Field manual for engineers

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Dbarvard College Library 


FROM 


GODFREY Li LIBRARY 


KFIELD MANUAL 


HENGINEHERS. 


4 BY 
PHILETUS H. PHALBRICK, C.E., M.S. 


M. Am. Marta. Soc., 


Chief Engineer, Kansas City, Watkins and Gulf Railway, 
'  N. Am, Land and Timber Co., etc., etc.: 
Sometime Professor of Civil Engineering at the 

Stute University of Towa. 


FIRST EDITION. 


FIRST THOUSAND. 


NEW YORK: 
JOHN WILEY & SONS. 
Lonpon: CHAPMAN & HALL, Limrrep. 
1901. 


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Copyright, 1901, 
BY 
P. H. PHILBRICK. 


ROBERT DRUMMOND, PRINTER, NEW YORK, 


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PREFACE. 


THE aim in this work has been: 

First. To present the subjects of the text in a mathematical 
and logical order. 

Second. To classify all problems presented so as to be easily re- 
ferred to. | 

Third. Especial care has been taken to express the resulting 
iormula of every problem in the form requiring the least numeri- 
cal computation. Many instances of this may be seen by com- 
paring these with similar formulas from other sources. 

Fourth. To furnish a large number of useful tables, more 
complete, more extended, and, where possible, with more ele- 
mentary and appropriate arguments than other similar tables 
possess. . 

Fifth. To treat the general problems of Railway Engineering 
more extensively than other similar works have done. The 
growth of the modern railway system during the past half-cen- 
tury has been so rapid that the mathematical needs of the 
subject have not kept pace with that growth. The technical 


books on this subject of some two decades ago are entirely | 


inadequate to supply the needs of to-day, and the most recent 
works are more restricted and elementary than the former. For 
these reasons the author, in attempting to supply a useful book, 
seeking utility rather than novelty, has found it necessary to 
modify some methods and formulas in use, and to introduce new 
subjects and formulas altogether. 

In the following the modified, extended, and extra topics 
referred to are, for clearness, considered in connection with the 
general matter of the text. 

Table II dispenses with all calculation in laying out curves, 

Hi 


lv PREFACE, 


as shown in Chapter IV; Table VII greatly simplifies the 
finding of the tangent and the external of any curve; and. 
Tables XVIII and XIX, and some others to which the remark 
can apply, are, it is believed, in terms of the proper arguments 
and in the best form. 

In Chapter IV the laws of errors in field-work are demon- 
strated and illustrated; and the best method of conducting a 
preliminary survey, introduced by the author a generation ago, 
is explained. | 

In Chapter V simple and exact formulas for determining the 
height of a mountain or other object by the dip of the horizon 
are substituted in place of the approximate formulas in use. 

For the stadii, as well as for the telemeter, new formulas are 
found which require no general computation; and the formula 
. for finding the proper elevation on curves, unlike other formulas, 
involves no large factors. 

In addition to a very general treatment of Compound Curves, 
Chapter VI includes the location of such curves of any number 
of branches (pp. 143-5), as well as easy and symmetrical 
formulas for finding their tangents (p. 176). | 

The reader interested in the philosophy of mathematics will 
find, it is hoped, an elegant and fruitful illustration of the prin- 
ciples of substitutions in determining general curves to fulfill 
required conditions on pages 145, 146. This is susceptible of 
general application. There has also been added a general treat- 
ment of the subject of Curves tangent to Curves, including the 
“Wye Problems”; and also that of Concentric Curves applic- 
able to Parallel Turnouts. 

The finding of the angles between the rails at the crossings 
of curved tracks is also thought to be a valuable addition. 

Chapter VII includes a variety of problems in Reversed Curves. 
The solution of Problems IV, V, and VI were first ziven to the 
Senior Class (1869) of the University of Michigan; while the 
author was In temporary charge. | 

Chapter VIII treats extensively of Turnouts. The distinction 
between connecting a straight line and a track, and two tracks, 
and what is required in each case, is shown on pages 199, 200. 
The impracticability of certain proposed turnout curves is 
demonstrated (p. 205), and a variety of methods of laying out 
turnout curves is shown, , 


PREFACE, Vv 


Chapter IX applies entirely to the author’s True Transition 
Curve. 

Chapter X shows that the formulas for the computation of 
earthwork may be abridged one half by supposing that the 
side slopes are produced in this intersection. A general relation 
between the “ end-area ” volume and the “ middle-area ” volume 
is shown by means of symbols, and new formulas are given for 
the volume of a frustrum of a pyramid and the frustrum of a 
eone. A formula showing the true correction of earthwork for 
curvature is also deduced, and the best method of computing 
earthwork tables explained and illustrated. 

Chapter XI contains only a brief exposition of the subject 
of approximate and abridged computations, which it is thought 
may serve to encourage the shortening of computations. 

Chapter XII describes the processes incident to construction 
and calls attention to two principles that aid very much in 
“staking out” earthwork. 

The logarithmic tables are not reproduced, for the reason 
that they are but little used and should not be used at all— 
and most emphatically so in this line of work. It is fair to 
observe that the space required for such tables is replaced by 
numerous useful tables, applying directly to the matter in hand, 
and also to enlarging the subject-matter of the book, thus sav- 
ing greatly in time and labor. Furthermore, it should be stated 
that there is no problem in the book requiring a computation 
more complex than to find the cost of 29 oranges (say), sup- 
posing that 17 oranges cost 43 cents. The author must believe 
that no person—much less an engineer—would think of apply- 
ing logarithms to the above example; and if so, he could not 
with any propriety apply them to any problem in the Manual, 
since the nature of the numerical computation to be made, and 
not the subject-matter of the problem, furnishes the test of 
methods. | 
- The author invites criticism, and, should another edition be 
called for, will make the best use possible of any suggestion 
that may in good faith be made to him. 

The Tables, all but three, were computed expressly for this 
book, and scrupulous care has been taken, by numerous readings 
and checks, to make all tables exact to the last figure. 

The book, as indicated in a few places, has been in prepara- 


v1 | PREFACE, 


tion several years and contains matter gathered all along the 
paths of the author’s experience; and is the result of a belief, 
on his part, of his ability to aid his professional brethren in this 
direction. 

While this delay has not been to the advantage of the author, 
it has nevertheless given opportunity for due reflection and re- 
consideration; and therefore, as a work of judgment based on 
experience, the volume is offered to his brother enquirers. If 
the book even partially accomplishes the object of the author’s 
aims, he will feel that the days he has devoted to it, though 
many, have not been spent in vain. 

P. H. PHILBRICK. 


CONTENTS. 


CHAPTER I. 


PRELIMINARY OPERATIONS. 


PAGE 
The Reconnoissance...........cccscccccccscccccccccccccvecccccctscecece s 
The Preliminary Survey.........ccccccccccsccccccccsccvccces secceeeeece S. 
The Location 2... ... ccc ccc cece cece cc ccec ccs neecsceercenessececccececse § 
The Organization of the Transit Party.................000 se ececceens 8 
The Compass: What kind to use and when to use it............00.: 8 
‘Requirements for a Successful Reconnoissance................0eeeeees 9 
Train Resistances........cccccccccccccccccrcceucceeccccescccctcteccesees 10 
Total Ascent the Main Test of Gravity Resistance..............sce08 II 
CHAPTER II. 
ADJUSTMENTS, Use, AND CARE OF INSTRUMENTS. 
The Transit. 
Adjust MENts ..cccccccccccccccccncccccscsceecc sees ee easseeaeesteescenees 12 
Use and Care of the Transit...........cccccccscsccccsscccccsscscncssscs 15 
Best Way to Set up the Instrument............. ccc cc cece nc cncceees 15 
To Measure the Angle between Two Lines or Objects cee cecccnseees 16 
Hints on the Care of Transits and Other Instruments.............. 16 
The Level. 

Adjustments ......ccccccccccnscscnssncecscncnccscceesssnsscecscvansscees 18 
Use of the Level... ccc ccc cece ce ccc eee een ccceceneeecceseeescsens 22 
The Compass. 

AdGjustMentS ....ccccccsccccccc ccc cenceccccesccccseneus once cence eeeceenes 22 
Use of the Compass........... ccc cece rece n cc cee cc careensecstecencnsecees 23 


Vill CONTENTS. 


CHAPTER III. 


PLANE TRIGONOMETRY. 


PAGE 
Definitions and Explanations..............cccccscccccevcrscceccevcseves 25 
Fundamental Relations.............ccccssesccsceccecceceseccesesecucees 26 
Solution of Plane Right Triangles........... ccc c cece e cence ccncccnsees 31 
Table for Plane Right Triangles............ 0. cc. cece eeeceececnseeeees 32 
Fundamental Relations for Oblique Triangles sce c cee e ence snscesccees 32 
Short Solution of ‘‘ Tangent Problem ”’...........0. cece cece eee ecens 34 
Solution of Plane Oblique Triangles............ccccseeccescecccecscees 35 
Table for Solution of Oblique Triangles..............ccecceeecnereeees 36 
List of Fundamental Formulas..............cccccccccccevccccvcscceces 37 
CHAPTER IV. 
SIMPLE Curves CONNECTING Ricut LINEs. 

Properties Relating to the Circle.......... ccc ccc cc cece ccc ccccccceee 39 
Some Elementary Relations..........c..ccccccccccccccevcscsccccccccece 4! 
Notation ..... cc cee cece cece cn cece cece ceceteec seven cesncereerecseeesccscees 41 
Degree of Curve Defined and Explained..............0.cee0- ececeee .- 4! 
Rational Treatment of Curves, Example IIlustrating................ 42 
Difference in Lengths of Arcs and Subtended Chords............... 43 
Huygens’ Formula for the Length of an Arc...........cccencccaecees 44 

Table Showing Excess of Arcs over Subtended Chords when the 
Arcs are Aliquot Parts of 100.........ceeeeeeeoes tec ecceterceccece os 44 

Table Showing Excess of Sub-chords over Aliquot Pasts of 100, 
when the Chord = 100.........cccceccceccncteccuseeececcececscesees 45 
Reason for these Large Excesses Pointed Out..........ccccccecsace 46 
Formulas for Radius, Tangent, External Secants, Offsets, etc..... 47 
Proper Course to Pursue in Locating a Curve...........ccceesseences 49 
Long Chords and Ordinates to Long Chords...........cccccseeeses - 50 
Approximate Value of Ordinates to Short Chords.............cceee. §I 
Offsets in Terms of the Degree of a Curve..............ccccccccccece §2 
Applications of Formula.........ccccccccccccccccccsccescccssecccesssess 53 

Laying Out Curves. 
A. By Deflection Angles....... Lec cece enc ceeeceececceteesceeseseeeseees 53 
B. By Tangent Offsets. New Method. Without Calculation....... 55 
C. By Ordinates from a Long Chord. Without Calculation........ 57 
D. By Chord Offsets. Without Calculation..............ccceecccecees 58 
E. By Middle Ordinates. Without Calculation................eeee0. 59 
F. By Radial Lines. Without Calculation.............ccccececcsceces 59 
Errors in Field-work—The Nature of. ° 

“A” Method of Laying Out........ ccc ccc cece eee ene eeceeececeens 60 
“B” Method of Laying Out............ cece eee n cece cere oeccccecccons 64 
“CC” Method of Laying Out........... cc cee cece erence eee e een eanees 64 


~ Method of Laying Out......... cece cc cece cece teen ceeeeccrcees 64 


CONTENTS. 1x 


PAGE 

“EE” Method of Laying Out............. ccc ccc cece eee seccecceesoeees 65 

Fourteen Problems in Simple Curves...............ccccccccccccccecs 65 
Obstacles in Surveying. 

To Erect a Perpendicular at Any Point of a Line..............0..0% 76 

Table and Forriulas giving Sides of Right Triangles................ 77 

To Drop a Perpendicular from a Given Point to a Given Line...... 77 


To Draw a Perpendicular to a Line from an Inaccessible Point.... 77 
To Prolong a Line past an Obstacle and to Measure its Length... 78 


Obstacles to Measuring a Line. 


When One End is Inaccessible............... cece cee ec cence eee eeeeee 79 
When Both Ends are Imaccessible................ ccc cece cece eee ence 80 
When an Inaccessible Space Intervenes................. ccc cccecccees 81 
Best Method of Making a Preliminary Survey.....................005 R2 
Table Illustrating the Same.............cccecce cee c eee ee cence cneeeeees 83 
To Replace a Broken Line between Two Points by a Straight _ 
DOSS « (aa 85 
To Find the Angle between Two Straight Lines when the Point of 
Intersection is Inaccessible...........c. ccc cece cece nce eceeesscecees 86 
To Connect Two Tangents by a Curve when the Vertex is Inac- 
or) 9 0) (- 87 
To Locate a Curve when the Vertex and Both Ends of the Curve 
are Imaccessible.............. cc cece cece eee e cece ete neeeee sec ceeeeenes 87 
To Pass from Any Point on the Curve to Any Point on the 
Tamgent ....cccc cece cence cece ccc c cence eee e eee e eee ease eee eeeeeneneee 87 
To Find any Desired Point on the Curve when Obstacles preclude 
tre Use of Ordinary Methods...........cccccccccceceneneceseeees 88 
CHAPTER V. 


LEVELING, STADIA MEASUREMENTS, ETC. 


Bench, or Bench-mark...........ccccccccccccccccccccccecscccesecseccce 90 
Form of Field-book for Level Notes...............e.00- bce eeeeccenee 92 
Proof for Level Notes...........ccecsccccccsce cence ssseccceereetnennes 93 
Benches: Where to Establish them.............. cc cece eee eee eeeee 93 
The Location of a Level Line.............. cece cece eee cece cet ecenes +s 94 
The Location of a Grade Line............ cc cece cece cnet cc nneenccnveees 94 
Correction for Curvature and Refraction...........cccccccccccccccece - 95 
Trigonometric Leveling.............cccccecccsececccrscccecceevsceeees »- 97 


True Simplified Formulas for Heights by the Dip of the Horizon.. 98 


The Stadia. 
Formulas for the Stadia—Simplified............ eeceeee eeeesecccccevese 102 


The Gradienter. 
Formulas for the Gradienter—Simplified..............cceeeeee ceseeeee 107 


x CONTENTS. ; 


Vertical Curves. 


PAGE 
When the Grades are both Ascending or Descending................ 109 
When One Grade is Ascending and the Other Descending.......... 110 
Elevation of Outer Rail on Curves...........cccccecccccccccccceccces Ii2 
Formula for the Elevation—Simplified............... ccc ccc ccccceccees 113 

CHAPTER VI. 
ComPouND CuRVES. 
General Formulas........... ccc cece cc ccc cen cccsccvccncccccccesccecsceers 116 
Proof that R,—R, > T,— T, in all Cases.......... cc cc cece eee ee een es 120 
Proof that Ks > 1s in all Cases...........ccceccccccccsccccecs veeeueaes 120 
Rk, 7, 


Table giving the Least Value of 2 — a for Different Values of V.. 120 
3 1 


Problems. 
Given T, and T,, also Either One of 0O,, O,, R,, and R,, to Find the 
8 Cs a: 121 
Given A, B, and C and Any One of O,, O,, R,, and R,, to Find the 
Others 2... ccc cece cece cece cece cece e cece cece ce ee eteereneceraseccescones 122 
The “ Engineering News’ Challenge Problem...................... 123 
Given R, and R, and Either Tangent, to Find the Other Tangent 
and the Central Angles.............ccceceece eee ce een eceseeeeseeees 123 
Given Either Tangent and Either Radius, also Either Central 
Angle, to Find the Other Tangent and Radius................66. 124 
Given the Tangents and a Central Angle, to Find the Radii........ 124 


Special Problems. 
To End a Simple Curve with a Shorter Radius so as to End in a 


Given Parallel Tangent............cccc cscs cece ccc cen ececccsccceeees 125 
To End a Simple Curve with a Longer Radius so as to End in a 

Given Parallel Tangent............. cc cece cccescnccccccccstcsveccons 126 
To Change a Compound Curve, the First Radius being the Shorter, 

so as to End in a New Given Parallel Tangent.................. 128 
To Change a Compound Curve, the First Radius being the Longer, 

so as to End in a New Given Parallel Tangent.................. 135 
To Change a Compound Curve so as to End at the Same Point as 

before, but on a New Tangent...........ccc cece ec ec ees ee eeenaes 140 


To Substitute for Radius R of a Curve, Two Radii R, and R,, the 
' Longer of which, R,, is to be used for a Certain Distance only 
at Each End of the Curve (Henck)............ccceseccssecsecves 142 
To Determine tke Distance between the Middle Points of a Simple 
Curve and of a Three-centered Compound Curve, Joining the 


Same Tangent Points.............ccccceec cree esse eeseceneseeseeracs 143 
To Connect Two Tangents by a Compound Curve of Four Branches, 
the First Two Consuming One-half of the Vertex Angle......... 143 


To Draw a Tangent to Two Given Curves...........cccc cece ewcneees 150 


CONTENTS. X1 


PAGE 
' To Locate a Curve Tangent Internally to Two Given Curves....... 151 
To Locate a Curve Tangent Externally to Two Given Curves...... 154 
To Locate a Curve Tangent to One Curve Externally and to 
Another Curve Internally................. Cece eee cece sccnncccas 156 
To Find a Curve Tangent to a Given Curve and to a Straight Line 
that Intersects the Curve..........cccc cece ccc ccnccccccsccceccecee 157 
To Find a Curve Tangent to a Given Curve Internally and to a 
Straight Line that does not Intersect the Curve...... bc nasceccens 163 
To Find a Curve Tangent to a Given Curve Externally and to a 
Straight Line that does not Intersect the Curve.............0000: 164 
“ Wye’ Problems. 
Example. A Curve Tangent to a Straight Line and Curve.......... 166 
Example. A Curve Tangent Internally to Two Curves.............. 167 
Example. A Curve Tangent Externally to Two Curves.............. 169 
Curves Tangent to Parallel Lines, Outer Curve Sharpened near 
Point of Tangency...........cccceccecceees Lecce ccc veces ccccecccecs 170 
The Same when Curves are Tangent to-One Line.............eee05. 171 
Curves Tangent to Parallel Lines, Outer Curve Flattened hear 
Point of Tangency.........cccccccc cece ccc encceeceseeseneeececeenens 173 
To Locate the Second Branch of a Compound or Reversed Curve 
from a Point on the First Branch.......... eee c eee ee eens eesecees 174 
Given a Sector ABO, to Find the Radius of another Sector, Center 
on AO, that will Cut BO at a Given Point..............ccceeeeeee 175 
To Find Lengths of Tangents of a Compound Curve of any Number 
Of Branche........ccccccccc ence cece cesses eceseceeeeeeeeeeeeeneenes 176 
To Find the Angles at Crossing of Rails of Curved Tracks se ceeees 178 
CHAPTER VII. 
ReveERsep CURVES. 
To Find the Reversing Point of a Reversed Curve .......c..scceeees 183 
Reversed Curve of Equal Radii between Parallel Tangents.......... 184 
Reversed Curve of Unequal Radii between Parallel Tangents....... 186 
Reversed Curve of Common Radius, Tangents not Parallel.......... 187 
Reversed Curve of Common Radius between Two Points on Two 
Diverging Tangents...........cscecccscccsscsccessccsesscescenececes 188 


Reversed Curve Connecting Diverging Tangents, Starting from a 
Given Point and Advancing toward or Receding from the Inter- 
section Of Tangents......ccccccccccccccccccccccccsescsnsctevscsscees 189 

Reversed Curve to Connect Two Given Points on Two Diverg- 
Ing Tamgents.......cccccccccccnccsccccccescceesssscscscceccescescens 192 

To Shift the P.R.C. of a Reversed Curve so that the Second Branch 
may End in a Given Parallel Tangent.................cceeceeseces 193 


CHAPTER VITI. 


Turnouts. 


Turnout and Frog Described................ ccc cece eccooerccee eeeeceee 195 
Turnout from a Straight Track....... 2... ccc ccc cece ccc ec cen enccccsecas 195 


xii | CONTENTS. 


PAGE 
To Connect a Straight Track and a Straight Line................... 199 
Two Connect Two Straight Tracks by a Curve of Three Branches, 
the Turnout Curve having Radii to Suit Given Frogs............ 199 
Formulas Supposing the Switch-rail to be a Part of the Turnout 
SF oh — 204 
A Simple Curve Cannot Meet the Conditions Required Above...... 205 
Several Methods of Laying Out Turnout Explained................. 207 
Double Turnouts from a Straight Track..........cc ccc cece ec eccccees 208 
To Fit a Curve to a Given Middle Frog...........cccscccccssscccees 210 
Turnouts from Curves. 
Turnout from the Inside of a Curve...........cccceccecccccsesscscecs 212 
Turnout from the Outside of a Curve... cc. cccecc cc ccccccccncccenees 213 
Double Turnout on Opposite Sides of a Curve............ccceeceenes 214 
To Find Degree of a Turnout from a Curve..........ccccccescecsess 215 
Other Turnouts from a Straight Track.............cccccceccsccececes 217 
To Fit the Turnout to a Given Middle Frog..................eeeee 219 
CHAPTER IX. 
THe True TRANSITION CuRVE. 
Reason for the Need of Such Curves.........cccccceccccccccccecereces 22 
Definition and Properties of Such Curves............. cc ccceccecccues 22 


Elementary Relations. 
To Find the Relative Length of the Offset and the Transition 


OF of - 225 
To Find Any Tangent Distance.........ccceccccccccccvcccccccccccces 226 
To Find Any Offfset........ccccccccc ccc cccn cece cccceceecccccecstecscces 228 


Having the Offset, t, Supposing the Offset Curve of the Same De- 
gree as the Original Curve, to Find the True Value, #’, of the 
Offset ....ccccccccc cece cence e cece ense cesses cece seceeee en eeesareseases 230 

To Find the Offset in Terms of the Cenrtal Angle and the Radius.. 230 

To Find the Angle between the Tangent and Any Chord Drawn 


0) 06 Ue: SPR 231 
Cubic Parabola Not Suitable for a Transition Curve................ 232 
To Find Point on Curve where the Tangent is Parallel to the 

Chord of the Curve.........ccccccccnccccsccccccccscnsscccccerecsees 232 
To Find Tangents at the Extremities of the Curve.................. 233 
To Find the Length of Any Radius Vector, or Chord, and the 

Angles between these Chords.............ccccecccsccccccccccnseees 233 
To Find Deflection Angle at Any Point, also Any Chord............ 235 
To Find the Exsec dV, also TV, etc.......... ccc cece cecec ne neccenees 236 
To Find the Radius of Curvature at Any Point..................08 236 
To Lay Out the Curve by Offsets from the Tangent AO............ 236 


Special Problems and Examples. 
Given Length and Degree of Kd, to Find the Offset AK, Tangent 
AO, and to Lay Out the Curve............ cc ccc cece cece eeenees 237 


CONTENTS. XH 


t 


PAGE 
Given the Degree of the Offset Curve and Offset AK, to Find the 
Length of the Transition Curve, etc...........ccccccecccccccceces 239 
Given the Degree (D') of Kd, and the Tangent AO, to Find the 
Length of the Transition Curve and the Offset AK............. 239 
Given the Degree of the Main Curve, and the Length of Kd, to 
Find the Offset, Tangent, etc..........ccccccccccccccccccccccceces 241 
Given the Degree of the Main Curve and the Length of Af Re- 
placed, to Find the Offset, Tangent, etc............ ccc cece eeees 242 
To Replace Each Half of a Simple Curve AfA! by a Transition 
Curve .... cc ccc cce cece e ce eeees Lecce eect eee een ene e ee aee ee eeecseeerns 243 
To Connect Two Tangents by Two Transition Curves, Each of a 
Given Length Si... cc ccc cece ccc cece ccc eee eee e eect ee teeseneeeeeseees 244 
To Connect Two Tangents by Two Equal Transition Curves 
having a Common Vertex Distance E..............c cence ccc ceees 245 
The Transition Curve Very General in Use................ccceceeees 245 
To Lay Out the Curve from Any Point on it........... cece ce eee 245 
To Substitute a Transition Curve for Each End Portion of the 
Main Curve without Changing the Rest of the Curve.......... 246 
The True Transition Curve Compared with Some Others and His- 
torical Note. ..rocccccencssvcccccccccesevecessescssscccuce ee cececcees 251 
CHAPTER X. 
CALCULATION OF EARTHWORK. 
Prismoid Defined.............ccccesccecccees ee acecees Deceneccesccesssces 253 
Area of Level Sections............. cece eee eeeee eee eees waeeeceeccecnees 254 
Area of Sections not Level..........cccccccccecccccecccncersvccceuceees 255 
Area of Irregular Sections............... cece cece cc eencsvencccsecrecese 256 
Formulas for Regular Excavations and Embankments.............. 256 
Error of the ‘‘ End Area Volume”? Always Twice the Error of the 
‘“* Middle Area Volume.” Demonstrated by Symbols............ 257 
The End-area Method Simplified..............cccccceccccececccceecens 258 
The End-area Volume Generalized..............ccccecccsccccccccecees 259 
Special Formulas and Cases.........ccccccee cece ce cccecesescccssceeacs 259 
Formula for the Volume of the Frustum of a Pyramid—Sim- 
plified ..............6 cece cece ce cec rece ee cone ceeees ccc e ec n ec eecenens 261 
Loaded Flat Cars, Piles of Stone, etc..........cccceccevccecceeecceues 263 
Ends of Embankments or ‘ Dumps ”’.........ccscccccccccsccccccecces 264 
Ground Irregular Laterally...........cccccccccccccsaveccuvccecesecccecs 264 
Mixed Work, Excavation, and Embankment..............cc.cceeeees 265 
Correction of Earthwork for Curvature............ccsscecccescesscceee 267 
Overhaul 2.2... ccc cc cect c ccc cee cence enone ene n ec areestneeesssescecereucece 268 
Monthly Estimates..............0.00- bee c cen n ccc ne et aceeeereetsneenees 269 
Final Estimates..........ccccccccccenccceccesreccccccceenesesseesscenaes 270 


Computation of Prismoids Level Laterally...,....... secececeaveseees 270 


XiV CONTENTS. 


CHAPTER XI. 


APPROXIMATE AND ABRIDGED COMPUTATIONS. 


PAGE 
Definitions and Notation............. sec cc cee ccescecscceceecseceseescs 293 
The Relative Erropr........ dace c cece ences ee saee tacts eeacsccncecseesees 273 
FAs Cob 8 9 Co) » Um 274 
Subtraction ....... cece cece rece cece tec e nce esccecccecenecsccassecceecesscs 276 
Multiplication and Division...........c.cccececccccecceececcccecescese 277 
Abridged Multiplication.........cccccccncccccccccccvccccscccccccevecces 279 
Abridged Division........ccccccsccccccccccccucccuscvcesveccscccccsecess 281 
CHAPTER XII. | 
CONSTRUCTION. 
Clearing and Grubbing, How to do it........cccccececeeeeesuesascees 284 
Grade-lime oo... cc ccc cece cnc cec eee c ec eceeeetsasessccccavaccsece ee eeeeee 284 
Surface Ditches—Importance Of........... ccc eec ee cescceceeeees seeeeees 285 
Cross-sections—Proper Places for..........cccccccccccssceccccscccecccs 285. 
Staking Out Earthwork when Ground is Level............../....... 286 
Staking Out Earthwork when Ground is Not Level.................. 287 | 
Two Principles to Aid in Laying Out Earthwork.................... 201 | 
BOrrow-pitS 2.2... cc cece ccc eee ccc e ence ee nnee cess ee eceeessentereeesceters 294. 
Shrinkage of Earthwork...........cccccecccccccccsncnscccccccecesnvecse 204) 
Retracing the Line............c. cece cc ccc cern e ener ceceescescssccecrccces 295 | 
Track-laying .....cccccccccecccc cece cece creme esecceeeenesssessseeseeneers 296 
Culverts .... ccc ccc cc ccc cece eee cere cere cases sseeeessnerecesseeene 297 
Location of Bridge Piers..............scccccceccecvecscs eee eescececesce 298 
Tunnels . ..cccccerecscccecsercscccncersrerscensscsnstesenceecessessecsess 298 | 
CHAPTER XITI. 
EXPLANATION OF TABLES AND MISCELLANEOUS ToPIcs. 
To Gauge a Stream Approximately............cccccccccccccccceccceces 301 
Transverse Strength of Beams...............eccecesseeecees eee cecenees 302 | 
Safe Bearing Power of Piles..........cccceccecncccccevcccecvers ceeees 302 | 
TABLES. - a 
Table for Right Triangles...........ccccccccccccccccccecccuscusceceens 32 | 
Table for Oblique Triangles................ccccccccncccvcecuccceecseas 36 
Table Showing Excess of Arcs over Subtended Chords when the 
Ares are Aliquot Parts of 100...........ccceccccccvecces bc eseceeeee 44 
Table Showing Excess of Sub-cords over Aliquot Parts of 100 
when the Chord == 100.........c.cscecccccccccccecccnccccvesveceeses 45 
Table Giving the Sides of Right Triangles..................cccseees 77 
Table for Traverse Survey...........ce cesses ccscccc cece enencescescucne 83 
Table for Level Notes. ...... 0. ccc ccc cece cence ee nce occ eecacessesepecs p2 
»—kR 
Table Showing the Least Value of T. =. eee ete tancneceseeeeas 120 
2 l 


CONTENTS, xv 


PAGE 
Table Showing Computation of Prismoids........ ce cceeences seecceese 271 
I. Degrees, Radii, etc.......... cc ccc cece cece enc cenaceccecscvcceees 304 
II. Tangent Offsets, 1 to 100 Feet..... dace ececeestecsecccesceces 310 
Til. Offsets for Arcs of 100 Feet...........ccccccccccccccccscccess 312 
IIIa. Middle Ordinate Arcs of 100 Feet............ccccccceneeees 313 
IIIb. Chords of Arcs of 100 Feet........ccccccescccccscceseccacece 313 
TV. Long Chords....... ccc ccc c ccc c cence nce nteecrensesceessccscrsees 314 
V. Middle Ordinates........ ccc cccc ccc cc ce sacsncacscceeeseeenes 316 
VI. Turnouts from a Straight Track........... ccc ec eseccccecrcees 318 
VII. Tangents and Externals of a 1° Curve...... occ ccccecenceeees 319 
VIII. Arcs of Degrees, Minutes, and Seconds for Radius — 1.... 323 
IX. Acres for Various Lengths and Widths...........sscscecees 323 
X. Total Grades.... ccc. cece ccccccccscccccecccncenseresssccsccesces "324 
XI. Correction for Curvature and Refraction.................... 325 
XIT. Elevation of Outer Rail..... cece erence eee eentacecesceeereences 325 
XIII. Coefficients for Stadia.......ccccccscccccccccsccscececvecccseses 32 
XIV. Coefficients for Gradienter........... cece cee cceseeeeescesoecces 327 
XV. Offsets for Transition Curves............ ceeees eee cccec cence 328 
XVI. Tangent Distances for Transition Curves.....ccsccsessecees 330 
XVII. Deflection Angles for Transition Curves..... eee ccc ccecenae 332 
XVIII. Earthwork Tables, Different Slopes and Bases.............. 333 
XIX. Earthwork Tables, Two Slopes and All Bases.............. 337 
XX. Sines and Cosines........... cece ccc cen secceneececeeeneseeeaees 338 
XXII. Tangents and Cotangents.......ccccccccvccccccevccscccsscccces 352 
XXII. Versines and Exsecants........cccccceccccscccccsccccccccceeese 359 
XXIII. Useful’ Numbers and Formulas..............ccceccscvcccscees 382 
XXIV. Conversion of Feet into Meters and Meters into Feet; 
also Miles into Kilometers and Kilometers into Miles..... sevee 383 


FIELD-MANUAL FOR ENGINEERS. 


CHAPTER I. 
PRELIMINARY OPERATIONS. 


1. THE engineering operations preparatory to the construction 
of a railroad are : 

The Reconnoissance ; 

The Preliminary Survey or Surveys ; and 

The Location. 


2. The Reconnoissance is a general but incomplete examination 
' of the country through which the proposed road is to pass, made 
for the purpose of acquiring data upon which surveys may be 
_ made and compared, and the best possible route for the road 
selected. 


3. A Preliminary Survey consists of the measurement of a line, 
including its angular deflections ; the elevations of various points 
upon it, the determinations of the topography along it and near 
it, for the purpose of furnishing the data from which the line 
may be definitely located; or the survey compared with other sur- 
veys, for the purpose of selecting one from which the location 
may be made. 


4. The Location consists in placing the line in the exact posi- 
tion in which it is intended to be. This nosition is called The 
Location. 


6. It is convenient to carry on these operations concurrently. 
The main points to consider in the location are the relative 
cost of grading and. bridging, and the relative grades and curva- 


6 FIELD-MANUAL FOR ENGINEERS, 


ture of the lines. In grading, the character of the soil for stabil- 
ity, in both ‘“‘cut” and ‘‘ fill,” should be considered. 

It is sometimes important to know the relative value of property 
traversed by different lines; and if the lines are far apart, the 
probable amount of traffic that the respective lines can command 
must also be taken into account, 


6. It is evident that the best possible location requires the least 
possible grading, bridging, curvature, etc., taken together, re- 
garding the cost and the expense of operating the road. We 
can afford, therefore, to increase the curvature, for example, if 
by so doing we can at the same time decrease the earthwork, and 
the line is bettered more by the latter than it is damaged by the 
former. 

The field-work of location has for its object to determine the 
exact position of the selected route on the ground, to establish the 
grade, to compute the amount of earthwork, decide upon the 
amount of bridging, etc. 


7. A railroad line usually follows the valleys of watercourses 
or the dividing ridges between watercourses, or crosses valleys 
and ridges more or less obliquely. 


8. The location on dividing ridges is perhaps the simplest of 
all. In this no bridges and few culverts are required ; and the 
elements governing the location are mainly the amount of earth- 
work and the curvature of the line. In this case asketch of the 
ridge, especially of its prominent features and governing points, is 
made while walking over it; and the preliminary line is accurately 
run, and made into a location, if the route is adopted. 


9. The location along the valley of a stream is usually more 
complex than the former. If the stream is so small that the cost 
of bridging it would be plainly less than the advantage to the 
alignment by crossing ; or if, on the contrary, the river is so large 
that the crossing of it is out of the question, the cost of bridging 
is not considered and the problem of location is reduced, in the 
main, to that of making the best alignment within the limits of 
the valley in the one case, or upon one side of the river in the other. 
The reconnoissance and surveys would be made as already de- 
scribed. 

Usually the most favorable ground both for alignment and con- 
struction alternates from one side to the other of the stream, and 


PRELIMINARY OPERATIONS, " 


only the results of careful and scientific surveys can tell how 
many and where the crossings must be in order to make the align- 
ment and grades the best possible, and to secure the most favora- 
ble ground. In this case the ground on both sides of the stream 
must be carefully examined, and if the stream, in consequence of 
banks or bottom, is not easily crossed at most points, the most suit- 
able crossings must be found. When this is done the engineer 
will mark out and survey one or more lines so as to fit the chosen 
crossings and other governing points. A comparison between 
different lines will point out the best, and generally the one from 
which the best location can be made. 


10. In locating a line across valleys and ridges, the engineer 
must find the best crossings of the streams, and the gaps or notches 
in the ridges, and must connect such of the former with such of 
the latter as will furnish the best line. 


11. Sometimes a railroad may occupy either valleys or dividing 
ridges for the greater part of its length ; in which case a choice 
must be made between the higher and the lower line. 

The higher line will require very much less drainage than the 
lower line, which is an important advantage; but, on the contrary, 
probably the curvature and the length of the higher line will ex- 
ceed that of the lower. 


12. It is not to be supposed that, in general, a railroad will 
follow either a valley ora ridge; or that it will cross valleys and 
ridges obliquely throughout its entire length; but parts of the 
line will generally do so, and to these, and therefore to the whole 
line, the preceding principles will apply. 


13. The regular ‘‘ reaches ” in a stream furnish the best cross- 
ings. Throughout these, compared with other points, the flow of 
the current is most uniform; the wash of the bottom, and the 
caving and the shifting of the banks, are least; while the security 
of the foundations of bridges and of approaches is greatest. If 
the bottom of the stream is variable, the best site on some given 
‘reach ” must be selected with a view tothe kind of foundations 
»uitable to the place. Sharp bends in streams should be studi- 

asly avoided. 


14. The engineer should freely consult the best maps of the 
country that he can command, and he should prepare a map on « 


8 FIELD-MANUAL FOR ENGINEERS, 


convenient scale, upon which he should copy the principal features 
of the country, such as streams and lakes, roads and towns, and 
fill in the details as he progresses. He should also locate on the 
map the governing points of the route, such as the best crossings 
of streams, the ‘‘gaps” in the ridges, mountain passes, etc. He 
may then sketch the line on the map. 

In a densely wooded country, the making of a thorough recon- 
noissance is comparatively difficult. In this case it will often be 
necessary to cross and recross the country many times before a 
comprehensive knowledge of it can be gained. 


15. The one almost indispensable instrument in making a re- 
connoissance is a pocket-compass. Field-glasses, hand-levels, 
telemeters, and other instruments are sometimes used, but are 
rarely needed, and cannot be used to advantage. Those who 
cannot make a proper reconnoissance without them would better 
employ some one who can, and turn their attention to other parts 
of the work, 


16. For preliminary surveys the corps of engineers may be 
constituted as follows: A chief engineer or engineer in charge, an 
assistant engineer or transitman, a levelman, a rodman, a stake- 
man, a rear flagman, two chainmen, and one or more axmen accord- 
ing to needs, The head flag should be carried by the head chain- 
man. 


17. Since a survey can be made more rapidly with the compass 


than with the transit, the compass may be used in preliminary 


work to a limited extent. Owing, however, to the inherent in- 
aecuracies of the compass, a line run with it is generally worthless, 
except as a guide to a transit preliminary, from which the location 
may be made. The compass is therefore of very little, if any, use 
in a comparatively level country, but is useful, if at all, for run- 
ning the first preliminary line in a hilly region, where several lines 
must be run. The compass should be light, should be mounted 
on a Jacob’s-staff, and should have a narrow slit in one sight, and 
a fine platinum wire, or its equivalent, stretched along the center 
of a wide slit in the other sight. A self-reading rod is best for 
this work because it saves much time over the sliding rod; and 
because, too, it enables the levelman to do his own reading. The 
ax for driving stakes should have a broad head. Stakes shouid 
be of a generous length, say from 80 to 36 inches, well driven into 


PRELIMINARY OPERATIONS. 9 


the ground, and projecting above the grass and other vegetation, 
so that the line may be easily followed or recovered, in fields or 
woods, by the engineer and others. Short stakes occasion the loss 
of much time, and those shorter than about 80 inches are generally 
a nuisance, except on streets, well-traveled roads, etc. 


18. To make a successful reconnoissance requires a good eye 
for distances, elevations, etc., and a quick and clear perception of 
the salient features of the country. One must be able to form a 
mental picture or image of the country along the proposed line; 
and of a number of lines crossing and recrossing each other within 
the limits of his mental map, Thus he may be able to sketch a 
proposed line, or several proposed lines, which at any time may be 
tested. The comparison of lines, to test their relative economy, 
is a question of science, aided by mathematics. 


19. It is plain that without skill in making the reconnoissance, 
the surveys would be at first more or less at random, and the 
reaching of a location roundabout and expensive. On the other 
hand, the greatest skill without a sufficient knowledge of mathe- 
matics and without a knowledge of the principles of engineering 
bearing upon the question, cannot produce the best location. 

A happy combination of the qualities necessary to the successful 
locating engineer is most assuredly found in comparatively few 
individuals, 

The two main obstacles to contend with in building a railway 
are grades and curves, Since both affect the cost of construction 
and the expense of operating the road, such grades and curves as 
will render the total cost of construction and operation of road a 
minimum are, as already suggested, the best. Since the resistances 
due to grades as well as to curves add to other resistances to the 
movement of trains, and since it is often possible to lessen grades 
at the expense of curvature, and vice versa, it becomes necessary 
to briefly consider the nature of the resistances that a moving train 
encounters, with a view to compensation. 

The resistances to overcome are, first, the friction of the moving 
parts of the engine and train; the friction of the wheels on the 
rails; impacts and oscillations, and the resistance of the air. 
These resistances vary with the condition of the rolling machinery, 
the road, and the weather, and are not accurately known. Fric- 
tion is nearly independent of the speed of the train; but the res’ 


10 FIELD-MANUAL FOR ENGINEERS. 


ances due to impact increase with the speed, and those due to the 
atmosphere increase in a still greater ratio. 

The sum of these resistances on a level track in fairly good con- 
dition and average fair weather is, according to Vose, 


= +8. wee ee ee DY 


in which 7 is the resistance in pounds per ton, and 9 is the velocity 
in miles per hour. It is not to be supposed that this formula is 
accurate, though perhaps it is as nearly so as any; and probably 
the resistances are no greater than the formula indicates, With 
a velocity of 20 miles per hour the formula gives 7 = 10.34 pounds 
per ton of the entire weight. Under unfavorable conditions, as a 
wet, soft, and rough track and high wind, the resistance might 
be 20 to 40 per cent more, or even greater still. 

The second resistance is due to grades of the track. The force 
necessary to overcome this is such a part of the total weight of 
the train as the vertical rise of the grade isto its length. Let 
7 = resistance per ton, and 4 = the ascent per 100 feet. Then 
y= a < 2000 pounds per ton. For a 1% grade (52.8 feet per 
mile) A = 1 and r’ = 20 pounds per ton, the force necessary to 
overcome the load. <A 4% grade would of course create a resist- 
ance of 10 pounds per ton, or approximately the same as the 
resistance caused by friction, impact, resistance of the atmos- 
phere, etc., at a speed of 20 miles per hour, as stated above. 

The third resistance is due to the curves of the track. This re- 
sistance is not accurately known. On Americanroads with American 
rolling stock it is probably about one half a pound per ton for each 
degree of curvature. Letting 7’ represent the resistance per ton 
per station, and D the degree of curvature, the resistance per sta- 
tion is 7’ = .5D. We observe that the resistance of a 1° curve 
is but the 7, part of that due to a 1% grade, or equivalent to 
vs = 1.82 feet per mile. The resistance of a 5° curve is equiva- 
lent to 6.6 feet per mile, and that of a 10° curve to 13.2 feet per 
mile, etc. ‘The resistance offered to a train in moving through 100 
feet of a 1° curve is one-half pound per ton moved 100 feet, which 
is equivalent to the total load lifted _ X 100 = zy = .025 ft.; 
and since the resistance varies as the product of the degree of 


PRELIMINARY OPERATIONS. 11 


curvature and length of the curve, or as the total curvature, 
tbe resistance on‘ any curve for each degree of change of direc- 
' tion is equal to the lifting of the train j, of a foot. Since an en- 
gine can haul over a road only what it can haul over the most 
unfavorable part, easy curves are necessary where heavy grades 
must occur, and vice versa ; the object being, of course, to keep at 
all points the sum of the resistances due to grades and curves be- 
low the allowable maximum. In equating for grades and curves, 
however, the ruling element is largely the total ascent and not 
the slope merely, as is too often considered to be the criterion. 
No more power is required to make an ascent of 20 feet, for ex- 
ample, on a 2% grade 1000 feet long than is required on a 3% grade 
4000 feet long. 

To compare grades simply, that is, the rate of ascent or descent 
especially on short grades, as is usually done, is meaningless at 
least—it is absurd. 

On light grades no allowance usually need be made for curva- 
ture, since the momentum of the train would carry it over the 
ascent even without aid from the engine. Grades situated near 
Stations, however, or other points where a train must stop or 
“slow up ’’ are greater obstacles than in other situations to the 
inovement of trains, and upon such grades an allowance for cur- 
vature, as above shown, should be made. 

Every locating engineer is presumed to make himself familiar 
with the principles involved in locations before making them; 
and to aid in this tle reader is referred to Vose and others, and 
especially to the elaborate treatise of Wellington on ‘‘ Railway 
Location.”? Wellington gives in Table 118 the ‘Total Energy, 
or Potential Lift, in Vertical Feet in Trains moving at Various 
Velocities,’ and discusses at, length all subjects connected with 
location. 


CHAPTER II. 
ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 


THE TRANSIT. 


The following are the usual adjustments of the transit : 

A. To Adjust the Levels (that is, to place the levels in a plane 
perpendicular to the upright axis),—Set up the instrument upon 
its tripod as nearly level as may be; unclamp the plates and bring 
the two levels above and on a line with the two pairs of leveling- 
screws. Then by means of one pair of screws bring the bubble 
of the level above them to the middle of the opening. Without 
moving the instrument bring the other bubble to the middle in 
the same way. Since in moving one pair of screws very far the 
other pair is liable to become cramped and the corresponding 
bubble somewhat disturbed, it is advisable to bring the bubbles | 
in succession near the middle, repeating if necessary, and ending | 
by bringing them exactly to the middle. 

When both bubbles are in place turn the instrument through 
about 180°; if the bubbles are now in position they need no cor- 
rection; but if not, turn the small nuts at the end of the levels 
until the bubbles are moved over one half of the error. Then 
bring the bubbles again to the middle by the leveling-screws and 
repeat the operation just described until the bubbles will remain 
in the middle during a complete revolution of the telescope. This 
shows the adjustment to be completed. 

B. To Set the Cross-wires Vertical and Horizontal.—Level the 
instrument. Move the telescope upward or downward and note 
whether the vertical wire traverses some fixed point or not. If 
not, loosen the four cross-wire screws and, by the pressure of the 
hands on their head outside the tube, move the cross-wire ring 
around, what seems to be sufficient, and repeat the operation, if 
necessary, until the vertical wire will traverse some fixed point. 

The cross-wire screws are near the screws of the centering-ring 


ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 13 


of the eyepiece (the heads of which are usually covered by an 
outside ring), but between them and the axis of the telescope. 

C. To Make the Line of Collimation Perpendicular to the Axis of 
the Wyes, so that tt will revolve in a plane.—Set up the transit 
at some point 0 near the middle of a level piece of ground 
and level it carefully. Let AOB be a straight line, AO and 
BO being nearly equal. Direct the line of sight to A, clamp 
the instrument and revolve the telescope, and if the line of 
collimation is not perpendicular to the axis, the line of sight 
will determine a point C, say, on one side of B. To test the mat- 
ter, loosen the upper clamp and turn the vernier plate almost or 
quite half-way around, so that the line of sight will again be on 


Cc 

A 5 8 
8) 

Fie. 1. \ 


A, and clamp the plates. Again revolve the telescope and note the 
point D, suppose, thus determined. If the point C coincides with 
B, D will also coincide with B and the line of collimation is in 
adjustment. If, however, C is on one side of B, as shown, D will 
be equally far on the other side, showing the line of collimation 
out of adjustment. 

To correct the error, use the two capstan-head screws on the side 
of the telescope, to move the ring to which the wires are fastened 
laterally, and with it, of course, the intersection of the wires... 

Having moved the vertical wire until by estimation one fourth 
of the space DC is passed over, return to A, clamp and revolve 
the telescope, and if the correction has been carefully made the 
line of sight will now cut the point B. It will, however, generally 
be necessary to repeat the operation, the adjustment not being 
perfected at first. 

Remember that the eyepiece inverts the position of the wires, 
and therefore in moving the ring the operator must proceed so as 
seemingly to increase the error. 

The wires may now be brought into the center of the field of 
view by moving the screws of the centering-ring of the eyepiece, 
which are slackened and tightened in pairs, the movement being 
now direct until the wires are seen in their proper position, 


14 FIELD-MANUAL FOR ENGINEERS. 


It is proper to observe that the position of the line of collima- 
tion depends solely upon that of the objective, so that the eye- 
piece may be moved in any direction, or replaced by another 
without at all deranging or in any way affecting the adjustment 
of the wires, 

D. To Adjust the Standards to the Same Height 0 that the Line 
of Collimation will Revolve ina Vertical Plane.—Set the transit 
as close as convenient to the base of a lofty spire, or other high 

object; level it carefully and clamp it, and 

A direct the telescope to the top of the spire or 

other elevated and well-defined point, as A 

in the figure. Bis in the same vertical plane 

with A and the instrument, that is, with the 
line of sight. 

Turn down the telescope to some good point 
on the ground, either found or marked, as C. 
Unclamp the plates or spindle, revolve the 
telescope and turn it half-way around, or far 
enough to again sight to the high point. 
Again clamp and turn down the telescope to 
C B D some point D opposite C. 

In setting C the left standard must have 
been too high, and in setting D the same 
standard (now the right by revolving the telescope) shows to be 
equally high, the errors BC and BD being equal. Correct the 
error by raising or lowering the sliding-piece at one end of the 
axis by means of the screw beneath and those above, so that when 
the telescope is directed to A and lowered it will cut B half-way 
between C and D. If the instrument is in adjustment it will, of 
course, cut B instead of C and D in the first two trials, 

E. To Adjust the Vertical Circle.—Set up the instrument firmly, 
level it carefully, bring into line the zeros of the circle and 
vernier, and with the telescope find some well-defined point, from 
two to five hundred feet distant, which is cut by the horizontal 
wire. Turn the instrument half-way around, revolve the tele- 
scope, fix the wire upon the same point as before, and observe if 
the zeros are again in line. If not, loosen the capstan-head 
screws which fasten the vernier, and move the zero of the vernier 
over half of the error. Bring the zeros again into coincidence 
and proceed as before, as many times as necessary, until the 
error is entirely corrected. 


Fia. 2. 


ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 15 


F. To Adjust the Level on Telescope.—First level carefully and 
camp the telescope approximately horizontal by the eye. 

Then, having the line of collimation previously adjusted, drive 
astake, say two or three hundred feet away, and note the height 
cut by the horizontal wire upon a staff set on top of the stake. 

Fix another stake in the opposite direction and at the same dis- 
tance from the instrument, and without disturbing the telescope 
turn the instrument upon its spindle, set the staff upon the stake, 
and drive the stake into the ground until the reading is the same 
as on the first stake. The tops of the two stakes are equally high 
however much the telescope may be out of level. 

Now set the instrument some twenty or thirty feet from one of 
the stakes and on the prolongation of the line joining the two. 
Level the instrument, clamp the telescope as nearly horizontal 
as may be, and note the readings of the staff on the two stakes, 
If they agree, the telescope is level. If they do not agree, then 
with the tangent-screw move the wire over fully the whole error, 
as shown at the distant stake; repeat the operation just described 
a- many times as inay be necessary, so that the wire will give the 
saine reading at both stakes, showing that the telescope is truly 
horizoutal. Taking care not to disturb the position of the tele- 
scope, bring the bubble into the middle by tke little leveling-nuts 
at the ends of the tube, which will render the adjustment com- 
plete. 

The above are all the adjustments usually required at the hands 
of the engineer; for any others see the Manual of W. & L. E. 
Gurley. 


Usk AND CARE OF TRANSIT, 


The instrument should be set up firmly with the plates nearly 
level, and thus save much time in turning the leveling-screws, 
besides the worse than useless wear upon them. 

To do this: Set up the transit approximately over the desired 
point. Hold firmly the leg of the tripod on the left with the left 
hand, and move the foot of the leg on the right with the right 
hand in any direction, so as to bring the lower plate of the tripod 
head approximately horizontal, determined by placing the eyes in 
the plane of its upper surface and sighting as nearly as may be 
to the distant horizon. It will do no harm to bend the body the 
{rifle necessary to do this. This sight is so long that, according to 


mae om 


16 FIELD-MANUAL FOR ENGINEERS. 


the principle of the ‘‘ division of errors,’’ the plate can be readil 
set very nearly horizontal in almost any position of the instru 
ment. | 

Now place the left hip against one leg; grasp the opposite le; 
with the left hand, and the leg on the right with the right hand 
and keeping the relative position of the legs unaltered, move th 
instrument laterally over the desired point and lower it to th 
ground. 

Of course the two movements of tilting and sliding may be don 
together and in much less time than required to describe thy 
movement, 


To MEASURE THE ANGLE BETWEEN Two LINES OR OBJECTS. 


Level the instrument carefully; bring the zeros of the vernier; 
and limb together by means of the upper clamp and tangent-screw 
and direct the telescope toward one of the objects by means o! 
the lower clamp and tangent-screw. Upon loosening the uppe! 
clamp and directing the telescope toward the second object, the 
angle desired is then shown upon the limb, | 

Before making an observation with the telescope, the eyepiece 
should be moved in or out until the cross-wires may be distinctly 
seen. This may be accomplished with the greatest precision by 
directing the telescope toward some white object. The sky 
will serve very well for this. The objective is then adjusted by 
moving it in or out until the object is seen clear and well defined, 
and the cross-wires appear as if fastened to its surface. ! 

Water and dust are very destructive to instruments ; indeed, 
the most destructive of all agents is dust. An instrument may be 
badly bruised, bent, or broken, by a fall or otherwise, and yet it 
may be repaired. If, however, it is allowed to stand on its tripod 
in a dusty office—and all offices are dusty—its life will be short, 
The author has known of instruments so choice that their owners 
would allow no one to handle them lest they might become 
slightly soiled, or some other mishap befall them; and yet in a 
few months they were ruined, in the way. pointed out. A minute. 
quantity of dust on the sockets will probably cause them to grind; 
and more of it will increase the grinding so that the instrument 
will soon become useless, 

Whenever not in actual use the eyepiece should be covered by 
the lid, and the object-glass by the cap, to protect them from dust, 
moisture, rain, etc, 


ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 17 


When an instrument is exposed to the hot rays of the sun for 
-sone time its parts are subjected to very unequal expansion, which 
throws the instrument more or less out of adjustment, thus tosome 
extent vitiating the work and damaging the finer parts of the in- 
sirament. Toprevent this itshould be shielded by an umbrella or 
screen of some kind supported on top of a high stake; or, if nothing 
better is at hand, a cloth should be placed over the telescope. 

When an instrument is in the office or in transport it should 
ve clamped and placed in a stable position, upon proper supports, 
ina tight box, well cushioned if possible. 

If the box cover does not fit upon the box closely it can be 
remedied by sticking suitable cloth on the top edge of the box or 
on the under side of the lid. 

When handling an instrument it should be supported by plac- 
ing the hand under the lower plate. 

Tangent and micrometer screws should be used equally on all 
portions of their length. 

Keep the tripod legs tight enough on the tripod head to 
secure stability by tightening the nuts on the bolts when 
necessary. Also tighten the shoes of the tripod if they become 
loose. Neglect of these things sometimes prevents an engineer 
from keeping his telescope on a point, and causes him to run 
1a zigzag line without knowing the cause of it. Secure the 
instrument well to the tripod head before using it; and bring 
all four leveling-screws to a bearing and cover the instrument 
with an oilcloth hood before carrying it. Use a fine camel- 
hair brush or a piece of old and soft linen to clean the glasses 
of the telescope. 

Dust, moisture, perspiration, ete., will sometimes cause a 
film to form on the lenses of a telescope which may greatly im- 
pair the sight through it. 

To remove the film, the lenses, after being carefully brushed, 
should be gently wiped with a piece of chamois-skin moistened 
with alcohol, and the lenses must be wiped dry by using fresh 
portions of the skin on separate parts of the lens. 

To remove dampness in the main tube of the telescope, take 
out the eyepiece, cover the open end with cloth, and leave the 
instrument in a dry room for some time. 

The centers of an instrument should always be lubricated 
“with fine watch-oil only, and after a careful cleaning. First 
wipe off all old grit and oil before applying fresh oil, 


18 FIELD-MANUAL FOR ENGINEERS. 


If dust settles on the cross-wires, unscrew the eyepiece and 
the object-glass, and gently blow through the telescope tube; 
cover up both ends and wait a few minutes before replacing 
the eyepiece and object-glass. 

Be sure to bring the object-glass cell to a firm bearing against 
its shoulder, and then examine the adjustment of the lines of 
collimation. | 

To clean the threads of a leveling or tangent screw use a 
stiff tooth-brush to remove the dust, then apply a little oil 
and turn the screws in and out with alternate brushing to 
remove dust and oil, until it moves freely and smoothly. Use 
such a brush to clean the object-slide. No screws should be 
strained more than necessary to insure a firm bearing, and 
this applies with special force to the cross-wire screws. : 

All straining of such screws beyond this impairs the accuracy 
of the instrument and the reliability of adjustment. | 

These remarks in reference to the care of the transit apply 
substantially, of course, to all instruments. 

If obliged to work with an instrument of faulty graduation, 
it is best to read each angle on different parts of the circle, 
and take the mean value. If we take the mean result obtained, 
by both verniers, we eliminate the errors due to eccentricity 
of the vertical axis, and also reduce the errors of graduation. 

For greater accuracy clamp the vernier-plate to zero and read 
the angle by both verniers. Then keep the vernier-plate 
clamped, point the telescope to the first object and proceed as 
before, any number of times. Then read the verniers, adding 
360° for each complete revolution which has been made, and 
divide this sum by the number of times the angle was read. 
The quotient is the required angle. 

The cross-wires can be illuminated easily by placing a piece 
of white cardboard, with a hole through it for the line of sight, 
in front of the telescope, and in an oblique position, so as to. 
reflect into the telescope the rays of a lamp or of a lantern 
placed back of the object-glass and near the telescope. 


THE LEVEL. 
The principal adjustments of the level consist in the follow- 
ing: 
], Bringing the cross-wires into the optical axis of the tele- 


ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS, 19 


wope and consequently parallel to the line of bearings of the 
“ye rings. ; 

2. Causing the line of bearings to be parallel to the plane 
it the level. 

3. Making either of these lines, and therefore all of them, 
rarallel to the bar and consequently perpendicular to the 


iris Of the instrument. 


]. To adjust the line of collimation, that is, to bring the 
‘ross-Wires into the optical axis, so that their point of inter- 
ction will remain on any given point during an entire revolu- 
tion of the telescope. 

Set the tripod firmly, remove the wye-pins from the clips, 
~v as to allow the telescope to turn freely; clamp the instru- 
ment to the leveling-head, and by the leveling and tangent 
«rews bring either of the wires upon the clearly defined edge 
of some object. Then carefully rotate the telescope half-way 
around. 

If the wire does not coincide with the line observed, bring 
it half-way back by means of the capstan-head screws at right 
angles with it, always remembering the inverting property 
of the evepiece. Then by the leveling and tangent screws 
hiring it again upon the “edge.” and repeat the above opera- 
tion if necessary, and continue to do so until the telescope may 
le rotated without changing the position of the wires. 

If both wires are much out of position, it will be well to 
approximately adjust the wires alternately before attempting 
tu make the adjustment of cither one complete, since an error 
in one somewhat affects the other. 

It may be advisable to center the eyepiece. To do so unscrew 
the covering of the eyepiece centering-screws, and move each 
pair in succession, with a screw-driver, until the wires are 
brought into the center of the field of view. 

The inverting property of the eyepiece does not affect this 
operation, and the screws are moved directly. To test the cen- 
tering, rotate the telescope, and if an object observed appears 
tu change position the centering is not perfect. 

In all telescopes the line of collimation is determined by the 
-ross-Wires and objective, and is not affected in any way by the 
eveplece. 

2. To make the line of bearings parallel to the bubble-tube. 


20 FIELD-MANUAL FOR ENGINEERS, 


so as to insure that it is horizontal, when the bubble is in the 
center. This adjustmenteembraces two parts: 

First, to bring the center-line of the bubble and the line of 
bearings in the same plane. 

Second, to make these lines parallel. 

To effect the first: Clamp the level and bring the bubble 
to the center by the parallel-plate screws. Now rotate the 


telescope in the wyes 20°, more or less. If the bubble runs . 
toward the end, it shows that the center-line of the bubble and | 


the axis of the telescope are not in the same plane; in other 
words, the bubble-tube lies crosswise of the telescope. 
To correct the error, bring the bubble by estimation half- 


wavy back, by the capstan-head screws, which are set in either | 


side of the level-holder. 


Again bring the level-tube under the telescope, the bubble | 
to the middle, etc., repeating the operation just described until | 
the bubble will keep its position, when the telescope is re-_ 


volved. 

For the second part: Bring the bubble to the middle of the 
tube by the leveling-screws, and take the telescope out of the 
wyes carefully and turn it end for end. 

If the bubble runs toward either end, lower that end, or 
raise the other by turning the adjusting-nuts on one end of the 
_ level until by estimation half the correction is made. Again 
bring the bubble to the middle by the leveling-screws, and 
repeat the whole operation just described until the reversion 
ean be made without causing «ny change in the bubble. 

3. Having made the previous adjustments, it remains to 
make the level-bubble (and therefore the line of bearings and 
line of collimation) parallel to the bar, and therefore perpen- 


dicular to the vertical axis of the level, so that the bubble will | 
remain in the middle during an entire revolution of the tele- » 


scope. 


of them bring the bubble to the middle. Turn the instrument 
half-way around horizontally. If the bubble runs toward either 


Place the level over a pair of leveling-screws, and by means © 


end, bring it half-way back by either pair of nuts, at the ends — 
of the bar. Then bring the bubble to the middle again, by the | 


Jeyeling-screws, ete., repeating the operation just described, 


| 


ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 21 


until the bubble will remain in the middle of the tube when 
the instrument is revolved. 

In making this adjustment it is best to use the opposite 
pairs of leveling-screws alternately, thus bringing the upper 
parallel plate of the tripod head into a position as nearly hori- 
zontal as possible, so that the error caused by not revolving the 
instrument precisely 180° may be the least possible. 

This adjustment is for convenience and not for accuracy in 
any appreciable degree. 

Now turn the telescope in the wyes until the pin on the 
clip of the wye will rest in the little recess in the ring to which 
it is fitted. 

Apply the horizontal wire to any level line, and in case it 
does not coincide with it, loosen two cross-wire screws at right 
angles to each other, and by their heads outside turn the cross- 
wire ring until the horizontal wire coincides with the level 
line. 

The line of collimation must then be adjusted again. In 
readjusting the line of collimation none of the lines referred to 
in making the adjustment is disturbed, and the adjustments 
are complete, 


ADJUSTING BY THE “ PEG” METHOD. 


By this method the main adjustments are effected at once. 

Drive two pegs several hundred feet apart, and set the in- 
strument midway between them. Read the rod on each, keep- 
ing the bubble in exactly the same position, preferably at the 
center. The difference of the readings is the difference of the 
heights of the pegs, no matter how much or in what way the 


_level may be out of adjustment. Then set over either peg and 


measure the height of cross-wires above top of peg. 

The difference of heights of pegs, added to this, or taken 
from it, according as the instrument is over the higher or lower 
peg, gives the height of the cross-wires above the other peg. 
Set the rod on that peg, and bring the horizontal wire to that 
height on the rod by the leveling-screws, keeping them at a 
bearing. Then bring the bubble to the center by raising or 
lowering one end of the level-tube. 

The first part of the second adjustment,—namely, to bring 


22 FIELD-MANUAL FOR ENGINEERS. 


the level-bubble and line of collimation in the same plane,— 
also the third adjustment, should be made as heretofore. 


USE OF THE LEVEL. 


The instrument should be set up firmly, with the top of the 
tripod as nearly level as may be, so as to save time in leveling 
and the wear of the screws, etc. | 

The setting up of the level is, of course, similar to that of the 
transit already described. 

The bubble should then be brought over each pair of level- 
ing-screws successively, and leveled in each position. | 

Bring the wire precisely in focus by the eyepiece, and the 
object distinctly in view by the objective, so as to avoid all 
“traveling of the wires” or parallax. 

It is best, where practicable, to take approximately equal 
fore and back sights, so as to eliminate any error due to a lack: 
of perfect adjustment, which is difficult to secure. 

For precise reading the rod should not be over 400 or 500 
feet from the instrument. 

If the socket of the instrument sticks in the leveling-head so, 
as to be difficult to remove, be sure that the instrument is un-' 
clamped and the leveling-screws are free. Then place the palms 
of the hands under the wye-nuts under each end of the bar, 
and give a sudden upward blow to the bar, and take care also 
to grasp it the moment it is free. 


To ADJUST THE COMPASS. 


The Levels.—First bring the level-bubbles into the middle 
by the pressure of the hands on different parts of the plate; 
then turn the compass half-way around. If either bubble runs 
toward one end of its tube, it indicates that that end is too 
high. 

Lower it by loosening the screw under the lower end, and 
tightening the one under the higher end, until the error is, 
by estimation, half removed. Level the plate again, and repeat 
the operation until the bubbles will remain in the middle dur- 
ing an entire revolution of the compass. 

The Sight-vanes.—The sights may next be tested by observ- 
ing through the slits a fine hair or thread made exactly verti- 
cal by a plummet. 


ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS, 23 


If either slit does not coincide in direction with the hair or 
‘tread, it must be made to do so by filing its under surface on 
the higher side. 

The Needle.—Having the eye nearly in the same plane with 
the graduated rim of the compass circle, observe whether or not 
the ends of the needle cut opposite points on the rim. If net, 
bend the center-pin (by means of a small wrench) about an 
eighth of an ineh below the point of the pin, so as to make the 
ends cut opposite points.on the rim. 

The needle now may be supposed to occupy -the position 
Yps, the pivot p not being 
in the center O. Now N’ AN 


keeping the needle in the 
same position, turn the 
compass half-way around. 


The needle will now oc- 
cupy the position N’p’S’. 


Correct half the error by £ Ww 
bending (that is, straight- ’ 

fring) the needle, making 

t cut points half-way be- 

tween its former positions, 

ko that it occupies the posi- 


tion Ap’B and is straight; SBS 
and correct the other half Fig. 3. 
by bending the pin, placing 
the point of the pin at O and giving the needle the position 
NOS. 

The operation should be repeated until perfect reversion is 
secured in the first position. 
_ Then try the needle on another quarter of the circle, and if 
an error is manifested, correct the center-pin only, the needle 
being already straightened by the previous operation. 

Do the same on other quarters of the circle until the needle 
will reverse in any position. 


To USE THE COMPASS. 


In using the compass keep the south end toward the person, 
and read the bearings from the north end of the needle. 
| Mark every station or point at which the compass is set, so 


24 FIELD-MANUAL FOR ENGINEERS, 

that it may be easily found for verification or a resurvey. It is 
much more important to have the compass level laterally, or 
crosswise of the sights, than in their direction; since if it is 
not so, on looking up or down hill through the lower part of 
one sight and the upper part of the other the line of sight will 
not be parallel to the N. and S. or zero line on the compass, 
and an incorrect bearing will be obtained. 

A continuous line thus run, to say nothing of other im-, 
perfections of the compass, would be a zigzag line probably, 
very much in. error. | 

The compass cannot be leveled by the needle, for the dip of, 
the needle is continually varying. If the needle touches the 
glass when the compass is leveled, balance it by sliding the coil 
of wire along it. 

The vibrations of the needle may be checked by gently rais-| 
ing it off the pivot, so as to touch the glass, and letting it, 
down again, by the screw on the under side of the box. | 

The compass should be smartly tapped after the needle has 
settled, to destroy the effect of any adhesion to the pivot or 
friction of dust upon it. 

The glass sometimes becomes charged with electricity by 
carrying it against clothing, or wiping it, etc., so that it at- 
tracts the needle to its under surface, preventing its free move- 
ment. 

The difficulty may be remedied by breathing on the glass, 
or touching it in different places with the moistened finger. 

Of course the chain and all other metals should be kept away 
from the needle. 


CHAPTER III. 
PLANE TRIGONOMETRY. 


1. Plane Trigonometry treats of the relations of the sides and 
angles, and of the solution of plane triangles. 

2. Some French writers divide the right angle into 100 degrees, 
the degree into 100 minutes, the minute into 100 seconds, etc. 
This centesimal system, in which reductions ‘are made by simply 
moving the decimal point, is altogether preferable to our sezagesi- 
mal system. 

3. Two angles whose sum is equal to 90° are complementary. 

Two angles whose sum is 180° are supplementary. 

4. Let us consider a series of right triangles ABC, AB’C’, etc., 
having the-common angle A. The triangles are equiangular and 
therefore similar, and we have 


BC BC’ B'Cc" (1) 
4B AB’ ~ AB™' 


BO BC’ _ Bo" @) 
40 AG’ “AG”? 


Fia. 4. 


Thus it appears that the ratios of the sides are the same in all 
right triangles having the same acute angles; and therefore if 
these ratios are known in any one of these triangles, they will be 
known in all of them. 

As any triangle may be divided into two right triangles, it is 
evident that the solution of oblique triangles may be made to de- 
pend upon the solution of right triangles. 

The above ratios, depending upon the angle alone and not at all 
apon the absolute lengths of the sides, may be considered as indices 

25 


Ww 


6 FIELD-MANUAL FOR ENGINEERS. 
of the angle, and have received special names, which we will ex- 
plain. 

5. Let us represent the sides and angles of a triangle in the 
usual way, shown in Fig. 5. The 
side opposite an angle, divided by 

c the hypothenuse, is called the sine of 
@ that angle. Thus 
A b C # — sin A, and = sin B. 
Fig. 5. ¢ 


a ,o 


The side opposite an angle, divided by the adjacent side, is called. 
the tangent of that angle. Thus | 


b 
? = tan A; ~ = tan B, 
b a 


The hypothenuse, divided by a side adjacent to an angle, is 
called the secant of that angle. Thus 


= sec B. 


Q/° 


c= sec A; 
b 

6. The terms cosine, cotangent, and cosecant are convenient 
abbreviations for the ‘‘sine of the complement,’’ ‘‘ tangent of the 
complement,’’ and ‘‘secant of the complement,” respectively. The 
reader must not suppose that there is such a thing or entity as co- 
sine, cotangent, or cosecant of an angle. 

Since the acute angles of a right-angled triangle are comple, 
mentary, that is, A = 90° — Band B = 90° — A, it follows that , 


cos A means the sine of (90° — A), or sin B; 
cot A means the tangent of (90° — A), or tan B; 
cosec A means the secant of (90° — A), or sec B, ete. 


Hence 

. a . b 

sin A = cos B= —; cos A = sin B= —; 
a b 

tan d= cotB=-~; cot A=tan B= ~; - (4) 
b a’ | 
C ¢ 

sec A = cosec B= =; conee A = see B =. | 


PLANE TRIGONOMETRY. 24 


Sin A = cos B is really an identical equation, since cos B is the 
sine (90° — B) = sin A; and so is tan A = cot B, etc. Furthermore, 


vers A = ane — cos A, 
c— bd 1 ‘1—~cosA_ versA 
exsec A = b =secA—L= Gg -1= aed wed’ 


7. Plane trigonometry, applied to plane triangles, is but the ap- 
plication of the one well-known proposition in geometry: ‘‘ Equi- 
angular triangles have their homologous sides 
proportional and are similar,” B 

Comparing mgs. » and 6, we observe that 
the above ratios © - os etc., change as the angles 
Aand B change. These ratios have been com- y a 
puted, however, for all values of the angles, 
differing by single minutes or less, and placed 
in tables under the corresponding headings, 
sine, tangent, etc., and opposite the correspond- 
' ing angles, 

S. From the above equations we observe that 


Fia. 6. 


1 
cosec A = — 7 (5) 
cos A = a 6 
—~ sec A’ ° e ® o ° ° (6) 


1 


cot A=: °- © «© @ 28 @ (2) 


Hence ; 
sin A cosec A = sec Acos A =tan AcotA=1. . (8) 


We also have, by division, 


sin A a 
cos A BT tn A; o e@ e e© @ @ (9) 
cos A b 
at a tA >. © © © @ @ (10) 


98 FIELD-MANUAL FOR ENGINEERS. 


Again, 
sin? A + cos? A = + o = “ =1. . (11) 
Again, 
sec? A =o = ope a14e a =1-+ tan? A, 
and ° 


seccA=W1+tan?A: ....., (12) | 


cosee! A = sec? B= © = ote =1+4 “= 1+oot' A, 
and | 
cosec A= 1+ cot?A. . . . . . (18) | 
From (9), 
sin A = cos A tana A = tan A __.tan A (14) 
sec 4 4/1 + tan? A 
From (10), 


cot A _ cot A 8) 
cosec 4 4/1 +. cot? A 


If one of the above functions of the angle A is given, all the 
others may easily be found. For example, if sin A is given, we 
have, from (11), 


cos A = sin A cot A = 


cos A = /1 — sin® A, 


Then 

PN (16) 

cos A 71 —sin? A’ 7 * 

cos cos A — ¥1— sin’ A — sin? A 
1 1 

sec A =e Cl ———====——" 5 e e e 18 
cos A 4/] — gin? A’ (18) 
A= . . 19 
cosec — sin A’ eo © «© © © © © © @ ( ) 


PLANE TRIGONOMETRY, 


If cos A is given, we have, from (11), 


sin A = 7/1 — cos* A 


Then tan A, etc., as above. 

Similarly when other functions of the angle are given. 

9. The sine and cosine of two angles being given, to find the 
sine and cosine of their sum, and the 
sine and cosine of the difference of their 


angles, 


In Figs. 7 and 8 let HOF = A, and 
FOH=B  ; then, in Fig. 7, HOG=A-+B, 
and, in Fig. 8, HOG=A—B. From 
any point #’ in OF draw FE perpen- 
dicular to OF; also EG and FAH per- 6 


pendicular to OH, and FX perpendicular 


to EG. Now the three sides of the tri- 
angle FHK are perpendicular to the three sides of the triangle 
FOU, and therefore FHK = FOH. 


Now, in Fig. 7, 
HG  FH4+ EK FH, EK 
sin (A + 5) = 55> no = 0+ Ko" 
But 
FH FH FO . 
Fo = Fo HO = 82 43 B, 
and 
EK EK EF : 
Fo > EF’ Ho = 084 sin B. 


Again, 


cos (A -++ B) = = 


sin (A + B) = sin AcosB+ cos Asin B. . 


oe OH — KF _ OH KF 
OE ~ 0H Of 


_ OH OF KF EF 
~ OF OK EF’ OF 


= cos A cos B — sin A sin B. 


29 


E 


GH 


(20) 


(21) 


30 FIELD-MANUAL FOR ENGINEERS. 


Then, in Fig. 8, 
sin (A — B) = EG HF— KE _ AF OF _ KE FE 
EO” HO — OF’ HO FE’ KO 
= sin AcosB—cosAsinB;. . . . « (22) 
a OH + FR FE _ OH OF . FK FE 
cos (A — B)= G5=—on = Or HOt FE’ OF 


= cos AcosB-+-sinAsinB.. . . « « (23) 


Fie. 8. 
In (20), make B = A and get 


sin 2A = 2sin A cos A. (24) 
In (21), make B= A and get 
cos 2.A = cos? A — sin* A 
= (1 — sin’? A) — sin? A= 1—2sin®?9A. . ,. . (25) 
== cos* A — (1 — cos* A) = 2cos*$A~—1.. . . (26) 
(20) and (21) give 


sin A sin B 
sin(A +B) _ sin AcosB+4+cosAsinB_ ‘cos A + cos B 
cos(A-+ 8)  cos.Acos B—sin Asin B sin A sin B’ 
~ ‘cos A cos B 
or 
tan A + tan B 
tan (A -+- B) = 1 — tan Atan B e e e e e e (27) 
Similarly, from (22) and (28), 
tan (A — B) = tan A — tan B r e ry e (28) 


1+ tan A tan B’ - 


A 


PLANE TRIGONOMETRY. 31 


We note some special values of the trigonometric functions. 
See Fig. 9. 

Let COP represent any triangle ; angle COP = O. Let O = 0. 
Then CP = 0, PO and BO coincide and are equal. 

Hence 

. 0 
sin 0 = 55 = tan0 = — =0, seccQ0=—.=1. 

Let O= 90°. Then PC and PO coincide with B’O, and CO = 0. 

Hence 
sin 90° = Bo =1, tan 90° = BO = 0, sec 90° = BO _ op. 


B'O 0 0 
Let O = 180°. Then PO and CO coincide with B’Oand PC = 0. 
Then 
0 0 B'"0 


sin 180° = BO = 0, tan 180° = Bo = 0, sec 180°= BO 


Let O = 270°. Then PO and PC coincide with B’’O and CO = 0. 
Hence 


; B’'O » BO ; _B’0 
sin 270° = Bon tan 270 =—7 =O, see 210° = 0 
The values of the functions of 360° are the same as those of 0°. 
We need not consider angles greater than 180°. 


= ©. 


SOLUTION OF PLANE RIGHT TRIANGLES. 


16. In order to solve a plane right triangle it is only necessary 
to select from equations (4) an equation containing the two given 
paris aside from the right angle and the part sought. By trans- 
posing, if necessary, so as to express the latter in terms of the 
former, it becomes known. There are two cases: 

CasE I.—A side and an angle given. Given A andc. See 
Fig. 5. 

Ezample.—Let A = 85° 23’, and c = 874.8. We have 


B= 54° 87’, 
Also table of sines and cosines gives 


sin 85° 23’ = .57904, and cos 35° 23’ = .81580. 


Hence a=csin A = 874.8 x .57904 = 506.54; 
b =ccos A = 874.8 x .81530 = 718,22, 


32 FIELD-MANUAL FOR ENGINEERS. 


CasE II.—Given two sides, 
Hrample,—Let a = 184.3, and c = 246. We have 


= .74919. 


. a 
sin A = — 
Cc 


We find in the table 


sin 48° 31’ = .74915, .°. A = 48° 31’ to the nearest minute; 
B= 90° — A= 41° 29; 6 =ccos A = 246 x .6624 = 162.95, 


or b= Vc? — a? = 162.95. 
TABLE FOR SOLUTION OF RIGHT TRIANGLES. 


Given. Required. Formulas. 


| 
| 
| , 
1. a, b | A, Bye tan A = 5, B= 90° — A, ¢=bsec A=asec B. 
a, ¢|A, B, 6 sin A ==, B = 90° ~ A, b= ccos A=atan B. 


A,a |B, b, ¢ | B= 90° — A, b= atan B, c=asec B=b sec A. 
A,b'B,a,c | B= 90° — A, a=btan A, c=a sec B=bsec A. 
. Ase |B, a,b | B=90° — A, a=csin A, d=a tan B=ccos A. 


Op 9 Bw 


If, in the second case above, dis given in place of a, then a 
and bas wellas Aand B change places in the formulas. However, 
A= 90° — B is the same as B = 90° — A. 

17. We will now deduce formulas for the solution of oblique 
triangles. 

Draw BD in Fig. 10 perpendicular to AC. Then, from the tri- 


B angle ABD, 
e a BD=csin A, . (29) 
| and, from the triangle BCD, 
C 
A b D BD=asinC@. . (80) 
Fra. 10. 


_ CD=AC—- AD=b—ccosA,. ... (81) 


PLANE TRIGONOMETRY. 3: 


Equating (29) and (80) gives 


. . a sinA 
csin A=asine, or —=-— . 
ec sind 
Similarly, or by analogy, ~ « « (82) 
a sinA and 6 sin B 
6” sin B’ c. sinG’ J 


From the figure, 
a? = (BD) + (CD). 
Substituting for BD and CD from (29) and (31), we have 
a? = c? sin? A + 0? — 26c cos A -} ¢ cos’ A; 


or, since sin? A + cos’ A = 1, 


2 2 2 
a’? = 0?+ ¢ —2becos A, oF cos A te | 
2be | 
Similarly, or by analogy, | (88) 
2 2 _ }? 2.14? — 2 
cos B= Oe and cos C= 2 .. 
2ac 2ab J 


From (29) and (31) we also have 
BD _ csinA sin A 


tanC= Gp =5-cosd 5b,’ 
~ — cos A 
sin 0 
tan B = —— 
® — cos O - (84) 
b 
tan A = - sin B 
—— cos B 
a 


18. Let ABO, Fig. 11, represent a plane triangle, the parts be- 
ing represented as usual. 

Take CH = CA, and draw AD and HH perpendicular to AZ. 
We have 


CAE + CEA = 180° — C=A+R. ... CAH=CHA =}(A+8), 
and BAEK = CEA — CBA = (A +B)-—- B= (A — B). 


34 FIELD-MANUAL FOR ENGINEERS, 


Also CAD =90°— CAH, and CDA =90° —-(AHC=CABE). 


Hence . CD= AC = b. 
Now 
a+ : _BD_AD_ AFtan}(A+B)_tan}(A+B) 35 
a— BE HH” AHtani(A—B)~tany(d—pBy © ©) 
D 
Fie 11. 
From triangle ABZ, 
BE _sinBAEF 
AB ~ sin AEB’ 
a—b_ sin}(A — B) ; 
or C — sin L(A + By e ee e e (36) 
In mane ABD, 
_@+6_ sin BAD _ cos}(A — B) 87 
a5 = c¢ sin ADB cos }(A + B) (37) 
Kq. (37) divided by (36) also gives 
bt 
a+6_ tan 4(A + B) (38) 


a—b tan}(A—By'*** 


which furnishes another demonstration for (35). 

A slight variation of the above solution of the tangent problem 
was given by the author in Vol. I, No. 1, of The American Math- 
ematical Monthly. It has since found its way into text-books on 
‘~igonometry. Still another solution by the author may be seen 

‘1. TIT, No, 11, of the same journal, 


PLANE TRIGONOMETRY. 3d 


SOLUTION OF PLANE OBLIQUE TRIANGLES, 
| 19. There are three cases. 
CasE I.—In this case two of the given parts are a side, and the 
angle opposite ; the other part being either a side or an angle, 


Example 1.—Let A, a, and B, Fig. 10, be given. 
C = 180 — (A + B). 


From (82), 


sin B similarly ¢ = a@ sin C 
sin A’ yY ¢= 47a’ 


b= 


Example 2.—A, a, and 0 given. 


(32) gives sin B = sin Az; then the table gives B, 


Now GO = 180° —(A-+ B); and ox ane 


sin 


CasE II.—Given two sides and the included angle, 
Let b, c, and A be given. We have, from (34), 

tan C= sin A A . 
——cosA 
c 


Then e 
sin A 


Or, from (33), 
a = (0° + c*? — 2bc cos Ay}. 
Then 


sin B = sin AZ, and (@ = 180° — (A +B). 


CasE III.—Given the three sides a, 6, and c. 
Eq. (88) gives 
b? + ec — q? 


A= 
cos She 


36 FIELD-MANUAL FOR ENGINEERS. 
Then 


sin B = sin A, and (@ = 180° — (A + B). 


The above formulas are all-sufficient for all practical purposes. 


This chapter constitutes a complete treatise on trigonometry, 


though the deductions from it are endless, as the examples in 
arithmetic are endless, 


TABLE FOR SOLUTION OF OBLIQUE TRIANGLES. 


(See Fig, 10.) 


Given. Required. Formulas. 


, _ asinB _asin(A+ B) 
6. A, B, a C, b, Cc b — sin A , Cc — sin A e 
7. A, a, b B, G, ¢ sin B — Ssin A o = sin (A+B) 
u sin A 
8. A, b,c | B,C, a| tan B = na q —? sin A 
. a cos A sin B 

_ @—(b—c) (a+b—c)(a+e—b) 

9. a, b, c A, B,C vers A = —3- — = De ’ 
sin B= b sin A 
— “a 


Use the first form for vers A with a table of squares, the second 
without; they are the best formulas known for this case. 

The following are the best formulas known for the area, 

10, Area iad sin C = jac sin B = joe sin A. 

It is never necessary to compute but one unknown part, and in 
the 3d case none at all, to have the required data for one of these 
equations; and the computation is shorter than by any other 
formula, 

Observe that asin Cis equal to the perpendicular from Bupon 4, 
6 sin ( is equal to the perpendicular from A upon «, ete. 


| 


UL. 


12, 


13. 


14. 


. exsec A = sec A —1 = 


. tan A + tan B= 


. cot A + cot B = 


PLANE TRIGONOMETRY. 


TRIGONOMETRIC FORMULAS. 


. vers A =1-— cos A = 2sin’? fA. 


vers A 
cos A~ 


. sin (A + B) = sin A cos B + cos A sin B. 

. cos (A + B) = cos A cos B F sin A sin B. 

. sin A + sin B = 2 sin 4(A + B) cos }(A — B). 
. sin A — sin B = 2 cos 3(A + B) sin 4(A — B). 
. cos A + cos B = 2 cos 4(A + B) cos 4(A — B). 
. cos B— cos A = 2sin 1(A + B) sin (A — B). 
. sin? A — sin? B= cos? B — cos? A =sin (A+ B) sin (A — B). 
. cos? A — sin? B= cos (A + B) cos (A — B). 


sin (A + B) 

cos A cos B° 

+ sin(A + B) 
sin Asin B * 


3 


~ 


wo a 1 
i = — cos? A = — = 2 sin . 
sin A Vi — cos? A <osee A 4A cos 14 
csA = Yi —sin? A= a = cos? 44 — sin? $f 
sec A 
= 2cos? 34 -1=1 — Ysin? $A. 
sin A 1 
= = - eee —_— t 2 
tan A cos A cord 7 CO8ee 2A — cot 2A 
_ 1—cos2A __ sin2Aa 
~~ gin2A4 ~— 1+ cos2A° 
cot A = cosec 2A + cot 2A = the reciprocal of any expres- 
sion for tan A. 
A = _— the reciprocal of any expression tu> cos A 
. sec -= oan p y exp ur cos A, 
1 , ; , 
. cosecA = = the reciprocal of any expression forsin A. 


38 FIELD-MANUAL FOR ENGINEERS. 


The above formulas contain the practical general relations exist. 
ing among the functions of an angle. By writing 4A, 2A, etc., 
in place of A, by repetitions, etc., the formulas may be greatly 
multiplied without producing any new relations. ‘This practice 
is too common in ‘Trigonometries, Field-books, etc, 


yas. 


CHAPTER IV. 
SIMPLE CURVES CONNECTING RIGHT LINES. 


Let ABCDE represent a circular arc joining the straight lines 
AV and. #Y, which are 
tangent to the curve at 
Aand £. 

AV and EY are tan- 
gents to the curve, A and 
Kare tangent points, and 
the angle A VH'is the an- 
vie of intersection, and 
shows the change of direc- 
tion in passing from one 
tangent to the other. 

V is the point of inter- 
section, or vertex. 


PROPERTIES RELATING 
TO THE CIRCLE. 


The following proposi- 
tions rest upon elementary 
geometrical _ principles, 
inay be regarded, for the most part, as axiomatic. 

(a) A tangent toa circle is perpendicular to the radius, at the 
point of contact. 

(5) Tangents drawn to the circle from the same point are 
equal, and the angle between these tangents, and the chord join- 
ing the tangent points, are equal. Thus, 


AV=HV, and VAH= VEA. 

(c) The central angle AOZ, subtended by a chord, is equal to 
the exterior angle K VH between two tangents to the curve at the 
extremities of the chord. 

(dq) The angle between a tangent and chord is equal to the 
angle subtended at the circumference by the same or by an equal 


chord. Thus, VAB= BCA = BAC, ete. 


Fig, 12. 


39 


40 FIELD-MANUAL FOR ENGINEERS. 


(¢) An angle between a tangent and chord, or an angle subtended | 
at the circumference by that chord, is equal to one half the central 
angle subtended by the same chord. ‘Thus, 


VAB = BCA = {AOB. 


(f) Equal chords subtend equal angles at the center of a circle, — 
and also at the circumference. Thus, if AB = BC, etc., 


ACB = BAC, etc., and AOB= BOC, etc. 


(7) A radius perpendicular to a cherd bisects the chord, and 
also the angle and the arc subtended | 
by the chord. Thus, if OC is per- | 
pendicular to"AZ, we have - 


AM= ME, AOV= EOP, 
and AC= CE, 


(X) Parallel chords, or a tangent | 
and parallel chord, intercept equal 
arcs. Thus, in Fig. 13, if P7, P’7’, 
and P’7'’ are parallel, then PP’ 
and 7”7” are equal, also PP’ and 

Fig. 18. TT’ are equal. 

(i) The exterior or deflection angle ABC, Fig. 14, between any 
two chords AB and BC is equal to half the central angle 4 OC, 
subtended by the chords. This 
is easily shown as follows: 

1. Join AC. BAC = }BOC, 
and BCA = 4A0OB, as stated 
above. But ; 


KBC = BAC + BCA; 
KBC = 1AOB + 4B00 
— 1400. 


2. Draw the tangent HBL. 
Then , 


KBC = KBL + LBC 
= HBA + LBC 
= }40B + 4B0C=}400. 


Fig. 14, 


SIMPLE CURVES CONNECTING RicH?T LinEs. 41 


3. Since the chords AB and BC are parallel to tangents drawn 
at the middle of the arcs AB and BC, it is evident that, in passing 
from one chord to the other, we turn through an angle measured 
by one half the arc AB -+- one half the arc BC, that is, through an 
angle equal to one half AOC. 


SoME ELEMENTARY RELATIONS. 


In Fig. 12 drop the perpendiculars Bb, Cc, etc., upon the tangent 
AY. 

Ab, be, etc., are called tangent distances, and Bb, Cc, etc., tan- 
vent offsets. 

Prolong AB, making Bd = AB. aC is called the chord offset. 

Then, from the preéeding article, we have 


CBd = 4A0C = AOB. 


Drop the perpendicular Bp upon Cd. This bisects the angle 
CBd and the base Cd. Hence 


CBp = dBp = }CBa = $AOB = BAD. 


Since, therefore, the triangles CBp, dBp, and BAb have an 
acute angle in each equal, and the hypothenuses also equal, they 
are equal in all respects; and therefore 


Cp = pd = Bb; also Cd = 2Bb. 


Since Ad is tangent to the curve at A, Bp is likewise tangent to 
the curve at B. me 

Represent the radius AO by R, the tangent, or vertex distance, 
AVby 7, and the angle of intersection K VZ, as v.ell as the central 
angle AOH, by V. Represent the chords AB, BC, etc., by ¢, a 
long chord, as AC, by C, the tangent distances Ab, bc, etc., by 
d,d,, etc., and the tangent offsets DB, cC, etc., by t, ¢,, etc. Then 
the chord offset Cd = 2t. Represent the vertex distance CV by 
E, and the middle ordinate of along chord by M. Let Z represent 
the length of the curve, P.C. (Point of Curve) the beginning of 
the curve, and P.7. (Point of Tangent) the end of the curve. 

The degree of a curve has been defined as the number of degree 


49° FIELD-MANUAL FOR ENGINEERS. 


subtended at the center by a chord 100 feet in length. This defi- 
nition of curvature is, however, awkward, arbitrary, and false. 
It is founded on error; it involves unnecessary labor and ends in 
anomalous and erroneous results. 

The degree of a curve may be defined as the change in its di- 
rection between one point, and another 100 feet from the first, 
measured on the curve. ‘This is the change in direction which 
one would make in moving 100 feet on the curve from one point 
to another. Or, as the angle subtended at the center of the curve 
hy an arc 100 feet in length. 

Let D = the degree of the curve. 

The circumference of a 1° curve is therefore 360 « 100 = 36000 


30000 == 5729.578 feet, almost exactly. 


feet, and its radius is 


21 


Since a 2° curve changes its direction 2° in 100 feet, its circum- 
ference is only half that of a 1° curve; and since the radius varies 
directly with the circumference, the radius, too, is only half that 
of ai° curve. For the same reason the radius of a 38° curve is 
precisely ¢ the radius of a 1° curve; and, generally, the radius of 
a D° curve is exactly equal to the radius of a 1° curve divided 

was 5729.57 
by D; it is therefore equal to — po 
Referring to Fig. 15, we see that for the same central angle 
AOE, the arc, the tangent, the 
chord, the middle ordinate, the ex- 
ternal secant, etc., vary directly witia 
£ the radius or inversely with the de: 


V 
A ZA |M oOo 
gree of curvature. For example « 
If a0 = 34A0, then ape = 1 APE, 
ai=4AV,etc. If APHisa 1° or 
\ PSS 
O 


2° or 8° curve of length Z, then ape 
is a 2° or 4° or 6° curve of lengtls 
41. 

Hence to compute any function 
of any curve, the tangent or exter- 
nal, for example, from the corre- 

Fra. 15. sponding function of another curve, 
it is only necessary to multiply or divide, as the case may be, by 
the ratio of their radii or degrees of curvature. 

“rample.—Find the tangent of a 7° 13’ == 438’ curve, the cen- 
ngle being 37° 50’. 


SIMPLE CURVES CONNECTING RIGHT LINES, 43 


The tangent of a 1’ curve, by Table VII, is 117812. Then 
T = 117812 +- 433 = 272.1. 
Using a table giving functions of a 1° curve, we have: 
Tangent of a 1° curve = 1963.6. 
Then Tangent of a 1' curve = 1963.6 x 60 = 117816. 


Finally, 117816 -+- 483 = 272.1. 
Or. 13’ = 60 = 0°.216, 
and therefore 7° 18’ = 7°.216. 
Then 1968.6 -- 7.216 = 272.1. 


With a table giving functions of a 1° curve there is no escape 
from dividing by 60, which division is obviated by giving the func- 
tions of a 1’ curve as in Table VII. 

It is often desirable to know the difference in the lengths of a 
chord and its subtended arc, and for this purpose we deduce the 
following formula : 


d = 001269289249”. — 0000000048820. - « (8) 
n n 
2 
= 001260, — ,0000000048 =, very nearly, . (9) 


,0012697, approximately. 2 2 6 e@ © 610) 


In these equations D = the degree of curvature, d = the differ- 
ence between any chord and the subtended arc, and 7 = arc of 100 
feet divided by this arc. 


For an arc of 100 feet n = 1 and @ = .001269D', nearly; . (11) 
For an arc of 50 feet n = 2 and d = .000158D', nearly; . (12) 
For an arc of 25 feet m = 4 and d = .00002D?, nearly. . (18) 


For n sub-chords per station the sum of the difference per 
station is 
dD? 


nd = .001269— (14) 
(L) 


44 FIELD-MANUAL FOR ENGINEERS. 


Representing the central angle of the curve by V,and the numn- 
. V 
ber of stations in the curve by NV, we have V = Dp} and hence the 


length of the curve exceeds the sum of the lengths of the sub- 
chords by 


| 2 VD 
E = Nnd = .on1269 x ¥ = .001269-;. . (14 
n D n 


D 
Hence, in laying out curves, ar should be nearly constant. 
Wt 


Since in a 4° curve the chord of an arc of 100 feet is 99.98 feet, 
curves from 0° to 4° can be properly laid out with chords of 100 
feet, aud with the same degree of accuracy we may lay out curves 
from 4° to 16° with chords of 50 feet, and those from 16° to G4° 
with chords of 25 feet. 

Very sharp curves can be easily laid out by swinging a chain 
around, while one end is held at the center of the curve. 

Let ¢ represent any arc, c its chord, and c, the chord of one half 
of the arc s. 

Then, from above, 


8c, —€ 
3 ° 


ad 


s—e=8(5 er) = 4s —8c,, or 8= 
This is said to be Huygens’ approximation to the length of an 
arc, 


The following table shows the differences between arcs of 25 
feet and of 50 feet. and the chords of those arcs. (See Fig. 16.) 


Deg. Are Are | Deg Are Are | | Deg. Are Are 
D 2A D 25 50 | D 2 5 
\ 
1 .000 .000 11 002 .019 21 009 .070 
2 .000 .001 12 008 .023 22 010 OTF 
3 .000 001 13 003 027 23 010 .084 
4 .000 003 14 Os 031 24 011 091 
5 .000 .004 15 004 036 25 012 .099 
6 .001 066 16 005 041 26 013 107 
vi 001 .008 17 006 .046 7 014 .116 
8 001 .010 | 18 C06 .052 28 Ole 124 
9 002 013 | 19 007 .058 2 017 . 138 
10 002 . 016 | 20 008 064 30 018 .143 


SIMPLE CURVES CONNECTING RIGHT LINES. 45 


II. Suppose the chord AB = 100 feet, and AOB = D, (Fig. 16). 
Then 
A om=—, MAan=*), etc. 


dence 


AM = AF sec MAE 


D, 
== 50 sec 7. 
Therefore 


AM — 50 = 50( see I7 1) 


dD, 
— 4 
= 00 exsec Tt 
Similarly, AK = AF'sec z = 25 sec + sec 3° etc., etc. 


The following table gives the excess of AK over 25 feet, and of 
AM over 50 feet, when chord AB is 100 feet. 


D, ~25 | —50 1 —25 | Dy —% | — 
1 000 | .000 1! .036 | .058 21 131 211 
2 oot | .002 12 .043 | .069 22 144 231 
3 003 | .004 13 050 | 1081 23 157 252 
4 005 | .008 14 058 |. 2 71 OTR 
5 oo7 | .012 15 067 |” 101 25 186 20¢ 
6 O11 | .017 16 076 | .192 26 201 323 
7 O15 | .028 17 .086 138 oF Q17 349 
8 o19 | .030 18 096 155 28 233 7 
9 024 | .039 19 107 172 29 250 403 
10 030 | .048 20 (119 181 30 268 431 


Comparing the preceding tables we learn that, when the chord 
AB is 100 feet, the chord AWM differs nearly four times as much 
from 50 feet as the chord of the arc of 50 feet differs from 50 feet. 
Furthermore, that the chord AK differs nearly sixteen times as 
much from 25 feet as the chord of the a7c of 25 feet differs from 
25 feet. 

Thus let us first suppose AMB (Fig. 16) to be a 10° curve, and 
the chord AB = 100 feet. Then by the table AK = 25.080 feet, 
and this would lead to an error of .030 « 40 =-1.2 feet in laying 
out a curve 25 x 40 = 1000 feet long, taking chord AK = 23 feet 
long. 


46 FIELD-MANUAL FOR ENGINEERS, 


Suppose, secondly, that the arc AMB= 100 feet and therefore 
arc AK = 25 feet, and the chord AX is but .002 of a foot less than 
25 feet. 

Hence in laying out the above curve, taking the chord AA = 25 
feet, the error committed would be only .002 x 40=.08 of a 
foot. 

Thus we see that when the chord of a station is taken = 100 
feet, the sub-chords AM, AK, etc., differ so much from 50 feet, 
25 feet, etc., as to largely vitiate the results, whereas such is by 
no means the case when the arc A MB is taken = 100 feet. 

Indeed, when the chord AB is 100 feet, the shorter the sub- 
chords used in laying out a curve, the greater the discrepancy in 
the measurement, on this basis, Thus fora 10° curve the difference 
bet ween four equal sub-chords and 100 feet is .080 x 4=—.120 of a 
foot; whereas the difference between to equal sub-chords and 
100 feet is .048 K 2 = .096 of a foot. The reverse is of course the 
case when the arc AMB is made the standard of measurement. 

These facts are evident ; for when the chord is made the stand- 
ard of measurement, the sum of the lengths of the sub-chords ez- 
ceeds more and more the length of the chord, the shorter they are ; 
whereas when the arc is the standard of measurement, the sum of 
the lengths of the sub-chords falls short of the length of the arc 
less and less the shorter they are. 

Most recent writers have endeavored to obviate the inconven- 
iences and inconsistencies above pointed out by inconsistent assum p- 
tions, such as basing the curves of different degrees upon chords 
of different lengths. This scheme gives the values of some of the 
radii quite correct; but it. causes sudden breaks in the values 
where the changes are made, 

For example, how can there be two different values (819.02 and 
818.64) of the radius of a 7° curve? 

And why should the radius of a 7° 10’ curve be given quite cor. 
rect, while that of a 6° 50’ curve is quite incorrect? 

Again, we are told by a recent author that, in practice, it is cus- 
tomary to take the radius of a 1° curve as 5730 feet, and to assume 
the radius to vary inversely as the degree. Thus for a 4° curve 
the radius would be = 1432.5 feet. This is rational and, 
moreover, is precisely what is here advocated, except that the true 
value of the radius of a 1° curve is used, viz., 5729.58 feet. 


SIMPLE CURVES CONNECTING RIGHT LINES. 47 | 


FORMULAS. 
From the triangle AOV, Fig. 12, we have 
AV= AOtan AOV, or T=RtaniV. . . (15) 


Transposing, we have 


T 
— — 1 
From the triangle A OF, we find 
AF $c 
0= Gn dor’ R= sin }D ~ dc cosec 3D. . (17) 
This value of # in (15) gives 
1 
_ fetangV 
~~ gin 4D (18) 


Measure equal distances VH and VZ along the tangents in 
Fig. 12 and draw HNL. 
Measure HW and VN. 


; AV _VN _ 4, VN 
Now, 40 ~ Hy’? % AV = 40pR: 
. VN 
that 18, T= Ry" ry e t r ry e . (19) 


If AM and VM are measured, then 


VM _AM | 
T= Ray or R= Tr 2 8 « e (20) 


Eqs. (19) and (20) serve to fix geometrically, without measuring 
angles, the tangent point of a curve of given radius that will unite 
two straight lines on the ground. 

In Fig. 12 draw VC’ perpendicular to A V to meet AC prolonged. 
Now the angle AVC = 3A VH = 4(180 — V) = 90—4V. Hence 


VCC’ = ACO = 90° — BOC = 90° —tV. 


¥ 
‘eX 
“4 
; 
v4 
t 
a 
* 
" 
» 
» 
ad 
~ 
beg 
“4 
f 
¢ 
4 
14 
wid 
i 


aro 


cE BO LB 
Bo Bre Oe 
on in nn ohn 
ze ( 
oc 
and Be=t=5,- me ee ee (RY 


We aso Lave 


Bb = ABsin BAX, or t=esniD, . 2. . (24) 


~ 
ig 


and Ato 2sineD 2. 2. 2 1... (25) 


Frample —Given R = 1909.9 feet and ¢ = 100 feet. to find the 
tangent and the chord offsets for 100°feet. | 

By (23). 
10000 


—_— ——_ — 2, . 2t — 5. . 
3819.8 618, and 2 5.236 


or Table J shows that 1909.9 is the radius of a 3° curve, and 


, 43D) = 1° 30', and sin 1D = .02618. 


Then, by (24), 
t = .02618 x 100 = 2.618, and 2¢ = 5.236. 


The tangent offsets are given in Table I, and the chord offsets | 
are twice the tangent offsets. | 


SIMPLE CURVES CONNECTING RIGHT LINES. 49 


In laying out curves, the chain is stretched from point to point 
~mn the curve, and coincides, therefore, with chords of the curve. 

Since the process is the same whatever the length of chain 
used, we will assume it to be 100 feet long. 


‘Fig. 17. 


The length of the curve is expressed in chains, in terms of the 
central .angle AO? = V, and the degree of the curve AOB = D. 


The number of chains is evidently equal to x 


Thus, in the figure, if V = 23° and D = 5°, the curve is 23 = 42 
chains, or 460 feet long. As the angle 4OH = 5° x 4 = 20°, 
HOF = 23° — 20° = 32°; and the arc HF is 2 x 100 = 60 feet long. 

It is usual, in laying out curves, to assume the radius R, and to 
find the degree of the curve D from it; or to assume D (usually 
in degrees and minutes), and to find R from it. Neither way 
is best. 

To assume a value of # or of D does not aid in the least in 
properly locating the curve. Generally the surface of the ground 
does indicate .approximately the position of the curve, and the 
proper course to pursue, therefore, is the following: Divide the 
tangent of a one-minute curve by the length in round numbers of 
the desired tangent, and neglect the decimal in the quotient. 
This gives the degree of the required curve in minutes. Wemay 


50 FIELD-MANUAL FOR ENGINEERS. 


change the quotient to an even number of minutes, or to some 
multiple of 10 if we wish, if there is sufficient latitude to be taken 
in the position of the P.C. 

Then divide the tangent of a one-minute curve by the corrected 
quotient for the tangent required, 

EHrample.— V = 46° 30’, and the tangent should be 1100 feet or 
over, Find the degree of the curve, and the length of the tan- 
gent. | 

Dividing the tangent of a 1’ curve by 1100 gives 


147697 + 1100 = 134’ = 2° 14’. 


Now 2° 10’ = 130’, and 147697 = 130 = 1186.13, the tangent re-, 
quired. | 


LONG CHORDS AND ORDINATES TO LONG CHORDS. | 


Let A, B, C, etc., represent stations upon a curve. | 

Draw the lines as_ repre- 
sented, HX being a perpendic-.: 
ular from the middle of the 
curve # upon the tangent 
AT. Let M= EM, M' = FY, 
etc. 

Draw AZ, and we have the 
angle HAM equal tothe angle 
HAX, and therefore the tri- 
angles HAM and HAX are 
equal in all respects. From 
this we learn that the tangent 
offset HX for any arc AZ i: 
equal to the versin HM af 
that arc, or to the middle cr- 
dinate HM of the chord AK 


of twice the arc. 

Or, draw the tangent Ht and the perpendicular Kt upon it. 
Then the tangent offset Ki of the arc HK =the versin HM of 
the same arc = the middle ordinate HM of the arc AHK. 

To find the middle ordinate, M. (See Fig. 18.) 


1°. M = HM = EX = AMtangent HAM =j3Ctanz?V. . (26) 
2°. M= EM = EX = HT cos TEX = Hoostv. . . . (26) 


SIMPLE CURVES CONNECTING RIGHT LINES. 51 


3°. M = HO — MO= R—- ReosiV = R(1—cos}V) 
= Rvers§V.. . . (26) 


4°, M = HO—MO=EO~— Y AO?— AM’°=R— ¥ R?—-3C?. . (26) 


To find any ordinate #N distant d from the center of the chord. 
Prolong FN, to meet OS, drawn parallel to 4K, and join FO. 


Now, FS = FO — 08 = VR? — d*. NS= MO= R- YM, 


o FN= Y/R? —a@—-(R-M)= M+ VR? -@-R. 


Other methods will be givén in connection with laying out 
curves by ordinates from a long chord, 


Approximate Values of Ordinates to Short Chords, 


Divide the chord into any number 
of equal parts, eight for example, at 
s, k, 7, etc., and erect the ordinates 
Em, Fn, etc., and prolong them to 
meet the curve in #', F’, etc. Let 
Em =m, Fn =m’, etc. We have, 
from geonetry, 


Similarly, 
Fa=m = eae — 2 & = {$m =m — 3m; (28) 
Gp =m" = exe ic Sn = 12m = m — pgm; (28’) 
Hq =m!" = or = a 5 = jn =m — Pym. . . (28”") 


For any other equal divisions of the chord we have similar re- 
sults. 


52 FIELD-MANUAL FOR ENGINEERS. 


To establish the points H, F, ete. 

Set # equally distant from A and B, and at the distance m from 
the point m; then F, equally distant from'k% and B, and } 
from 2; then G, equally distant from m and B, and $m trom 2 D; 
lastly, H, equally distant from p and B, and s,m from q. 

This method involves much less labor than that of drawing sub- 
chords to find the points 7, G, etc. If we draw a tangent at £, 
the offsets to F, G, H, etc., will be, according to the preceding 
formulas, 4m, 34m, 23m, etc. This shows a convenient way of 
finding the points on the curve. 

To compute tangent offsets and middle ordinates by means cf a 
series. 

In a way similar to that pursued in finding the difference be- 
tween an arc ard its chord we find 


t = .872664625997.D — .00002215240389.D* 


+ .00000000022498360386.D' —, . (29) - 


or 


t = .872664626D — .000022152404.D3-+-.000000000224933604 D>. (30) 


We may find the tangent offset for m stations by multiplying 
the successive terms of (30) by m?, m‘, etc. 


. m . 
We thus find tang offset for arc of 50 feet, or 3 stations, 


t’ = .218166156499.D — .00000138452524D* 
+ .00000000000351458756D® —. . (31) 


For 1 foot m? = (.01)? = .0001, m= .00000001, etc. Then, 


The tang offset for 1 ft. 
= t, = .0000872664626D — .00000000000022152404.D*-+ . (82) 


From (30), we have ¢ = .87D, approximately. 
For 7 stations in = .8777D.. . . 2 1 we (88) 
Wellington in Ratlway Location, ch. xxx, recommends the 


equation t=iwD,. . 1. ee ee (33’) 


SIMPLE CURVES CONNECTING RIGHT LINES. 53 


which is practically the same as (33). (383) is a trifle more accurate 
than (33’), but either is sufficiently accurate for all cases in 
which n?D does ne+ much exceed thirty. 

The same formu:a gives the middle ordinate, » representing the 
natmnber of stations on either side of the center, or 22 the number 
of stations in the arc. Any other ordinates desired are then given 
by eqs. (27), (28), ete. 

Erample.—Find six offsets of a 3° curve at points 50 feet apart. 
(See Fig. 18 ) : 


Forn=34, t=ix*t X8= _ .66; 
n = 1, ={xX1xkK3= 2.62; 
n=% t=2xX2 xX38= 5.91; 
n=2 t=ix4 X3= 10.50; 
n=24,¢=%x 28x 3 = 16.41; 
w=8, t=iX*K9 XK 3 = 23.62, 


These results are of course equal to the middle ordinates for 
1,2, 3... 6 stations of the same curve. 


LAYING OUT CURVES, 


Since in laying out curves the operation or method is the same: 
whatever the length of the chain or chord, we will here assume it 
to be 100 feet long. 


A. By Deflection Angles. 


Let A in Fig. 17 be the P.G. Set the instrument at A, and 
turn off from the tangent AV the given deflection angle VAB= 
1D, D being the degree of curve. This will give the direction, 
AB, and measuring 100 feet in this direction, the point B will be 
determined. Turn off the additional angle BAC = 4D, the tele- 
scope being now directed toward C, and set C in the line AC and 
100 feet from #. Turn off the additional angle CAD = 4D, and 
set D in the line AD and 100 feet from C. Proceed in the same 
way for other stations to the end of the curve, or so far as the 
stations can be seen from A. 

It is usually impossible, on account of obstructions likely to be 
met with, to lay out the whole of a curve from the first station. 
When such is the case, we determine as many stations as conven- 
ient, remove the instrument to the last station so determined, ar-’ 


54 FIELD-MANUAL FOR ENGINEERS. 


proceed from that as from the first station. For example : Suppose 
RB, C, and D to be found with the instrument at A. Remove the 
instrument to D, sight to A, turn off the angle ADV = DAV= 
£D, and the line of sight will be in the direction of the new tan- 
gent DV at D. Reverse the telescope, and the line of sight will 
point forward along the same tangent VDG. Now set ZH, F, etc., 
from the tangent DG, as B, C, and D were set from the tangent 
AV. | 

In setting the first stake from the new position of the instru-. 
ment, as # from D, no notice need be taken of the tangent at J), 
it being necessary simply to turn off from the line ADH the angle. 
HDE = HDG+ GDHE= ADV+ GDEH= 4(iD). | 

In the new position of the instrument we observe that, in all 
cases, the defiection from the line pointing to the back station to 
the line pointing to the forward station is as many times the de- 
flection angle 4D as there are chains in the curve between the 
back and the forward station. Of course this applies to simple. 
curves only. ! 

The beginning of a curve, as well as the end, usually falls: 
between regular stations, giving short chords atthe ends. The 
deflection angle for a short arc is such a part of the full deflection 
angle as the short arc is of the full are. 

Let @ represent the arc AB or BC, etc., and a, represent the| 
arc HF. Also HOF = Dy. 

Hence DFH=14D, and HDF=%4D,, and therefore 


3D 
i 
2a 


a 
—- = —, or 4D, = 4D-, or “=D . 


In beginning a curve, the instrument being at the first station, 
it is convenient to place the zeros of the instrument plates to-| 
gether, and direct the line of collimation along the tangent to the 
curve. The reading on the limb for any station will then be 
equal to the total deflection angle for that station. 

If the vernier is not disturbed while laying out the curve, it is 
plain that when the instrument is moved to its second position D, | 
Fig. 17, and the line of sight directed to A, the reading will be. 
equal to the total deflection from A to D; and that, an additional | 
angle equal to this deflection being turned off, the reading will be | 
equal to the central angle AOD, and the line of sight will be in 
the direction of the tangent at D. The same is true for all posi-— 


SIMPLE CURVES CONNECTING RIGHT LINES. a5 


tions of the instrument. Since the deflections for the last section 
of the curve, that is, from the last position of the instrument to 
the end of the curve, is turned off but once, the reading on the © 
instrument, when the curve is finished, will be equal to the total 
central angle V, less the deflection for the last section, This fur- 
nishes a convenient check for the work, 


B. By Tangent Offsets. New Method. 


Let ABCDEFGH represent acurve having a short chord AB 
::¢’ subtending an angle AOB 
=: D, at one end of the curve. 

Define the tangent A V by set- 
ting stakes upon it. Draw Be, 
id, ete., parallel to A V, and BX, 
icX’, etc., perpendicular to AV, 
or suppose such lines to be 
\irawn, Now 


BX=ABsin BAX=c'sin} D,=bt, 


and is given by Table II. 

Since the chords BC, CD, etc., 
are parallel to the tangents at the 
middle of the arcs BC, CD, etc., 
we have 


CBe= D+ 3D, DCd=D,+ 3D, HDe = D, + 8D, ete. 
Hence 
Ce=csin(D, + 3D), Dd =c sin (Di + 3D), 
He =csin(D, + §D), etc. 
We observe that Cc, Dd, He, etc., are respectively the tan- 


gent offsets for curves of the degrees (D, + 4D), (D: + $D), 
(D, + §D), etc., and may be taken from Table III. Then 


CX' = BX + Cc, DX" = CX'+ Dd, HX" = DX" + He, ete. 


Set B at a distance c’ from A, and ¢, from AV; then C a dis- 
tance c from B, and CX' from AV; D adistance c from C, and 
Dr’ from A V, etc. 


56 FIELD-MANUAL FOR ENGINEERS. 


~ 


It will be observed that in this method we avoid constructing 


a tangent at B, in consequence of the short chord AB; we avoid, © 


secondly, the finding of AX, XX’, etc.; thirdly, the errection of 
perpendiculars, at X, X’, etc.; and, lastly, we avoid the use of 


the radii which are large and fractional. Since ¢ is generally 100, 
though sometimes an aliquot part of 100, no computation is gener- | 


ally required, and none to mention in any case. Thus we see how 
simple and short this method is, compared with the method of 
tangent offsets in general use. 

When the curve begins at a station there is no short chord. 
Then 


AB=c,D =D; BX=csin}D, Ce=csingD, Dd =csin §D, 
etc. 


It is best to lay out the curve from each end, so as to avoid off- 


sets inconveniently long. Long offsets may be avoided by drawing | 
a tangent at any station and continuing the curve from that station 


precisely as from A, when there is no short chord. 

To draw a tangent at any station, draw a line through that sta- 
tion and at a perpendicular distance from an adjacent station equal 
to the tangent offset for a station = csin3D. Or, draw it parallel 
to the chord joining adjacent stations. 

It will be noticed tliat stations on the curve are not opposite 
stations originally set on the tangent. 

The length of the curve gives the number of the station at H. 

This is perhaps the best of all methods with out a transit; but 
a combination of methods is sometimes advisable, as will be shown 
further on. 

We also have 


AX = Ccos34D,, XX' = Be=ccos(D, + {D), etc., 


or XX’, X'X”, etc., are respectively equa! to the tangent distances 

for one station of curves of the degrees D, + 4D, D, + 3D, etc. 
These quantities are not needed, however, and it is to be ob- 

served that the points X, X’, etc., are not established or used. 


SIMPLE CURVES CONNECTING RIGHT LINES. 59 


C. To Locate a Curve by Ordinates from a Long Chord. 


Let Fig. 21 represent a curve having an odd number of full 
chords, and a short chord 
AB =c’ subtending an 
angle AOB = D,, and a 
short chord KZ = cc" sub- 
tending an angle KOL 
=D,. Draw the tan- 
gent AZ’, and drop the 
perpendiculars BX and 
CX’ upon the tangent. 
Draw Be parallel to AT. 

Since Be is perpendic- 
ular to AO, and BC is 
perpendicular toa radius 
_ bisecting the angle BOC, the angle CBc = D,-+ 4D. 

Now BX is found in Table II; and Cec may be found from the 
same table or Table III, supposing the curve to be of the degree 
D,+4D. Or, 


BX=c'’sinjD,, and Cco=c sin (D,+ 4D). 


Suppose ec =80, D=4?. 
D, = 3° 12’, 
and (D, + 4D) = 5° 12. 
Table II gives BX = 2.23, 
and Ce = 9.06. 
CX’ = 11.29. 


Place B 80 feet from A and 2.23 feet from the tangent A7’; then 
Ca chain-length from Band 11.29 feet from the tangent AT. 

Locate K from the tangent at Z in the same manner as B is 
located. This gives the line CK. 

Sappose the stations already located ; the long chords HG, DH, 
etc., drawn, and also the perpendiculars Dd, Eel, etc. 

Now Cd = Cm — Di = 4(CK — DQ) = one half of the difference 
between the chords of six and of four stations. See Table lV. 

d= Di — Hf = }(DH — HG) = one half of the difference be- 
tween the chords of four and of two stations. 

im = one half the chord of two stations. 


58 FIELD-MANUtAL FOR ENGINEERS, 


Again, Dd = Fm — Fi = the difference between the middie 
ordinates of chords of six and of four stations. (See Table V.) 

El = Fm — Ff = the difference between the middle ordinates of 
chords of six and of two stations. 

Finally, #m is the middle ordinate of a chord of six stations. 

We note that points and lines on the right of Wm are of course 
symmetrical with corresponding points and lines on the left. 


Hence Cn = CK ~ nK = CK — Cl 
Ch = CK —nrK = CK — Ca, etc. 


Now lay off Ca, Cl, Cm, ete. 

Set Dastation from Cand a distance Dd from d, or CK; ther 
Ea station from D and a distance #/ from J, or CK; then F&F : 
station from Z and a distance #’m from m, or CK, etc., etc. 

We also have 


Cd =ccos§D, dli=ccos$D, and lm=ccosiD; 
Dad=csin§D, He=csin§D, and Hf=csin ZED. ° 


These quantities may be taken directly from a table of sines 
and cosines, as already pointed out. It is not necessary, as shown 
under the preceding method, to compute and lay off Cd, Ci, etc. 


D. To Locate a Curve by Chord Offsets. 


Let ABCDEF be the curve, having short chords AB = c’, sub. 
tending an angle AOB = D, 
Di p, Ey and HF = c”, subtending an 

angle HOF = Dy. 
Fz Locate B as in the las? 
F method. Prolong AB, mak 
ing BC, =c. Theangle CBU. 
= CAB+ ACB =3D+44D,. 
Hence C,C is the chord offse’ 
av’ for a curve of 4D+ 4D 

degrees, 


y 
o C,C = 2csin (4D + 4D,) 
= 2¢sin}(D, + D). 


Place C, therefore, at a distance c from B, and 2¢’ from Q,, 2¢’ being, 
*wice the tangent offset which is given by Table III opposite thw 


SIMPLE CURVES CONNECTING RIGHT LINES. 59 


gree 4(D -+- D,). Prolong BC to D,, making CD, = c. Now 
D,D = 2t is given by Table III, being twice the tangent offset. 
Place D, therefore, at a distance c from C, and 2¢from D,. Place 
all regular stations similarly. ( may be placed, also, as in the 
preceding method. is easily placed from the tangent at H, as B 
was placed from the tangent at A. 

It may be observed that by the above method we are able, by 
means of a table giving offsets for full stations only, to locate any 
station, as C, by chord (or tangent) offsets, though the chord, as 
AB, preceding the adjacent chord may be of any length. 


E. To Locate a Curve by Middle Ordinates. 


Let ABCDEFG be the curve, having a short chord AB = ce’, 
subtending an angle AOB = D,, 
and a short chord #G =c’", sub- 
tending an angle FOG = D,. 

Locate the stations Band C from 
the tangent Az, or by some other 
method as already explained. 
Then set off on CO the distance 
Ce = t=csin 4D = the middle 
ordinate of a chord of two stations; 
and set Din the prolongation of 
Be, and at a distance c from C. 
Set off Dd the same as Cc, and set # in the prolongation of Cd, 
and at a distance c from D. Locate all subsequent stations in the 
same way. To test the accuracy of the work, measure the per- 
pendicular Fy from the last regular station upon the tangent at @. 
This distance ought to be 


t, = c’ sin 1D. 


F. To Lay out a Curve by Radial Lines from the Center. 


Consider the case of a half-mile race-track, having two parallel 
sides, each 600 feet long, connected at the ends by semicircles, as 


QO — 
shown in Fig. 24. Now ae = 720, the length of each 
oe 180° 
semicircle. Hence the degree of each curve = 739 > 25°, and 


the radius is 229.18 feet. 


60 FIELD-MANUAL FOR ENGINEERS. 


Set the instrument at O, and run radial lines O01, O02, etc., making 


Fia. 24. 


angles of 30° with each other, for example, and set stakes at 1, 2, 
etc., 229.18 feet from O. These stakes are 100 §2 = 120 feet apart, 
measured on the arc. 

Determine other points on the curve by middle ordinates. 

The inside of the track is three feet inside of the line measured 
above. 


Errors 1N FIELD-WORK. 


In laying out curves, as well as in all field operations, errors 
will sometimes occur, and it is important to know the immediate 
and the ultimate effect of such errors; to know when such errors 
are increasing, from one stage of the work to another, and when 
they are Cecreasing; when they are too small to be of importance, 
and when they are so large as to vitiate the result if not cor. 
rected. 

It is important, also, to know the law of increase or of decrease 
of such errors; for in that case the error at any point may be 
found from that at any other point, the end of the curve, for ex- 
ample; thus making it possibie to properly, that is, really, correct 
the curve without rerunning it. 


I. Curves Laid out by Deflection Angles. 


A. To find an approximate value of the error at the end of a 
curve due to an error in the length of a chord.—Suppose A, B, C, 
etc., to represent stations on the true curve, and A’, B’, C’, ete., 
stations on the false curve. 

1°. Suppose the first stake from A to be set at B’,a distance 
BB’ = ¢ beyond B, its true place. 


StMpPLi CURVES GONNECTING RIGHT LINES. 61 


Then C’ will be set on AC prolonged and a chain from B’, D’ 
will be set on AD prolonged and a chain from (’, etc. 

To prove that CC’ is less than BB’. Draw CH equal and paral- 
lel to BB’. Then BB'CA is a parallelogram, and B’H = BC = 
BC" = achain. Hence C” is on the arc whose center is B’ and 
radius B’H. With Caseenter and radius CH = e describe the 


A O 
Fig. 25. 


arc HK. HE lies outside of the arc HC", since the arcs at H are 
perpendicular to CH and B’H respectively, and are limited by AC 
prolonged. Hence 


CC’ < CK = CH = BB’ =e. 


For the same reason DD’ is less than ('C’, and so on to the end 
of the run, that is, to the last stake set with the instrument at A. 

Suppose the instrument moved to D’. 

Since, in setting H’ from D’, we turn off from D’A the same angle 
that we would turn off from DA in order to set H, we have D'EH’ 
equal and parallel to DH, and therefore HH’ is equal and parallel 
to DD’. For the same reason FF” is equal and parallel to HZ’ 
or DD’. S>on to the end of the curve. 

2°. Suppose an error to occur in some other chord than the first— 
in the second chord, for example. B is supposed, therefore, to be 
set correctly. Suppose the next stake set at C’ instead of at C 


62 FIELD-MANUAL FOR ENGINEERS. 


Let BC’ =c +e Now BC+ CC’ > BC, or 
e+ CC’ >c+e. “. CO’ > e. 


-From C’ the errors will follow the same law as in the former 
case. 

With reference to these two cases we remark that if the error 
occurs in the chord adjacent to the instrument, the errors in the 
stations will decrease slightly to the end of the run, and from 
that point remain constant to the end of the curve; but if the 
error occurs in a chord not adjacent to the instrument, the error 
in the station, at the end of that chord, is slightly greater than 
the error in the chord, though the errors slightly decrease from 
this point to the end of the run, and remain constant from the 
end of the run to the end of the curve. In all cases the error at 
the end of the curve may be regarded as equal to the error in the 
chord, whether adjacent to the instrument or not. 

3°. Suppose the chain is 
in error, in which case add 
the chords will be in error. 

Let the curve AB(R = 
AO = radius, and D = de- 
gree) be run with a chain 
100 feet long, and the curve 
AB (R’ = AO’ = radius) 
with a chain 100+ e feet 
long. The number of 
chords is equal to t = Nn. 
Then for the difference in 
length of the curves we 
have AB’ = (100 + e)n, 
AB=100n, and therefore 


AB’ —-AB=en. ..... « (84) 


For the chord we find, since &’ = a(t), 


100 +e 


AB = 2x 100 


sin} V, AB= 2Rsin}V, 


and therefore 
2Re sin 3V 


SIMPLE CURVES CONNECTING RIGHT LINKS. 63 


180 
But R = 100 xD’ 
, 860e8ingV  114.60e sin" dV or 
BB = rs) —_— po e (35) 
Also 
a 180 , _ 1800e _ 57.36 ; 
R= (100 +- 6)” and R’— R= xD = Dp” . (36) 
We also have 
x3 1 2 
hh = BB’ sin B'Bh = BB' sin4V = wate Cn gh) , . (87) 
~ , 
tnd Bh = BB’ cs }V = Si feein Y (38) 
Ezample.—Let e = .02, V = 60°, and D = 6°. 
The error of the curve = .02 x a = .200. 
1 
The error of the long chord = ae aX = .191. 
Also R’'-—-R= $7.3 xe = .191, 
114. .02 
Bh = AES KO KE = .0955, 
9) 
and Bh = wEeK X_-886 = .169. 
B, Suppose that the first stake is set at B’, Fig. 27, instead of 
it B, the error in the angle being m4 
BAB’ = A. 


Then it is evident that C’, 
D’, etc., will be set at the same 
ingular distance from the chords 
AC, AD, ete., as B’ is set from 
4B. B’, C’, etce., are on the 
‘urve, having O’ as center and 
AO’ = AO as radius. OAO’ 
= BAB’. 

To find the error at the end of 
the curve. 

Suppose £# the end of the true 
ourve, 


64 FIELD-MANUAL FOR ENGINEERS. 


Now AK’ = AB, and the angle HAK’ = the angle BAB’. 
Hence HH’ = 2AM sin }HAEH’ = 2A sin } BAB’. 


If the error in the angle is corrected at the end of a run, say at 
H’, then, for reasons given above, the error in the position of the 
stations is constant from #’ onward to the end of the curve. If 
the error is corrected during a run, say just before D’ is set, 
then that station will be set at D, on AD and 100 feet from C’. 
Similarly H, will be set on AH and 100 feet from D,. 

It may be shown that HZ, is less than DD, precisely as it was 
shown that CC’ is less than BB’ in Fig. 25. Hence in this case 
the error will decrease to the end of the run, and remain constant 
‘from that point to the end of the curve. 


II, Curves Laid Out by Tangent Offsets. 


If in Fig. 22 the tangent at B is swung through an angle A, sav, 
then all stations following B will be misplaced, the error increas- 
ing to the end of the ‘‘run” in the manner shown in Fig. 27. Ili 
a new tangent is drawn before reaching tne end of the curve, as 
at H’, Fig. 27, it is easy to see that such tangent would make an 
angle A with the tangent to the true curve at #, and that the 
error, therefore, would go on from #’ forward precisely as from 
B to H’, and hence the error would increase regularly from B to 
the end of the curve. | 


III. Curves Laid Out by Ordinates from a Long Chord. 


In this case, if the end C of the chord CK, Fig. 21, is misplaced, 
then all stations from D to Z inclusive will be misplaced in pro- 
portion to their distances from K, the greatest displacement being 
less than that of C. If K is also misplaced, the same stations 
will likewise share that error, in proportion to their distances | 
from C. * | 


IV. Curves Laid Out by Chord Offsets. (Fig. 22.) 


If in this method any station, as B, is set at one side of its true 
position, then all subsequent stations will be in error in the same 
direction ; the errors increasing regularly to the end of the curve, 
in the manner shown in Fig. 27. 

If, owing to an error in some chord, some statiun is set forward 
--h--kward from its true position, then all subsequent stations 

. error the same amount and in the same direction. 


SIMPLE CURVES CONNECTING RIGHT LINES. 65 


V. Curves Located by Middle Ordinates. 


In Fig. 23 suppose (to be placed a distance a to the right, say, 
that is, along CO,) of its true position. Then c will be a distance 
: to the right; D and d will be 2a to the right ; Hand e will be 
4uto the right of their true positions, etc. 

If C is correct, but c a distance a too far to the right, then D 
‘nd d will be 2a@ to the right, H and e will be 4a to the right, F 
und f will be 6a@ to the right of their true positions, etc. 


PROBLEMS IN SIMPLE CURVES. 


I. Given the tangent distance AB = d and the tangent offset 
BD = t, to find the radius of a curve that will pass through J 
and be tangent to AB at A. 


let AD=c. We have _ A BB 
OD = OU? + CD’,' . D’ 
or R= (R—d4+ a. C 8) 


From this we have 
@? 
k= iF +4). ° (a) 


From (@), 


t= R— /R?—d, . . (89) Fia. 28. 


and d= yt(2R—t). ....... (40) 
Second Solution.—Let AOD = A; then 
BAD = ADC = jA. 
Now the triangle ABD gives 


d 

r =cot#A. . ......, (41) 
Also = rers A 42 

R —— V 7 8 9 e ry e e e e bd ( / 


and p= cose aL - oe 2 «we (48) 


66 FIELD-MANUAL FOR ENGINEERS. 


If d and ¢ are given, find A from (41), then I? from (42) or (48). 
If ¢ and Rare given, find A from (42), then d from (41) or (43) 
Finally, if d and J? are given, find A from (48), then ¢ from (41) 
or (42). 

zample 1.—Given ¢ = 12 and d = 171.6, to find R. | 


tan }dA = 171.6 7 .06993. 


4#1= 4°, and A=8., 


t 12 


=- —~-- —  — = 1233.3. 
~~ vers . 00973 1233.3 


Then 


Erample 2.—Given R = 1233.3 and d = 171.6, to find ¢. 


1283.3 _ 


cosec A = - Tre 7 7.187. .. A=8?’. 


Then ¢ = 1233.3 « .00973 = 12. 


This problem is useful in finding points on a curve beyond an’ 
obstacle. 

II. To find the distance to a curve in a given direction from a: 
A given point on a tangent. 


NP We have 


AO=R, AB=d, and ABP= B. 


tan ABO = .. PBO = ABO — ABP: 
| OB 
sin OPT = sin OPB = R x sin PBO 
_ sin PBO 
~ gin ABO 
Then 
0) PB = Ret (OPI — PBO) (44) 


Fia. 29. sin PBO — 


This problem furnishes a general method of finding any desired 
point on a curve when obstacles preclude the usual methods. 
(See Problem 12.) 

II. Having run the curve AD, radius AO = Rf, Fig. 28, to 
find the radus RF’ of a curve that will pass through D’, given by 
angle BDD’ = D, and DD’ = #. 


ns 


SIMPLE CURVES CONNECTING RIGHT LINES, 6% 


Let AB=d, and BD=t. 
Draw D’H parallel to AB. Now 


HD’ =FEsinD, and HD = EKeosD. 
AB’=—d+Hsin D=ad,, and B'D'=t— Hes D=t. 


1 /d,? 
Then Pr = 5(¢ + a). 


It will be observed that when AB’ < AB, # sin D must be 
subtracted from d to give d,; and that when B’D’ = BH > BD, 
Ecos D must be added to ¢ to give 4. 

IV. Having run a curve of radius R and tangent 7, to find the 
ew tangent 7” corresponding to a new radius R’, or to find a new 
radius Ji’ corresponding to a new tangent 7’, the central angle re- 
maining constant. 

Eq. (15) gives 


T’ = Rh’ tantVv; T= RtantV; 


. 7’ -T=(R'—R)tangVvy, . . . . (48) 

or R’'—-R=(T’—T)cothV. . . . . (46) 
Similarly, from eq. (17), we get 

C'—C=2%AR — R)singvV. . . . . (4%) 


These equations are of advantage for computing the change in 
one element, 7’ — JZ for example, from the change in another, 
R’ — R for example, when the given change is small, or is an 
aliquot part of the element changed. 

EHrample 1.—Having run the curve of radius R = 1910, and the 
central angle V = 46° 12’, and tangent distance 7 = Rtan 4V 
=1910 x .42654 = 814.7, to find 7’ when #& is made equal to 1900. 

Eq. (45) gives 


TT’ =T+(R — R)tan3V 
= 814.7 — 42654 x 10 = 814.7 — 4.97 = 810.4. 


Example 2.—Given V = 52° 04’, & = 5730, and 7’ = .48845 
x .5780 = 2798.8, to find R’ corresponding to 7’ = 7 + = 
We have 


R=R+ a = 2798.8 + 2383.2 = 3032.0. 


68 FIELD-MANUAL FOR ENGINEERS. 


V. Given a curve joining two tangents, to change the position | 

of the P.C. so that with the same radius the curve may end in 
| a given parallel tangent. 

Let AB be the given curve, | 
and B’ V’ the parallel tangent. 

V V'=a shows the distance | 
and direction that all points of | 
the curve are moved. The 
curve will therefore begin at 
A’andendat B’; AA’ and BB’, 
as well as OO’, being equal 
and parallel to VV’. It is. 
- not necessary to run the tan-— 
0 0 gent BV’ in order to find the 

Fig. 30. distance VV’, To find this 

distance run a line, such as BB’, parallel to _AV from any point 
on BV to meet DYV’. Then make 4A’ = BB’. 

If the perpendicular offset Bh = h is measured, we have 


h 


A ‘= a — oT 
A BB sin V’ 


V being the vertex angle. 
If B’V’ were on the other side of BV from that shown, the | 
new tangent point A’ would fall on the opposite side of A from _ 
that shown in the figure. 
If the new curve is required to end at a given point on B’ ae 


we have, then, the length of the A V y’ 
new tangents, : = 


A'V’ — B’ V’ = T’, 


which gives the pusition of A’ (and 
B’), and the corresponding radius, 


R’ = T" cot dV, 
or, by (46), 


R’ = R+(T’ — 1) cot bV. 


Fig. 31. 


VI. Given acurve AB joining two tangents, to find the radius 
of a curve that from the same P, C, will end in a given parallel 
rt, 


SIMPLE CURVES CONNECTING RIGHT LINES, 69 


Let AV=BV=T7, AV’ = B’VW’=T7", AO=R, ACR’. 
We have, from the figure, 

R’ T’ , vyyt 

R = ike or R = Roi 
Also, from eq. (46), 

R’=R+(T’ — T) cot hV. 


Or, prolong AB to B’ and measure BB’. Let AB =c, and 
AB’ =e, BB’ =¢’—c. Then, from the figure or from (17), 


co ¢ 
2sin}V" 


Ri=R+ 


If the parallel tangent is defined by a perpendicular offset, as 
B’p = h, draw BC parallel to AO. Then 


Cp = BC cos BCp = (R’ — R) cos V. 
. CB’=(R'—R)=(R'—R) cos V+-A, or (R’—R)(1—cos V)=A, 
or (2’ — R)versin V=h, or R’=R+ h 


versin V’ 

The quantity added to & in the above equations must be sub- 
tracted from it to find ’ in the case in which V’ falls between A 
and V, that is, when 7’’ is less than 7’. 

Example 1.—V = 78°, R= 954.9. JZ’ may be computed or found 
by Table VII to be 773.3. 

Let VV’ = 20 feet. Then 


R’ = R+(T’ — T) cot 4V = 954.9 + 20 x 1.2349 = 979.6. 


Kazample 2.—R = 1909.9, V = 46° 88’. ZT may be computed or 
found by Table VII to be 823.2. 
It is desirable to move the vertex from Vto V’ about 100 feet. 
Find the new radius 2’. 
823.2 + 8 = 102.9; 
1909.9 + 8 = 238.7. 


Hence AV’ == 8238.2 + 102.90 = 926.1, 
and the new radius 


AVY = RR’ — 1909.9 -+ 288.7 = 2148.6, 


70 FIELD-MANUAL FOR ENGINEERS, 


VIL. Given a curve joining two tangents, to find the new tan- 
gent points, corresponding to the 
same radius, after each tangent has. 
been moved any distance in the di-. 
rection of the other. 

Let AVand BV be the given and 
A’V’ and B’V’ the required tan- 
gents. Let H be at the intersection 


of AV anl B’V’. Let VT =a, 
B VV’ = b, and VV’ = «. 
We observe that 
KX VHV' = 180° —V. 
.. sin ViZV’ = sin V, 


— cos V. 


Fra. 32. and cos VHV’ 
Chap. II, formula No, (8), gives 
tan HVV’ = Ww sin 0 
a 
iv’ +- cos - + cos V 

bsin V 
sin HVV" 

We observe that the directions of VHand HV’ are the same 
as that in which the tangents BV and A V are moved, and there- 
fore there can be no ambiguity about the direction of these lines 
or of VV’, which is the line joining V and V’. 

Since Rand V are not changed, it is evident that all parts of | 
the curve are moved in the direction VV’ and a distance equal to 
VV’. 

Ilence make the angle VAA’ = HVV’, and AA’ = VV’. | 

The curve will begin at A’ and end at B’, BB’ as well as OO’ 
being equal and parallel to VV’. 

If the distances the tangents are moved are given by perpendic- 
ular offsets V/A’ = hh’, and Vh=h, the triangles VHh and, 
V’Hh’ are similar and give 


Then VV’= 


HV eh 
HV’ / 
1 sin VHV’ 


HY’ — cos VHV 


SIMPLE CURVES CONNECTING RIGHT LINES. wl 


! . V 
or tan HVV’ => -. 
i + cos V 
Now the triangle h VV’ gives 
, h' 
VY" = sin VV 


VV’ and VA are on the same side of BV; also, VV’, and V’h' 
are on the same side of A V. 

VIII. Given a curve AB joining two tangents AV and BY, to 
change the curve so as to end at the 
same point as before, but in a tan- A __A V_Vv'_OD 
gent inclined at a given angle A ——~\ 
with the original tangent. 

Let AV = BV = T, and the new 
tangents A’V’ = BV’ = T’, 


Let AO= R, and A’'O = R’, 


B 
Draw BD = p perpendicular, and BB 


BMN parallel to AV. 

Let 
V’=BV’'D=BVD- VBV'= V-+-A. 

Now An = R versin V, O 
and A’m = R’ versin V’. Fia. 38. 
But A’m = An; .. Rversin V = RB’ versin V, 

,  RKversin V 
or = ———, 
versin V 


With this value of R’ run the curve back from Bthrough the 
angle V’, and it will end at A’, tangent to AV, A’V’ being equal 
to BV’. 

If the length of the new tangent is desired, we have, from the 
triangle VBV’, 


qr — LsinV _ BD a 
~ sin V’ ~ sin V’ sin V” 
Then 
R’ r = Pp = P 


~ tangV’ sin V’tan3V’~ vers V" 


02 FIELD-MANUAL FOR ENGINEERS. 


If we wish to run-the curve from A, we have 
DV=pcot V, and DV' =pecot V’. 
'. VV’ = p(cot V — cot V’), 
and AA’= AV+ VV' —A’V'’ = VV'’+T— T’. 


When V < 90°, 7 increases as V decreases, and vice versa ; 
and when V > 90°, 7’and V increase and decrease together. In 
all cases R increases as V decreases, and vice versa. 

IX. Given a curve, radius AO = R, joining the tangents 

A V AV and BY, to find the radius 

AO’ = Ff of a new curve start- 
ing from A when the forward 
tangent VB’ takes a new direc- 
tion from the vertex. 

We have 


VA = RtanjiV, 


and 
R’ = VA cot }V’. 
. BR’ = Rtan $V cot $V’ 
tan 4V 
o" ~ "tan iy" 
3 
Fig. 34. 

X. Given a curve AB, radins AO = R, joining the tangents 
AV and VB, to find the change a A’ y 
in the P.C., the radius remain- 
ing the samme, when the forward 
tangent takes a new direction 
from the vertex. 


We have 
VA = RtaniyV; B’ 
VA’ = Rtan $V’. 
*, AA'=Ri(tan 4 V—tan 30 ). B 


XI. Given the angle of in- 
tersection V of two tangentsQ Oo 
1V and BY, to find the radius Fie. 3. 


SIMPLE CURVES CONNECTING RIGHT LINES. v3 


| R and tangent distance 7 of a curve joining the tangents and 
passing through the point Z. 
1°, Let # be given by VH¥ =1, 
and angle HVO = A. 
Lett VEO=H, VOH=0O, 
and AO=R, AOV= UV. 


Now 
os t Pa AO KO _ sin A . 
~t" " VO" VO ~ sin Kk’ 
. sin A 
sin #H = cos FV’ 


This gives H. Then 
O = 180° — (A + £). 


Moreover, 
| 
HO _&_ sin A ir _ ,sin A 
EV 7% sno’ ‘ = "gin O° 
Finally, T= Rtan} 


2°. If His given by VIZ and HE perpendicular to each other, 


EH 
EVF = FVH — EVH = 90° — }V — EVH; 


VH 


VE = cos EVH’ 


With these values proceed as above. 


3°. Let EH be given by VD = a, DE = b, the angle VDF being 
eqialto VOA =14V. Produce DH to F'and G. 


Let DF = ec. Then ¢c = acos}V, and AD = YDE x DG. 
Bt DG = DF4 FG = DF + EF =2DF — DE = 2c — b. 
-, AD= Vb(2e—b),, and VA=VD4+DA=T. 
Now R= T cot $V, 
o HH= DHKsin EDH= bsin tV. 


04 FIELD-MANUAL FOR ENGINEERS. 


DH = b cos } V. 
Then VH = VD — DH = a — bcos $V. 


With these values of VH and H# proceed as above. 

XII. Given a tangent and curve (Fig. 36), to find the distance 
from a given point on the tangent to the curve in a given direc- | 
tion. 

Let V be the point, and suppose the direction defined by the 
angle HVA=B. Let AV=T7. We have 


T 
tan 1V= R’ 


Then EVF = DVF — EVA = (90° — 3V) — B=A, say. 


Now equation under Problem XI gives 


. sin A 
sin # = - iv 
Then O = 180° — (A+ #), 
sin O . 
and VE= t= sin A’ 
or, 
tan FVD = af and EVF = FVD—EVD=FDV—B= 4A. 


Then find sin Z, then O and / as before. 
Erample.—R = 954.98, T = AV = 860, AVH = 40°. We have 


» 


tan A VO = : = 2,72837; .*. AVO = 69° 52’, 


and AOV = 4V = 20° 8’, and HVF = 29° 52 = V. 


sin A .49798 
* -— — — e ; oe EH —_— 8 re 
sin H = =08 1 17 “93880 7 58039 = 147° 58 


O = 180° — 177° 50’ = 2° 10’; 


954.93 x .08781 _ 
0 gg TR. 


SIMPLE CURVES CONNECTING RIGHT LINES. 


75 


XIII. To locate a tangent to a curve of given radius R from a 


' given point V. (Fig. 37.) 

1. If the curve is marked 
by stakes visible from the 
given point, a tangent can 
be sighted in at once. 

2. If the curve is not 
visible from the point, run 0) 
a trial tangent VB, and Fig. 37. 
measure VB = A and the angle VBO = B. 

Chapter III, formula (8), gives 


E 


tan BVO — 


sin B EO 


Then OV = OB. and sin HVO = 


Lastly, BVE = EVO — BVO. 


K oH This gives the angle to be laid off from 
the trial tangent to give the true tan- 
gent AZ. 

c XIV. Given two curves ABand AC, 
radii AO = R, AO, = R,, subtending 
the central angles AOB=9O and 
AO,C = O,, to find the length of the 
line BC. (Fig. 38.) 

Let A, B, and (represent the angles 
of the triangle ABC, and a, d, and cthe 
sides opposite. We have 


c= 2Rsin40, and b= 2R, sin 40). 


Also, 
BAO = 90° — 340, CAO, = 90° — 30: 
Fie. 28. . BAC= A= 4(O — Or). 
tan B= _Sin A (See Chap. III, formula (8).) 
© — cos A 
b 
_ ,sin A 


a —— e es 
sin B 


"6 FIELD-MANUAL FOR ENGINEERS. 


If the curves are run through an integral number of stations, 


Tables IV and V give at once the tangent distances AK and A/l | 


and the tangent offsets BK and CH. Then, drawing CUD parallel 
to AlITto meet BA in D, we have 


CD = AH — AK=4d,say, and BD= BK — CH=, say. 


Then BO= Vda? 4+ &, 
CD d 
or tan CBD = BD == t? 
,.. BD 
and BC = OID 


OBSTACLES IN SURVEYING. 


It is often necessary to draw lines parallel and perpendicular to 
other lines. Hence the follow- 
ing problems : 

I. To erect a perpendicular at 
any point of a line. (Fig. 39.) 

1. Let A be the point, and BC 
the line. Make AB = AC, and 
with B and C as centers and 
any radius greater than AB de- 
scribe arcs intersecting at D or 
at ZZ, or (with a different radius) 
at F. Any two of the points A, 
D, H,and F' determine the per- 
pendicular required. * 

2. Fix any two points of the 
chain at Band at C. Take hold 
of the point midway between B 
and C and stretch the chain, the 
middle point being at D. AD is the perpendicular required. 
We may find, similarly, other points #, F, etc. Any two of these 
points A, D, H, F, etc., determine the perpendicular required. 

3. Let C be the point. Take any point D as a center, and with 
a radius DC describe an arc BC. Prolong BD, making DH = BD. 
CH is the perpendicular required. For DA (A being at the mid- 
dle of BC) is perpendicular to BC, and, by construction, CH is 
parallel to AD. 

4, A right angle may be obtained by laying off on the ground 
the three sides of any of the triangles represented in the following 

‘ble, or any equimultiples of these sides, making one of the 


SIMPLE CURVES CONNECTING RIGHT LINES. rar 


sides adjacent to the right angle (@ or 0) coincide with the line. 
Let ¢ = the hypothenuse. 


ener fe | ff | Ll 


I 
1/ 38 | 4] 5 6 | 20 | st | 29 
2) 6 | a | as || 7 | a | a | ae 
3 | 8 | 1] 17 |} 8) 9 | wo | 41 
4 | 7 | 24 | 2s 9 | 11 | 6 | 61 
a | 
| 


Thus, in Fig. 40, using 70 links of the chain, hold the first 
end, also the end of the 70th link of the chain, at if, the end of 
the 21st link at B, and the end of 
the 50th link at C. C 

If in the three expressions 
am? —n?, 2nn, and m? + 2? we 
assign to m and 27 any values at 
pleasure, m being greater than n, 
we will have sets of numbers yy *”) 
representing the sides of right- 
angled triangles. In that way 
the above numbers were found. 

Equimultiples of any set of the 
above numbers will represent the B a1 A 
sides of a right-angled triangle. Fia. 40. 

II. Zo let fall a perpendicular from a given point to a given 
line.—Let Zin Fig. 89 be the point. Measure any line HB to 
the given line. At the middle of HB take Das a center, and 
with a radius DB describe an arc BC. HC is the required per- 
pendicular. For continuing the arc to H, we see that the angle 
BCG is inscribed in a semicircle. 

p II. To let fall a perpendicular to a 
‘line from an tnaccessible point.— Let 
BC (Fig. 41) be the line, and P the 
point. Let p represent the perpen- 
dicular PA. Then 


BK=pcot B, and CK=pcot C. 
Bk _ctB 


” CK ~ cot @ 
C ad BK _  ctB 
Fia, 41. ee BE + CK ~ cot B+ cot C’ 


"8 FIFLD-MANUAL FOR ENGINEERS. 


Since BK + CK = BC, we have 


_ an cot B 
BK = BCot B+ cot O" 
This gives the foot of the perpendicular A. If BC is taken equal 
to 100 or some small multiple of 100, BA is very easily found. 

This problem is particularly useful in locating important ob- 
jects, such as mills, warehouses, bridges, etc., on one side or the 
other of a railway survey, 

In this case B and C represent stations or points on the survey, 
and the angles at B and C can be measured and recorded while 
the instrument is set at B and at C. The simple divi-ion required 
to find the position of K can be made at any time. Of course the 
point Pis located graphically by drawing BP and CP. 

IV. To prolong a line AB (Fig. 42) past an obstacle and to 
measure its length.—This is easily done by perpendicular offsets, 
a method to. familiar to need description, but not the best way. 


A B H D E 


Fia@. 42. 


Or, measure BC in any convenient direction, and at C deflect any 
angle ¥CD. Draw BD and the perpendicular CH. The angle 
CDB = FCD — CBD. Hence 


sin CBD 
CD = BCT BDC ' 
sin FCD 
also BD= sin BDC’ 


If the angle BCD is made equal to 90°, then 


CD = BC tan CBD, 
and BD = BC ~ cos CBD. 


SIMPLE CURVES CONNECTING RIGHT LINES. V9 


If the angle CD is made equal to 2CBD, then 
CDB = FCD — CBD = CBD. 
Hence CD= BC, and BD=2BH = 2BC cos CBH. 


CH is the departure of the line BC, or of DC, from the line 
ABD. It is also the approach of the line CR, or of CD, to ABD. 

If necessary more than one course may be run away from the 
main line ABD, and more than one in returning to it. 

‘lo recover the main line it is only necessary to make the sum 
of the approaches equal to the sum of the departures, 

The distance measured on the main line is obtained as above. 

V. Obstacles to Measurement.—Methods have been pointed out 
' in connection with Fig. 42 for finding the length of obstructed lines 
when the ends are accessible. When inaccessible the following 
problems apply. 

A. When one end of the line is inaccessible, (Fig. 43.) 

1. Let AB be the line to be measured, across a river for ex- 
ample. Measure AC’ in any 
convenient direction, and the 
angles at A and (. Then 


AC sin C 
AB= gin B 


sin C 
sin (A + C)’ 
2. If the angle ACB is 


made equal to half of DAC, 
then 


CBA = CAD — BCA 
= 2BCA—BCA= BCA. 
“ AB= AC. Fra. 43. 


8. Or, in Fig. 48, make the angle BAC = 90°. Then 
, AB = AC tan ACB. 


If, in this, AC = 100, or some simple multiple of 100, which is 
usually easy to effect, the formula requires no computation 
whatever. 

4. If ACB in Fig. 44 is made equal to 45°, AB = AC, 


80 FIELD-MANUAL FOR ENGINEERS, 


5. If at C we make the angles ACB and ACD equal, we have 
AB = AD. 


‘When the river or other obstruction occurs on a continuous 
survey, as a railway survey, AD is a measured line, and this 
method gives AB = AD without any computation whatever. 


Fia. 4. Fig. 45. 


6. In Fig. 45, AB being the distance required, run and measure 
any line AC and measure the angle BAC = A. Make 


ACB = 90° —-A= 0. 
Then AB = ACsin C. 


B. When both ends of the line are inaccessible. (Fig. 46.) 
Let AB be the line to be meas- 
ured. Find the distancesfromthe A B 
point C to each end of the line A 
and 8B by preceding methods, and 
measure the angle C. Then 


nC D E 
sin 
tan A = 40 4 
BC C 
(see Chapter II, formula (8).) Fia. 46. 
sin C 
AB= BC snd: 


C. To erect, at a given point A (Fig. 47), a line AX perpendicu. 


SIMPLE CURVES CONNECTING RIGHT LINES. 81 


lar to an inaccessible line 
BC, and to draw a parallel 
AH tothe same line.—Find 
AB =cand AC = b by pre- 
ceding methods. Then 

tan B= _sin A : 
— cos A 


Now draw A, making 
BAK = 90° — B. 
AK will be the required per- 


A 
Fig. 47. 


pendicular, and AH, making BAH = B, will be the required 


parallel. 


D. To find the length and relative position of an inaccessible line, 
AB, from an accessible line, CD, separated 
from the former by an inaccessible space. 

Measure CD and the angles at Cand D. 
Fzample.—Let CD = 4000; 


BCD = 
ACB = 


126° 257’; 
3°10’; ADB= 38°01’. 


ADO = 47° 53}'; 


-- ACD = 129° 85)’; BDC = 50° 54y’. 


Also, 


CAD = 180° — ACD — ADC = 2 31’; 
CBD = 180° — BCD — BDC =2° 40. 


Hence 


AC= 40005 


sin 47° 584’ 


nO ay > 67581.9; 


BC = 4000 sn 50” 544" _ g6798.8. 


Cc tan CAB = 


0 Finally, AB = BC —. 


Fie. 48, 


AC_. = 70143115 
BC 
. CAB = 75° 282". 


2° 40 


sin C .0552406 —3 85987, 


— cosC’ 


sin 3° 10’ 
sin 75° 283’ = 3807.8, 


82 FIELD-MANUAL FOR ENGINEERS, 


If AB is accessible, it can be measured as a check on the com- | 


putation, Such is the case when it is a tangent of a railway sur- 
vey adjacent to tiie inaccessible space. 

The data of this exainple are taken from an actual night survey 
across an inaccessible sea-marsh. 

Rockets were thrown and lights then exhibited at A and B. 
which were observed with transits from the tops of towers at C 
and D. 

The computed and the measured length of AB agreed within a 
few inches. 

The best method of making a preliminary railway survey 
through a wooded region is by a suitable adaptation of the method 
of traversing, which we will now explain. 


So far as known to the author, this was first devised by him in | 


1869, aud used for him by his assistant, Prof. J. B. Davis (now of 
Michigan University) in making the preliminary surveys of the 
Owossvu and Northwestern Railway. 

Suppose we wish to run from A in the direction of ABF, which 
we will call the base line; and upon which numerous obstacles, such 


as trees, occur, aking it necessary to run the line ABC, D,Z, F, 
called a traverse. The deflection angles at B,, C1, D,, end H, are 
supposed to be small. 

From B,, C,, D,, and #, draw perpendiculars to AB, and from 
B,, C,, aud D, draw parallels B,K, C,L, and DiP to AB as 
shown. Prolong AB, to R, and B,C, to 8. 

The course of a line is its direction with reference to the 
base line. 

The departure of a line is the distance that a point recedes 
from or approaches to the base line in moving from one end of 
the line to the other. 


amt 


We have D,L = C.D, sin D,C,L. Now since the sines of | 


small angles vary nearly with the angles or with the number of 


minutes in the angles, we see that the departure of a line varies | 


SIMPLE CURVES CONNECTING RIGHT LINES. 83 
. ® 

nearly as the product of the length of the line by the number of 
minutes in the course. 

The departure of a point is its distance from the base line, 
Thus the departure of D, = DD,. 

The departure of the end of a line, as B,C,, inclining from the 
' base, is equal to the departure of the beginning of the line plus 
the departure of the line. Thus CO = BP, + CAH. The 
departure of the end of a line, as C,D,, inclining toward the base, 
is equal to the departure of the beginning of the line minus the 
departure of the line. Thus DD, = CC, — DL. The departure 
of the end of the line D, #, which crosses the base line is equal 
to the departure of the line minus the departure of the beginning 
of the line. Thus HZ, = £,P — DD,. 

The record of the survey can be conveniently kept, as shown 
in the following table, the columns of the transit-book serving 
the purpose perfectly. 


; h Departures. 
Angles wit 

Angles turned. Main Line. 

Stations. Each Course. Total. 


Left. | Right. 


Left. | Right.’ Left. Right. Lefe. “Right, 


fa ee NT 


Cy=+] 47 10’ 7040 10040 


D,= 19] 3¢’ 40’ 3800 6240 


E,= 24 1° 20’ 40 20000 13760 


Path 13760 00 .00 


The deflection at station 10 is 10’ R., and at station 13 it is 
22' R., etc. The course from 10 to 13 is evidently 10’ R.; from 
13 to 15 + 20 it is 10 + 22 = 387 R.; from 15 + 20 to 19 it ig 
42 — 32 = 10’ L., etc. 


Ld 


84 TIIELD-MANUAL FOR ENGINEERS, 
e 
From 10 to 13 the departure is 30010 = 3000 foot-minutes, 
‘* 13 “ 15-+-20 the departure is 22032 = 7040 ‘‘ 


‘© 15-420 to 19 * “ ‘* 38010 = 3800 ‘ 
‘19 to 24 ¢ “ ** 50040 = 20000 ** etc. 
The aggregate departures are : 

7 2: Se 3000 R. 

At 15 + 20..... 3000. -+ 7040 = 10040 R. 

At l19............ 3000 + 7040 — 3800 = 6240 R., etc. 


The distance necessary to run from a given station on any given 
course to reach the base line is found by dividing the departure 
at that station by the course. Thus from station 24 forward the 
course is 40’ R. Then 13760 -+ 40 = 344 feet, showing that the 
auxiliary line (#,¥F in the figure) will reach the base line 344 feet 
beyond station 24, or at 27 + 44. 

The figure represents a main angle at F, the forward tangent 
being #77, and which may be followed approximately the same as 
«1 F was followed. 

Let / = the length and d = the departure of any line, and 7 = 
the number of minutes in the course. Then 


ad = JTsin n’ = in sin 1’ = .0002909/n. 


Since Jn is given in the last two columns, the departures in feet 
are found by multiplying the quantities in these columns by 
.0002909 or .00C29 nearly. We observe that 


1 -- .0002909 = 3438 nearly. 


Hence the departures given in the table (in foot-minutes) 
divided by 3488 will give the departures in feet. 

A rough approximation for the purpose of keeping sufficiently 
near the base line on sideling ground is generally all that is 
needed. This being the case, it is not in general necessary or 
advisable to find the total departures, except when it is desirable 
to ‘‘run for the base” preparatory to turning a main angle. 

Thus to find the departure at station 24. The sum of the 


product to the left is... .....00. 008 one 3800 + 20000 = 23800 

and the same to the right is........... 3000 + 7040 = 10040 
The difference i8....-. 0... cece ce cee ce cnevees ee ceee . 13760 L. 
We have AB = AB, cos BAB,, 

or 


AB, — AB = ABI — cos BAB,) = AB vers BAB. 


SIMPLE CURVES CONNECTING RIGHT LINES. 85 


For BAB, = 2° 34 this becomes AB, — AB = .00LAB nearly. 
This shows that AB, exceeds the true distance measured along 
'the base by only one thousandth part of its length for an angle 
of 2° 34’. If greater accuracy than this is desired, the angles 
between the auxiliary line and the base line, or the ‘‘ courses,” 
may usually be made smaller than 2° 34’. Since the error is 
approximately as the square of the number of minutes in the 


angle, 
for 1°17 it is nearly .00025, or nearly 1 in 4000; 
and for O° 3s «ss «00006, “ = ** 1 ** 16000, ete. 


To find the angle in minutes between the base line and a line 
joining any two stations. 

Divide the difference or the sum of their departures, according 
as they are on the same or on opposite sides of the base line, by 
their distance apart. 

Thus the line B,D, makes with the base line the angle 


DD, — BB, _ 6240-3000 4g oy 


BD ~ 600 

The line AH, makes with the base an angle 
EE, 18760 , 
ae = Tog = 9-88 = 9 50” 


In platting, the auxiliary lines are penciled only, so as to plat 
observed objects in proximity to the line necessarily observed 
from the auxiliary lines. When these objects and the base lines 
are mapped the auxiliary lines need not be retained. 

This method yields quite accurate results when the angles be- 
tween the lines of the survey are 2° or 3°, as we have seen. 

For perfect accuracy, however, use the following method : 

Problem.—Having run a broken line ABCD, to find the angle 
between the first course and the direct course AD. Represent the 
lines run, in their order, by a, b, and c, and the deflection angles at 


B H 
Fia. 50. 


Band at C by B and C respectively, Draw CR parallel to AB, 
and CH and RDEK perpendicular to AB, 
Angle DCR = DCE — RCH = C— B 


86 | FIELD-MANUAL FOR ENGINEERS, 


Now 
BH=bcsB,; Hk=ccos(C—B) AK= AB+ BH+ AK. 
CH = bsinB; DR=csin(C—B); DK = CH — DR. 


DK 
N ° t = 
ow an DAK = AK’ 
Also AD = AK ~ ccs DAK. 


Of course the method is applicable whatever the number of lines 
run, All but the last line could usually be taken equal toa whole 
number of chains, which 
would reduce the required 
computation to a simple mul- 
tiplication, 

E. To find the angle of 
defiection, JV, between two 
straight lines AV and V&, 
when the point of intersection 
is inaccessible; and the dis- 

A B M tances of the intersection from 
Fia. 51. given points on the lines. 

1. Run and measure a perpendicular PK to one of the lines. 

Measure the angle VPK = P. Then 


V = 90° + P; 
KV = KP tan P, 
and PV = KP — cosP. 


. Run and measure any line PZ from one line to the other. 
Measure alsu the angles VPL = P’ and PLV = I. Then 
Ve P+. 

Hence, in the triangle PVL, PZ and the angles are known, to 
find VP and VZ, 

3. If obstructions prevent the use of the former methods, run 
and measure any broken line ABCD. Prolong AB and BC te 
meet VM at M and N. 

Measure the deflection angles CBM = B, DCN = C, and CDN 
= D. 

Let AMV = M,andCNV=WN. Then 


sin D _ 7 sin D | 


BN=BC+CN. Also M=N-B=C4+D-B: 


SIMPLE CURVES CONNECTING RIGHT LINES, 87 


sin V 
BM = BN——; AM = AB+ BM. 

Now we have AM and the angles at A and M, to find AV, MV, 
and the angle V = .1+ M. A similar explanation will apply to 
any case, 

When a broken line must be used, the above method involves 
fewer computations than any other. 

F. To locate a curve joining two tangents when the vertex is 
inaccessible. A 

Find by the last problem the 
distances Va and V6 to convenient a 
points on the tangents, and the V 
angle V. Then assigning or com- 
puting the tangent 7 = AVor BV b 
from the radius, we have 


aA = T7’— ay, 
and bB= T — bY. 
O B 

We now have the tangent points Fie. 52. 
and can run in the curve as usual, 

G. To locate a curve of radius 2 or tangent Z' when the vertex, 
the beginning, and the end of the 
curve are inaccessible, Find, as 
shown with Fig. 52, the angle V 
and the distance Va to any point, 
a, on AV. Then 


n A 


aA = T — aV, 
and 
. cd aA 
0 B sin AOe = 1 = 
Fic. 58. Also 


ac = Ad = Rversin AQc. 
Drawing the tangent cb, we have 


. abe = AOc, or ach = 90° — AOc. 
This gives the direction of the curve atc, and it may be run in 
each way from c. 
To pass from any point con the curve to any point on the 
tangent. 


ac 
We have tan (den = anc) = ~ 


88 FIELD-MANUAL FOR ENGINEERS. 


Set the instrument at cand turn off from the tangent cb an angle 
ben = bed — den 
= abc — ane, 
an 
cos ane 


and measure cn = 


H. To find any desired point on a curve when obstacles preclude 
the use of ordinary methods. 

(1.) In Fig. 28 measure any convenient tangent distance AB = d. 
Then, as shown in Problem 1, eq. (43), 


cosec A = a; then ¢=d tan 3A. 


d and ¢ give the point D on the curve. 

It is important to note that if AB is made equal to one half the 
long chord for any number of stations given by Table IV, BD = 
AC is the corresponding middle ordinate and may be found in 
Table V. 

Hzample.—Let AD be a 4° curve, and AB = one half the chord 


of four stations = 398.70. = 199.35. Then Table V gives 


2 
BD = 18.94. 


(2.) Problems 2 and 12 of this chapter furnish general methods 
of overcoming obstacles on curves. 

(3.) We can find points on the curve as follows: 

Let 5 be a station near the obstacle. Deflect from the tangent 
at 6 some small multiple of the deflection 
angle for one station 3D, giving the line 
bd. The length of dd may be taken at 
once from Table IV and measured off, 
giving d, a station on the curve beyond 
the obstacle. Taking bm = }$bd, and 
measuring off tle middle ordinate me 
taken from Table V, gives also a point c 
on the curve. 

If more convenient, make ba = cm, and 
ac = bm, which also gives c. 

Again, run the tangent DV’ =d’ any 
Fia. 54. convenient distance. Then 
bo =k  R 
ov’ ov’ 


. 


cot 0OV'’ = 


\ 


{ 


SIMPLE CURVES CONNECTING RIGHT LINES. 89 


Defiect at V’ an angle equal to 2b0V', and muke V’d = dV’. 
a@ will be a point on the curve. The number of stations from } 
is equal to . 
60d 2b0V’ 
Dp =D 
bV’ should usually be taken equal to a whole number of chains, 


. . EK 
in which case a 

The lines }V” and Vd lying on the inside of dd may be run 
instead of DV' and V’'d. 


is very readily found. 


CHAPTER V. 
LEVELING, STADIA MEASUREMENTS, ETC. 


THE field operations in connection with the level are more 
simple than those required with the transit, but they require 
greater skill and facility in manipulation in order to produce 
correct results. 

It is to be observed that the elevation of points is a relatire 
matter. The elevation of some point, from which all others are 
to be found, is arbitrarily assumed to be 100, or some other 
number sufficiently large, so that the elevation of all points 
to be considered will be greater than zero. 

Near the coast, and in fact wherevcr practicable, it is im- 
portant to refer the levels to the mean level of the sea, calling 
this zero, or 100, or some other number, taking care to estab- 
lish from it some convenient and permanent reference-point 
ealled a bench-mark, or bench. 

All points having the same height as this bench are some- 
times said to be on a level surface called the datum. This, 
however, makes no difference with the work and serves no use- 
ful purpose, and need not be considered. 

All elevations thus found become of much importance in 
determining the relative elevations of the country, and in the 
construction of physiographical maps, ete. 

Having established the first bench, and recorded its elevation, 
the rodman stands squarely on both feet behind the rod, and 
rests it on the bench as nearly in a vertical position as possi- 
ble, which is best done by simply steadying it with the thumbs 
and fingers, taking care not to grasp it. 

The levelman sets up his level, preferably in the direction 
that the line extends, in any position from which he can well 
see the bench, as well as points to be afterwards observed. 

He then makes sure that the instrument is in adjustment, 
and is focused; levels it carefully and sights to the rod. He 

90 


LEVELING, STADIA MEASUREMENTS, ETC. 91 


may plumb the rod laterally by means of the vertical cross- 
wire of the level, and the rod may be waved gently on each 
side of the vertical toward and from the instrument, the short- 
est reading being the true reading. 

The line of sight on the rod covered by the horizontal cross- 
wire is then on a level with, or at the same height as, the wire 
itself, and the latter is therefore higher than the bench by the 
distance intercepted on the rod between the line of sight and 
the bottom of the rod. This is called the reading of the rod, 
or simply the reading. Adding this reading to the height of 
the bench, we obtain the height of the cross-wire, technically 
ealled the height of instrument, and designated by the initials 
H. I. 

Having obtained the height of instrument, the elevation of 
any other point upon which the rod can be read can be found 
by taking a reading of the rod upon it. Of course the point is 
below the instrument an amount equal to the reading, which 
must therefore be subtracted from the height of instrument to 
give the elevation of the point. The elevations of any number 
of points may be thus obtained. 

In order to obtain the elevation of points above the instru- 
ment, or below it more than the length of the rod, the instru- 
ment must be moved from its present position to one higher 
or lower as the case may require. 

Before the instrument is moved to a new position a temporary 
bench, called a turning-point (and designated by T. P. or 
“Peg ”’), must be established and its elevation ascertained with 
care, since any error in its elevation is carried forward through- 
out the whole line of levels. A turning-point must be firm 
and definite and not easily disturbed or lost. A small stake 

r “peg” driven with its upper surface about flush with the 
surface of the ground is generally used. The top of a rock may 
well serve the purpose. 

Benches and turning-points are of course the same in prin- 
ciple, but the more or less permanent point taken as the basis 
of the elevations of the Survey, and also those made usually 
along and near the line, for future reference, whether used 
as turning-points or not, are usually called benches. 

From this new turning-point we proceed precisely as before, 
by getting a new height of instrument, etc., and it is important 


92 FIELD-MANUAL FOR ENGINEERS. 


to note that the operation just described, of obtaining a height 
of instrument from a bench or turning-point, and then obtain- 
ing the heights of any number of desired points within range 
of the instrument, including a new bench or turning-point, 
includes the whole subject of leveling. | 
Since the cross-wires must be higher than any point upon: 
which a reading is taken it must be remembered that: | 
1. The reading on a point, added to its elevation, gives the 
height of instrument. 
2. The reading on a point subtracted from the height of in- 
strument gives the elevation of the point. | 
In other words: We must add a reading (to the height of | 
some point) to get a height of instrument, and must subtract | 
a reading (from a height of instrument) to get the height or | 
elevation of some point. 
The theory of leveling requires, therefore, only a simple | 
application of addition and subtraction, and it is not easy, it 
would seem, to go wrong in it. 


Station. +8 H. I. -~ S | Elevs. Remarks. 
BM 200.00 | W. Oak 60 ft. R. of Station O | 
0| 8.46 | 203.46 | 7.29 | 196.17 
1 5.34 198.12 | 
Peg = 2 |......00].- «--- 0.81 202.65 
+40; 4.17 | 206.82 | 1.12 | 205.70 
38 8.16 | 203.66 
4 6.09 | 200.78 
5 4.14 202. 
Peg + 60 1.07 | 205.75 
6.18 | 211.938 | 3.13 | 208.80 


The accompanying table shows a convenient form of field- | 
book for keeping the level notes of a railway or other survey. 
The first column contains the stations and benches. The second 
the plus readings taken on points whose elevations are assumed 
or already determined. The third column contains the heights 
of instrument recorded one line below the elevation of the turn- 
ing-point (or bench) from which it is calculated. The fourth 
column contains the minus readings. The fifth column con- 
tains the elevations of all points observed. The right-hand 
page is reserved for remarks describing the benches and their 
location, also objects crossed by (or near) the line, as roads, 

treams, ditches, ete. 


"LEVELING, STADIA MEASUREMENTS, ETC. | 93 


It is to be observed that for any series of levels the sum 
of the plus sights less the sum of the minus sights (omitting 
those for determining intermediate points on the ground) is 
equal to the difference between the first and last elevation. 

Thus to prove station 3, we have 


3.46 + 4.17 — 0.81 — 3.16 — 203.66 — 200 — 3.66. 
To prove the H. I., 211.93, we find 
3.46 + 4.17 + 6.18 — 0.81 — 1.07 = 211.93 — 200 — 11.93. 


In practice it is best to check each page of the field-book by 
comparing, as above, the first turning-point or height. of instru- 
ment (brought over from the preceding page), with the last 
turning-point or height of instrument on the page. 

To facilitate this work some engineers use two columns for 
the minus sights, placing those which determine the turning- 
points in a column by themselves. 

This practice is commendable. 

Benches should be established at short distances apart along 
the line, taking care to locate them, so far as possible, near the 
crossings of roads, streams, railways, etc., and at all points 
where their need can be foreseen, in the location of cattle- 
guards, culverts, bridges, etc. Of course an extra-good bench 
should be established at the end of the survey. An extra-good 
turning-point or bench should also be established at the end 
of each day’s work. 

The object of obtaining a line of levels is to furnish a profile 
of the line surveyed, showing the undulations of the surface 
over which it passes. 

The elevations are platted on profile paper, the horizontal 
scale being about 400 feet to an inch, and the vertical scale 
about 25 feet to an inch. This distortion of scale magnifies the 
vertical measures about *°/,,—16 times, so that the slight 
changes in the elevation of the surface may be distinctly seen. 

In running a line of “flying” levels no readings are taken 
except on turning-points. If the difference of levels of the ex- 
treme points only is desired, it is necessary to find the differ- 
ence only between the sum of the plus and of the minus read- 
ings, as already explained. This is very convenient for testing 


94 FIELD-MANUAL FOR ENGINEERS. 


a line of levels already run; in which case it is best to touch 
on the benches only, and if found correct, the intermediate ele- 
vations may be regarded as correct also. 

No line of levels should be taken as correct, and so used. 
without first being carefully checked. 

The Philadelphia rod is the most convenient and best rod 
in use. It is plainly lettered and easy to use, and may be 
read by the levelman when desirable, and at a distance of sev- 
eral hundred feet. 

To Locate a Level Linre.—Set a peg at the desired height, as 
a starting-point, and take a reading of the rod thereon. Send 
the rod forward in the desired direction, and have it moved up- 
ward or downward along the slope of the ground until a point ~ 
is found which gives the same reading as before. 

Of course the reading is taken on a peg. This second peg 
is of the same height as the first. Find in the same way a 
third peg from the second, ete. 

In this way stakes may be set at points on the ground level 
with the top of a proposed dam, or with the supposed top 
of water flowing over the dam. Then joining these stakes by 
lines~the area thus inclosed may be nteasured. 

The water behind a dam is not level, but is curved con- 
cavely upward and so increases in height back of the dam, 
and sets back farther than if level. 

For the subject of backwater, Works on Hydraulics must be 
consulted. 

Other applications of the level line are to obtain “ contour 
lines’ for topographical maps, for levees in irrigated rice- 
fields, ete. 

To Run a Grade-line.—This consists in setting a series of 
pegs so that their tops shall be points in a line, which shall 
have any required slope ascending or descending. 

First drive pegs at each end of a line to the heights required. 
These heights may differ by a given amount, or this difference 
may be undetermined. 

Set the level over one of the pegs and measure the height, a, of 
the cross-wires above the top of the peg. 

Set the rod on the other peg, and make the reading on the rod 
equal to the height a. 

. Without disturbing the level drive any desired number of pegs 


¢ 


LEVELING, STADIA MEASUREMENTS, ETC, 95 


along the line, so that the reading on each will also be equal 
to a. 

A line of uniform grade or slope is not a straight line. 

Calling the globe spherical, this line when traced in the plane 
of a great circle would be a logarithmic spiral. On a length of 
six miles the distance of its middle point from the middle of its 
straight chord would be six feet almost cxactly. 


CORRECTION FOR THE EARTH’S CURVATURE AND FOR 
REFRACTION. 


This is necessary for long distances. 

Let AB (Fig. 55) represent a portion of a section of the earth's 
surface. Then if a level be set at A, the line of sight of the level 
will be the tangent 4D, while the 
true level will be the arc AB. The 
difference BD between the line of 
sight and the true level is the cor- C 
rection for the earth’s curvature for 
the distance AB. This must be sub- 
tracted from the reading of the rod 
at £B, or, what is the same thing, 
added to the height of B, as given 
by the reading of the rod. 

Let AX = R, AB = D, and BD 
= KH. By geometry, 


D 


AD’ = BD(BD + 2R), E 
_ AD Fie 55. 
ot BD= BD FRR 


Omitting BD in the right-hand member, since it is small com- 
pared with 2R, and supposing AD = AB = D, we obtain 


dD D? 


a 2, 
K= aR = 3 >< 20013650 -000000023908D*. . . (1) 


This formula gives a result or value for # slightly too small; 
but the relative error is only about one in 24,000 for a distance of 
R 57°.29578 


7, = 39.6 mi —.——— = 84’ 22”.65, 
100 39.6 miles, or arc of 100 34’ 22/'.65, 


96 FIELD-MANUAL FOR ENGINEERS, 


In observing distant objects, a ray of light traversing the air 
fron an object to the eye or instrument is refracted, and takes a 
curved path which, for points near the surface of the earth, is 
practically the arc of a circle, concave downward, and whose 
radius is 72. 

Thusa point at ( (Fig. 55) would appear at D higher than it 
really is by an amount CD, This may be found from the above 
formula by substituting 7R for BR. 

Hence the correction for refraction is 

D? 


‘= —S- = ° 2, ° ° e ° 
H’ = = = -000000008415D (2) | 


The correction for curvature and refraction is 


W _ _ D? dD _ 38D  _ 
E" = BC=BD-—-CD= 3R 14k FR = .000000020492.D?. (3) 

This must be added to the apparent elevation of the observed 
object to give the true elevation. 

Table XI gives the value of the correction for the value of 
AR = 20911790 feet. 

When it is possible to set the level midway between the points 
whose heights are required, the corrections will balance each 
other and may be omitted. 

The above equations may be put into a form sometimes more 
convenient as follows : 

The length of arc on the earth’s surface subtending angle of 
one minute is 


2x X R 
360 >< 60 = 6083 feet = D,, say. 
D,? 
Then 14R = 12638 a ry e e s e e (1’) 


= the correction for refraction for distance 6083 ft. or are of 1’. 


Di}? 


Also sp 7 886. ee ee 


= the correction for curvature for the same distance or arc 1’, 


LEVELING, STADIA MEASUREMENTS, ETC. 97 


3D;? 
TR 


and 


= .75828 . . ... .. (8) 
= the correction for curvative and refraction, 


TRIGONOMETRIC LEVELING. 


First Method.—When the point C (Fig. 56) can be seen from 
two points A and B on the same level, then 


AD=CDcot CAD, and BD= UD cot CBD, 


Subtracting gives 


AB 


AB=CD(cot CAD— cot CBD), or CD=— AD —cot CBD" (4) 


Second Method.—Let A and B (Fig. 57) occupy any positions 


C 


A B D 
Fia. 56. Fia. 5%. 


except in line with C. Measure AB and the angles at A and B; 
also the angle of elevation CAD. 


C=190°—A—B, AC= ABA? and ('D = AC sin CAD. (3) 


If A, B, and C are in the same vertical plane, the solution is in 
no wise affected. 

Unless the distance AD = D is short it is necessary to add to 
the correction found by the preceding formulas the correction 


D? . 
= 2 2 
TR .0000000204927)?, 


for curvature and refraction, namely, 


93 FIELD-MANUAL FOR ENGINEERS. 


To find the height of instrument by an observation of the horizon 
(Fig. 58). 

First Method.—Let C be the place of the transit, and BAA’ a 
portion of a section of the 
earth’s surface. 

Were there no refraction 
the line of sight would be 
the tangent CA; ACH = C 
would be the angle of de- 
pression or dip; and we 
would have 


BC = BO X exsec COA 
= R exsec C. 


E 


Owing to refraction, how- 
ever, the line of sight would 
be a curve concave down- 
ward whose radius = 7K; 
and it would therefore ex- 
tend from C' to a point A’, 
say a distance AA’ beyond 
A. 
Draw CDF tangent to this 
curve at ( to meet the ra- 
dius AO prolonged in F. 
Draw also the tangent A’t. 

Let COA=H, COD=H', 


Fig. 58. . and COA’ = O. Then 
, 
CO=rsec H# = cos Ht’ 


Let H, on A’O prolonged but not shown, be the center of the 
arc A’U. Now 


Git’ — 0" — co 
cos O = — cos COZ = CH — HO — CO 


2C0. HO 
137? — r?sec? H — 18cos H — sec H 
-12risecH 12 ; 


Clearing, substituting 1 — vers O for cos O, 1 — vers H for 
cos H, and 1 + exsec A for sec H, we find 


12 vers O = 18 vers H + exsec H. 


LEVELING, STADIA MEASUREMENTS, ETC. 99 


Or, writing versines for exsecants or vice versa, we have 


vers O = 3 vers H, 


or, approximately, 
exsec O = j exsec H. 


In the triangle U'ZO, 


__ 497? + 367? ~ CO" 857? — r*sec* H 85 — sec? H 


eS S56 XK Tr ””~C« B4 
CO sec H 
_ . E =— ¢ ° Oo. -— . O 
Also sin sin a sin : 
..sin #sin O = sim? oc # = (1 — cos? oe 


144 “4 


_ 170 — 169 cos? H — sec? H sec 7 


Now since QA’ is the prolongation of HO, and HC and OD are 
perpendicular to CD and therefore parallel, we have 


COA’ — DOA' = COD, or O- H= H'; 
-, cos H’ = cos(O — #) = cos O cos # + sin Osin £Z. 


Substituting in this the above general values of cos 0, cos E, 
and sin # sin O, and expanding and reducing, we find 


18 cos H + sec H — cos H+ (sec H — cos Hf) 


cos H’ = 14 14 


Patting cos H=1— vers H, sec H=1-+ exsec H, etc., we 
find 
(exsec H — vers H) 


vers H’' = $ vers H — id ; (a) 
or vers H’ = $ vers H, very nearly,. . . . . . (6) 
or exsec H' = $exsec H, very nearly. . . . . « (7% 


exsec H’ — vers H’ 


Hence vers H’ = § vers H — 3 


From this we have 


4 g(exsec H’ — vers H’) 


vers H = j vers H' 12 —, . (0d) 


100 FIELD-MANUAL FOR ENGINEERS. 


or vers H=jvers H', very nearly, . .... (6) 


or exsec T= jexsec H', very nearly, . . . . . (7) 


Since the versines of small angles are very nearly in the ratio 
of the squares of the magnitudes of the angles, we have, from (6), 


H = ¥iH' = 1.081’, approximately. . . . (7’) 


Supposing 7 = 4000 miles and AB = 5 = 50 miles, then it is 
easy to show that the error of eq. (7) is Jess than .00000002, and 
the error of (6) is about .0000000004. 

Example 1.—The observed dip of the sea horizon is H’ = 24’, 


What is the height of the instrument above the sea? 
We have 
.BC=r exsec 1 = rj exsec H’ = 20914000 X .00028467 


= 595.35 feet, exactly. 
Example 2.—Let = 4000 miles, and AB = 50 miles. What is 


the observed angle of depression H’, and what is the height of 
the observer above the sea? 


We have 
wa -. x 57°,29578 = 0°.7162 = 42’ 58” 
4000 ~ - ° 
- H’ = 0.716 ~~ 1.08 = 0°.668 = 89’ 47". 
Also, 


h=vrexsec H = 4000 x .0000781 = .8124 miles = 1649.47 feet. 


The exact relations between H and H’, shown above, would 
seem to be more satisfactory than the approximate equations in 
general use even if these were regarded as sufficiently accurate. 


THE STADIA. 


The stadia is a compound cross-wire ring or diaphragm having 


three horizontal wires. 
The two outer ones are called stadia wires, and distances deter- 


mined by means of them are called stadia measurements, 


LEVELING, STADIA MEASUREMENTS, ETC, 101 


The stadia wires are adjusted so as to intercept a certain space 
on a rod at a given distance from the transit and perpendicular to 
the line of sight. 

Let C (Fig. 59) = the distance of the object-glass from the axis 
of the transit, and f = the focal length of the object-glass. 


Fie. 59. 


This focal length is equal to the distance of the cross-wires 
from the object-glass when this is focused for a distant object. 

This focal length may be found also by removing the ol:ject- 
glass, exposing it to the rays of the sun, and noting at what dis- 
tance from the center of the glass tle rays form a perfect and 
minute image of the sun on a smooth surface. 

Let Cm=l, Cn=l, DH=S8', and HK=8S. 

The focal distance OF is constant, but Co = ¢ varies with the 
position of the object-glass, and hence CF is also variable. 

Let Co = c’ when the rod is at DH, and Co =c when the rod is 
at HK. 

Now, from the figure, 


Fn AK tats) 8 


Fm DE? —(c +f) — ig” 


or TH+ f= ell -C@+P))- pe. 8) 


S’ is usually assumed = 1 foot, and Fim = l’ — (c’ + f) = 100 
feet; and the stadia wires are then adjusted accordingly. 

c’ is measured on the telescope when the object-glass is focused 
on the rod at the assumed distance. 

To measure any other distance the rod is again observed at thie 
desired point and the space S noted, which placed in (8) gives 
l—~(e+f)=4, say. We may then measure c¢ on the telescope. 


Then 
¢t=e 4-f + dy. 


102 FIELD-MANUAL POR ENGINEERS. 


Since, however, c has but a small range of values, it will usually 
be sufficient to assume it to be constant and equal to some mean 
value. 

Suppose that in (8) ¢ = c’ = ¢,, and solving we find 


S’ — 81 
C1 +f= s— 3-3 - © © © © «© @ (9) 


If we observe S’ and S corresponding to any two distances U | 
and / and substitute in (9), we have c, +f. : 
Having found ¢, +f, lay off Cn = 100 + ¢, + f, or Fm = 100, 
and adjust the stadia wires to sabtend DH = 

distance. 
Then from (8), writing c, for c and c’, we have 


~—(atf)=V —(e, + fer = 1008, or / = 1008S + (¢.+/S), 


or, omitting accent, ! 
¢= 1008 + (c+ f). 2 6 © « © (10) 


Hxample.—Suppose at U’ = 100 we find S’ = 1, and at? = 500 
we find S = 5.0453. 
Then eq. (9) gives 
_ 504.53 — 500 


c+f= 40453 = 1.12. 


Then, from eq. (10), /= 1008 +- 1.12, provided the stadia wires 
are spaced so as to intercept 1 foot at 101.12 feet distance from 
the center of the instrument. | 

The foregoing formulas must be modified when the line of 

p collimation is oblique to the 

n horizon, which is usually the 

P case. Thus, in Fig. 60, let 

DE’ = the space intercepted 

. on the rod when the line of 

Cc F a collimation #'H is horizontal, 

and let DH = S = the space 

intercepted when the line of 
collimation 7’n makes an angle n¥H = a with the horizon, 

Let DFE = DFE’ = §, 


Fia. 60. 


LEVELING, STADIA MEASUREMENTS, ETC, 103 
In Fig. 60, 
S= DH — HH = HF [tan (a + 36) — tan (a — 36)]. 
The horizontal reading desired is D’/H’ = 2HF tan 46. 


Dividing gives DEY tan 
gs S ~ tan (a + $6) — tan (a — $6) 

_ 2 sin 49 

Bat 2 tan $6 = cos 40" 


Also, by Chapter III, formulas (26), (27), 


sin @ 
tan (a + 40) — tan (a ~ 4) = cos (a 4 4p) cos (@ — 6) 
— ~ sin 46 cos 30 
~ cos? @ — sin? 46° 


Substituting these values, we obtain 


Dk’ cos? @ — sin? 46 , 
3 = ote wee ee (11) 


If we neglect sin? 36, or, what is equivalent, add sin® 36 to the 
numerator, we introduce a relative error of 


sin? 46 sin? $0 


im? 1 2 
2 —_—_—=-- = sin? 19 sec? a, very nearly. 
cos? @ — sin? 16 ~ cos? * y y 


Again, if we add sin’ $9 to the  orominaler, making it unity, 
46 

we introduce a relative error of = 3 cost 46 = = sin? 40 sec? 46. 

The first of these errors increases the fraction, and the second 

; . 57°. 29578 , aan 

decreases it. Moreover, since 6 = joo = 34" 22.65, we 
have in al] practical cases a > $9, or sec? a > sec? 36. 

Hence the fraction is, by the double approximation, increased 
slightly more than it is decreased. Hence 


4 U / 


——— | . 
5 < cos'a; but 3 


The total relative error is 


= cos’ a, almost exactly. . (12) 


6 = sin? 46 (sec? a — sec? 46) 


= sin’ 46 (tan? a — tan’ 49), 


104 FIELD-MANUAL FOR ENGINEERS. 


Since tan’ 46 is very small compared with tan’ a, we have 
é = sin’ 36 tan’ a, very nearly, 


.000012 tan® a, very nearly, 


*{{ 


z~ = cos’ a(1 — .000012 tan’ a) 
— cos? am .000012 sin® a. ° ° ° e ° (13) 


Since sin $9 = tan $9 = .005, very nearly, the quantity 


200 
neglected above is only (.005)* = .00000000625 and does not come 
within the range of the table. 

The above is the coefficient of reduction by which to multiply 
the observed space DH = S in order to produce the true space 
DE’ which would be observed at the same distance if the line of 
collimation were horizontal. 

Hence we have, from (10), 


L = (1008 + ¢ +f )(cos? a — .000012 sin? a). 
By eqs. (25) and (26), Chap. III, 


cos’ ¢ = 1 + cos 2a =f vers 20 
2 2 

a ain’ a _.1 — cos 2a __ vers 2a 
an = = 2 . 

Hence \ 

vers 2a vers 2a 

[= (1008+ ec +1 _ 9 — .000012 —). - (14) 
or 

l= (10S8+¢+/) (1 - == ), very nearly. . . . (14’) 


These coefficients may be read from a table of versed sines 
without any computation whatever. 

The last equation is quite accurate enough, but the coefficients 
of Table XIII are calculated by the exact formula. 

Ezample.—Find the coefficient for a = 7° 20’. 

We write half the vers 14° 40’............ eeeeee = .01629 
and subtract from unity and find coefficient......... = .98871 

Avother method of procedure is that in which the rod is held 
perpendicular to the line of collimation. 


LEVELING, STADIA MEASUREMENTS, ETC. 105 


To secure this position of the rod a bar is attached to it having 
sights upon it, through which the rodman watches the instrument 
during an observation, the line of sights being perpendicular to 

| the rod, 


Fia. 61. 


The horizontal distance of the point B from the insti ument is 
TH = 1K+ KH= Imcosa+ Bm sina, 
or IH = (1008 +¢+ f)cos a + rsin a. 


(See Fig. 61, in which 7 is the reading of the rod by the line of 
collimation. ) 
The elevation of B above J is 


BH = mK — Bm cos a, 
or BH = (1008-+¢+/f)sina—rcosa .. . (15) 


When the distances are sufficiently great correction, must be 
made for curvature and refraction as already pointed out. 


THE GRADIENTER. 


This attachment consists of a screw working against a clamping- 
arm suspended from the horizontal axis, on the opposite end from 
the vertical arc. 

A strong spiral spring presses the arm against the end of the 
screw. The large silvered head of the screw is usually divided 
into 100 equal spaces. When the screw is turned the head moves 
along one division of a small silvered scale for each revolution of 
the screw. 

When the regular clamp of the telescope is free the telescope 
may be revolved; but when the telescope is held by this clamp it 


106 FIELD-MANUAL FOR ENGINEERS. 


can be moved only by the gradienter screw, which thus takes 
the place of the vertical tangent-screw. 

The screw is cut with such a ‘‘ pitch’ as to cause the horizon- 
tal cross-wire to move over a space of 1 foot on a rod 100 feet dis- 
tant when the screw is turned through one complete revolution. 
The wire will therefore move ;3, of a foot on the rod 100 feet 
distant when the screw is turned through 4, of a revolution, 
which is equal to one of the small spaces on the head. 

Grades may be established with great facility with this screw. 
First, set the screw at zero and level and clamp the telescope. 
Then move it upward or downward as many spaces of the screw 
as there are hundredths of a foot to the hundred feet of the re- 
quired grade. Thus for a grade of 1.25 of a foot turn the screw 
through 125 spaces, or one revolution and one fourth of a revolu- 
tion, 

This screw is also very convenient for measuring short dis- 
tances. Since one revolution of the screw will move the wire over 
one foot at the distance of 100 feet, it is only necessary to turn the 
screw through one revolution and observe the number of feet the 
wire moves on the rod at the required distance, and multiply that 
by 100. 

Thus suppose the space on the rod is 2.57 feet, then the dis- 
tance = 2.57 & 100 = 257 feet. 

When the distance to be measured is not approximately hori- 
zontal some modifications are necessary, which we will now ex- 
plain. 

In Fig. 62 let O be the position of the transit; DOB = H= the 

angle of elevation to the foot of 

CLA the rod; AOB = 6 = the angle 
subtended by any number of rev- 

B olutions of the gradienter screw; 

AB = S = the space on the rod 

subtended by the angle 6 when 

the rod is vertical. BCisdrawn 

‘e) O perpendicular to BO. Then 


Fig. 62. ABC = BOD—E; 
OAB = 90° — (E+ 8); 
BC _ sin [90° —(#+-6)] _ cos Hcos 6 — sin Hsin 8 
AB” sin(90° +6) cos 6 
° BC = SAcos H — sin # tan A); 


LEVELING, STADIA MEASUREMENTS, ETC. 107 


BO= BO = 6 8 _ ain wr). 
tan 6 tan 6 - 
DO = BO cos H = ane a ~— }sin 2n 


These formulas are applicable to any gradienter screw. 
Let tan 6 = +4, for one revolution of the screw. Then we have 


BO=S(100 cos H— sin #);. . . . 1. we ee (16) 


100(1+- cos 27) 


DO= &100 cos? # —4sin 2)= si 3 


—sin 2H), (17) 


This last equation is decidedly the simplest formula for compu- 
tation, since it involves 2 only { 
and no exponent. 

When the angle 6 is an angle of 
depression the point B is the 
upper end of the part AB of the 
rod used. (See Fig. 63.) The 
horizontal distance only is usual- 
ly required; and Table XIV gives, 
for different values of XH, the co- 


200 F008 7) _ gsin on 


D 


efficient 


Fig. 63. 


by which to multiply the read- 
ings of the rod in order to find the horizontal distances, 


VERTICAL CURVES. 


Vertical curves are used to round off the angles formed by the 
meeting of two grades. 

Let AC and CB (Fig. 64) be two grades meeting at C. ‘shese 
grades are determined by the rise or fall per station in going in 
either direction. Thus, in the figure, let g represent the vise per 
station of the grade of AC, and g’ represent the fall per station of 
the grade CB. In going in the opposite direction BC would, of 
course, be a rising grade and CA a falling grade. 

The parabola furnishes a very suitable vertical curve, and it is 
very easily computed and laid out. 

Let AZB represent the required parabola. Draw the vertical 
line C7, also a vertical line through B to mect AC prolonged in 


108 FIELD-MANUAL FOR ENGINEERS, 


Ff, and the horizontal lines AK and CG to meet BF in K and G. 
Join AB, and drop the perpendiculars aim,b, and aamabe. Let 
m = the number of stations in AC or CB. Then, since these dis. 


Fig. 64. 


tances are measured horizontally, we have AH = HK, and there 
fore AD = DB. | 

The vertical line CD is therefore a diameter of the parabola, 
and the distances of points on the curve in a vertical direction | 
from corresponding points on the tangent AF are in the ratio of 
the squares of the distances of these points from A, Also, 


CD=}3FB; CHE=}FB=j3CD; 
.. CH= ED. 
Now, since CF = CA, 
FG = Cif= the rise of AC in n stations = ng. 
Also, GB = the fall of GB in n stations = ng’ ; 


FB=n(g +9). 
Now, since B is 2n stations from A, we have 
. FB g+ 9 
Offset at first station from A = aym, =a = in? = adn (18% 


The value of a,m, being determined, the distances of the curve 
at all points from the tangent AY are also known. Thus 


&gm_, = 4a, Ce = Qa, etc. 


It is easy to calculate the heights of the points of the curve 
above AK, though these heights are scarcely needed. Thus 


mb, = ab, — Qam = g — a, 
Mabe = Qgby — Aem, = 29 _ 4a; 
HH = CH — CH = 3g — Qa, etc., ete, 


LEVELING, STADIA MEASUREMENTS, ETC. 109 


Finally, 
BK = FK — FB = 2ng — 4n?a = 2ng — n(g +9) =nlg —-9g), 


or 


BK = CH — GB= 1g — ng = n(g — g') as before. 


The successive grades are found by taking the successive differ- 
ences of the heights just found. Thus 


m,),—0 =9—@; Mobg—™M,b, = g—38a;, HH— mb, = g—Sa, ete. 


The change in the grades from station to station, which is the 
same as the chord deflection of the curve, is, we observe, equal to 
the constant quantity 2a. 

Second Proof.—The curve must change its direction in its 
length of 2n stations an amount equal to g + g’, and therefore 
gig 

2n 


in each station it must change an amount equal to ; and 


this is equal to 2a by eq. (18). 
Third Proof.—The ordinates to the curve are a, 4a, 9a, 16a, etc. 
The differences of ordinates are 3a, 5a, Ta, etc. 
-. the differences of grades are 2a, 2a, etc., a constant. 
Fourth Proof.—Let m, represent the nth station, and prolong 
the chord m,m, to meet CH in z (not shown). 


Now aim, = (n — 1)*a, dsm, = n'a, and CH= (n + 1)°a. 
But | 


Cr+ aim, = 2asmo, or Cx = 2am. — Am, = [(n + 1)? — Qa. 
Hence Hz = CH — Cz = 2a. 


In finding the value of a, etc., it is necessary to know when we 
are to take the sum of the grades, and when the difference; for 
there may be four combinations of two adjacent grades. 

It is only necessary to observe: 

(1) That when the grades are both ascending or descending FB 
is equal to n(g — g’), g being the steeper of the two grades, and 
therefore 


ng- 9) _ 9-GF 


o= 4n®* ~ 4n 


110 FIELD-MANUAL FOR ENGINEERS. 
(2) That when one of the grades is ascending and the other de- 


scending, 


FB =n(g+qQ'), and a= ate 


If one of the tangents is horizontal, let g = grade of the other 


and then a = TF 
4n 


In all cases it is necessary to consider only the change of grade 
at C; which we will represent by G. Then 


G=g-g, or G=g+79, 


according as the grades are both rising or falling, or one rising 
and the other falling. Then 


FB=nG, and a= 


These formulas apply without change to points between stations, 
as the following examples will illustrate. 

Example 1.—A .9% up grade joins a .8% down grade at station 
76 at an elevation of 94.0. Find ordinates, etc., for a curve 6 sta- 
tions long. 


a= 2t-3 _ 4, (See Fig. 64.) 


12 
Stations on AF’,....... 73 74 15 %6 C7 %8 79 
Elevations on AF..... 91.3 92.2 93.1 94.0 94.9 95.8 96.7 
Corrections...........- 00 1 4 9 #16 2.56 38.6 


Elevations on curve.... 91.8 92.1 92.7 93.1 93.8 93.8 93.1 


It is necessary to compute the ordinates a, 4a, etc., for one half 
the curve only, since they are the same for corresponding points 
on each tangent. Thus we have: 


Stations on AC andCB 73 74 5 76 #77) )~6=6©6%8)~—OD 

Elevationson ACandCB91.8 92.2 93.1 94.0 98.7 93.4 93.1 
CorrectionS........000 0.0 1 4 0.9 4 1 0.0 
Elevations on curve.... 91.3 92.1 92.7 938.1 93.3 93.8 93.1 


LEVELING, STADIA MEASUREMENTS, ETC. 111 


Example 2.—A .2% up grade is continued beyond station 17, 
whose elevation is 46.4, by an .8% up,grade. Find ordinates, ete., 
for a curve 6 stations long. 


8 — .2 
a= —- = .05. 


Stations on AC and CB.. 14. 15 16 17 18 19 20 

Elevationscen AC and CB 45.8 46.0 46.2 46.4 47.2 48.0 48.8 

Ordinates ... .......... 0.0 .05 .20 .45 .20 .05 0.0 
| Elevations on curve..... 45.8 46.05 46.4 46.85 47.4 48.05 48.8 
To find the grade of the parabola at any point. 

Let the curve be that of Example 1, and let the point be at a, 
+ 40, or 140 feet from A. 

Elevation at a, + 40 = 92.2-+ .40.9 = 92,2 + .36 = 92.56. 


Also, (1.452°@ = 1.960 = .196 — .20, nearly. 


Therefore elevation of curve at a, + 40 = 92.86. 

The special need of these curves is in sags in the grades to pre- 
vent the breaking of the train. 

According to Wellington, Railway Location, page 365, vertical 
curves in sags should be at least 200 feet long, or 100 feet on each 
side of the vertex, for each tenth in the rate of change of grade. 

This would call for a curve 1200 feet long in the last example. 

It is not always practicable to meet this requirement, but such 
curves could te and should be much longer than they are gener- 
ally made, 


} 


113 FIELD-MANUAL FOR ENGINEERS: 


ELEVATION OF THE OUTER RAIL ON CURVES. 


A car of weight w moving on a curve of radius Ft with a velocity. 
of » feet per second develops a centrifugal force in the direction 
ab (Fig. 66) expressed by 


wv? 
I= 32.15R° 


To counteract this force the 
outer rail on a curve is raised 

€ above the inner rail an amount 
b be = e, so that the car may 


f. rest on an inclined plane. 
Fia. 66. Let ac = g, the gauge of the 
track. 
The component of fin the direction of ac is 
ab ab , wv? = ab 
Wve =S5 =I" = 30 75R g. 


The component of w in the direction of ca is 


Now f' and w’ are opposite in direction, and in order to satisfy 
the mechanical conditions they must be equal. 
Equating their values, therefore, we find 


aby? 


¢ = 30 15R" ° r) ° = 8 ° e (19) 


But ae = distance between rail centers = gauge -+ one rail 
head = 4.708 +- .202 = 4.91. For an elevation of 6 inches 


ab = ac — .03 = 4.88, nearly. 


This is a good average value for ad. 
Again, if V = the velocity in miles per hour, we have 


5280 22 
V= 35 


° = 3600 Vz 


LEVELING, STADIA MEASUREMENTS, ETC. 113 


Furthermore, & = i in which D is the degree of a curve of 


D 
| radius R, and R, = the_radius of a one-degree curve = 5729.58. 
- Substituting these values in (19), we have 


488 x 484DV* __ ; 
= ST 3c BBB Se 5729.58 — 000056809) V 


= .000057D V?, almost exactly. . . . . . . (20) 


If we substitute in (19) ab = g, and » = =v, we have, approx- 


imately, 
. 066889 V ? 
é = DO ° es 


a 


TLis is the formula in general use. 
To give the most favorable view possible of this formula, + ub- 


stitute in it g = 4.708 and find 
2814907 


é = ra . 2 s s s e . e . (22) 


Erample.—Find the elevation, e, for a 7° curve and a velocity 


of 40 miles per hour. 
Ea. (20) gives 


e = .000057 « 7 x 1600 = .638 feet = 7.66 inches. 


Kq. (22) gives 


é= “oe = .616 feet = 7.39 inches. 

It will be observed that the two large factors appearing in (22) 
do not occur in (20); and, moreover, eq. (20) is very much mere 
accurate than the formula (22) generally used. 

The author has elsewhere pointed out the evil of elevating the 
outer rail on sharp curves sufficiently to meet the requirements of 
too high velocities. 

The elevation should be much less than required for the speed of 
the fastest passenger-coaches ; for it is better, on such a curve, 
for a coach to hug the outer rail somewhat or to slacken speed, or 
both, than to pull all the freight-cars against the inner rail. 

-The best conditions are realized when the speeds of fast trains 


114 FIELD-MANUAL FOR ENGINEERS. 


are lessened so as to exceed only slightly the speed for which 
the elevation of the curve is suited. 

Thus a train moving at a speed of 50 miles per hour should be 
decreased to a 35-mile rate or less in passing over a curve elevated 
for a 30-mile rate. Under such conditions the cars would slightly 
press the outer rail around the curve; their motion would thus 
be steadied, and the movement woud be the steadiest possible. 

The maximum elevation should probably not exceed eight 
inches, except, possibly, at some special point, under
…[truncated]