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REESE>LIBRARY
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UNIVERSITYTDF CALIFORNIA.
*
ci(ecevved
, Occasion No.
Sj — imi —
.92418 • C/jss
r-u-u-ll1
FIELD MANUAL
FOR
ENGINEERS.
BY
PHILETUS H. PHILBRICK, C.E., M.S.
M. AM. MATH. Soc.,
Chief Engineer, Kansas City, Wat\ins and Gulf Railway,
.A'. Am. Land and Timber Co., etc., etc.;
Sometime Professor of Civil Engineering at the
State University of Iowa.
FIRST EDITION.
FIRST THOUSAND.
:
extended, and extra topi*,
,10r clearness, considered in connection with the
icrai matter of the text. ..xpi<.
Table II dispenses with all calculation Dy numerous readings
the last figure.
\j & 4- .L>aces, has bden in prepara-
IV 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 stadia, 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 given 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 \^-^ iuired in each case, is shown on pages 199, 200.
of certain proposed turnout curves is
and a variety of methods of laying out
PREFACE. V
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
cone. 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 be"en in prepara-
VI 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 lias 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, lie will feel that the days he has devoted to it, though
many, have not been spent in vain.
P. H. PHILBKICK.
CONTENTS.
CHAPTER I.
PRELIMINARY OPERATIONS.
PAGE
The Reconnoissance 5
The Preliminary Survey 5
The Location 5
The Organization of the Transit Party 8
The Compass: What kind to use and when to use it 8
Requirements for a Successful Reconnoissance 9
Train Resistances 10
Total Ascent the Main Test of Gravity Resistance n
CHAPTER II.
ADJUSTMENTS, USE, AND CARE OF INSTRUMENTS.
The Transit.
Adjustments 12
Use and Care of the Transit 15
Best Way to Set up the Instrument 15
To Measure the Angle between Two Lines or Objects 16
Hints on the Care of Transits and Other Instruments... . 16
The Level.
Adjustments
Use of the Level.
The Compass.
Adjustments 22
Use of the Compass 23
vit
Viii CONTENTS.
CHAPTER III.
PLANE TRIGONOMETRY.
PAGE
Definitions and Explanations 25
Fundamental Relations 26
Solution of Plane Right Triangles 31
Table for Plane Right Triangles 32
Fundamental Relations for Oblique Triangles 32
Short Solution of " Tangent Problem " 34
Solution of Plane Oblique Triangles 35
Table for Solution of Oblique Triangles 36
List of Fundamental Formulas 37
CHAPTER IV.
SIMPLE CURVES CONNECTING RIGHT LINES.
Properties Relating to the Circle 39
Some Elementary Relations 41
Notation 41
Degree of Curve Defined and Explained 41
Rational Treatment of Curves, Example Illustrating 42
Difference in Lengths of Arcs and Subtended Chords 43
Huygens' Formula for the Length of an Arc 44
Table Showing Excess of Arcs over Subtended Chords when the
Arcs are Aliquot Parts of 100 44
Table Showing Excess of Sub-chords over Aliquot Parts of 100,
when the Chord = 100 45
Reason for these Large Excesses Pointed Out 46
Formulas for Radius, Tangent, External Secants, Offsets, etc 47
Proper Course to Pursue in Locating a Curve 49
Long Chords and Ordinates to Long Chords 50
Approximate Value of Ordinates to Short Chords 51
Offsets in Terms of the Degree of a Curve 52
Applications of Formula 53
Laying Out Curves.
A. By Deflection Angles 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 58
E. By Middle Ordinates. Without Calculation 59
F. By Radial Lines. Without Calculation 59
Errors in Field-work — The Nature of.
"A" Method of Laying Out 60
"B" Method of Laying Out. 64
"C" Method of Laying Out 64
" D " Method of Laying Out 64
CONTENTS. IX
PAGE
"E" Method of Laying Out 65
Fourteen Problems in Simple Curves - 65
Obstacles in Surveying.
To Erect a Perpendicular at Any Point of a Line 76
Table and Formulas 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 79
When Both Ends are Inaccessible 80
When an Inaccessible Space Intervenes 81
Rest Method of Making a Preliminary Survey KJ
Table Illustrating the Same 83
To Replace a Broken Line between Two Points by a Straight
Line 85
To Find the Angle between Two Straight Lines when the Point of
Intersection is Inaccessible 86
To Connect Two Tangents by a Curve when the Vertex is Inac-
cessible 87
To Locate a Curve when the Vertex and Both Ends of the Curve
are Inaccessible 87
To Pass from Any Point on the Curve to Any Point on the
Tangent 87
To Find any Desired Point on the Curve when Obstacles preclude
ti.e Use of Ordinary Methods 88
CHAPTER V.
LEVELING, STADIA MEASUREMENTS, ETC.
Bench, or Bench-mark 90
Form of Field-book for Level Notes 92
Proof for Level Notes 93
Benches: Where to Establish them 93
The Location of a Level Line 94
The Location of a Grade Line 94
Correction for Curvature and Refraction 95
Trigonometric Leveling 97
True Simplified Formulas for Heights by the Dip of the Horizon.. 98
The Stadia.
Formulas for the Stadia — Simplified 102
The Gradient er.
Formulas for the Gradienter — Simplified 107
Xll CONTENTS.
PAGE
To Connect a Straight Track and a Straight Line ................... 199
Two C'onnect Two Straight Tracks by a Curve of Three I tranches,
the Turnout Curve having Radii to Suit Given Frogs ............ 199
Formulas Supposing the Switch-rail to be a Part of the Turnout
Curve ................................................................ 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 ............................. 208
To Fit a Curve to a Given Middle Frog ............................. 210
Turnouts from Curves.
Turnout from the Inside of a Curve .................................. 212
Turnout from the Outside of a Curve ................................ 213
Double Turnout on Opposite Sides of a Curve ....................... 214
To Find Degree of a Turnout from a Curve ......................... 215
Other Turnouts from a Straight Track ............................... 217
To Fit the Turnout to a Given Middle Frog ........................ 219
CHAPTER IX.
THE TRUE TRANSITION CURVE.
Reason for the Need of Such Curves
Definition and Properties of Such Curves
Elementary Relations.
To Find the Relative Length of the Offset and the Transition
Curve ................................................................ 225
To Find Any Tangent Distance ...................................... 2_>6
To Find Any Offset .................................................... 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 ................................................................ 230
To Find the Offset iri Terms of the Cenrtal Angle and the Radius.. 230
To Find the Angle between the Tangent and Any Chord Drawn
from A ............................................................... 23 1
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 ................................................. 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 ...................................... 233
To Find Deflection Angle at Any Point, also Any Chord ............ 235
To Find the Exsec dV ', also TV, etc ................................. 236
To Find the Radius of Curvature at Any Point ..................... 236
To Lay Out the Curve by Offsets from the Tangent AQ ............ 236
Special Problems and Examples.
Given Length and Degree of Kd, to Find the Offset AK, Tangent
AO, and to Lay Out the Curve ................................... 237
CONTENTS. xiii
PAGE
Given the Degree of the Offset Curve and Offset AK, to Find the
Length of the Transition Curve, etc 239
Given the Degree (Z)1) 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 241
Given the Degree of the Main Curve and the Length of Af Re-
placed, to Find the Offset, Tangent, etc 242
To Replace Each Half of a Simple Curve AfA1 by a Transition
Curve 243
To Connect Two Tangents by Two Transition Curves, Each of a
Given Length 5 244
To Connect Two Tangents by Two Equal Transition Curves
having a Common Vertex Distance E 245
The Transition Curve Very General in Use 245
To Lay Out the Curve from Any Point on it 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 251
CHAPTER X.
CALCULATION OF EARTHWORK.
Prismoid Defined 253
Area of Level Sections 254
Area of Sections not Level 255
Area of Irregular Sections 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 258
The End-area Volume Generalized 259
Special Formulas and Cases 259'
Formula for the Volume of the Frustum of a Pyramid — Sim-
plified 261
Loaded Flat Cars, Piles of Stone, etc 263
Ends of Embankments or "Dumps" 264
Ground Irregular Laterally 264
Mixed Work, Excavation, and Embankment 265
Correction of Earthwork for Curvature 267
Overhaul 268
Monthly Estimates 269
Final Estimates : 270
Computation of Prismoids Level Laterally 270
XIV CONTENTS.
CHAPTER XT.
APPROXIMATE AND ABRIDGED COMPUTATIONS.
PAGE
Definitions and Notation 273
The Relative Error 273
Addition 274
Subtraction 276
Multiplication and Division 277
Abridged Multiplication 279
Abridged Division 281
CHAPTER XII.
CONSTRUCTION.
Clearing and Grubbing, How to do it 284
Grade-line 284
Surface Ditches — Importance of 285
Cross-sections — Proper Places for 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 291
Borrow-pits - 291
Shrinkage of Earthwork 294
Retracing the Line 295
Track-laying 296
Culverts 297
Location of Bridge Piers 298
Tunnels 298
CHAFTEB 'Mil.
EXPLANATION OK TARI.KS AM. MISCEI.LAKKOUS TOPICS.
To Gauge a Stream Approximately jo i
Transverse Strength of Beams 302
Safe Bearing Power of Piles 302
TABLES.
Table for Right Triangles 32
Table for Oblique Triangles 36
Table Showing Excess of Arcs over Subtended Chords when the
Arcs are Aliquot Parts of 100 44
Table Showing Excess of Sub-cords over Aliquot Parts of 100
when the Chord = 100 45
Table Giving the Sides of Right Triangles 77
Table for Traverse Survey 83
Table for Level Notes 92
R., — R,
Table Showing the Least Value of T~-^—^-- 120
CONTENTS.
PAGE
Table Showing Computation of Prismoids 271
1. Degrees, Radii, etc 304
II. Tangent Offsets, i to 100 Feet 310
III. Offsets for Arcs of 100 Feet 312
Ilia. Middle Ordinate Arcs of 100 Feet 313
Illb. Chords of Arcs of 100 Feet 313
IV. Long Chords 314
V. Middle Ordinates 316
VI. Turnouts from a Straight Track 318
VII. Tangents and Externals of a i1 Curve 319
VIII. Arcs of Degrees, Minutes, and Seconds for Radius = i 3.3
IX. Acres for Various Lengths and Widths 323
X. Total Grades 324
XL Correction for Curvature and Refraction 325
XII. Elevation of Outer Rail 325
XIII. Coefficients for Stadia 3^6
XIV. Coefficients for Gradienter 327
XV. Offsets for Transition Curves 328
XVI. Tangent Distances for Transition Curves 330
XVII. Deflection Angles for Transition Curves 332
XVIII. Earthwork Tables, Different Slopes and Bases 333
XIX. Earthwork Tables, Two Slopes and All Bases 337
XX. Sines and Cosines 33'-$
XXL Tangents and Cotangents 352
XXII. Versines and Exsecants 35<;
XXIII. Useful Numbers and Formulas 382
XXIV. Conversion of Feet into Meters and Meters into Feet;
also Miles into Kilometers and Kilometers into Miles 383
FIELD-MANUAL FOE 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 \vhich 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 position is called The
Location.
5. 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-
K
6 ITELD-MAKFAL 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 a sketch 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. 7
only the results of careful and scientific surveys can tell how
many and where the crossings rnust 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 or a 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
ihe bottom of the stream is variable, the best site on some given
" reach " must be selected with a view to the kind of foundations
f/.ui table to the place. Sharp bends in streams should be studi-
ously avoided.
.14. The engineer should freely consult the best maps of the
( (Kuitry that he can command, and he should prepare a map on a
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. lie
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 transitinan, a levelman, a rodman, a stake-
man, a rear flagman, two chainrnen, 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-
accuracies 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 billy 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 shotild
be of a generous length, say from 30 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 30 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 resist-
10 I'lKI.iJ-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,
in which r is the resistance in pounds per ton, and v 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 r = 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 is to its length. Let
r' — resistance per ton, and U = the ascent per 100 feet. Then
?•' = — ; • X 2000 pounds per ton. For a \% grade (52.8 feet per
1 UU
mile) h = 1 and r' — 20 pounds per ton, the force necessary to
overcome the load. A \% 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 American roads with American
rolling stock it is probably about one half a pound per ton for each
degree of curvature. Letting ?•'' represent the resistance per ton
per station, and D the degree of curvature, the resistance per sta-
tion is r" = .5D. We observe that the resistance of a 1° curve
is but the ^ part of that due to a \% grade, or equivalent to
K<) Q
-^- = 1 .32 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 = ^ = .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,
the resistance on any curve for each degree of change of direc-
tion is equal to the lifting of the train ^ 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/c grade 1000 feet long than is required on a \% grade
4000 feet long.
To compare grades simply, that is, the rale 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
movement 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 the 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 hring
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
12
ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 13
of the eyepiece (the beads 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 tlie Axis of
tlie Wyes, so tliat it 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 0, say, on one side of J3. 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
FTG. 1.
A, and clamp the plates. Again revolve the telescope and note tbe
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.
4 tftELft-MANUAL FOR ENGINEERS.
It is proper to observe that the position of the line of collirna-
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.
I). To Adjust the Standards to the Same Height so that the Line,
of Collima'.ion icill Revrtve in a 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
direct the telescope to the top of the spire or
other elevated and well-defined point, as A
in the figure. B is 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.
Unclainp 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
~Q 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 EG 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.
ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. 15
F. To Adjust the Level on Telescope.— First level carefully and
clamp the telescope approximately horizontal by the eye.
Then, Laving the line of collhnation previously adjusted, drive
a stake, 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 may be necessary, so that the wire will give the
same reading at both stakes, showing that the telescope is truly
horizontal. Taking care not to disturb the position of the tele-
scope, bring the bubble into the middle by the 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
ofthe engineer ; for any others see the Manual of W. & L. E.
Gurley.
USE AND CAKE OP TRANSIT.
The instrument should be set up firmJy 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
trifle necessary to do this. This sight is so long that, according to
16 FIELD-MANUAL FOR ENGINEERS.
the principle of the " division of errors," the plate can be readily
set very nearly horizontal in almost any position of the instru-
ment.
Now place the left hip against one leg; grasp the opposite leg
with the left hand, and the leg on the right with the right hand;
and keeping the relative position of the legs unaltered, move the
instrument laterally over the desired point and lower it to the
ground.
Of course the two movements of tilting and sliding may be done
together and in much less time than required to describe the
movement.
To MEASURE THE ANGLE BETWEEN Two LINES OR OBJECTS.
Level the instrument carefully; bring the zeros of the verniers
and limb together by means of the upper clamp and tangent-screw,
and direct the telescope toward one of the objects by means of
the lower clamp and tangent-screw. Upon loosening the upper
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 CAKE OF INSTRUMENTS. 17
"When an instrument is exposed to the hot rays of the sun for
some time its parts are subjected to very unequal expansion, which
throws the instrument more or less out of adjustment, thus to some
extent vitiating the work and damaging the finer parts of the in-
strument. To prevent this it should 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
be clamped and placed in a stable position, upon proper supports,
in a 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
a 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, etc., 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 aluavs 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 ahd 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. Xo 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 Are 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 positio'n, 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:
I, Bringing the cross-wires into the optical axis of (he lele-
ADJUSTMENTS, USES, AND CARE OF INSTRUMENTS. ID
scope and consequently parallel to the line of bearings of the
wye rings.
%2. Causing the line of bearings to be parallel to the plane
of the level.
3. flaking either of these lines, and therefore all of them,
parallel to the bar and consequently perpendicular to the
axis of the instrument.
1. To adjust the line of collimation, that is, to bring the
cross-wires into the optical axis, so that their point of inter-
section 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,
so as to allow the telescope to turn freely; clamp the instru-
ment to the leveling-head, and by the leveling and tangent
screws bring cither of the wires upon the clearly defined edge
of some object. Then carefully rotate the telescope half-way
around.
! f the wire docs 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 eyepiece. Then by the leveling and tangent screws
bring it again upon the "edge," and repeat the above opera-
tion if necessary, and continue to do so until the telescope may
be 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
to make the adjustment of either 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
to change position the centering is not perfect.
In all telescopes the line of collimation is determined by the
cross- wires and objective, and is not affected in any way by the
eyepiece.
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 adjustment* embraces 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 screw*. 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-
way 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
can be made without causing tiny 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.
Place the level over a pair of leveling-screws, and by means
of them bring the bubble to the middle. Turn the instrument
half-way around horizontally. If the bubble runs toward either
end, bring it half-way buck by either pair of nuts, at the ends
of the bar. Then bring the bubble to the middle again, by the
level ing-screws, etc.. 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.
Xow 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, Athich 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-ranes.— 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 CAKE OF INSTRUMENTS. 23
If either slit does not coincide in direction with the hair or
thread, 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 not,
bend the center-pin (by means of a small wrench) about an
eighth of an inch 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
Npti, the pivot p not being
in the center 0. Now
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 El
bending (that is, straigJit-
CH'UH/) the needle, making
it cut points half-way be-
tween its former positions,
so that it occupies the posi-
tion Ap'B and is straight;
and correct the other half
by bending the pin, placing
the point of the pin at 0 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 FOll 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 sexagesi-
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'G', etc.,
having the common angle A. The triangles are equiangular and
therefore similar, and we have
BC BG' B"G"
AB ~ AB'
AB"
BG
AG
B'C' B"C'
AG
AB _ AB'
AC ~ AC'
AC'
AB"
FIG.
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
upon the absolute lengths of the sides, may be considered as indices
25
2G 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
the hypothenuse, is called the sine of
that angle. Thus
— = sin A,
c
and - = sin B.
c
The side opposite an angle, divided by the adjacent side, is called
the tangent of that angle. Thus
The hypothenuse, divided by a side adjacent to an angle, is
called the secant of that angle. Thus
v .
— = sec A;
b
- = sec 3.
a
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° — B and 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, etc.
Hence
sin J. =
cos 5=;
c
tan A — cot B =• — >
cos A = sin B = — ;
cot A = tan B = — ;
a
(4)
sec A — cosec B = r ; cosec A = sec B = - .
h a
PLANE TRIGONOMETRY.
Sin ,4 = cos B is really an identical equation, since cos B is the
sine (90°— B) = sin A; and so is tan A = cotB, etc. Furthermore,
c — b
vers A = = 1 — cos ^L.
exsec A = - — ; — = sec A — 1 =
cos A
1 — cos A vers A
cos J.
cos A'
B
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"."
Comparing Figs. 5 and 6, we observe that
the above ratios -, -, etc., change as the angles
A and B change. These ratios have been com-
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.
8. From the above equations we observe that
b
FIG. 6.
Hence
(6)
(7)
sin A cosec A = sec A cos A = tan A cot A = 1. , (8)
We also have, by division,
sin A a
cos A b
3 FIELD-MANUAL FOR ENGINEERS.
Again,
sin' 4 + cos' ;t=^+i' =<;=!. . . . (11)
Again,
6» 62
and
sec A = 4/1 -f tan2 A ; (12)
cosec2 A = sec2 jB = — = — = 1 -f- — = !-}- cot2 ^4,
and
cosec A =
From (9),
tan
sin A = cos A tan J. =
From (10),
cos J. = sin J. cot J. —
cosec .
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 = |/1 — sin5 A.
Then
sm A sin A
tan J. = T= - = J • . = • (16)
cos J. /i _ sins j.
sm A sin
—— = -— : — ...
cos A 4/1 _ sin2 A
(18)
PLANE TRIGONOMETRY.
If cos A is given, we have, from (11),
sin A — 4/1 — 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 EOF = A, and
FOE=B; then, in Fig. 7, EOG=A+B,
and, in Fig. 8, EOO = A - B. From
any point F in OF draw FE perpen-
dicular to OF; also EG and FH per-
pendicular to OH, and FK perpendicular
to EG. Now the three sides of the tri-
angle FEK are perpendicular to the three sides of the triangle
FOH, and therefore FEK = FOH.
No\v, in Fig. 7,
FIG.
!„ (A
EK
EO
But
and
FH FH FO
Ji/Ji
= cos A sin B.
. '. sin ( A + B) = sin A cos B -f- cos A sin B, . .
Again,
. OG OH-KF OH KF
. (20)
OR OF KF EF
= C°S A Cos 3 ~ sin A
30 FIELD-MAKUAL FOR ENGINEERS.
Then, in Fig. 8,
_ EG _ HF-KE_ HF OF _ KE FE
sm(A-B)=~gQ- EQ ~ OF' ^EQ~ ~YE'~EO
= sin A cos B — cos A sin B\
(22)
OG
cos (A-B} = =
FK_OH OF^ FK^ FE
1 ~ OF' E0~^~ FE ' OE
— cos A cos B + sin A sin B.
(23)
0 H G
FIG. 8.
In (20), make B = A and get
sin 2A = 2 sin A cos A. (24)
In (21), make B = A and get
cos 2 A = cos2 A — sin2 A
= (1 — sin2 A) — sin9 A — 1 — 2 sin8 A.
= cos2 J. - (1 - cos2 A) = 2 cos2 ^. - 1.
(20) and (21) give
sin A sin B
sin (J. -j- B) _ sin .4 cos 7? -f cos AsinB_ cos A cos B
cos (A + B) ~ cos .4 cos B — sin .A sin B
1 -
cos
X
sin B '
or
tan ( A -f B) =
tnn
tan B
I — tan A tan
Similarly, from (22) and (23),
tan (A ~ B) = ~
tan A — tan B
1 -|- tan A tan B
(27)
(28)
PLANE TRIGONOMETRY. 31
We note some special values of the trigonometric functions.
See Fig. 9.
Let COP represent any triangle ; angle COP = 0. Let 0 = 0.
Then CP = 0, PO and BO coincide and are equal.
Hence
SinO = ^ = 0, tanO = JL = 0, «c 0 = ™ = I.
Let 0 = 90°. Then PC and PO coincide with B'O, and CO = 0.
Hence
~£)f f\ T> ' /") T>/ /^)
sin 90° = —r =1, tan 90° = — — — oo , sec 90° = ~- = oo .
_O U 0
Let 0 = 180°. Then PO and CO coincide with B"0 and PC = 0.
Then
sin 180° =^L = 0, tan 180° = ^ = 0, sec 180°= |^ = 1.
Let 0 = 270°. Then PO and PC coincide with B'"0 and CO = 0.
Hence
7?"'O 7?//7O Ti"fO
sin 270° = 57^=1, tan 270°= — -- =00 , sec 270° =^r-^= oo .
1> U 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
parts 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 and c. See
Fig. 5.
Example.— Let A = 35° 23', and c = 874.8. We have
B = 54° 37'.
Also table of sines and cosines gives
sin 35° 23' = .57904, and cos 35° 23' = .81530,
Hence a = c sin A = 874.8 x .57904 = 506.54;
b = c cos A - 874.8 X .81530 = 713.22.
FIELD-MANUAL FOR ENGINEERS.
CASE II.— Given two sides.
Example.— -Let a = 184.3, and c = 246. We have
sin J. = _= .74919.
c
We find in the table
sin 48° 31' = .74915, .•. A = 48° 31' to the nearest minute;
B = 90° - A = 41° 29'; & = c cos A = 246 X .6624 = 162.95,
b = Vc2 - a8 = 162.95.
TABLE FOR SOLUTION OF RIGHT TRIANGLES.
Give-u.
Required.
Formulas.
1. a, b
A, B, c
tan A = -,
b
B =
90° - Ay
c = b sec ^1 =a sec .Z?.
2. a, c
A, B, b
a
sin A = — ,
B =
90° - A,
5 = c cos .4 =a tan B.
3. A, a
B, b, c
B = 90° -
Ay b
= a tan B
, c— <zsec j5=&secu4.
4. A, b
B, a, c
B = 90° —
A, a
= b tan A
, c=a sec _/?:=& sec A.
5. Ay C
B, a, b
B = 90° -
A, a
= c sin A
, b=at&n. B=ccosA.
If, in the second case above, b is given in place of a, then a
and 5 as well as A and 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-
angle ABD,
BD = c sin A, . (29)
and, from the triangle BCD,
BD = a sin C. . (30)
Also CD •=. AC — AD — b — c cos A. . . . (31)
PLAKE TRIGONOMETRY.
Equating (29) and (30) gives
a sin A
c sin A = a sin c, or — = -: — ~.
c sin 6
Similarly, or by analogy,
a _ sin A
b~ sin B'
From the figure,
(32)
b sin B
and c=^C' J
+ (CD)*.
Substituting for BD and CD from (29) and (31), we have
a'2 = c1 sin2 A + 62 — 2bc cos -4 -|- c2 cos2 A\
or, since sin2 A -(- cos* 4 = 1,
a». — £2 _|_ c-> _ 2£c cos ^} or COB A ~
Similarly, or by analogy,
cos B = ^C , and
From (29) and (31) we also have*
BD c sin A
tan G = 7=^ = T-
sin
6 - c cos A b
cos A
tan 5 =
tan J. =
sin (7 .
a
- — cos U
0
sin B
(34)
c
a
18. Let ABC, Fig. 11, represent a plane triangle, the parts be-
ing represented as usual.
Take GE = CA, and draw AD and EH perpendicular to AE.
We have
CAB + OEA = 180° - (7 = 4 + 7?.
and £4# = (7^4 - GBA = l(A +B) - B = ±(A - B).
FIELD-MANUAL FOR ENGINEERS.
Also GAD = 90° - CAE, and CD A - 90° - (AEC '= CAE).
Hence CD = AC = b.
Now
a + b BD AD AEi&n \(A +B) tan
JEET
. . (35)
FIG 11.
From triangle ABE,
or
BE _ sin BAE
AB~ jsin AEB'
a — b _ sin \(A — B)
c ~ sin \(A + B)
In triangle ABD,
BD _ a -f- b sin BAD cos |(^4 — B)
AB ~ c ~ sin ^IJW? ~~ cos ^(^4 + B)'
(36)
• (37)
Eq. (37) divided by (36) also gives
tan
+ B)
tan
— B)'
(38)
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
trigonometry. Still another solution by the author may be seen
jn Vol, III, No, 11, of the same journal,
FLAKE TRIGONOMETRY. 35
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 (32),
sin B sin C
b = a —. — — ; similarly c = a — : — -T-.
sm A sm A
Example 2. — A, a, and b given.
(32) gives sin B = sin A — ; then the table gives B.
Now C = 180° - (A + B) ; and c = a --.
CASE II. — Given two sides and the included angle.
Let b, c, and A be given. We have, from (34),
sin A
tan C =
Then
cos A
c
= ISO8 — (4 + <7) ; and a = b ~.
Or, from (33),
a = (&' -f e* - 2bc cos
Then
sin B = sin A — , and C — 180° — ( A -f- B).
CASE III. — Given the three sides a, b, and c.
Eq. (33) gives
COS J. =
36 FIELD-MANUAL FOK ENGINEERS.
Then
sin B = sin A-, and C = 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 FOB SOLUTION OF OBLIQUE THIANGLES.
(See Fig. 10.)
Given. Required.
Formulas.
6. A,B,a C, b,c
7. A, «, b B,C,c
8. A, b, c B,G, a
9. a, b, c A,B,C
asinB
a sin (A -f B)
sin A
b sin A
sin A
a sin (A + B)
a
sin A
sin A
b sin A
tan B — •
- — cos A
b
a*-(b-cy
~ sin B '
(a+b-c}(a+c-b)
2bc
b sin A
2bc
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 lab sin C — %ac sin B = \bc 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 a sin C is equal to the perpendicular from Bupon b,
b sin G is equal to the perpendicular from ^1 upon <(, etc.
PLANE TLUGONOMKTRY. 37
TRIGONOMETRIC FORMULAS.
tl. sin A = 4/1 — cos'J A — — 2 sin ^A cos ±A.
cosec A
12. cos A — \'\ — sin* A = — - = cos2 IA - sin'2 \A
sec 4.
= 2 cos2 ^ - 1 = 1 - 2 sin2 \A.
13. tan A — ; = r = cosec 2 A — cot 2 A
cos A cot .4
/ 1 - cos 2.4 sin 2/1
sin 2 A 1 + cos 2^ '
14. cot ^4. — cosec 2 A + cot 24 — the reciprocal of any expres-
sion for tan A.
15. sec A — -- — the reciprocal of any expression for cos A.
cos A
16. cosec A — — - - - — the reciprocal of any expression for sin A.
sin A
17. vers A = 1 — cos A — 2 sin2 \A.
18. exsec A — sec J. — 1 = — -p.
cos 4
19. sin (.1 ± B) = sin .1 cos 5 ± cos A sin 5.
20. cos (A ± B) = cos A cos .5 T sin A sin Z>.
21. sin A. + sin Z? = 2 sin |(4 + 5) cos \(A — B).
22. sin A — sin I? — 2 cos |(/1 -f B) sin |(.l — 7?).
23. cos A + cos B = 2 cos |(A -f .#) cos |(4 - 1?).
24. cos B — cos A = 2 sin \(A + 5) sin £(J. - B).
25. sin2 4 - sin2 B = cos2 B - cos2 /I = sin (A + 5) sin (.4 - 1?).
20. cos2 A — sin2 5= cos (4 -f- B) cos (4 — B).
27. tan A ± tan B=sin
28, cot 4 ± cot 7? =
COS 4. COS 5
si" <A ± B>
38 FIELD-MANUAL FOR ENGINEERS.
The above formulas contain the practical general relations exist-
ing among the functions of an angle. By writing \A, 2A, etc.,
in place of A, by repetitious, etc., the formulas may be greatly
multiplied without producing any new relations. This practice
is too common in Trigonometries, Field-books, etc.
CHAPTER TV.
SIMPLE CURVES CONNECTING RIGHT LINES.
LET ABODE represent a circular arc joining the straigLt lines
A V and E V, which are
tangent to the
C'
curve at
A and E.
AV and JSV are tan- V
t^ents to the curve, A and
Z£are tangent points, and
the angle KVE is the an-
gle of intersection, and
shows the change of direc- c
tion in passing from one
tangent to the other.
Vis the point of inter-
section, or vertex.
PROPERTIES RELATING
TO THE CIRCLE.
The following proposi-
tions rest upon elementary
geometrical principles,
may be regarded, for the most part, as axiomatic.
(a) A tangent to a circle is perpendicular to the radius, at the
point of contact.
(&) 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 - EV, and VAE - VEA.
(c) The central angle AOE, subtended by a chord, is equal to
the exterior angle KVE between two tangents to the curve at the
extremities of the chord.
(d) 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 = BAG, etc.
39
40
FIELD-MANUAL FOR ENGINEERS.
(e) 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,
(/) Equal chords subtend equal angles at the center of a circle,
and also at the circumference. Thus, if AB -- BC, etc.,
ACS = BAG, etc., and AOB = BOG, etc.
7
FIG. 13.
radius perpendicular to a chord bisects the chord, and
also the angle and the arc subtended
by the chord. Thus, if OG is per-
Px_^_ xN.T pendicular to'AE, we have
AM = ME, AOV = EO V,
and AC=GE.
(h} Parallel chords, or a tangent
and parallel chord, intercept equal
arcs. Thus, in Fig. 13, if PT, P'T',
and P"I " are parallel, then PP"
and T'T" are equal, also PP' and
TT' are equal.
(i) The exterior or deflection angle KBG, Fig. 14, between any
two chords AB and BG is equal to half the central angle AOG,
subtended by the chords. This
is easily shown as follows :
1. Join AC. BAG = %BOC, / \i
and BGA = %AOB, as stated
above. But / / \C
KBC= BAG+BCA-
.-. KBC = ^AOB + iBOC
= \AOC.
2. Draw the tangent EBL.
Then
KBC = KBL + LEG
= NBA + LBG
Fio. H.
SIMPLE CURVES CONNECTING RIGHT LINES. 4l
J3. Since the chords ^47? and BC are parallel to tangents drawn
at the 'middle of the arcs AB and B C, 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-
gent offsets.
Prolong AB, making Bd = ' AB. dC is called the chord offset.
Then, from the preceding article, we have
CBd = ^AOC = AOB.
Drop the perpendicular Bp upon Cd. This bisects the angle
CBd and the base Cd. Hence
CBp = dBp = iCBd = \AOB = BAb.
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 Ab is tangent to the curve at A, Bp is likewise tangent to
the curve at B.
Represent the radius AO by R, the tangent, or vertex distance,
A Fby T, and the angle of intersection K VE, as well as the central
angle AOE, by V. Represent the chords AB, BC, etc., by c, a
long chord, as AC, by C, the tangent distances Ab, be, etc., by
d, di , etc., and the tangent offsets bB, cC, etc., by t, ti , etc. Then
the chord offset Cd = 2t. Represent the vertex distance GVby
E, and the middle ordinate of a long chord by M. Let L represent
the length of the curve, P. C. (Point of Curve) the beginning of
the curve, and P.T. (Point of Tangent) the end of the curve.
The degree of a curve has been defined as the number of degrees
FIELD-MANtJAL FOR
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
liy an arc 100 feet in length.
Let D == the degree of the curve.
The circumference of a 1° curve is therefore 360 X 100 = 36000
OfjAAA
feet, and its radius is ~— — — 5729.578 feet, almost exactly.
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 a 1° curve. For the same reason the radius of a 3° 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
57*^9 57S
by 1); it is therefore equal to 1~ — .
Referring to Fig. 15, we see that for the same central anglo
AOE, the arc, the tangent, the
chord, the middle ordinate, the er.-
ternal secant, etc., vary directly with
the radius or inversely with the de-
gree of curvature. For example •.
If aO = %AO, then ape - \APE,
ai = \AY, etc. If APE is a 1° or
2° or 3° curve of length L, then ape
is a 2° or 4° or 6° curve of length
Hence to compute any function
of any curve, the tangent or exter-
0 nal, for example, from the corre-
FIG. 15. spending 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.
Example. — Find the tangent of a 7° 13' — 433' curve, the cen-
tral angle being 37° 50',
SIMPLE CUHVES CONNECTING EIGHT LINES. 43
The tangent of a 1' curve, by Table VII, is 117812. Then
T= 117812 H- 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, :i!7816 •+• 433 = 272.1.
Or, 13' T*| 60 = 0°.216,
arid therefore 7° 13' = 7°. 216.
Then 1963.6 -4- 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 = .001269239249^* - .0000000048329—. (8)
nA nb
IP D4
= .001269 .0000000048^, very nearly, (9)
ns n5
Z>2
= .001269—, approximately (10)
In these equations D = the degree of curvature, d = the differ-
ence between any chord and the subtended arc, and n = arc of 100
feet divided by this arc.
For an arc of 100 feet n = 1 and d = .001269Z*2, nearly; . (11)
For an arc of 50 feet n = 2 and d = .000158D2, nearly; . (12)
For an arc of 25 feet n = 4 and d = .00002Z)'2, nearly. . (13)
For n sub-chords per station the sum of the difference per
station is
nd -= .001269 — (14)
44
FIELD-MANUAL tOlt ENGINEERS.
Representing the central angle of the curve l>y V, and tlie num-
ber of stations in the curve by A", we have N ; and hence tin-
length of the curve exceeds the sum of the lengths of the sub-
chords by
7)2 V VD
E= Nnd= .001269=^ X -~ = .001269—,.. . (14')
Hence, in laying out curves, — should be nearly constant.
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, and 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 04°
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 s represent any arc, c its chord, and et the chord of one half
of the arc s.
Then, from above,
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.)
De*
D
Arc
25
Arc
50
Deg.
D
Arc
s>5
Arc
50
Deg.
D
Arc
25
Arc
50
1
.000
.000
11
.002
.019
21
.000
.070
.000
.001
13
.003
.023
22
.010
.077
3
.000
.001
13
.003
.027
23
.010
.084
4
.000
.003
14
.004
.031
24
.011
.091
5
.000
.004
15
.004
.036
25
.012
.699
6
.001
.000
16
.005
.041 .
26
.013
.107
7
.001
.008
17
.006
.046
27
.014
.116
8
.001
.010
18
.COS
.052
28
.01C
.124
9
.002
.013
19
.007
.058
29
.017
.133
10
.002
.010
20
.008
\m
30
.018
.143
bIMI'LE CURVES CONNECTING KKIHT LINES.
45
II. Suppose the chord AB = 100 feet, and AOB = D, (Fig. 16).
Then
AOM=^, MAtt = — , etc.
2 4
Hence
AM = AEsec MAE
= 50 sec ~i.
4
Therefore
AM - 50 = 50 (sec -£ - 1 j
= 50 exsec — .
Similarly, AK = AFsvc — = 25 sec — - sec — , etc., etc.
o 4 o
The following table gives the excess of AK over 25 feet, and of
AM over 50 feet, when chord AB is 100 feet.
Deg.
AK
AM
Deg.
AK
AM
Deg.
AK
AM
D!
-25
-50
A
-25
-50
DI
-25
-50
1
.000
.000
11
.036
.058
21
.131
.211
2
.001
.00-2
1-2
.043
.069
22
.144
.231
3
.003
.004
13
050
.('81
23
.157
258
4
.005
.008
14
.058
.093
24
.171
.275
5
.007
.012
15
.067
.101
25
.186
.29?.
6
.011
.017
16
.076
.122
26
.201
.323
7
.015
.0:23
17
.086
.138
27
.217
.349
8
.019
.030
18
.096
.155
28
.233
.375
9
.024
.039
19
.107
.172
29
.250
.408
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 AM 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 arc of 25 feet differs from
25 feet.
Thus let us first suppose A MB (Fig. 16) to be a 10° curve, and
the chord ^47? = 100 feet. Then by the table AK = 25.030 feet,
and this would lead to an error of .030 X 40 — 1.2 feet in laying
out a curve 25 X 40 = 1000 feet long, taking chord AK = 25 feet
long.
46 FIELD-MANUAL FOR ENGINEERS.
Suppose, secondly, that the are AMD — 100 feet and therefore
arc A K — 25 feet, and the chord AK is but .002 of a foot less than
25 feet.
Hence in laying out the above curve, taking the chord AK ' — 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 AMB 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 for a 10° curve the difference
between four equal sub-chords and 100 feet is .080 X 4 = .120 of a
foot ; whereas the difference between two equal sub-chords and
100 feet is .048 X 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 ex-
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 assump-
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 value?
where the changes are made.
For example, how can there be two different values (819.02 an<*
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 — j- = 1432.5 feet. This is rational and,
4
moreover, is precisely what is here advocated, except that the true
value of the radius of a 1° curve is used, vi/., 5729.58 feet.
SIMPLE CURVES CONNECTING RIGHT LINES. 47
FORMULAS.
From the triangle AOV, Fig. 12, we have
AV=AOt&nAOV, or T=Rtau^V. . . (15)
Transposing, we Lave
From the triangle AOF, we find
A0 = .Ani;, or R = -rn = & cosec
sin ^107^ sin \J)
This value of R in (15) gives
Measure equal distances FIT" and VL along the tangents in
Fig. 12 and draw HNL.
Measure #JVand VN.
A V VN VN
that is, T=R-. ... ..... (19)
If ^4J/ and FJf are measured, then
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 VG' perpendicular to A V to meet A C prolonged.
Now the angle AVG — \AVE = |(180 - F) = 90 - |F. Hence
VCC' = AGO = 90° - BOO = 90° - 4LF.
48 FIELD-MANUAL FOR EXGJNKEkS.
Also, VG'C = 90° - CA V = 90° - i V;
.-. VC'C- VCC', and VC' = VG = #.
Hence
7^7 /"fr TT
^- = -jy = laniF, or E = T \&\\ ±V. . . ("21)
Substituting T — ft tan IF for T, from (15), gives
^= tftan^Ftan £F. ...... (21')
Since the triangles BCd and 7?0(7are similar, \ve have
Gd BC BC*
or
and Bb = t = ~ ......... (3:.)
We also have
56 = AB sin 7?^4&, or i = c sin |D, . . . (24)
and 2t = 2c sin \D ........ (25)
Example. — Given R = 1909.9 feet and c — 100 feet, to find the
tangent and the chord offsets for 100 feet.
By (23),
1 0000
= 2.618, and 2t = 5.236,
or Table I shows that 1909.9 is the radius of a 3° curve, and
\D = 1° 30', and sin \D — .02618.
Then, by (24),
t = .02618 X 100 = 2.618, and 2t = 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
>n 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 long1.
FIG. 17.
The length of the curve is expressed in chains, in terms of the
central angle AOF = V, and the degree of the curve AOB = D.
V
The number of chains is evidently equal to — .
Thus, in the figure, if V— 23° and D = 5°, the curve is -2/ = 4f
chains, or 460 feet long. As the angle AOE = 5° X 4 — 20°,
EOF— 23° — 20° = 3° ; and the arc EF is f X 100 = 60 feet long.
It is usual, in laying out curves, to assume the radius 11, 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 R 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 give.s the degree of the required curve in minutes. We may
..nzrociTV 11
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.
Example. — 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 = 1136.13, the tangent re-
quired.
LONG CHOKDS AND ORDINATES TO LONG CHORDS,
Let A, B, C, etc., represent stations upon a curve.
Draw the lines as repre-
sented, EX being a perpendic-
ular from the middle of the
curve E upon the tangent
AT. Let Jlf = EM, M' = FN,
etc.
Draw AE, and we have the
angle EA M equal to the angle
EAX, and therefore the tri-
angles EAM and EAX are
equal in all respects. From
this we learn that the tangent
offset EX for any arc AE h
equal to the versin EM of
that arc, or to the middle or-
dinate EM ot the chord AK
of twice the arc.
Or, draw the tangent Et and the perpendicular Kt upon it.
Then the tangent offset Kt of the arc EK = the versin EM of
the same arc = the middle ordinal e EM of the arc AEK.
To find the middle ordinate, M. (See Fig. 18.)
1°. M = EM = EX = AM tangent EAM = \C tan
3°. M = EM = EX- ET cos TEX - #eos V,
. (26)
. (26' )
SIMPLE CURVES CONNECTING RIGHT LINES. 5l
3°. M = EO - MO = U - R cos | F = #(1 -cos £ F)
= R vers | F. . . . (26")
4°. J/= EO-MO=EO- i/AO*-AM*=R- \/ltr-iC-. . (26'")
To find any ordiuate FN distant d from the center of the chord.
Prolong FN, to meet OS, drawn parallel to AK, and join FO.
Now, FS =
- OS" = 4/-R8 - d*. JVS = MO = R - M .
= M+
- - d* - R.
Other methods will be given in connection with laying out
curves by ordinates from a long chord.
Approximate Values of Ordinates to Short Chords,
^FG,
Divide the chord into any number
of equal parts, eight for example, at
z, k, ly etc. , and erect the ordinates
Em, Fn, etc., and prolong them to A
meet the curve in 13', F', etc. Let
Em = m, Fn = m', etc. We have,
from geometry,
_ Am X mB __ \c X $c
or, approximately,
»»= 2r-=55- • • <27>
6.K oil
E'F
FIG. 19.
^G
JW
Similarly,
c X |c 15 c*
^L
WSR
= m =- = . ~ = m = m -
Hq = m'" =
(28)
(28')
(28")
For any other equal divisions of the chord we have similar re-,
suits.
FIELD-MANUAL FOR ENGINEERS.
tablisli the points E, F, etc.
Set E equally distant from A and 13, and at the distance m from
the point m ; then F, equally distant from k and 13. and \im
from n ; then G, equally distant from m and B, and £w* from p;
lastly, H, equally distant from^) and B, and T7gw from q.
This method involves much less labor than that of drawing sub-
chords to find the points F, G, etc. If we draw a tangent at E,
the offsets to F, G, H, etc., will be, according to the preceding
formulas, fam, yV^, ~i%m> e^c- This shows a convenient way of
finding the points on the curve.
To compute tangent offsets and middle ordinates by means of a
series.
In a way similar to that pursued in finding the difference be-
tween an arc and its chord we find
t = .872664625997Z> - . 000022 15240389.D3
-f.00000000022493360386Z>5 -, . (29)
or
t = .872664626,0 -.000022152404£3-f-.000000000224933604Z)5. (30)
We may find the tangent offset for m stations by multiplying
the successive terms of (30) by m1, m4, etc.
We thus find tang offset for arc of 50 feet, or — stations,
2
t1 = .2181G6156499Z) - . 000001 38452524D3
+ .00000000000351458756Z)5-. . (31)
For 1 foot m* = (.Ol)'2 = .0001, m4 = .00000001, etc. Then,
The tang offset for 1 ft.
= t, = .0000872664626Z) - .00000000000022152404D3 + . (32)
From (30), we have t = .SID, approximately.
For n stations tn = .87/iJX> (33)
Wellington in Railway Location, ch. xxx, recommends the
equation t = |»8A , , , (33')
SIMPLE CURVES CONNECTING RIGHT LINES. 53
\vliicli is practically the same as. (33). (33) is a trifle more accurate
than (33'), but either is sufficiently accurate for all cases in
which ii-D does no* much exceed thirty.
The same formuia gives the middle ordinate, n representing the
number of stations on either side of the center, or 2n the number
of stations in the arc. Any other ordinates desired are then given
by eq.-?. (27). (28), etc.
Example.— Find six offsets of a 3° curve at points 50 feet apart.
(See Fig. 18 )
For
n
n
n
n
n
n
=
1,
1,
t.
2,
2i,
3,
^ — 7
t =
t =
t —
• x
X
X
\s
x
i
1
9
•r
4
L' 5
"T"
6
X
x
X
X
x
X
3
:j
3
8
3
8
1=
9
5
10
16
2:'
.66;
.62;
91;
.50;
.41;
.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.C. Set the instrument at A, and
turn off from the tangent AVthe given deflection angle VAB =
\D, D being the degree of curve. This will give tlie direction,
AB, and measuring 100 feet in this direction, the point B will be
determined. Turn off the additional angle B AC = \T), the tele-
scope being now directed toward C, and set C in the line AC &nd
100 feet from B. Turn off the additional angle CAD = ^D, 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, and
54 FIELD-MANUAL FOR ENGINEERS.
proceed from that as from the first station. For example : Suppose
B, C, and D to be found with the instrument at A. Remove the
instrument to Z), sight to A, turn off the angle ADV ' = DA V —
\D, and the line of sight will be in the direction of the new tan-
gent DV Sit D. Reverse the telescope, and the line of sight will
point forward along the same tangent VDG. Now set E, F, etc.,
from the tangent DG, as B, C, and D were set from the tangent
In setting the first stake from the new position of the instru-
ment, as E from D, no notice need be taken of the tangent at D,
it being necessary simply to turn off from the line ADUihe angle
HDE = HDG -f GDE = AD V+ GDE - 4(iZ>).
In the new position of the instrument we observe that, in all
cases, the deflection from the line pointing to the back station to
the line pointing to the forward station is as many times the de-
flection angle \D 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 at the 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 arc.
Let a represent the arc AB or BC, etc., and a, represent the
arc EF. Also EOF =Di.
Hence DFE—^D, and EDF=$D1, and therefore
— = — , or
\D a
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 EIGHT LINES. ^
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 a curve having a short chord AB
-- c' subtending an angle AOB
= - Di at one end of the curve.
Define the tangent A V by set-
ting stakes upon it. Draw Be,
TX etc., parallel to A V, and BX,
iJcX', etc., perpendicular to AV,
or suppose such lines to be
•drawn. Now
BX=ABsm£A X=c'sm \Di = t^
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., FIG. 20,
we have
CBc = D, + \D, DCd - D, + \D, EDe = A -f- ID, etc.
Hence
Cc = c sin (D1 4- |£), Dd = c sin (D, + |Z>),
Ee = c sin (D, -f |Z>), etc.
We observe that Cc, Dd, Ee, etc., are respectively the tan-
gent offsets for curves of the degrees (Di -(- ^D), (Z>i -f- %D),
(2>, _|_ |/))> etc., and may be taken from Table III. Then
CX' = BX + Cc, DX" = CX' + Dd, EX"' = DX' + Ee, etc.
Set B at a distance c' from A, and ti from A V; then C a dis-
tance c from B, and CX' from AV; D a distance c from C, and
Dx" from A V, etc.
FIELD-MANUAL FOR ENGINEERS.
be observed that in this method we avoid constructing
t at B, in consequence of the short chord AE\ 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 c 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
A B = c, D! — D ; BX = c sin $D, Cc = c sin |Z>, Ed — c sin f Z>,
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 = c sin \D. Or, draw it parallel
to the chord joining adjacent stations.
It will be noticed that 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 without a transit ; but
a combination of methods is sometimes advisable, as will be shown
further on.
We also have
AX— CCOS^D! , XX' — Be = c cos (Dl -f ^D), etc.,
or XX ', X'X", etc., are respectively equal to the tangent distances
for one station of curves of the degrees D\ -f- \D, D\ -f- f Z), 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 EIGHT LINES.
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 = Dl} and a
short chord KL — c" sub-
tending an angle KOL
= D<I. Draw the tan-
gent AT, and drop the
perpendiculars BX and
CX' upon the tangent.
Draw Be parallel to AT.
Since Be is perpendic-
ular to AO, and BG is
perpendicular to a radius
bisecting the angle BOG, the angle GBc = J)l-\- \D.
Now BX is found in Table II ; and Ce may be found from the
same table or Table III, supposing the curve to be of the degree
A + \D. Or,
BX — c' sin IDj , and Cc = c sin (Dl -f- %D}.
Suppose c' = 80, D = 4°.
i=3° 12',
and
Table II gives
and
Cc = 9.06.
Y' = 11.29.
Place B 80 feet from A and 2.23 feet from the tangent A T ; then
C a chain-length from B and 11.29 feet from the tangent AT.
Locate K from the tangent at L in the same manner as B is
located. This gives the line CK.
Suppose the stations already located ; the long chords EG, DH,
etc., drawn, and also the perpendiculars Dd, Eel, etc.
Now Cd= Cm - Di = l(CK— DH) = one half of the difference
between the chords of six and of four stations. See Table IV.
dl — Di — Ef= \(DH— EG] = one half of the difference be-
tween the chords of four and of two stations.
lm =• one half the chord of two stations.
58
FIELD-MANUAL FOR ENGINEERS.
Again, Dd — Fm — Fi = the difference between the middle
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, Fm is the middle ordinate of a chord of six stations.
We note that points and lines on the right of Fm are of course
symmetrical with, corresponding points and lines on the left.
Hence
On = CK- nK = CK - Cl
Ch = CK- hK = CK - Cd, etc.
Now lay off Cd, Cl, Cm, etc.
Set D a station from C and a distance Dd from d, or CK; ther
Ea. station from D and a distance El from I, or CK; then F i
station from E and a distance Fm from m, or CK, etc., etc.
We also have
Cd = c cos f D, dl — c cos f D, and fan, = c cos \D ;
Dd - c sin %D, Ee = c sin f D, and .FJf = c sin £Z>.
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, Cl, 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\
and EF — c", subtending an
angle EOF = D.,.
Fa Locate B as in the las'
T method. Prolong AB, mak
ing BCi = c. The angle CBC,
= CAB + ACB = \D-\-\Di.
Hence CiCis the chord offse
2t' for a curve of \D -)- \D
degrees.
= 2c sin i(Z>, + D}.
Place C, therefore, at a distance c from B, and 2t' from Ci, 2t' beinjv
twice the tangent offset which is given by Table III opposite the.
SIMPLE CURVES CONNECTING RtGHf LINES. 59
degree $(D + DI). Prolong BC to D, , making <7A = c. Now
D1D — 2t is given by Table III, being twice the tangent offset.
Place D, therefore, at a distance c, from (7, and 2t from DI. Place
all regular stations similarly. C may be placed, also, as in the
preceding method. J^is easily placed from the tangent at E, 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 = <•',
subtending an angle AOB — D\ ,
and a short chord FG = c", sub-
tending an angle FOG = D«.
Locate the stations B and C from
the tangent A x, or by some other
method as already explained.
Then set off on CO the distance
Cc = t — c sin \D — the middle
ordinate of a chord of two stations ;
and set D in the prolongation of
Be, and at a distance c from C,
Set off Dd the same as Cc, and set E 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 G.
This distance ought to be
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
204Q 1900
shown in Fig. 24. Now - — — — '• - = 720, the length of each
180°
semicircle. Hence the degree of each curve — «-™ = 25°, and
the radius is 229.18 feet.
66 FIELD -MAX UAL FOR ENGINEERS.
Set the instrument at 0, and ran radial lines 01, 02, etc., making
6
FIG. 24.
angles of 80° with each other, for example, and set stakes at 1, 2,
etc., 229.18 feet from 0. These stakes are 100 f <? = 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 IN 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 decreasing; 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 possible 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", etc.,
stations on the false curve.
1°. Suppose the first stake from A to be set at B' , a distance
BB' = e beyond B, its true place.
SIMPLE CUliVK-i COXXECTiXG 1UGHT LltfES. 6'1
Then C' will be pet on AC prolonged and a chain from B', D'
will be set on AD prolonged and a chain from C', etc.
To prove that CC' is less than BB'. Draw GH equal and paral-
lel to BE'. Then BB'CII is a parallelogram, and B'H — BC —
B'C' = a chain. Hence C' is on the arc whose center is B' and
radius B'H. With C as center and radius Ctl = e describe the
FIG. 25.
arc HK UK lies outside of the arc HG' , since the arcs at // are
perpendicular to CH and B'H respectively, and are limited by AC
prolonged. Hence
CC' < CK = CII = BB' = e.
For the same reason DD' is less than CC', and so on to the end
of the i'U)i, that is, to the last stake set with the instrument at A.
Suppose the instrument moved to D'.
Since, in setting E' from D', we turn off fromD'A the same angle
that we would turn off from DA in order to set E, we have D E'
equal and parallel to DE, and therefore EE' is equal and parallel
to DD' . For the same reason FF' is equal and parallel to EE'
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,
FIELD-MANUAL FOR ENGINE
Let BCr = c + e. Now BC -f- CC' > BC', or
c + <7C" > c + e. .-. (7(7' > 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 all
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 -f- e feet
long. The number of
V
chords is equal to — = n.
Then for the difference in
length of the curves we
have AB' — (100 + e)n.
AB = ICOn, and therefore
AB' - AB = en.
(34)
For the chord we find, since R' — lit—
AB' =
and therefore
AB' - AB = BB' =
', AB = 21i
ZRes'm ^V
~100 '
SIMPLE CURVES CONNECTING RIGHT LINES.
But
BB'
360tfsin^y 114.60esin"-|-y
(35)
TtU D
Also
11' = (100 + <
< 180
180. 57.3,
(36)
We also have
TV 7»7.
-BB'~'m * V 114-6^(sin*F)2
(37)
ul
TV,
57.3^ sin y
— 7? 7? ' cos i y — •.
(38)
Example.— Let e = .02, T7 = 60°, and D = 6°.
The error of the curve = .02 X — TT = -200.
o
114 6 X 1 X 02
The error of the long chord = — — = .191.
Also 11' - R - - '—£— ~ = -19! •
114.6 X .02 X
([
= .0955,
B. Suppose that the first stake is set at B', Fig. 27, instead of
at B, the error in the angle being
BAB' = A.
Then it is evident that Cf,
D', etc., will be set at the same
angular distance from the chords p,
AC, AD, etc., as B' is set from
AB. Bf, C', etc., are on the
curve, having 0' as center and
AO' = AO as radius. OAO'
= BAB'.
To find the error at the end of
the curve.
Suppose .Z£the end of the true
curve.
C4 FIELD-MANUAL FOll ENGINEERS.
Now AE' = AE, and the angle EAE' = the angle BAB'.
Hence JE7i" = 2^1# sin \EAE' = 2AE sin
If the error in the angle is corrected at the end of a run, say at
IS', then, for reasons given above, the error in the position of the
stations is constant from Er 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 Di on AD and 100 feet from C'.
Similarly Ei will be set on JJ^and 100 feet from 2)lf
It may be shown that EEi is less than DDl precisely as it was
shown that CO' 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, say,
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. If
a new tangent is drawn before reaching the end of the curve, as
at E', Fig. 27, it is easy to see that such tangent would make an
angle A with the tangent to the true curve at E, and that the
error, therefore, would go on from E' forward precisely as from
B to E', and hence the error would increase regularly from B to
the end of the curve.
III. Curves Laid Out by Ordmates 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 H inclusive will be misplaced in pro-
portion to their distances from K\ the greatest displacement being
less than that of G. 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 station is set forward
or backward from its true position, then all subsequent stations
will be in error the same amount and in the same direction.
SIMPLE CU-RVES CONNECTING RIGHT LINES.
65
V. Curves Located by Middle Ordinntes.
In Fig. 23 suppose G'to be placed a distance a to the right, say,
(that is, along CO,} of its true position. Then c will be a distance
ft to the right ; D and d will be 2a to the right ; ^and e will be
'*i to 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, E nud e will be 4« to the right, F
ind/ will be $a 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 D
ind be tangent to AB at A.
Let AD = c. We have
From this we have
1 - d\ . . (39)
FIG. 28.
and d = 4/t(2JZ - t). . .
Second Solution. — Let AOD = A ; then
BAD = ADC = %A.
Now the triangle ABD gives
d- = cot $A. . .
(40)
Also
and
- = vers ^4,
— = cosec A.
a
(41)
(42)
(43)
66 FIELD-MANUAL FOR ENGINEERS.
If d and t are given,. find A from (41), then R from (42) or (43)
If t and .7? are given, find A from (42), then d from (41) or (43).
Finally, if d and R are given, find A from (43), then t from (41)
or (42).
Example 1. — Given t = 12 and d = 171.6, to find It.
tan §A = p^-g = .06993.
.;. \A = 4°, and A = 8°.
Then R = _ - _!_ - --^_ = 1233.3.
vers .4 .00973
Example 2.~ Given 72 = 1233.3 and d = 171.6, to find t.
cosec A = -^8j|- = 7.187. .-. 4 = 8°.
Then t = 1233.3 X .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
given point on a tangent.
We have
AO = 11, AS = d, and ABP = B.
tan ABO = ? . PBO = ABO - ABP;
d
sin OPI = sin OPB = ~ X sin
ji
sin
•• sin ABO'
Then
1 sm(OP2-PBO)
FIG. 29. —faTPBO
This problem furnishes a general method of finding any desired
point on a curve when obstacles preclude the usual methods.
(See Problem 12.)
III. Having run the curve AD, radius AO = It, Fig. 28, to
find the radus R' of a curve that will puss through D', given by
angle BDD' = D, and DD' = E.
SIMPLE CURVES CONNECTING RIGHT LINES. 67
Let AB = d, and BD = t.
Draw D'H parallel to AB. Now
HD' = E sin D, and 1W = E cos D.
AB' = d+ #sin D = dlt and B'D' = t - E cos D = ti.
It will be observed that when AB' < AB, E sin Z> must be
subtracted from d to give d* ; and that when B'D' = BH > BD,
E cos D must be added to t to give £,.
IV. Having run a curve of radius R and tangent T, to find the
new tangent T' corresponding to a new radius R' ', or to find a new
radius h' corresponding to a new tangent T' ', the central angle re-
maining constant.
Eq. (15) gives
T' = R' tan \V\ T = R tan % V\
.'. T' - T— (11' - R) tan * F, . . . . (45)
or 72' — R = (T' — T) cot \V. '..'.. (46)
Similarly, from eq. (17), we get
C' — C = 2(72' — R) sin ^F. .... (47)
These equations are of advantage for computing the change in
one element, T' — T 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.
Example 1. — Having run the curve of radius 72 = 1910, and the
central angle V= 46° 12', and tangent distance T = R tan \ V
= 1910 X- 42654 = 814.7, to find T' when 72 is made equal to 1900.
Eq. (45) gives
= 814.7 - .42654 X 10 = 814.7 - 4.27 = 810.4.
Example 2.— Given F = 52° 04', 72 = 5730, and T = .48845
X .5730 = 2798.8, to find 72' corresponding to T' — T + •-- .
We have
R' = R + ^ = 2798.8 + 233.2 = 8032.0.
68
FIELD-MANUAL FOR ENGINEERS.
V. (riven 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 ^47? be the given curve,
and IV F'the parallel tangent.
W' —a shows the distance
and direction that all points of
the curve are moved. The
curve will therefore begin at
A ' and end at Bf ; A A and BB',
as well as 00', being' equal
and parallel to YV. It is
not necessary to run the tan-
gent B' V in order to find the
distance VV. To find this
FIG. 30.
distance run a line, such as BB', parallel to A V from any point
on BV to meet B'V. Then make A A' = BB'.
If the perpendicular offset Bh — 7t is measured, we have
AA = BB' =
V being the vertex angle.
It B'V were on the other side of .BFfrom 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'V,
we .have, then, the
new tangents,
length of the
A'V = B'V = T',
which gives the position of A (and
B'), and the corresponding radius,
It' = 7" cot IF,
or, by (46),
R' =, R + (T' - T) cot *F.
FIG. 31.
VI. Given a curve AB joining two tangents, to find the radius
of a curve that from the same P, C. will end in a given parallel
tangent.
SIMPLE CURVES CONNECTING RIGHT LINES. 09
Let^F=£F = T, AV' = B'V' = T't AO = R, A0f = R'.
We liave, from the figure,
R' T' 7"
R=-T> °r ll =ET-
Also, from eq. (46),
R' = R + (Tf - T) cot£F.
Or, prolong AB to B' and measure BB'. Let ^45 = c, and
A#' = c', BB' — c' - c. Then, from the figure or from (17),
If the parallel tangent is defined by a perpendicular offset, as
B'p — Ji, draw BG parallel to AO. Then
Cp = BCcos BCp = (R1 - R) cos V.
.-. CB' = (R'-R) = (R'-K)coz V+U, or (1? -£)(!- cos V)=7i,
or (R' - R) versin V = h, or R' = R + - A
The quantity added to IMn the above equations must be sub-
tracted from it to find R' in the case in which V falls between A
and V, that is, when T' is less than T7. •
Example 1. — V -- 78°, R = 954.9. T.7 may be computed or found
by Table VII to be 773.3.
Let VV = 20 feet. Then
R' =R+ (T' - T) cot £F= 954.9 + 20 X 1.2349 = 979.6.
Example 2.—R — 1909.9, F= 46° 38'. T may be computed or
found by Table VII to be 823.2.
It is desirable to move the vertex from Fto V about 100 feet.
Find the new radius R'.
823.2 -* 8 = 102.9;
1909.9 H- 8 = 238.7,
Hence A V - 823.2 + 102.90 = 926.1,
and the new radius
AO' = If _ 1909.9 + 238,7 = 2148.6.
70
FIELD-MANUAL FOR ENGINEERS.
VII. Given a curve joining two tangents, to find the new tan-
gent points, corresponding to the
same radius, after eacli tangent Las
been moved any distance in the di-
rection of the other.
Let A Fand .Z?Fbe the given and
A' V and B'V the required tan-
gents. Let Hbe at the intersection
of AV &ul B'V. Let VH = a,
VII = b, and VV = c.
We observe that
B
VHV = 180° - F.
.'. sin VII V — sin V,
FIG. 32. and cos VHV = — cos F.
Chap. Ill, formula No. (8), gives
sin F sin F
HV
r-f -j- COS V
cos V
Then
VV =
b sin V
sinhVV
We observe that the directions of F//and HV are the same
as that in which the tangents B V and .4 Fare moved, and there-
fore there can be no ambiguity about the direction of these lines
or of FF', which is the line joining F and V.
Since R and F are not changed, it is evident that all parts of
the curve are moved in the direction VV and a distance equal to
FF'.
Hence make the angle VAA' = HVV, and A A' = VV.
The curve will begin at A' and end at B' , BB' as well as 00r
being equal and parallel to W .
If the distances the tangents are moved are given by perpendic-
ular offsets Vh' = hf, and Vh = h, the triangles Vllh and
V ' Hh' are similar and give
HV
HV
sin VHV
HV
HV'
or
SIMPLE CURVES CONNECTING RIGHT LINES.
sin V
~ h
71
Now the triangle hVV ' gives
vv- -7^-
smHW
VV and Vh are on the same side of B V; also, VV, and Vh>
are on the same side of A V.
VIII. Given a curve AB joining two tangents AV and BV, to
change the curve so as to end at the
same point as before, bat in a tan- A. A' V V' D
gent inclined at a given angle A
with the original tangent.
Let A V- BV = T, and the new
tangents A'V = BV = T'.
Let AO = /?, and A'O' = II'.
Draw BD = p perpendicular, and
BMN parallel to A V.
Let
V'=BV'D=BVD+VBV'=V+A.
Now An = It versin V,
and A'm — 11' versin V.
But A'm = An',
B
FIG. 33.
. '. R versin V = R' versiu V,
R versin V
R' =
versin V
With this value of R' run the curve back from B through the
angle V, and it will end at A', tangent to A V\ A' V being equal
to BV.
If the-length of the new tangent is desired, we have, from the
triangle VBV,
Then
sin V
T'
sin V ~ sin V
P
tan
sin V tan \V ~ vers V
72 FIELD-MANUAL FOR ENGINEERS.
If we wish to run the curve from A, we have
D V = p cot V, and D V = p cot V.
.-. VV =p(cot V- cot V),
and AA' = AV+ VV - A'V = VV -\-T-T'.
When V < 90°, T increases as V decreases, and vice versa;
and when V> 90°, I7 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 and BV, to find the radius
AO' = R' of a new curve start-
ing from A when the forward
tangent VB' takes a new direc-
tion from the vertex.
"We have
and
VA = R tan \Vt
R' = VA cot£F.
tan
FIG. 34.
tan | V
X. Given a curve AB, radius AO — R, joining the tangents
A Fand VB, to find the change
in the P.C., the radius remain-
ing the same, when the forward
tangent takes a new direction
from the vertex.
We have
VA = 5 tan £F;
VA' = Btan^F.
FIG- 35.
XL Given the angle of
tersection V of two tangents 0
4 Fand BV> tP find the radius
SIMPLE CURVES CONNECTING RIGHT LINES.
R and tangent distance T of u curve joining the tangents and
passing tb rough the point E.
1°. Let E be given by VE=l,
and angle EVO = A.
Let VEO = E, VOE = 0,
and AO = 21, AOV = $V.
No\v
i v- A0 - E0 - ^A.
-Jd~'VO~
sin
sin E —
sin A
cos \ V
This gives E. Then
0 = 180° - (A 4 E).
Moreover,
EO _ R _ sin A
~EV ~~ T " ~ sin 0
sin A
or 7? — I -. — —
sin 0
Finally, T=
2°. If E is given by VZZ" and HE perpendicular to each other,
EH
then
tan E VII =
VII'
EVF = FVH - EVH = 90° - £F - EVH;
VII
~~ cos EVH '
With these values proceed as above.
3°. Let Ebe given by VD — a, IXfiJ = ft, the angle VDF being
rial to FO^l = 4F. Produce i>^to 2^ and (7.
Let DF = c. Then c — a cos ^F, and AD = \/DE X DO.
But DO = DF+FG = DF -\- EF=2DF- DE=2c-b.
~1>), and VA= VD + DA = T.
.-, AD=
Now It= Tcot \V,
or HE = DE sin
= b sin i F.
74 FIELD-MANUAL FOR ENGINEERS.
D1I = b cos £ V.
Then VH= VD — DH = a - b cos \V.
With these values of F/iTand HE proceed as above.
XII. Given a tangent and curve (Fig. 36), to find the distance
from a given point on the tangent to the curve in a given direc-
tion.
Let V be the point, and suppose the direction defined by the
angle EVA = B. Let AV = T. We have
T
taniF=-.
Then EVF = D VF - EVA = (90° -$V)-B = A, say.
Now equation under Problem XI gives
sin A
sin E = -- T^pf.
cos^F
Then 0 = 180° -(A + E),
or,
^, and EVF = FVD - EVD = FDV - B = A.
Then find sin J^, then 0 and 7 as before.
Example.— R = 954.93, T7 = A V = 350, ^L VE= 40°. We have
tan AVO = - 2.72837; .'. J[FO = 69° 52',
and AOV=±V= 20° 8', and #FF = 29° 52' = F.
0 - 180° - 177° 50' = 2° 10';
_ 954.93 X -03781 _ 70 4
.49798"
SIMPLE CURVES CONNECTING RIGHT LIKES.
75
XIII. To locate a tangent to a curve of given radius R from a
given point V. (Fig. 37.) v c
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
a trial tangent VB, and FIG. 3T.
measure VB — A arid the angle VBO = B.
Chapter III, formula (8), gives
sin B
tan^FO =
Then OV=OB.
Lastly,
sin B
— — cos B
and sin EVO =
sin B VO"
BVE = EVO - BVO.
EO
or
sin A
- — cos A
This gives the angle to be laid off from
the trial tangent to give the true tan-
gent AE.
XIV. Given two curves AB and AC,
radii AO = R, AOi = R\, subtending
the central angles AOB = 0 and
AO^C = 0,, to find the length of the
line EC. (Fig. 38.)
Let A, B, and G represent the angles
of the triangle ABC, and «, b, and c the
sides opposite. We have
c = 2R sin \0, and & = 2R} sin |0,.
Also,
BAO = 90° - \0, CAOv = 90° - £0,
.-. BAC= A = 1(0 - 0i).
-, (See Chap. Ill, formula (8).)
and then
,
a = b
sin A
- — ~.
sin B
FIELD-MANUAL Foil EXGIXKEKS.
If the curves are run through an integral number of stations,
Tables IV and V give at once the tangent distances AK and AH
and the tangent offsets BK and CH. Then, drawing CD parallel
to A //to meet BK'ni D, we have
CD = AH - AK = d, say, and BD = BK - Ctf= t, say.
Then BC = f <
and
BC =
BD
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 cf a line. (Fig. 39.)
1. Let J. be the point, and EC
the line. Make AE = AC, and
with B -and C as centers and
any radius greater than AE de-
scribe arcs intersecting at D or
at E, or (with a different radium)
at F. Any two of the points A,
D, E, and F determine the per-
pendicular required. "!
2. Fix any two points of the
chain at B and at C. Take hold
of the point midway between B
and G and stretch the chain, the
middle point being at D. AD is the perpendicular required.
We may find, similarly, other points E, F, etc. Any two of these
points A, D, E, 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.
CIlis 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
table, or any equimultiples of these sides, making one of the
FIG. 39.
8IMPLK CURVES CONNECTING RIGHT LINES.
77
sides adjacent to the right angle (a or b) coincide with the line.
Let c = the hypothenuse.
No.
u.
b.
c.
No.
a.
b.
C.
1
3
4
5
6
20
SI
29
2
5
12
13
7
12
33
37
3
8
15
17
8
9
40
41
4
24
25
9
11
60
61
5
10
24
26
10
13
84
85
B
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 A, the end of
the 21st link at B, and the end of
the 50th link at C.
If in the three expressions
ra2 — ir, 2mn, and m2 -j- ifi we
assign to m and n any values at
pleasure, in being greater than n,
we will have sets of numbers
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
sides of a right-angled triangle.
II. To let fall a perpendicular from a given point to a given
line. — Let //in Fig. 39 be the point. Measure any line HB to
the given line. At the middle of ///> take D as a center, and
with a radius DB describe an arc BC. HO is the required per-
pendicular. For continuing the arc to II, we see that the angle
BCIIis inscribed in a semicircle.
III. To let fall a perpendicular to a
line from an inaccessible point. — Let
BC (Fig. 41) be the line, and P the
point. Let p represent the perpen-
dicular PK. Then
BK — p cot B, and CK = p cot C.
BK _ cot B
•'' CK ~
21
Fm. 40.
FIG. 41.
and
BK
cot C'
cot B
cot B -j- cot C '
78 FIKLD-MAXUAL FOR EXGTXEKRS.
Since BK -\- CK = BC, we have
BK = BC
cot B -f cot 0 *
Tills gives the foot of the perpendicular K. If BC is taken equal
to 100 or some small multiple of 100, BK 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 G 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 .STcan be made at any time. Of course the
point Pis located graphically by drawing .Z?Pand 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.
B
FIG. 42.
Or, measure BC in any convenient direction, and at C deflect any
angle FCD. Draw BD and the perpendicular CH. The angle
CDS = FCD - CBD. Hence
sin FCD
also <P-L> = ^^ sin BBC'
If the angle BCD is made equal to 90°, then
CD = BC tan CBD,
and BD = BC -f- cos CBD.
SIMPLE CURVES CONNECTING RIGHT LINES.
If the angle FCT) is made equal to 2CBD, tlien
CDS = FCD - CBD = CBD.
Hence CD - BC, and BD — 2BH = 2BC cos CBH.
OH is tlie departure of the line BC, or of DC, from the line
ABD. It is also the approach of the line CB, 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.
To 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 rinding 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 C. Then
AC sin C
AB = —
B
sin B
= AC-.
sin C
sin (A + C)'
2. If the angle ACS is
made equal to half of DAC,
then
OB A = CAD - BCA
.'. AB - AC. FIG. 43.
3. Or, in Fig. 43, make the angle BAG = 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-MAXUAL FOR EXGIXEERS.
5. If at G we make the angles ACB and ACD equal, we liavo
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.
FIG. 45.
6. In Fig. 45, AB being the distance required, run and measure
any line AC and measure the angle BAG — A. Make
Then
ACB = 90° - A = C.
AB = AC sin C.
B. When both ends of the line are inaccessible. (Fig. 46.)
Let AB be* the line to be meas-
ured. Find the distances from the A-
point G to each end of the line A
and B by preceding methods, and
measure the angle G. Then
sin G
tan A =
AC
(see Chapter III, formula (8).)
C
FIG. 46.
.
sin A
C. To erect, at a given point A (Fig. 47), a line ylZf perpendicu-
SIMPLE CURVES CONNECTING RIGHT LINES.
81
\
lar to an inaccessible line
BC, and to draw a parallel
AH to the same line. — Find
.47? = c and AC — b by pre-
ceding methods. Then
pin A
tan B = - .
- — cos A
I)
Now draw A K, making
BA K = 90° - B.
A K will be the required per-
pendicular, and AH, making BAH — B, will be the required
parallel.
D. To find the length and relative position of an inaccessible lino,
AB, from an accessible line, CD, separated
n from the former by an inaccessible space. -
Measure CD and the angles at C and D.
Example.— Let CD = 4000;
BCD = 126° 25V; ADC = 47° 53 V;
ACB= 3° 10'; ADB= 3° 01'.
.-. ACD = 129° 35V; SJ)C = 50° 54 1'.
Also,
CAD = 180° - ACD — ADC — 2° 31';
CBD = 180° - BCD - BDC = 2° 40'.
Hence
40= 400o5!2*™a.' = 67«81.»;
BC = 4000
tan CAB =
sin 2° 40'
sin C
= 66728.8;
'AC
FIG, 48,
Finally,
CAB = 75° 28V.
^sin 3° 10'
-, - 3807.8.
82 FIELD-MANUAL FOH ENGINEERS.
If AD 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 the inaccessible space.
The data of this example 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, and used for him by his assistant, Prof. J. B. Davis (now of
Michigan University) in making the preliminary surveys of the
Owosso 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
FIG. 49.
as trees, occur, making it necessary to run the line ABiCiDiE[F,
called a traverse. The deflection angles at Bi, C\t Diy pud E^ are
supposed to be small.
From Bi , d , D\ , and Ei draw perpendiculars to AB, and from
Si, C>, and Dl draw parallels BiK, dL, and D,P to AB as
shown. Prolong ABi to R, and B} Ci 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.
We have D1L — CiDi sin DiCiL. 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 CUHVES CONNECTING 1UGHT 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 Z>, = 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 Cd = #77, + f,7i. 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 DDi = CC\ — DiL. The departure
of the end of the line D\E\ which crosses the base line is equal
to the departure of the line minus the departure of the beginning
of the line. Thus EEl = E,P - DDt.
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.
Angles turned.
Angles with
Main Line.
Departures.
Stations.
Each Course.
Total.
Left.
Right.
Left.
Right.
Left.
Right.
Lcfr.
Right.
A = 10
10'
10'
11
12
£,= 13
22'
32'
3000
3000
14
15
Ci = + 20
4'2X
10'
7040
10040
16
17
18
Z>!= 19
3C'
40'
3800
6240
21
E,= 24
1°20'
40
20000
137GO
25
26
27
F =+44
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 -f- 20 it is 10 -f 22 = 32' I*.; from 15 + 20 to 19 it ig
42 - 32 = 10' L., etc.
84 FIELD-MANUAL FOR ENGINEERS.
From 10 to 13 the departure is 300x10 = 3000 foot-minutes.
" 13 " 15+20 the departure is 220x32 = 7040
" 15+20 to 19 " " "380X10:= 3800
'< 19 to 24 " "500X40=20000 " etc.
The aggregate departures are :
At 13 ................................ 3000 R.
At 15 + 20 ...... 3000 + 7040 = 10040 R.
At 19 ............ 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 -4- 40 — 344 feet, showing that the
auxiliary line (E^Fin 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 FIT, and which may be followed approximately the same as
AF WAS followed.
Let I = the length and d = the departure of any line, and n —
the number of minutes in the course. Then
d = I sin n' = In sin 1' = .0002909fo.
Since In is given in the last two columns, the departures in feet
are found by multiplying the quantities in these columns by
.0002909 or .00029 nearly. We observe that
1 4- .0002909 = 3438 nearly.
Hence the departures given in the table (in foot-minutes)
divided by 3438 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 ____ ............. 3800 + 20000 = 23800
and the same to the right is ........... 3000 + 7040 = 10040
The difference is .................................. 13760 L.
We have AR = AB, cos
~ 4W - <-«s #4#.) = AB vers BA
SIMPLE CURVES CONNECTING RIGHT LINES. 85
For BABl = 2° 34' this becomes AB, - AB = .WlAB nearly.
This shows that ABi 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° 1?' it is nearly .00025, or nearly 1 in 4000;
and for 0° 38£' " " " .00006, " " 1 " 16000, etc.
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 BiDi makes with the base line the angle
PA - BB, _ 6240-3000 _ _
BD WO
The line AE, makes with the base an angle
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 tbe 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 JIT). Represent the
lines run, in their order, by a, &, and c, and the deflection angles at
FIG. 50.
B and at C by B and C respectively. Draw CR parallel to AB,
and (7/7 and KDK perpendicular to AB.
Angle DCR = DCE - RCE ~ C - B,
86
FIELD-MANUAL FOR ENGINEERS.
Now
Now
BH=bcosB; HK = c cos (C - B) ; AK - AB + BH+ HK
CH = b sin B; DR = c sin (C - B); DK = CH - DR.
DK
4jr
Also AD = AK -*- cos DAK.
Of course the method is applicable whatever the number of lines
run. All but the last line could usually be taken equal to a whole
number of chains, which
would reduce the required
computation to a simple mul-
tiplication.
E. To find the angle of
deflection, i", between two
straight lines A V and VM,
when the point of intersection
is inaccessible; and the dis-
A M tances of the intersection from
FlG- 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= A'Ptan F,
and PV - KP • t- co* P.
2. Run and measure any line PL from one line to the other.
Measure also the angles VPL = P' and PL V =i /,. Then
V= P' + L.
Hence, in the triangle PVL, PL and the angles are known, to
find FPand VL.
3. If obstructions prevent the use of the former methods, run
and measure any broken line ABCD. Prolong AB and BC to
meet VM at M and N.
Measure the deflection angles CBM = B, DCN = C, and CDN
= D.
Let AMV = M, and CNV = N. Then
sTiHv ' tto.(0+J5'
= BC -f CN. Also M = W - B- C -f D - /?;
SIMPI.I-; CURVES CUNXKCTI.NI.T RIGHT LINES.
BM =
sin N
AM = AB + BM.
Now we have ^43/arid the angles at A and M, to find vl F, MV,
and the angle F = A 4- Jf. A similar explanation will apply to
any case.
Wlie.n a broken line must he used, the above method involves
fewer computations than any other.
F. To locate a curve joining two tangents when the vertex is
inaccessible.
Find by the last problem the
distances Va and Vb to convenient
points on the tangents, and the / \ \, V
angle V. Then assigning or com-
puting the tangent T = A For BV
from, the radius, we have
and
aA = T - a V,
bB = T- bV.
We now have the tangent points
and can run in the curve as usual.
GK To locate a curve of radius U or tangent T when the vertex,
the beginning, and the end of the
curve are inaccessible. Find, as
shown with Fig. 52, the angle F
and the distance Fa to any point,
«, on A V. Then
a A = T - a F,
and
cd a A
FIG. 53.
Also
ac = Ad = J?versin AOc.
Drawing the tangent cb, we have
abc = AOc, or acb = 90° - AOc.
This gives the direction of the curve at c, and it may be run in
each way from c.
To pass from any point c on the curve to any point n on the
tangent.
We have
1an (dcu = an<<) =
88 FIELD-MANUAL FOR ENGINEERS.
Set the instrument at c and turn off from the tangent cb an angle
ben = bed — den
= dbc — anc,
, Us IV
and measure en = — — .
cos anc
H. To find any desired point on a curve when obstacles precl udo
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 = — ; then t — d tan ±A.
d and t 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.
Example. — Let AD be a 4° curve, and AB — one half the chord
of four stations = — ^— = 199.35. Then Table V gives
BD = 13.94.
(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 b be a station near the obstacle. Deflect from the tangent
at b some small multiple of the deflection
angle for one station ^D, giving the line
bd. The length of bd 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 the middle ordinate me
taken from Table V, gives also a point c
on the curve.
If more convenient, make bx = cm, and
xc = bm, which also gives c.
Again, run the tangent b V = d' any
jrIG 54 convenient distance. Then
SIMPLE CURVES CONNECTING RIGHT LINES. 89
Deflect at V an angle equal to 2bOV, and make V'd = bV.
d will be a point on the curve. The number of stations from b
is equal to
bOd 2bOV
bV sliould usually be taken equal to a whole number of chains,
7)
in which case —^-f is very readily found.
The lines bV" and V"d lying on the inside of bd may be run
instead of bV and V'd.
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 relative
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 wherever 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
called 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, etc.
Having established the first bench, and recorded its elevation,
the rod man 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
tho bench, we obtain the height of the cross-wire, technically
called 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
henfJt, tailed 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
or " 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
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.
EM
0
1
Peg 2
3.46
203.46
7.29
5.34
0.81
200.00
196.17
198.12
202.65
W. Oak 60 ft. R. of Station 0
+ 40
8
4
5
Peg + 60
6
4.17-
6.18
206.82
211.93
1.12
3.16
6.09
4.14
1.07
3.13
205.70
203.66
200.73
202.68
205.75
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,
streams, ditches, etc.
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 -f 4.17 — 0.81 — 3.16 = 203.66 — 200 — 3.66.
To prove the H. I., 211.93, we find
3.46 -f- 4.17 4- 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 ease 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 Line.— 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 :i
third peg from the second, etc.
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 measured.
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, etc.
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, KTC. 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 exactly.
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 AD, while the
true level will be the arc AB. The
difference BD between the line of
sight and the true level is the cor-
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 AE = R, AB = 1), and BD
= E. By geometry,
AD* = BD(BD 4-
A&
BD =
FIG 55.
BD +
Omitting BD in the right-hand member, since it is small com-
pared with 2R, and supposing AD — AB = J), we obtain
22t 2 X 20913650
rx = .0000000239087)5. .
(1)
This formula gives a result or value for E slightly too small;
but the relative error is only about one in 24,000 for a distance of
^Q = 39.6 miles, or arc of5- m
= 34' 22".65.
96 FIELD-MANUAL FOR ENGINEERS.
In observing distant objects, a ray of light traversing tlie air
from 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 7.K.
Thus a point at C (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 2L
Hence the correction for refraction is
E' = ~_ = . 00000000341 5D2 ..... (2)
The correction for curvature and refraction is
IP D2 3 /)2
E" = £C=BD-CD=: — - ~—rt = - - = .000000020492D8. (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
R = 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
Then ~ = .12638 (!')
= the correction for refraction for distance 6083 ft. or arc of 1'.
Also ~~ = .8846 (2')
= the correction for curvature for the same distance or arc 1',
LEVELING, STADIA MEASUREMENTS, ETC,
97
and
i 1 .75828 (3')
7/t
= 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 - CD cot CAD, and BD = CD cot CBD,
Subtracting gives
AB
AB=OD(cot CAD- cot CBD), or ^=calCAD_^cB1)- W
Second Method. — Let A and B (Fig. 57) occupy any positions
C
B
FIG. 56.
except in line with C. Measure AB and the angles at A and B\
also the angle of elevation GAD.
C=1SQ°-A-B, AC =
sin C
, and CD = AC sin CAD. (5)
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
O 7^0
for curvature and refraction, namely, — — — .0000000204927)2,
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\ ACE— C
would be the angle of de-
pression or dip ; and we
would have
BC = BOX exsec CO A
= It exsec C.
Owing to refraction, how-
ever, the line of sight would
be a curve concave down-
ward whose radius = 7/£;
and it would therefore ex-
tend from C to a point A',
say a distance A A' beyond
A.
Draw CDF tangent to this
curve at C to meet the ra-
dius AO prolonged in F.
Draw also the tangent A't.
Let COA=N, COU=I1',
and CO A' = 0. Then
FIG. 58.
CO = r sec // =
cos H'
Let E, on A'O prolonged but not shown, be the center of the
arc A'C. Now
#0V_ CO9
cos 0 = - cos COE = -
2CO . EO
- ___
TsT^eclT^ 12
Clearing, substituting 1 — vers 0 for cos 0, 1 — vers H for
cos II, and 1 -f exsec 7?" for sec H, we find
13 vers 0 — 13 vers // + exsec Jf.
LEVELING, STADIA MEASUREMENTS, ETC. 99
Or, writing versines for exsecants or vice versa, we have
vers 0 = £ vers II,
or, approximately,
exsec 0 = £ exsec H.
In the triangle CEO,
49 r5 -f 36 ?•* - ~U6l _ 85r2 - r2 sec9 H _ 85 - sec2 H
2~x~ 6r"x~7V * 84?" ~~84~~
Also sin E = sin
.-. sin #sin 0 = sin2 0' ---- = (1 — cos'2 Oj— — —
_ 1 70 - 169 cos2 H - sec2 H sec H
~l44~ ~~T"~
Now since OA' is tlie prolongation of EO, and JS'C'and OZ) are
perpendicular to CD and therefore parallel, we have
CO A - 1)0 A' = COD, or 0 - E = II';
.-. cos H' = cos (0 — E) = cos 0 cos E -\- sin 0 sin E.
Substituting in this the above general values of cos 0, cos E,
and sin 7? sin 0, and expanding and reducing, we find
13 cos E + sec H , (sec H — cos //)
Putting cos H= 1 — vers //, sec 7/= 1 + exsec IT, etc., we
find
(exsec // — vers H)
vers /f = f vers 77 -- — — — — , . . (a)
or vers 77' = | vers 77, very nearly, ...... (6)
or exsec //' = f exsec //, very nearly ...... (7)
exsec H ' — vers //'
Hence vers //' = f vers H -- — .
i&
From this we have
, £(exsec H' — vers H'}
vers H — \ vers //' -f ^ — , . (ft)
100 FIELD-MANUAL FOR ENGINEERS.
or vers //= | vers //', very nearly, ..... (6')
or exsec H= | exsec II', 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— 4/JZT = 1.08//', approximately. . . . (?')
T
Supposing r = 4000 miles and AB = — — 50 miles, then it is
80
easy to show that the error of eq. (7) is less than .00000002, and
the error of (6) is about .0000000004.
Ki'<n/'iple 1. — The observed dip of the sea horizon is //' .— 24'.
What is the height of the instrument above the sea?
We have
BC = r exsec H- r| exsec //' = 20914000 X .00028467
'= 595.35 feet, exactly.
Example 2.— Let r — 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
Kf\
H= j^ X 57°.29578 = 0°.7162 = 42' 58".
.-. H' = 0.716 -4- 1.08 = 0°.663 = 39' 47".
Also,
7i = r exsec H= 4000 X .0000781 = .3124 miles = 1649.47 feet.
The exact relations between II 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/ = the focal length of the object-glass.
FIG. 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 ohject-
glass, exposing it to the rays of the sun, and noting at what dis-
tance from the center of the glass the rays form a perfect and
minute image of the sun on a smooth surface.
Let Cm = I', Cn = I, DE = S', and HK = 8.
The focal distance OF is constant, but Co = c varies with the
position of the object-glass, and hence CF is also variable.
Let Co = c' when the rod is at DE, and Co = c when the rod is
at HK.
Now, from the figure,
*
'
— -
Fm ~ DE' I' - (c' +7)
8' is usually assumed = 1 foot, and Fm = I' — (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 the
desired point and the space 8 noted, which placed in (8) gives
I — (f' +/) — ^i» sav- We may then measure c on the telescope.
Then
102
FIELD-MANUAL FOE 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 = c' = Ci, and solving we find
SI' - S'l
If we observe S' and 8 corresponding to any two distances I'
and I and substitute in (9), we have Ci -(-/.
Having found c, +/, lay off Cm = 100 + c, -f /, or Fm = 100,
and adjust the stadia wires to subtend DE = just one foot at that
distance.
Then from (8), writing c, for c and c', we have
or, omitting accent,
(10)
Example. — Suppose at £' = 100 we find S' = 1, and at I = 500
we find S = 5.0453.
Then eq. (9) gives
504.53-500 _
~
Then, from eq. (10), I = 100$ -f- 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
collimation is oblique to the
horizon, which is usually the
case. Thus, in Fig. 60, let
jyE' = the space intercepted
on the rod when the line of
collimation FH is horizontal,
and let DE = S = the space
intercepted when the line of
collimation Fn makes an angle nFH = a with the horizon.
Let DFE = D'FE' = 0.
Ff0
LEVELING, STADIA M-EASUREMENTS, ETC. 103
In Fig. 60,
8 = DH - EH = HF [tan (a + |0) - tan (a - £0)].
The horizontal reading desired is D'E' = 2JBFtan £0.
D'^' 2 tan 40
Dividing gives — ^— = — — r^.
8 tan (a -f- £0) — tan (a — £0)
2 sin 40
But 2 tan 40 = -
COS ^U
Also, by Chapter III, formulas (26), (27),
sinO
tan (a -f £0) — tan (a — £6) =
cos (a -j- -|0) cos (a — i0)
2 sin |0 cos 4.0
cos8 a — sin2 0*
Substituting these values, we obtain
D'E' cos- a — sin2
cos2
(11)
If we neglect sin2 |0, or, what is equivalent, add sin8 ^0 to the
numerator, we introduce a relative error of
sin2 40 sin2 4,0
— — — — = sin* |0 sec2 a, very nearly.
cos8 a — sm- ^^ cos^ a
Again, if we add sin* |0 to the denominator, making it unity,
we introduce a relative error of — — ^-r = sin2 40 sec2 40.
cos2 |0
The first of these errors increases the fraction, and the second
decreases it. Moreover, since 0 = — — — — 34' 22". 65, we
J(JO
have in all practical cases a > |0, or sec2 a > sec2 10.
Hence the fraction is, by the double approximation, increased
sliglitly more than it is decreased. Hence
TV W T)' W
— — < cos8 a ; but — — — cos* a, almost exactly. . (12)
o o
The total relative error is
e — sin2 ^0 (sec2 a — sec2 4;0)
= sin3 £0 (tan2 a — tan2 £0).
104 FIELD-MANUAL FOR ENGINEERS.
Since tan1 ^Q is very small compared with tan2 a, we have
e = sin5 -|0 tan- a, very nearly,
= .000012 tan2 a, very nearly.
Hence
D'E'
= cos5 a(l - .000012 tan2 a)
o
=: cos2 a — .000012 sin8 a (13)
Since sin £9 = tan ^0 = — — = .005. very nearly, the quantity
neglected above is only (.005)4 = .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 DE — S in order to produce the true space
D'E' which would be observed at the same distance if the line of
colliination were horizontal.
Hence we have, from (10),
I = (1005 + c +/)(cos2 a - .000012 sin2 a).
By eqs. (25) and (26), Chap. Ill,
1 4- cos 2a
cos2 a = ~ — SB . J •
and Bin' a =
Hence
I = (100S+ « +/)l - ~- - .000013 , . (14)
I = (100/8 + c +/) (l - ver*2(l \ 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.
JSxample. — Find the coefficient for a = 7° 20'.
We write half the vers 14° 40' = .01629
and subtract from unity and find coefficient — .98371
Another 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 rodina
…[truncated]