angle C' 3 = 180- [(F2 -  Fr)
+ (82+ Sw;)] and C'3: ch:: F: r' + - h               (54)
Then, by subtracting I h, we have r'.
Having thus obtained our formula, we now give an example of
calculation.




TURNOUT FROM                INSIDE       OF CURVE.                 101
To find r' from    the frog angle,
Fr                     =     7~ 14' 42"/27
Sw                     =   - 1 08' 12"'
Fr + S                 =     8~ 22' 54"/ 27
2)171~ 37' 05"'73
~ [1800 - (Fr + Sw)]   =    85~ 48' 32" 86
r +   A h-d        =   5731-531
r - j h            =   5727-247
Sum                = 11458-778      co. ar.  log. = 5-9408619
Difference               4-284               log. = 0-6318495
i [1800- (Fr + Sw)] = 85~ 48' 32"'46         tan.,= 1*1350427
X                  _= 00 17' 32"/ 38         tan. = 7-7077541
2
C                  =-  00 35' 04"'76
2) 179~ 24' 55"'24
i (1800- C2)       = L890 42' 27"'62
r+ — h-d           =  5731-531
r - _h             =  5727-247
11458-778      co. ar. log. = 5-9408619
4-284               log. - 0 6318495
1 (1800- C2)       = 890 42' 27" 62          tan. = 2-2922459
X,- =                  40 11' 27"/14         tan. = 88649573
F2                 =  930 53' 54/" 76
qS                 =  850 31' 00"48
C2                 =  000 35' 04"/ 76
1800 00' 00"'00
PF,      = 930 53' 54"/76 co. ar. sin. = 0-0010062  S  =- 850 31' 00"'48 co. ar. sin. = 0-0013309
r +h- d= 5731-531 feet      log. = 3-7582707  r -  h = 5727-247 feet  log. = 3-7579458
C2       =  00 35' 04"76    sin. = 8-0087699                          sin. = 8-0087699
ch2      =   58-620 feet    log. = 1-7680468                          log. = 1-7680468
In triangle No. 3, we have
F2               =  930 53' 54"/76         S2  =  850 31' 00"48
Fr               =   70 14' 42" 27         Sw  =   10 08' 12"
F3               =  860 39' 12"/49         S3  =  860 39' 12"/48
S                =  860 39' 12" 48
173  18' 24" 97
1800 - (Ff + Ss) =   60 41' 35"/ 03




[FIo. 18.]
C
c
___ i__ _*z!__^^-=




TURNOUT     FROM   INSIDE    OF  CURVE.        103
By the problem, F3 and S should be equal. Then,
1800- (F3 + S,) =  60 41' 35'03 co. ar. sin. = 0-9334859
ch2         =                 log. = 1'7680467
F3          =  860 39' 12"'48   sin. =  9-9992587
r' +  i   h  = 502-101        log. =  2-7007913
h          =  2 350
r'          = 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   FORMULE.
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. Representing the mouth of the switch by S; the
length of the switch by s, and the switch angle by Sw; the centre
of the table by 0; 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 r'; 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 Sw C, which line we represent by q,
R: r' + d:: cos. Sw: q =(r' + d). cos. w  (55)
And in the triangle C O T, (by assuming C ( as radius,) we have
this analogy:': T (:: (C 0 =radius,): tan. C2= T —     (56)
And therefore, cos. C2: r':: R: C O =        (57)
Representing the distance of the centre of the turning table
from the main track (viz., Q 0) by n, and the line C ( byp, we
then draw the line 0 A, parallel with the centre line of main track,
and in the triangle A 0 C we have
p:R:: q —n:cos. C         q- I-        (58)
If we now deduct from the angle C, the angles Sw and C2, we
shall have left the angle C of the triangle S C T, which we represent by C3, then will I (180 - C3) = S = T; and
Sin. T: r':: sin. C3  S T =  r'n. 3      (59)
sin.T




TURNING     TABLE     AND   MAIN    TRACK.        105
To give an example of calculation, we will assume n =   36'4
feet; r' =  499-725 feet; s =  21 feet; T O =60 feet; d= 0 416
feet; Sw=   1 08' 12". Then,
r' +- d   = 500'141          log. = 2'6990925
Sw        = 10 08' 12"       cos. = 9-9999145
q         -= 500 043         log. = 2-6990070
T 0       =  60 feet        log. = 1-7781513
r         -= 499-725   co. ar. log. = 7-3012689
Ca        = 60 50/ 47"'41   tan. = 9-0794202
C,          60 50' 47" 41 co. ar. cos. = 0-0031080
r'        =  499-725         log. =  2-6987311
p         = 503-314         log. = 2-7018391
Sw = 10 08' 12"  q - n  = 463-643          log. = 2-6661837
C2 = 60 50' 47"-41 C     = 220 54' 02"/78  cos. = 9-9643446
70 58' 59"1141 (Sto + C2)  = 70 58' 59"-41
C3        = 140 55' 03"'37
2)1650 04' 56"-63
T = i (1800 — C) = 820 32' 28"'31 co. ar. sin. = 0-0036904
r'        = 499-725         log. = 2-6987311
C,        = 140 551 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 C2 the complement of C will leave
the angle A 0 T. Having thus obtained A 0 T, we place our
instrument at 0 and lay off said angle from A, and measure the
P




106               USEFUL FORMULA.
distance 0 T to T; then, moving the instrument to T, and pointing it to 0, we lay off the angle 0 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 been correctly prepared, we shall be
the distance d from the centre line of the main track, upon the
side towards the turnout curve. The curve may now be 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 convenient 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 S - the distance S Q, (as measured,) and a = S, we
have, by taking a as radius, the following analogy:: n-d:: (rad.): tan. S=   nn-d         (60)
wherein S will be equal to the angle at S in the triangle Q S 0;
then,           cos.::: R: a                      (61)
then, representing the angle S in the triangle C S O by S2, we
have 90~ + Sw - S = S2, and
r' + a: r' c a:: tan. - (180~ - S2): tan. X   (  a). tan. (180- S,)
2                           r/+V h'~l ~ V~ZII1~ 1L   Tt -4




TURNING     TABLE   AND    MAIN   TRACK.       107
and X + i (180~ - S2) =the angle 0; and X -    (180 - S2)
the angle C                                          (62)
We then have sin. 0: r':: sin. S2: C (); or, sin. C: a:: sin.
S2: C (; then, putting p  C (0, we have
p: R:: r': cos. C2= r                (63)
And cos. C2 r':: sin. C2: T O = tan. C2 r'    (64)
Then, deducting C2 from C leaves C3 = the angle C in the triangle
S C T; and 2 (180~ - C3) == the angle S = the angle T; and
Sin. T: r':: sin. C3: S T -  sin. C. r      (65)
To give an example of calculation, we assume n = 36*4; r'
499-725; d = 0416; Sw = 1~ 08' 12"; 6 - 200 feet.
6       =  200 feet  co. ar. log. =  7-6989700  900 00' 00"'00
n -  d  =  35-984         log. =  1-5561094      1~ 08' 12"
S       = 100 11' 58" 32  tan. = 9'2550794 90~ + Sw = 910 08' 12"
S      - 100 11' 58"/ 32
S       =            co. ar. cos. = 0-0069179 S2  = 80~ 56' 13"'68
a       = 200 feet       log. = 283010300     2(990 03' 46"'32
a       = 203'211 feet   log. = 2-3079479 i (1800- S) - 490 31' 53"-16
(60)
r'      =  499 725 feet
a       = 203-211
Sum     = 702-936    co. ar. log. = 7-1530842
Difference = 296-514      log. = 2'4720452
I(1800 - S2) =  490 31' 53"'16  tan. =  0'0689836
X       =  260 18' 34"'57  tan. =  9'6941130
0       = 750 50' 27"'73
C       = 230 13' 18"'59
Sa      = 800 56' 13"'68
1800 00' 00'00




108                USEFUL    FORMULAE.
0   =  750 50' 27" 73 co. ar. sin. =  0 0133981  C  =  230 13' 18"'59 co. ar. sin. =  0-4041819
r'   =  499 725 feet  log. =  2-6987311  a  =  203-211 feet  log. =  2 3079472
S2  = 80~ 56' 13"'68  sin. = 9'9945442           sin. = 9'9945442
p   =               log. = 2-7066734             log. = 2-7066733
r'  =               log. = 2-6987311
Ca  = 10~ 55' 27" 50  cos. = 9 9920577  C = 230 13' 18" 59  (63)
Ca = 100 55' 27"/50
C2  = 10~ 55' 27" 50  tan. = 9-2855789  C3 = 120 17' 51" 09
r' -=              log. = 2-6987311    2)1670 42' 08" 91
T   = 96-452 feet  log. = 19843100 - (1800-Ca)= 830 51' 04"'45  (64)
T   =  830 51' 04"'45 co. ar. sin. =  0'0025057
r'  =  499 725 feet  log. =  2-6987311
C3  =  120 17' 51"-09  sin. =  9-3283555
S T = 107 051 feet  log. = 2 0295923                    (65)
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,) C2 = 10~ 55' 27"-5, the complement to which = 79~ 04' 32" 5 =S the angle C 0 T; we have also
found, formula (62,) 0 = 75~ 50' 27"-73 = the angle C 0 S.
We now place our instrument at 0, and lay off from S the difference 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 instrument to T, and lay off the angle S T 0   - - (1800  C3) + 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 0'416 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
(4 9 ) 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 complicated 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.




[Fia. 10.]
A..............- -cc




TRACK    OVER    WATER.                 ll
side of the river, marked 0 on the diagram. Then, by considering
the tangent line from 0 to A to bear due east, (whatever may be its
direction,) we measure the angle A 0 1, by which we determine the
line from 0 to 1 to bear S.E. 75~ 08' 35" 37, and by measurement
we find the distance 152*5 feet. We then place a signal at station
2, and by a triangulation we ascertain the bearing from station 1
to 2 to be 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 bearing of this
line we determined to be S. 72~ 59' 36"-41 E., and its length =
1604 *264 feet; then, removing to station 3, we ascertain the bearing of the tangent line to be S. 60~ E., or (which is the same,)
N. 600 W.
Having thus connected the straight or tangent lines, by a traverse
running through the point selected as the most suitable for the
location of the curve, our next step will be to prepare our data to
ascertain its radius.
Upon examination of the foregoing, we find the courses and distances noted in the following table, viz.,
BEARINGS.   DISTANCES.
o 0   II
Station 0 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. = 1604.264 
and the bearing of the tangent lines from 0 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 westings, according to the requirements of the case.




112                USEFUL    FORMULE.
Representing 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 departures, which we represent by D. Then, putting R for the radius
of the tables, B for the bearings, and d for the distance, we have
R: d:: sin. B: D-sin. B. d;
and R: 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 = 391017 log. = 1-5921960
No. 2. B =  S 850 02' 37" E  sin. =  9'9983730  cos. =  8-9365008
d =  350 feet   log. = 2-5440680       log. = 25440680
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. = 32052779
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 distances, with the southings and eastings belonging to each; and
having summed them up, we proceed to compute the bearing and
distance from 0 to station 3. Thus, by making use of the symbols
of the preceding formula, with the addition of 6, by which we represent the distance from station 0 to 3, we have
Tan. B   —; and sin. B: D::: 6   -   -in;
or, Cos. B: L:: R:     -  -                       (N)




BEARINGS      OF   STATIONS.                113
BEARINGS.     DISTANCES.  SOUTHINGS.  EASTINGS.
Station 0 to 1 S 75~ 08 35!I.07 E  152.5 feet  39.102    147.401
"   1 to 2 S 85~ 021 37. 00 E  350.0  "   30.239      348.691
"  2 to 3 S 72~ 591 36/.41 E  1604.264 "  469.219     1534.119
L - 538.560 D   2030.211
D = sum of eastings  = 2030'211          log. = 3-3075411
L = sum of southings  = 538 560          log. = 2-7312341
Bearing from station 0 to 3 = B = S 750 08' 35" 37 E  tan. = 0'5763070
B = 750 08' 35"'37 sin. = 9-9852331      cos. = 9-4089261
D = 2030'211   log. = 3-3075411  L = 538 560 log. = 2-7312341
= 2100-429 feet log. = 3-3223080  Proof  log. = 3-3223080
Having thus obtained the bearing from station 0 to 3, viz.,
S 75~ 08' 35"'37 E, and distance =  2100 429 feet, our next step
will be to ascertain the distances 0 A and A 3. In the triangle
A 3 0, we have to find the several angles. The bearing from
0 to A      -  due East        )
> which gives angle at 0 -  14~ 51' 2411.63
O to3       -   S 75~ 08t 351.37E )
A to 0         due West        )
"     "   "  A = 150~ 00, 0 11.00
A to 3          S 60~ 001 001.00 E 
3 to A          N600 00 0000.OOW)
" ~    "   "   3 =  15~ 08 35U1.37
3 to       =-   N 75~ 081 3511.37 W)
Having found the angles, we have
Sin. A::: sin. 0: A3=. sin. 
sin. A;
and Sin. A: 6: sin. 3: 0A=. sin. 3  Thus,
sin, A




114                USEFUL    FORMULA.
A =  150~ 00' 00"'00 co. ar. sin. =  0'3010300  A  =  co. ar. sin. =  0 3010300
0 =  140 51' 24" 63.  sin. =  9'4089261  d  =   log. =  3;'3223080
=               log. =  3-3223080  3  =  150 08' 35t" 37  sin. =  9 4170259
A 3 = 1077'12 feet  log. = 3-0322641  A 0 = 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 0 G, the angle 0 and the
sides 0 A and 0 1, to find the side A 1 -= A G, and the angles A'
and G. Putting a for the side 0 A, and b for the side 0 1 = 0 G,
we have
a +b: a' b:: tan. I (180~ - 0): tan.  (A' c G) =
(a c  b). tan. 4 (1800 -0)
a+b b
and 2  (180 - 0) + 2 (A' c G)= G;
and I (180~ - 0) -   (A' cG)     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:: sin. 0 m''in 
sin.  l'
or, Sin. G: a:: sin. 0: m=  a.sin.0          (P
Having obtained m, and putting G' for the angle G in the
triangle A C G, we have
Sin.A::: R:AC=      i;
sin. 4A
then, representing A C by n, we have
1: sin. ( A -A'):: n: sin. G' == n. sin. (A A  A')  (Q)
And sin. (I A- A' + G'): m:: sin. (i A- A'): r
m. sin. (0 A - A')              - a
sin. ( A - A' + G')              \ 




BEARINGS      OF  STATIONS.                115
We now introduce an example of computation [formulae (O)
and (P.)]
a        = 1097'4 feet                            180o 00' 00" 00
b        =  1525 "                     0        = 140 51' 24"'63
a + b    = 1249-9 co. ar. log. = 6 9031247      2)1650 05 35/" 37
a r b    =  944-9     log. = 2-9753858  i (180~ - 0) = 82~ 34' 17"'68
(1800 - 0) = 820 34' 17"'68 tan. = 0'8847831
b (A' co G) = 80~ 12' 52".64 tan. = 0-7632936
G        = 1620 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
0        = 140 51' 24"'63     0 = 140 51' 24" 63  sin. = 9-4089261
Proof    = 1800 00' 00"'00    m = 950'8 feet     log. = 2'9780902
By formulae (Q) and (R) we have
3 A       = 750 00' 00" 00  co. ar.  sin. = 0-0150562 = n
i A - A'  = 720 38' 34" 96       sin. = 9-9797599
Ambiguous G' - 810 09' 53"/75 *  sin. = 9 9948161
True G'   = 980 50/ 06".25
i A - A'  = 720 38' 34". 96
i A - A + G'-1710 28' 41"-21  co. ar.  sin. = 0-8291893
m         = 950'8 feet           log. = 2-9780892
~ A - A'  = 720 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 =    h; hence
r   =  6124'05 feet  co. ar.  log. =  6-2129616
i ch =  25   "            log. = 1-3979400
D    = 00 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    FORMULA.
The deflection thus found being an awkward sum to add or subtract in the field, we may assume one more convenient without
materially changing the location of the curve; we therefore assume
0~ 14' as the measure of a deflection; then, by formula (5) we'   ch
have r = - si.; and, for the purpose of ascertaining by what
amount this change in the length of the radius will affect the location 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. 1 C =tan. 1 C. tan.  C. r-   tan..tan. C  ch
THE VALUE OP b, COMPUTED FROM THE RADIUS THE VALUE OF b, COMPUTED TO CORRESPOND TO
ALREADY OBTAINED.      A RADIUS BASED UPON A DEFLECTION OF 00 14'.
* C = 150 00' 00"  tan. = 9-4280525  D  = 0O 14' 00" co. ar. sin. = 2'3901470
* C = 70 30' 00"  tan. = 9 1194291  i C = 150 00' 00"  tan. = 9'4280525
r   = 6124-05    log. = 3 7870384  i C = 70 30' 00"  tan. = 9-1194291
b,= 216-033 feet  log. = 2 3345200  ~ ch = 25 feet  log. = 1'3979400
b  = 216-555 feet  log. = 2-3355686
b   =.......               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 proceed to determine a radius which shall correspond with the desired
deflection, viz., of 0~ 14', as explained in the foregoing.  By
formula (5) we have r= -    inD  Thus,
D   = 00 14'   co. ar. sin. = 2'3901470
j ch  =  25 feet     log. =  1-3979400
r   =  61386853 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 0 and 3, which have been already determined. By
formula (6) we have t = tan. i C. r. Thus,
r   =  6138-853 feet  log. =  3'7880870
f 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 = (AA-A') =750 - 20 21' 25"' 04 = 720 38' 34"'96
G'                       =  980 50' 06"*25
C = the supplement       = 80 31' 18"'79
Proof                      1800 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 G C T'. Thus,
i C               -  150 00' 00"-00
Supplementary C   = 80 31' 18"-79
Hence the C sought   60 28' 41"'21




m  [FIG.  20.]
~:~~ ~,  ~,-7~  a, 
6/,I 
\ ~II
~yA i I II B
cK
I       1~~~




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,"
C" = 23321'21   log. = 4-3677511
r =              log. = 3-7880870
r" =        co. ar. log. = 46855749
a =   694-09 feet  log. = 2-8414130
If we now consider T' as numbered 3-06, in the stations of location, 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 + 6 94,
which increases the number of the locating stations to 10. For the
purpose of avoiding fractions, we ascertain the point T' by measurement 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
A =         830 00'
B =         32~ 00'
10 =        650 00'
1800 00'




120                USEFUL    FORMULA.
Then, putting 6 for the distance between 10 and A, we have
Sin. B::: sin. 10: AB;     andsin. B: 6:: sin. A: 10 B  (S)
Thus,
B  -  320 00'  co. ar. sin. -   0-2757903
6  = 200 feet      log. = 2'3010300
10 - 650 00'       sin. = 9 9572757
A B - 342-05 feet  log. = 2'5340960
Sin. B =           log. = 25768203
Sin.-B =
A     =  830 00'    sin. =  9-9967507
10 B  =  374-61 feet  log. =  2'5735710
(5 0 ) 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 d, (which represent 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" =   r- a
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 =0~ 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 angle0~ 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 0~ 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.
R




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
a b ed. 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 a b e d; and upon
these principles we construct the following table, viz.,
RELATIVE       RELATIVE          RELATIVE
BEARINGS OF    BEARINGS OF       BEARINGS OF
PRIMITIVE STATION  b AND d           a AND C
N.E.           N.E.              N.E.
Station 10 to 10 50 angle =  281 + 21 1511 =  301 15/"
"   <" 11.00 " =    561 +  "  =    58' 151 - 4 3011 =  531 4511
"   "  11.50  " =   841 +  "  = 10 261 1511 -   = 10 211 451
"   "  12.00  " = 1121 +   "  = 10 541 15 —  "  = 10 491 4511
"  "  12.50 " = 1401   "  = 2~ 22I 151-  "  - 2~ 171 4511
"   "  13.00 " = 1681 +       = -"             = 2~ 45f 45
"'" 13.50 " = 196 = -N.E. 30 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 EASTINIGS, OF STATIONS,
FROM C, OR THE CENTRE OF THE CURVE.
NORTHINGS EASTINGS
PRIMITIVE STATIONS.  BEARINGS.   DISTANCES.
IN FEET.  IN FEET.
10-00 from C to 10  Due North  r + 0 feet =  6138-85 feet  6138-85  00-000
10-50 " " " d  N.E. = 00 30/ 15"  r-9 " =6129-85 "  6129-61  53-938
"(  "M  4" " b 6 "  " i "  r + 9  " =6147-85  "(  6147-61  54-096
11'00 " "< "  N.E. = 00 53'45"  r- 9 " =612985 "  6129 10  95-838
t"  "  " " a  c   cc "    r -+9  " =6147-85  "  6147-10  96-119
4"  "(  " " d  N.E. =  00 58 45"  r-9  " =6129-85  "  6128-97  103-860
"cc  I  " "6 b  "  "  "   r -+9  "  = 6147-85  "  6146-97  104-165
11'50 " "<" c  N.E. = 1021' 45"  r-9 " = 6129-85 "  6128-12  145-755
"  "4  " " a  "c  "  "   r r+9 " =6147-85 "  6146-11  146-183
<( (< " " d  N.E. = 1026'45"  r-9 " =6129-85 "  6127-92  153-176
s"  "  "cc" cb  it  "  "  r +r 9  "  =6147-85  "  6145-92  154-228
12-00  "  " " c  N.E.=  1049145/  r-9  " =6129-85  "  6126-73  195-662
"     " "cc  a    " "c "  r+ 9  " =6147-85  "  6144-72  196-237
t"  "4  " cd  N.E. =  10 54' 15"  r-9  "  =6129-85  "  6126-46  203-682
"  " " " b    t"  "' "   r-+ 9 " =6147-85 "  6144-45  204-280
12-50 " " " c  N.E.= 2017'45"  r-9," = 6129-85 "  6124-93  245-557
"  " " " a    "   "  "    r-+l9 " =6147-85 "  6142-92  246-177
"  "  " " d  N.E.  20 22' 15"  r-9  " =6129-85  "  6124-60  253-574
it  cc cc"  b  "  "  cc  r + 9  "  = 6147-85  "  6142-59  254'318
13-00 " " " c  N.E.= 2045'45"  r-9 " =6129-85 "  6122-81  295-435
"<  "  " c a  C(  "  "    r +9  " =6147-85  "  6140-79  296-302
13-50  "  " " 0  N.E. =  30 16' 00"  r-0  " =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 6 for d, we have, by
formula (M,)
D = sin. B 6; and L = cos. B 6.




124                              USEFUL           FORMULAE.
10-50  d   =  N.E. 00 30' 15"    sin. =  7-9444459                      cos. =  9-9999832
6  =         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
6  =         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. 0~ 53' 45"    sin. =  8-1940869                      cos. =  9-9999469
6  =         6129-85      log. =  3-7874499                      log. =  3-7874499
D  =         95838       log. =   1-9815368    L =   6129-10     log. =  3-7873968
11-00  a   =  N.E. 00 53' 45"    sin. =  8-1940869                       cos. =  9-9999469
6  =         6147-85     log. =  3-7887233                      log. =  3-7887233
D  =         96-119       log. =  1-9828102    L =   61471 0     log. =  3-7886702
11-00  d  =   N.E. 00 58' 15"    sin. =  8-2290013                      cos. =  9-9999377
6  =         6129-85      log. =  3-7874499                      log. =  3-7874499
D  =         103-86       log. =  2-0164512     L =  6128-97     log. =  37873876
11-00  b   =  N.E. 00 58' 15"    sin. =  8-2290013                      cos. =  9-9999377
6 =          6147-85     log.    37887233                        log. =  3-7887233
D  =         104-165     log. =  20177246      L =   6146-97     log. =  3-7886610
11-50  c   =  N.E. 10 21' 45"    sin. =  8-3761729                      cos. =  9.9998772
6d            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. =  83761729                       cos. =  9-9998772
j  =         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. 10 26' 15"    sin. =  8-3994397                      cos. =  9-9998633
j  =         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. =  9-9998633
6=           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. 10 49' 45"    sin. =  8'5040569                      cos. =  9-9997786
6  =         6129'85      log. =  3'7874499                      log. =  3-7874499
D  =         195-662      log. =  2-2915068    L =   6126-73     log. =  37872285
12-00  a   =  N.E. 1~ 41' 45"    sin. =  8-5040569                       cos. =  99997786
d  =         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
6 =          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. 10 54' 15"    sin. =  8-5215024                      cos. =  9-9997601
6  =         6147-85      log. =  3-7887233                      log. =  3-7887233
D  -         204-280     log. =  2-3102257     L =   6144-45     log. =  3-7884834
12-50  c   =  N.E. 20 17' 45"    sin. =  8-6027015                      cos. =  9-9996513
6  =         6129-85     log. =  3-7874499                       log. =  3-7874499
D  =         245-557      log. =  2'3901514     L =  6134-93     log. =  3-7871012
12-50  a   =  N.E. 2~ 17' 45"    sin. =  8-6027015                      cos. =  9-9996513
6  =         6147-85      log. =  3-7887233                      log. =  3-7887233
D  -         246-177      log. =  2-3914248     L =  6142'92     log. =  3-7883746
12-50  d  =   N.E. 20 22' 15"    sin. =  8-6166545                      cos. =  9-9996281
6  =         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
6 =          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. = 8-6830114       cos. = 9'9995011
6 =     6129-85  log. = 3-7874499          log. = 3-7874499
D =      295-435  log. = 2'4704613  L = 6122-81  log. = 3-7869510
13-00 a = N.E. 20 45' 45' sin. = 8-6830114       cos. = 9-9995011
6 =     6147-85  log. = 3-7887233          log. = 3-7887233
D =      296-302  log. = 2-4717347  L = 6140-79  log. - 3-7882244
13-50 station N.E. 30 16' 00" sin. = 8-7557469   cos. = 9'9992938
6 -      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 0 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. 450 = 374-61 feet  264-884  264-884
"  10 to A = S.W. 200 = 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. = 2-4230560  L = 264-884 log. = 2-4230560
S.W. 200 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.  770 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-   68'404 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, distinguishing 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     FORMULAF.
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 =  59.50914 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 easting,
we then have
AN oN bN  1:: A    bE: tan. B  A   b=
wherein B expresses the bearing sought.




BEARINGS OF STATIONS.                                              129
EXAMPLE OF COMPUTATIONS.
Station 10-50  AN c bN =       5950-914 co 6147'61 -  196-696        log. = 2-2937956
AE CO bE =    -   68'404 co   54-096 =  122-500        log. = 2-0881361
B' or bearing from A to 10-50 b  =  N.E. 310 54' 51" * tan. = 9-7943405
Station 10-50 AN co dN =       5950-914 co 6129'61  =  178-696        log. = 2'2521148
AE co dE     =   -  68-404 co  53'938 _ 122-342        log, = 2-0875756
B' or bearing from A to 10-50 d  =  N.E. 31~ 23' 49"  tan. = 9-8354608
Station 1100 AN co aN =       5950-914 co 6147-10 =   196-186        log. = 2-2926680
AE co aE =    -  68-404 co   96-119     164-523       log, = 2-2162266
Bearing from A to 11-00 a     =  N.E. 390 59' 00"     tan. =  9-C235586
Station 11'00 AN co bN =      5950-914 co 6146-97  -  196-056        log. = 2-2923801
AE co bE =    -  68-404 c,   104-165;- 172-569        log. = 2-2369628
Bearing from A to 11'00 b     =  N.E. 41o 21' 15"     tan. = 9-9445827
Station 11')0  AN co CN =      5950-914 co 6129-100 =  178-186        log. = 2-2508736
AE co CE =    -  68-404 co   95-838 =  164-242        log. = 2-2155842
Bearing from A to 11-00 c     =  N.E. 42~ 40' 29"     tan. = 9-9647106
Station 11'00 AN co dN =      5950'914 co 6128'97  =  178-056        log. = 2'2505566
AE co dE =-   -  68-404 co  103-860 =  172-264        log. =  2-2361945
Bearing from A to 11-00 d     =  N.E. 440 03' 10"     tan. =  9-9856379
Station 11-50 AN co aN =      5950-914 co 6146-11 =    195-196       log. = 2-2904709
AE c aE =    -  68-404 co  146-183 = 214-587         log. =  2-3342064
Bearing from A to 11-50 a     = N.E. 470 52' 48"      tan. =  0-0437355
Station 11-50 AN co bN =      5950.914 co 6145-920 =   195-006        log. = 2-2900480
AE co bE -    -  68-404 co  154'228 = 222'632         log. = 2-3475876
Bearing from A to 11-50 b     = N.E. 480 47' 04"      tan. =  0-0575396
* The northing of A being less than the northing of b, and the easting of A being less than b,
the bearing must of course be northeasterly.
S




130                             USEFUL           FORMULiE.
Station 11-50 AN co CN =        5950-914 co 6128-120 =   177-206        log. =  2-2484784
AE c   CE =   -   68'404 co  145-755 =   214-159        log. =  23307364
Bearing from A to 11-50 c      =  N.E. 500 23' 38"       tan. =  0'0822580
Station 11-50 AN co dN =       5950-914 co 6127-920 =    177-006        log. =  2'2479880
AE co dE =     -  68-404 co  153-176 =   221-580        log. =  2'3455306
Bearing from A to 11-50 d      =  N.E. 510 22' 56"      tan. =  0'0975426
Station 12-00 AN co aN =       5950-914 co 6144-720 =    193-806        log. =  2-2873672
AE co aE =     -  68-404 co   196-237 =  264'641         log. =  2'4226571
Bearing from A to 12-00 a      =   N.E. 530 47' 00"      tan. =  0-1352899
Station 12-00  AN co bN =      5950-914 co 6144-450 =    193-536        log. =  2-2867617
AE co bE =     -  68'404 co  204'280 =   272-684        log. =  2-4356597
Bearing from A to 12-00 b      =  N.E. 54~ 38' 06"      tan. =  01488980
Station 12'00   AN co CN =      5950-914 co 6126'730 =   175'816        log. =  2'2450584
AE o CE =     -  68'404 c. 195'662 =    264-066         log. =  2'4217125
Bearing from A to 12'00 c      =  N.E. 560 20' 39"      tan. =  0'1766541
Station 12-00  AN c   dN- =    5950-914 co 6126-460 =    175.546        log. =  2'2447125
AE co dE =     -  68-404 co   203-682 =  272'086         log. =  2'4347062
Bearing from A to 12'00 d      =   N.E. 570 09' 04"     tan. =  0'1899937
Station 12'50 AN       aN =     5950-914 co 6142-920 =   192'006        log. =  2'2833148
AE c   aE =    -  68-404 co  246-177 =   314'581        log. =  2'4977325
Bearing from A to 12 50 a       =  N.E. 580 35' 37"     tan. =  0-2144177
Station 12-50   AN co bN =     5950-914 co 6142-590 =    191'676        log. =  2-2825677
AE co bE =     -  68'404 co  254-318 =   322'722        log. =  2'5088286
Bearing from A to 12-50 b       =  N.E. 590 17' 33"      tan. =  0-2262609
Station 12-50  AN co CN =      5950-914 co 6124 930 =    174'016        log. =  2'2405892
AE co CE =     -  68-404 co  245-577 = 313'961          log. =  2-4968757
Bearing from A to 12-50 c       =  N.E. 61~ OG' 08"     tan. =  0-2562865




BEARINGS            OF STATIONS.                               131
Station 12-50 AN co dN =     5950-914:o 6124-600 = 173-686       log. = 2-2397648
AE    dE =   - 68'404 3o 253-574 = 325'978         log. = 2'5078262
Bearing from A to 12'50 d    = N.E. 61~ 39' 22"    tan. = 0-2680614
Station 13-00 AN o aN =      5950'914 co 6140-790 = 189'876       log. = 2-2784701
AE     a  =  - 68'404 o    296'302 = 364'706       log. = 2-5619430
Bearing from A to 13-00 a    = N.E. 620 29' 50"    tan. = 0'2834729
Station 13-00 AN o2 CN =    5950-914 o 6122-810 = 171-896         log. = 2-2352658
AE    v CE =  - 68'404 o   295-435 = 363-839       log. = 2-5609065
Bearing from A to 13'00 c    = N.E. 640 42' 41"    tan. = 0'3256407
Station 13-50 AN co 13-50N =    5950-914 -' 6128-870 = 177-956       log. = 2'2502980
AE o 13'50E =    - 68'404 c   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 BN o bN    =  5873-969 co 6147'610 = 273641         log. = 2-4371812
BE o- bE =     264'884 c   54"096 =210'788         log. = 2-3238459
Bearing from B to 10-50 B    -= N.W. 370 36' 27"' tan. = 9-8866647
Station 10,50 BN co dN   =  5873-969 co 6129-610 = 255-641       log. = 2-4076305
BE co dE =     264-884 co  53-938 = 210-946        log. = 2-S241713
Bearing from B to 10-50 d    = N.W. 390 31' 41"    tan. = 9-9165408
Station 11'00 BN co aN   =  5873-969 co 6147-100 = 273-131       log. = 2-4363710
BE co aE =     264-884 co   96-119 = 168-765       log. = 2-2272824
Bearing from B to 11-00 a    = N.W. 310 42' 41"    tan. = 9-7909114
Station 11-00 BN co bN  =   5873-969 co 6146-970 = 273-001       log. = 2-4361642
BE o bE   =    264-884 co  104-165 = 160'719       log. = 2-2060672
Bearing from B to 11 00 b    = N.W. 300 29' 09"    tan. = 9'7699030




132                             USEFUL FORMULA.
Station 11-00  BN co CN    =   5873-969 co 6129-106 =   255-131         log. =  2-4067632
BE co CE    =    264-884 co    95'838 =  169-046         log. =  2-2280049
Bearing from B to 11-00 b       =  N.W. 330 31' 40"     tan. =  9-8212417
Station 11-00  BN co dN      =  5873-969 co 6128-970 =   255-001        log. =  2-4065419
BE co dE    =    264'884 co   103-860 =  161-024         log. =  2-2068906
Bearing from B to 11-00 d       =  N.W. 32~ 16' 15"      tan. =  9-8003487
Station 11-50  BN c    aN  =   5873-969 co 6146-110 =    272-141        log. =  2-4347940
BE cO aE    =    264-884 co   146-183 =  118-701         log. =  2-0744544
Bearing from B to 11-50 a       =   N.W. 230 S3' 56"     tan. =  9'-396604
Station 11-50  B.N co biN  =   5873-969 co 6145-920 -    271-951        log. =  2-4344907
BE co bE   =     264-884 co  154-228 =   110'656         log. =  2-0439751
Bearing from B to 11-50 b       =   N.W. 220 08' 29"     tan. =  9-6094844
Station 11-50  BN co C_    =   5873-969 co 6128-120 =    254'151        log. =  2-4050918
BE co CE    -    264-884 co   145-755 =  119-129         log. =  2-0760175
Bearing from B to 11-50 c          N.W. 250 06' 51"     tan. =  9'6709257
Station 11-50   BN co dN   =    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. 233 44' 38"     tan. =  9-6433384
Station 12-00  BN c   aN   =   5783-969 co 6144-720 =   270'751         log. =  2-4325701
BE co aE    =    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  BN   -o bN  =    5783-969 co 6144-450 =   270-481        log. =  2'4321368
BE co bE    =    264-884 co  204-280 =    60-604         log. =  1-7825013
Bearing from B to 12'00 b       =  N.W. 12~ 37' 45"     tan. =  9-3503645
Station 12-00   BN co CN   =    5873-969 co 6126-730 =   252-761        log. =  2-4027100
BE co CE    =    264'884 co  195-662 =    69'222         log. =  1-8402441
Bearing from B to 12-00 c       =  N.W. 150 18' 56"      tan. =  9-4375341




BEARINGS OF STATIONS.                                              133
Station 12'00  BN c dN    =   5873-969 co 6126-460 = 252-491          log. = 2-4022459
BE o- dE   =    264-884 co  203-682 =   61'202        log. =  1-7867656
Bearing from B to 12'00 d      =  N.W. 130 37' 31"    tan. =  9-3845197
Station 12-50 BN co aN    =   5873-969 co 6142-920 =  268'951         log. = 2-4296731
BE co aE E      264-884 co   246 177 =   18-707        log. =  1-2720041
Bearing from B to 12'50 a      =  N.W. 30 58' 44"      tan. = 8'8423310
Station 12'50  BN co b    =   5873-969 co 6142'590 = 268 621          log. = 2'4291399
BE co bE   =    264'884 co   254-318 =   10'566        log. =  1'0239106
Bearing from B to 12'50 b      =  N.W. 2~ 15' 09"      tan. = 8'5947707
Station 12-50 BN co CN     =   5873-969 co 6124-930 = 250'961         log. = 2-3996062
BE c cE    =    264-884 co  245-557 =    19-327        log. =  1-2861644
Bearing from B to 12-50 c      =  N.W. 40 24' 13"     tan. = 8-8865582
Station 12'50  BN co dx   =   5873-969 co 6124-600 = 250-631          log. = 2'3990348
BE co dE   =    264'884 co  253-574 =    11-310        log. =  1'0534626
Bearing from B to 12 50 d      =  N.W. 2~ 35' 02"      tan. = 8'6544278
Station 13*00  BN co aN   =   5873'969 co 6140'790 =   266'821        log. = 2'4262200
BE co aE   =    264'884 co  296'302 =    31'418        log. =  1-4971785
Bearing from B to 13'00 a      =  N.E. 6~ 42' 56"      tan. =  9'0709585
Station 13'00  BN co CN   =   5873-969 co 6122'810 = 248-841          log. = 2-3959220
BE co CE   =    264'884 co  295'435 =    30'551        log. =  1'4844564
Bearing from B to 13'00 c      =  N.E. 60 59' 25"      tan. =  9'0885344
Station 13'50  BN co 13'50N =   5873969 Co 6128'870 =    254'901        log. = 2-4063715
BE co 13'50E =    264'884 eo   349'811 =   84'927        log. =  1-9290458
Bearing from B to station 13-50 =   NE. 18o 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     FORMULA.
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.
ANGLES     ANGLES',
BEARINGS FROM  BEARINGS FROM  AT STATION A AT STATION B
PIERS
STATION A TO THE  STATION B TO THE  WITH B, AND  WITH A, AND
AND
POINTS INDICATED  POINTS INDICATED  THE POINTS  THE POINTS
CORNERS.
IN COLUMN 1.  IN COLUMN 1.  INDICATED IN INDICATED IN
COLUMN 1.  COLUMN 1.
0  /,,       0  I  I,   0  I  I'0  I 
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      "  34 23 49    "  39 31 41    68 36 11  37 28 19
11-00  a     "  39 59 00    "   31 42 41   63 01 00   45 18 19
"     b     "  41 21 15    "  30 29 09   61 38 45    46 30 51
f   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
"(    b     "  48 47 04    "  22 08 29   54 12 56   54 51 31
"(    c     "  50 23 38    "  25 06 51    52 36 22   51 53 09
"     d     "  51 22 56    "  23 44 38    51 37 04  53 15 22
12-00  a     "  53 47 00    "   14 13 38  49 13 00    62 46 22
it    b     "  54 38 06    "  12 37 45   48 21 54   64 22 15
o' c        "  56 20 39    "  15 18 56   46 39 21    61 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
"     b     "  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     "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 09
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 alinements 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




[FIa. 21.]
l
a ---- -- ^^^^ s^ ^ -- -- - - -li~I —
A^^     ^^f^.
IV                I
a~~~~~~~~~~l




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    FORMULAE.
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 + D S a' = D S A'; and the following
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 R, we have
this analogy, cos.: sin.  R: tan. =  -; which,in practice,will
stand thus,
S D: (D a' +  a' A'):: R: tan.A' S D =  (D a'+a A
and 180- A' S D       A S D;
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   = 00 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. I C: radius = cot. I C 260  (64)
Thus,  260           log. =  2'4149733
i 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




[FIG. 22.]
S
m IT.,
\                      4
\    I Iw
IN         I
Ii
i I III I
~ i~iiizl!




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  c'
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 T C 1, the
angle at T = 90, consequently the angle at 1 is equal to the
complement of the angle C; and the angle C =-  T C T=  00 21' 10"
13        13
0~ 01' 37" 7; hence the angle at 1= 89~ 58' 22" 30, which
is common to the triangles T 1 C and T 1 a; then, in the triangle
T 1 a, we have T -   001' 37" 70-  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, b 2, 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~ + T: T 1:: sin.
T: a 1; but, since the sin. of 90~ + T is equal to the cos. T,
the formula may be rendered thus, cos. T: T:: 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.




142                 USEFUL    FORMULA.
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
TS, 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 t, 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 r, and the length of the curve by c, we shall have: r:: C":e =    r.="    cot.; t. C"        (66)
Thus,  C = 0~ 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
2 t = 520
Difference  =  0'0066
We find thus, that the arc and the tangent lines agree within
1600- of a foot, which is a little larger than ~ — of an inch, a
quantity quite too small to be considered an error in laying down
a railroad 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 anything 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 I we have T  = 0 00' 48"'85  tan. = 6 3744395
T 1  = 20 feet        log. = 1-3010300
a 1  =   0'00473663 feet log. = 7-6754695
In the triangle T b 2 we have T  = 00 01' 37"'07  tan. = 6-6754695
T 2  - 40 feet       log. = 1-6020600
b 2  =   0'0189465 feet  log. = 8'2775295
In the triangle T c 3 we have T  = 0~ 02' 26"'55  tan. = 6-8515608
T 3  = 60 feet        log. = 1'7781513
c 3  =   0'0426297 feet  log. = 8'6297121
In the triangle T d 4 we have T  = 0 03' 15"'40  tan. = 6-9764996
T 4  = 80 feet       log. = 1-9030900
d 4  =   0'0757861 feet  log. = 88'795896
In the triangle T e 5 we have T  = 00 04' 04"'25  tan, = 7-0734097
T 5  = 100 feet      log.  20000000
e 5  =   0 118415 feet  log. = 9-0734097
In the triangle Tf 6 we have T  = 00 04' 53"'01  tan. = 7-1525910
T 6  = 120 feet      log. = 2-0791812
f 6   =  0'170518 feet  log. = 9 2317722
In the triangle T g 7 we have T  = 00 05' 41"'95  tan. = 7-2195379
T 7  = 140 feet      log. = 2-1461280
g 7  =   0'232095 feet  log. = 9'3656659




144                  USEFUL      FORMULE.
In the triangle T h 8 we have T  = 00 06' 30"'08  tan. = 7-2775300
T 8  = 160 feet     log. = 2-2041200
h 8  = 0-303144 feet  log. = 9-4816500
In the triangle T i 9 we have T  = 00 07' 19"-65  tan. = 7-3286827
T 9  = 180 feet      log. = 2-2552725
i 9  =  0383668 feet  log. = 9-5839552
In the triangle T j 10 we have T  = 00 08' 08"'50  tan. = 7-3744403
T 10 = 200 feet      log. = 2-3010300
j 10 =   0-473664 feet  log. = 9-6754703
In the triangle T k 11 we have T  = 00 08' 57"'35  tan. = 7-4158332
T 11 = 220 feet      log. = 2-3424227
k 11 =   0-573134 feet  log. = 9-7582559
In the triangle T 1 12 we have T  = 00 09' 46"/20  tan. = 7-4536218
T 12 = 240 feet      log. = 2-3802112
1 12 =   0-682076 feet  log. = 9-8338330
In the triangle T m S we have T  = 00 10' 35" 05  tan. = 7-4883842
T S  = 260 feet      log. = 2-4149733
m S  =   0-800493 feet  log. = 9-9033575
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, but 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 be 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 l,'in the triangle T a 1, which we call y; 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 remainder of the ordinates, b 2, c 3, d 4, etc., to m S, according to
the expressions given in the following table.
COLUMN      EXPRESSION                   COMPUTED RESULTS
OF           OF           COMPUTED    BY PREVIOUS FORMULAE,
ORDINATES.    FORMUL2E.      RESULTS.      FOR COMPARISON.
No. 1.        No. 2.         No. 3.          No. 4.
No. 1 or a 1    =    y     =   0.00473663 feet  0.00473663 feet
2 " b 2      =   22 y    =  0.01894652 "    0.01894650 
3 " c 3      =   32 y    =  0.04262967 "    0.0426297  "
4    d 4     =   42 y    =  0.07578608 "   0.0757861 
5 " e 5      =   52 y    =  0.11841575 "   0.1184150  "
6 " f 6      =   62 y    =  0.17051868 "   0.1705180 
7 " 9 7      =   72 y    =  0.23209487 "   0.2320950  "
8 " h 8      =   82 y    =  0.30314432 "   0.303145
9 " i 9      =  92 y     =  0.38366703 "   0.383668
10 " j 10     = 102 y    =   0.47366300 "   0.473664  "
11 " k 11     = 112 y    =   0.57313223 "   0.573134 
12" 1 12      - 122 y     -  0.68207472 "   0.682076 
13 " m S      - 132 y    =   0.80049047 "   0.800493:(67)
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 the notation expressing the
TU




146                USEFUL    FORMULAE.
method of computation. Third column contains the resulting computations. 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
methols 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 obliterated, 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 line k k' to represent a level, (See Fig. 23 (a);)
then, draw the line n t, 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, let fall the perpendicular n'', until it intersects the line n 1'.
Then, from the intersection of n t with n'', draw the line' m
parallel with kc'; and, from the intersection of n 1 with n' 1, draw
the line 1 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 1 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 c' S t' will be equal to the grade
or inclination S n. Then, representing by
d, the distance between the stations, (usually 100 feet;)
g, the difference in heights between the stations in grade n S;'I      64                "           ",, "6'  n S;
n and n', the numbers of the stations at the points they represent;
A, the height of station n;
h', the  "      "    n'; we have
d   g:: n     n: n m;
and d:g':: no': n m'                 (68)
Then, by substituting p for n m, and p' for n' m'; 6 for n S,
and 6' for n' S; and, subtracting p from A, (which gives the height
of the line m T above the datum line, and which we will represent
by d;) then, subtracting p' from h', (which gives the height of m' I
above the datum line, and which we will represent by o,) we have




[rIIG.    23 (a).]
117.3g5
kin
*.992 m                          0e.a
fFrI.     23(b).]
n-12.452
I. — 5                               ~ -1234
__-.234:
4.452m                            0                              452__  o  S




VERTICAL    CURVES.                  149
g+g': d:: (h-o)     6    (h=-o) d
g +g'   d:: (h'- o):'=      (h'.-       (69)
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 it 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
d          =  100 feet   co. ar.  log. =  8'0000000
g          =  0-7575 feet       log. =  9-8793826
n  no n= 10 stations, or 1000 feet  log. =  3'0000000
n m = p    = 7575 feet         log. = 0'8793826
h          =  10-567
p -h = o      2-992
Again,  d        =  100 feet   co. ar.  log. =   8'0000000
AT'        =  0-3787 feet      log. =  9-5782953
n cn n'    =  10 stations, or 1000 feet  log. =  3'0000000
n' m' = p'  =  3-787 feet       log. =  0-5782953
h'         =  7-385
h'-p=o     = 3-598
g' +  g   =  1'1362 feet  co. ar.  log. =  9-9445452
d         =  100 feet           log. =  2-0000000
h - o     = 6969               log. = 0-8431705
6         =  6'1336            log. =  2-7877157
g +  g'   =  11362       co. ar.  log. =  2'9445452
d         =  100 feet           log. =  2-0000000
h -  o'   =  4'395             log. = 0-6427612
= 3-866              log. = 2-5873064
Having thus ascertained the distances n S, and n' S = 6 and 6';
if we now take n +- 6 =  the number of the station, represented
by S, (the point of the intersection of the grades;) then will n - 6
- the same number S, if the computations be correctly prepared.




150                USEFUL     FORMULE.
STATIONS.             STATIONS.
Thus,    n    = 10             n'   = 20
= 6-1336          6'    = -8664
n + 6 = 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 n' to S. If our computations are correct, the
results should be alike.  Thus, we have h -.g = the height of
S; and h' -'. g' =  height of S.
6     -= 6-1336           log. = 0'7877155
g     = 0 7575            log. = -98793826
6 g   = 4-6462                 0-6670981
h      = 10-567
Height of  S      - 5-9208  above datum line.
6'    =  3-8664            log. =  0-5873068
g'    =  0-3787            log. =  9-5782953
6'. g'  =  1-4642              0-1656021
hl'   =  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 0 800 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 copy verbatim the description of Fig. 23 (a); we therefore 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   n':n m; andd:g':: n c    n':n' m'  (70)
Then, as before, substituting p for n m, and p' for n' m'; 6 for
n S, and 6' for n S; and then, subtracting p from h, we get the
height of m 1' above the datum line, (which height we represent by
o';) and then, by adding p' to h' 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:
and g' g': d::'-  d: 6'         (71)
EXAMPLE OF COMPUTATION.
Let d = 100 feet; g =   0 800 feet; g' =  0 400 feet; and n=
10; n' =20; h=   12-452; h'-      8-234. We then have
Firstly, d: g:: n  n': n m;
Secondly, d: g'::n  n' n m'.
Thus,   d      =  100 feet  co. ar.  log. =  8-0000000
g      =  0-800 feet     log. =  9'9030900
n = n'  =  10 stations, or 1000 feet  log. =  3 0000000
n' m = p =  8-00 feet   log. = 0-9030900




[FIG.  24.]
D  1  2 I 3  5  7  8  9 10
c




CURVING     BOARDS.                   153
Secondly,  d    =  100 feet       co. ar.  log. =  8'0000000
g'     =   0400 feet             log. = 9-6020600
n co n'  =  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
(p - h) = o' = 4-452   (p + h') = o = 12-234
h'       =  8-234       h          12-452
(ht- o')  = 3-782       (h - o)  = 0-218
Again, we have
g   g': d:: h-o: 6;
and g   g': d::h'-':                (71)
g a g =  -0,400             co. ar. log. = 0-3979400
d     - 100 feet                  log. = 2'0000000
h - 0    0218 feet                log. = 9-3384565
6     =  0-545 stations, or 54-5 feet  log. =  1.7363965
g  o gt  0-400              co. ar.  log. =  0-3979400
d     = 100 feet                  log. = 20000000
h' - o'  83782 feet               log. = 0-5777215
6'    -.  945,5 feet, or 9-455 stations  log. =  2-9756615
(56) We conclude our remarks upon tracklaying with a formula 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     FORMULAE.
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 ordinate
corresponding to No. 1, by y; 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 = -  and cos. I C: a:: sin. I C: y =tan. I C'. a
1
tan. 1 C'. 1; it is now obvious that         r-; wherefore,
2~ 2-2      r- 2
tan. IC.1=y=        -                                     (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       =  22 y
3 =   c 3       =  32y
4 =   d 4       =  42y
5 =   e5        =  52y
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 computations the ordinate D d =    (10) (10.)  This ordinate will
correspond to 102 y, 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; R the radius of the tables; 2 ch the half of the chord AB;
Cthe angle D C B; then will r: R::    ch: sin. C= -   h;and
Cos. I C:  ch:: sin. i C:D d = tan. i C. - ch        (74)




156                 USEFUL     FORMULAE.
which expression corresponds to the greatest ordinate of the
calculations; and which, in the practical examples we shall give,
will be represented by 102 y.  So that the difference between the
ordinate found by the exact formula and 102 y will constitute the
error.
Example of computation, according to the approximate formula.
Assuming r -   300 feet; I ch =  10 feet; and a =1 foot; we shall
then have
r  -  300 feet   co. ar.  log. =  7-5228787
2  =             co. ar.  log. =  9-6989700
y -   0'0016666 feet    log. = 7-2218487
Having found  No. 1 =  a 1       =    y         =  0-0016666
we have       2 =   b 2        =   22 y       =  0.0066666
3 =   c 3        =   32 y       =  0.0149999
4 =   d4         =   42 y       =  0.0266666
5 =   e 5        =   52 y       =  0-0416666
6 = f 6          =   62 y       =  0.0599999
7 =   g 7        =   72 y       =  0.0816666
8 =   h 8        =   82 y       =  0.1066666
9 =   i 9            92 y       =  0.1349999
o  =  j 10        = 102 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.  C.  ch= D d =     10. 10
r    =  300 feet  co. ar.  log. =  7-5228787
~ ch = 10 "              log. = 1-0000000
C    - 10 54' 36/"767    sin. = 8-5228787




CURVING      BOARDS.                    157
i C  = 0O 57' 18"'383    tan. = 8-2219693
i ch - 10 feet                1-0000000
D d  = (10). (10) = 0-16671 feet  9-2219693
102 y =        0-16666
Difference   = 0'00005
It will be seen, by comparing D d, with 102 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 = 102 y.)  Thus,
102 y -    0     =  D d      =  0.1666666
102 y -    y     =  a I      =  0.1650000
102 y -  22 y    =  b 2'     =  0-1500000
102 y -  32 y    =  c 3      =  0.1516666
102  -   42 y    =  d 41     =  0.1400000
102 y -  52 y    =  e 51     =  0.1250000
102 y -  62 y    =  f 6      =  0.1066666
102 y -  72 y    =  g 7t     =  0.0851111
102 y -  82 y    =  h s8     =  0 0600000
102 y -  92 y    =  i 91     =  0.0316666
102 y - 102 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 R; we
have, in the triangle A B T', c: R:: b:cos. A =h; and
Cos. A:  c:: R: r=   co --                 (75)
2             cos. A    b
For an example of computation, we will suppose c = 100 feet;
and b -  S feet. Then,
c2 =  1002         log. =  4'0000000
b  =  8     co. ar.  log. =  9.0969100
2      co. ar.  log. =  9'6989700
r  =  625 feet    log. =  2-7958800
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 b =  8 feet;
as above. We then, by dividing a by 2, have in the triangle
A B T', (taking b as a cosine, and 1 a as a sine,) the following
analogy,
Cos.: sin.:: R: tan. A=  b-                 (76)
Then, Sin. A: l a:: R: c =     ACos. A:c:: R: r        c _ =    l            (77)
cos. A  2 sin. A cos. A
EXAMPLE OF CALCULATION.
a = 99 6795            log. = 1-9986059
b     8          co. ar.  log. =  9-0969100
A   - 850 24' 45Y-17    tan. = 1-0955159
A   -  850 24' 45"'17  co. ar.  sin. =  0-0013942
A   =             co. ar.  cos. = 1-0969102
2   co. ar.  log. =  9'6989700
a = 99-6795            log. = 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= 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
a2 + 62 =  2
C2   a2 +  b2 C2
and b: c:: c: d =   =     -      = -  + b   2 r    (78)
EXAMPLE OF COMPUTATION. Let a - 99-6795 feet; b - 8 feet.
Then,




[FGa. 25.]
D'  A
a      {n]
B




ELEVATING      OUTSIDE    RAIL    FOR   CURVE.     161
a-  = 99-67952 feet      log. = 3-9972118
b   =  8 feet      co. ar.  log. =  9-0969100
1242               log. =  3-0941218
+3b =    8
2)1250
r   =   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 Professor 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.
[A\/            B
[wi




162                USEFUL    FORMULaE.
"Let C A B 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 perpendicular 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
D A B 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. l 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      -d; but, by




ELEVATING        OUTSIDE      RAIL     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 be 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.t       It is evident
that a car moving 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 I),
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 FORMULA.
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         considerably 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 upon 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, R = 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 =
d     4   7 feet.
Formula E =          d V
Rg
R  = 4000 feet         co. ar.  log.   6 6 3979400
g  =    32'2           co. ar.  log. = 8-4921441
V2 =    51'33                   log. = 3-4207990
d        4-7                    log. = 0-6720979
E  _     0'09615                log. = 8-9829810
Again, for the      purpose of showing the changes of E, consequent
upon    the  changes     of R, we      assume R        -  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
V2 =    5133                    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. But, it is not our
intention to discuss generally the principles which should govern the construction of cars. 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     FORMULE.
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 formulae for the computation of the cubic
contents of earth, excavations, embankments, masonry, etc., in
constructing railroads.
(59) An article quoted from Silliman's 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 parallelopipeds; therefore, if you add the area of both ends of the section, 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 contained in the section would be just sufficient to fill the hollows,
and make the surfaces planes between them.
( 6 0 ) 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, parallelopipeds, wedges,
and surfaces of the cross sections of the excavations of a railroad;
we shall, therefore, only allude to the most common formule 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.




[FIG. 26(1).]
THE PYRAMID.
\........,\j
a
[FIG. 26 (2).]
THE WED G'E.
a................... _'.'.............. 1,,,"""'/....
C Z_____^         ^  -I




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 i a; if then we represent the height by
h, we have for the area of the base a X a = a2; for the area of
the middle section D a X  a = 1 a2; then, by the rule, ( a + i a 4) h
2 a6 h  ~ a2 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.
(6 2) 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 ~ 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-a2;
For middle section  a X I a —     a2.
Then, ( a+~6 a —4 ) h            2 a  a2 h =  the solid contents;
which corresponds with the formula in common use for determining
the solid contents of the wedge (2.)
x




170                USEFUL    FORMULAE.
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 somewhat 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 b 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 foursided one; and the pyramid a b B C being a triangular or threesided one.  We may further cut the four-sided pyramid in the
plane b B D, which divides that pyramid into two three-sided pyramids. 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 investigations 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 sum, 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 = a2, the middle area I a
X   a=4 a2; then, (a + a). h       2a   (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) i a X i a = a2; and (a4-). h        =   4^ (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 pyramid is exact, according to the rule. As a further proof, we may




[FIG. 27.]
IN WHICH THE TWO SEPARATE DIAGRAMS REFERRED TO BY THE TEXT MAY BE TRACED
~~II'   ~~    I,.     =,,,. ot.24s farea ^c',,o. o.ar."'"a.  \,'t,\,~  
\'''''. - "          "
\'"", \'^'^ 
X   L     a \  \ Jro B.sf.aree,'\
^\  r'i:o. ft.are\d  -u, f.am,.2  /




EXCAVATIONS      AND    EMBANKMENTS.           173
now 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
a h + ( a X i ) 4 ).a  a2 h;
a  h
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 parallelopiped by the rule, the simplest manner of proceeding 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 a2, and four times the
middle area of one of the blind pyramids, will be a X a = a2 each;




174                USEFUL    FORMULA.
there being two of them, we have, therefore, 4 a2 + 2 a2 = 6 a2,
and 6 a' h   a2; the same result as by the ordinary method of
6
determining the cubic contents of a parallelopiped.
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.
(6 3 ) 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
trapezoids 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 be 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 + K L, upon the supposition that the slopes of the
cuttings are as three to two. Thus,
The height of A= 4 feet
"i      "...=                   2
KL.                           — 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 + 2 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
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




176                 USEFUL     FORMUL E.
To find the area of the trapezoid a B L I, we have found
a  B.........       21 feet
L I........           9
a B +   L I).......  =  2)30
(a B  +   L  I).......       15
Height of B.......        8
Area.......  =    120 feet
To find the area of the triangle A B c, we have found
A c........       = 15 feet
- 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
I 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 g, equal the height of F.   =8 feet
+  half the height of F.....      4
+ GM......               9
Wherefore, F.....        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 g E  21 X 2-8
58'8 feet.
Secondly. In the triangle D E f we have
The length of the line D f equal the height of D.   10 feet
+ "      D.      5
+MH    =.          9
Wherefore f D..           24 feet
Then we have half the difference in height of f and E equal to
half the difference of D and E; thus,
The height of D...=           10'00 feet
"  "   E.......= 1360
Difference...                2) 3-60
Half difference.....         1-80
We now find the area of the triangle DEf - 24 X 1'8
43-2 feet.
Thirdly. To find the area of the trapezoid F G M g; we found
in the foregoing,
F  g.........    21'00 feet
GM.........  =  900
Fg+   M.......  2)30-00
Mean length....             15-00
Then, M G = height of F        = 8-00
Area.....               120-00 feet
Y




FIG. 28.]
B
A~ 15tf,I~~~~~~~~C
21       g




EXCAVATIONS       AND    EMBANKMENTS.            179
Fourthly. Then, to find the area of the trapezoid D H Mf; we
find'in the foregoing,
Df.....=                 24-00 feet
MH.=                             9'00
D f + M H.....        2)3300
Mean length.....        =  16-5
Then, M f= height of D            = 10 00
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 cubic contents of that portion of the half
section lying above the lines F g, g c, c A, and A F.
(6 4 ) It will be evident, from an inspection of the drawing, that
the upper portion of the half section may be divided into two apparent, 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 FORM.                   SECOND FORM.
A  c....    15-  feet  F g..=          21-00 feet
g  E....  =  56        c B....   -'4'00
9'0        4 times middle area  =,,84-00
754 times middle area.  = 84'00




~~~~~3   ~ ~ ~ ~ ~, -M —-- ^
^   ^*^^ ^.l   / > -""".... —-""""  ^  /~~~c,...<s l "'       ^ ^ ^^ -^
rs^ ^.........................^ ^ ^




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 cg F G M.    The lines Fg, 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.
A c........                  = 15 feet
Mg..........  =  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,
L c.=                                 4 feet
GM.....                              = 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,
L  c........=             4 feet
Fg.= 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 + 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 convenient 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
t6 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    FORMULAE.
liarly adapted to this method of computing cubic contents, and
affords a very convenient, as well as an accurate method of ascertaining 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 Bf 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 f D a having all of its surfaces 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 D E B, and intersecting in the line D B, a
plane passing through D B C, the solid will then contain only two
apparent pyramids, with no blind pyramid; viz., the apparent
pyramids D fE B and B a C D; the measurement of which is as
follows:
For the area of base of D E f B = 43-2 feet, and + 4 times middle area = 43-2  =  86'4 feet
"     ('(   a B C D = 42  " " -+ 4  " "  " = 42  =  84'0
Dividing by.....6)1704
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 f C,
and the blind pyramids C a f D and E f 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    = 84'0
" 4 times middle area of blind pyramid C af D = C a X f D = 4 X 24.  = 96 0
"    "   "   "   "       "   aBEf= aB X Ef= 21 X 36.  = 75'6
Dividing by...6)3420
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
"  "    "    "    "  "   ".....             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 frequently occur in practice, as the engineer would be likely to divide
z




186                USEFUL    FORMULE.
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 be
noticed when the field work is performed, so that the proper form of
dividing the cross section may be applied, large sections may
frequently be computed with as good a degree of accuracy, or even
better, than smaller sections without such notice.
To complete the computation of the cubic contents of the figure,
we have
The double area of the base of pyramid H Mf D a.....   330 feet
Also  "  "  "  "     "  aBLIH.......  =  240
Then, 4 times the middle area of the blind pyramid =fM X B a = 21 X 10.. = 210
And 4 " "   "   "  "   "   "  = B L X M H =9 X 8..  72
Dividing by...........6)52
142
Multiplying by....                                  100
Cubic contents of a B L I H MfD.....= 14200
Having thus discussed the operations necessary to obtain the
cubic contents of every portion of the section, we now add an example of summing up the contents after the manner in common
practice. Before entering on our work, we remark that we contemplate two summations; the first containing the computation of the
upper portion of the second half section, noticed in the foregoing as
containing no blind pyramid; and the second containing the computation of the upper portion of said second half section, noticed
as containing two blind pyramids.
(6 6) 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  =  60'00 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 gM XAc=15X8 = 120-00                    120-00
3 L XGM=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 "  = Mf X B a = 10 X 21 = 210-00       210-00
" 2 = B L X M I=8 X 9 - 72-00  Blind pyramid  72-00
Dividingby.....           6)1776-00  ac XfD=4 X 24 = 96 00
296'00  EfX 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 be
noticed, between the slope and centre stakes; but 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 measurement of the middle area of the blind pyramid C a f D in the second




188                USEFUL     FORMULJE.
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.