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fully to it. Depress the telescope until a pin can
be set in the ground at its base, as at B. Loosen
the clamp and turn the plate through 180 with-
out touching the telescope. Clamp the instru-
ment and sight again to the high point A. Again
Fic7264. depress the telescope and set another pin, which
it will be found is at some distance from B, as at C. The
vertical plane is the line A D, and it will be seen that the
error is doubled. The adjustment is made by raising or
lowering one end of the telescope axis by means of a small
screw placed in the standard for that purpose.



1236. Care of the Transit. The transit, though it
will bear a lifetime of legitimate service, will not stand
neglect or banging. The bearings are^ delicate and easily
marred by particles of dust or sudden blows. Moisture
clouds the lenses, and, when combined with dust, is doubly
injurious. Little advantage is gained from working in the
rain, and, unless the stress of work requires it, both instru-
ment and men are better off under cover. If the instru-
ments should encounter a wetting, carefully wipe the object
glass, eye-piece, and verniers with- a piece of chamois skin,
as moisture soon clouds them so as to prevent further work.
As soon as the party returns to office or camp, complete the
drying process by thoroughly rubbing with a piece of
chamois skin, which every engineering party should carry.
When a party rides to and from work, the instruments
should be carried in their cases, and they should always be
kept in their cases when in the office. The common cus-
tom of leaving an instrument on its tripod and standing on
a board floor can not be too severely condemned.

1 237. Setting Up the Instrument. As much of
the work of an engineering party is suspended while the
instrument is being set up, it is highly important to acquire
facility in setting it up. The following suggestions will be
of use, although practice alone will make one expert.

In setting up a transit, three preliminary conditions
should be met as nearly as possible, viz. :

1. The tripod feet should be firmly planted.

2. The plate on which the leveling screws rest should be
level; and

3. The plumb bob should be directly over the given

The third condition must be met to a nicety, and this is
rendered comparatively easy by means of a " shifting head "
with which most modern transits are provided. When these
three conditions are approximately met, the completion of
the operation is quickly performed with the leveling screws.


1 238. How to Prolong a Straight Line. Let A B,
in Fig. 265, be a straight line, and it is required to prolong
or produce it 400 feet to C.


The line can be prolonged in two ways by means of
foresight and backsight.

1. By foresight, set up the transit at A and sight to B ; let
the chainman measure 400 feet from B in the direction in
which the line is to be prolonged. Then, by means of signals,
move the flag to right or left until the vertical cross-hair
shall exactly divide the flag held at C. Then, the line B C
will be the prolongation of the line A B.

2. By backsight, set the transit at B and sight to A.
Reverse the telescope, and, having measured 400 feet from
B in the opposite direction from A, set the flag at C, then
the line B C will be the line A B produced.

1 239. Double Centers. In prolonging lines, a device
known as double centering is sometimes used. It is un-
necessary when using an instrument that is in proper
adjustment, but it is a good check, and a knowledge of the
method is valuable.

Let A B, in Fig. 266, be a given line which it is re-
quired to produce 1,000 feet. Set up the transit at B;
B 500' D 600' G

backsight to A, and reverse the instrument. Set a point C
500 feet from B. Unclamp the upper plate and revolve the
telescope through 180, backsighting again to A. Reverse
the telescope. If the line of sight does not come at C, then
the point C is not in line with the points A and B, and the
line of sight will be at some point, as I), on the opposite side
of the true line. Measure the space C D and mark its mid-
dle point E. The point E will be in the prolongation of the


line A B. Move the transit to E, and, backsighting to B,
determine the point H by the same means used in fixing
the point E.

1240. Horizontal Angles and Their Measure-
ment. A horizontal angle is one the boundary lines of
which lie in the same horizontal plane. Let A, B, and C,
in Fig. 207, be three

points, and let it be re-
quired to find the hori-
zontal angle formed by
the lines A B and A C
joining these points.
Set up the instruments FlG - 26r -

precisely over the angular point A, and carefully level it.
Set the vernier at zero, and place the flag at B and at C.
Sight the flag at B and set the lower clamp. Then, by
means of the lower tangent screw cause the vertical cross-
hair to exactly bisect the flag at B. Loosen the upper
clamp. With a hand on either standard, turn the telescope
in the same direction as that of the hands of a watch until
the flag at C is covered or nearly covered by the vertical
cross-hair. Clamp the upper plate and with the upper tan-
gent screw bring the line of sight exactly on the flag at C.
The arc of the graduated circle traversed by the zero point
of the vernier will be the measure of the angle BAG, equal
to 143 30'. The points A, B, and C are not necessarily in
the same horizontal plane, but the level plate of the instru-
ment projects them into the horizontal plane in which it

1241. A Deflected Line. A deflected line, or "angle
line," is a consecutive series of lines and angles. The direc-
tion of each line is referred to the line immediately pre-
ceding it, which preceding line is, in imagination, produced,
and the angle measured between it and the next line actually
run. The angles are recorded R f or L l , according as they
are turned to the right or left of the prolongation of the
immediately preceding line. An example of a deflected line



is shown in Fig. 2G8. Here the start-
ing point, A, of the line is a point in
the Jiead-block of the switch at Benton
\ Station, O. & P. R. R. The point A
is, of course, in the center line of the

Set up the transit at A with the
vernier at zero. Sight to a flag held
at F on the center line of the track,
O. & P. R. R. Loosen the vernier
clamp, and turn the telescope sight
to a flag held at B, the next point on
the angle line ; clamp the vernier, and,
by means of the tangent screw, ac-
curately sight to the flag held at B;
the angle reads 32 30', and is record-
ed R 1 32 30', with a sketch showing
the connection in which the term
3 head-block is designated by the abbre-
2 viation H. B. The bearing of the line
A B can not be taken at A on account
of the attraction of the rails. The
instrument is now moved to B, the
vernier set at zero and backsighted
to A ; the bearing of A B, N 75 00'
E, is taken, and the number of sta-
tion B, 2 + 90, together with the bear-
ing of A B, recorded. The telescope
5 is then reversed, pointing in the di-
! rection B B '. The point C being de-
5 termined, the upper clamp is loosened
| and the telescope turned to the right
and sighted to C. The reading of the
angle is found to be 14 30', and re-
corded R 1 14 30'. It measures the
angle B' B C. The bearing of the
line B C, N 89 20' E, is then re-
corded. The instrument is next set



up at C, the vernier set at zero, backsighted to B, and then
reversed ; the deflection to D, R' 10 00', is then read and
recorded, together with the number of the station at (7,
6 + 85. This deflection measures the angle C C D, and


Deflection. \ Mag. Bearing.

Ded. Bearing.



End of Line.


L'3000 f

N. 6925'E.

N. 6930'E.



R T 10OO'

S. 8030'E.





R T 143O'

N. 8920'E.

N.8930'E. \ l^b

JT R nf Switch



at Benton Sta.

gives the direction of the line CD. A good form of notes
for such a survey is that given above.

1 242. Checking Angles by the Needle. In spite of
the greatest care, errors in the reading and recording of
angles will occur. The best check to such errors is the
magnetic needle. And though it is not an exact check,
owing to the lack of precision in reading the needle and to
local attraction, yet it is the only reliable one, and in
universal use.

In Fig. 269, we have an example of the use of the needle



in checking angles. The bearing of the line A J5, which
corresponds to A B in Fig. 208, is N 75 00' E, and is
assumed to be correct. The bearing of the line B C, as read
from the needle, is N. 89 20' E. Its deduced or calcu-
lated bearing is obtained as follows: To the bearing of
the line A B, N 75 00' E, we add the R deflection 14 30';
the sum is 89 30', which is recorded in the column headed
Ded. Bearing. (See Art. 1241.) The deduced bearing,
it will be seen, is ten minutes greater than the magnetic
bearing as read from the needle and recorded in the column
headed Mag. Bearing. Had the deflection angle been re-
corded D instead of R, the deduced bearing would have
been the difference between 75 00' and 14 30', which is
60 30', and would be recorded N 60 30' E. The magnetic
bearing being N 89 20' E would have at once revealed the
error. The confusion of the directions R* and D is the com-
monest source of error in recording deflections, though
sometimes a mistake of ten degrees is made in reading the
vernier. It is a wise precaution to read both angle and
bearing after they are recorded and compare them with the
recorded readings.

1243. Simple Triangulation. Triangulation is

an application of the principles of trigonometry to the

measurement of in-
accessible lines and
angles. A common
occasion for the use
of trigonometry is
illustrated in Fig.
270, where the line
of survey crosses a
stream too wide and
deep for actual
measurement. Set
two points A and
B on line, one on


each side of the stream. Estimate roughly the distance
A B. Suppose the estimate is 425 feet. Set another point
C, making the distance A C equal to the estimated dis-
tance A B = 425 feet. Set the transit at A and measure
the angle B A C= say 79 00'. Next set up at the point
C and measure the angle A C B = say 50 20'. The
angle A B C is then determined by subtracting the sum
of the angles A and C from 180 ; thus, 79 00' + 56 20' =
135 20'. 180 00' 135 20'= 44 40'= the angle ABC.
We now have a side and three angles of a triangle given,
to find the other two sides A B and C B. These sides
may be easily found by the methods given for the solution
of triangles (see Arts. 759, etc.) by drawing a line from
the vertex of one of the angles A or C so as to divide
the triangle ABC into two right-angled triangles. A
simpler and easier method, however, is the following: In
higher works on trigonometry, it has been demonstrated
that, in any triangle, the sines of the angles are proportional to
the lengths of the sides opposite to them. In other words,
sin A : sin B:: B C : A C; or, sin A : sin Cl'.BC'.AB, and
sin B : sin C::A C : A B.

Hence, we have sin 44 40' : sin 50 20' ::425 : side A B.
Sin 50 20' = .83228;
.83228 X 425 = 353.719;
sin 44 40' = .70298;
353.719 ^.70298 = 503.17 ft. = side A B.

Adding this distance to 76 + 15, the station of the point
A, we have 81 + 18.17, the station at B.

Another and frequent occasion for the use of trigonome-
try is the following: Two tangents, A B and C D, Fig. 271,
which are to be united by a curve, meet at some inaccessi-
ible point E. Tangents (which will be more fully described
later on) are the straight portions of a line of railroad. The
angle C E F, which the tangents make with each other, and
the distances B E and C E are required. Two points A and
B of the tangent A B, and two points C and D of the tan-
gent C D, being carefully located, set the transit at B, and,



backsighting to A, measure the angle B C 21 45'; set
up at C, and backsighting to >, measure the angle
C B 21 25'. Measure the side B C = 304.2 ft.

Angle C E F being exterior to the triangle E B C is equal
(see Art. 1 194) to the sum of EB C and E C B =21 45'+
21 25' = 43 10'. The angle B E C= 180 - C E F =
136 50'.

From the principle stated we have sin 136 50' : sin
21 45' :: 304.2 ft. : side C E.

Sin 21 45' = .37056;

.37056 X 304.2 = 112.724352;

sin 136 50' = .68412;

side C E = 112. 724352 -f- .68412 = 164. 77 ft.


FIG. 272.

Again, we find B E by the following proportion :

Sin 136 50' : sin 21 25' :: 304.2 : side B E;

sin 21 25' = .36515;

.36515 X 304.2 = 111.07863;

sin 136 50' = .68412;

side B E = 111.07863 + .68412 = 162.36 ft.



A building H, Fig. 272, lies directly in the path of the
line A B which must be produced beyond H. Set a plug at
B and then turn an angle D B C = 60. Set a plug at C in
the line B C, at a suitable distance from /?, say 150 feet.
Set up at C, and turn an angle B C D = 60, and set a plug
at D, 150 ft. from C. The point D will be in the prolong-
ation of A B. Then, set up at D and backsighting to C,
turn the angle C D D' = 120. D D' will be the line re-
quired, and the distance B D will be 150 feet, since BCD
is an equilateral triangle.

A B and C D, Fig. 273, are tangents intersecting at some

inaccessible point H. The line A B crosses a dock O P, too


wide for direct measurement, and the wharf L M. F is a
point on the line A B at the wharf crossing. It is required
to find the distance B H and the angle F H G. At /?, an
angle of 103 30' is turned to the left and the point E set
217' from /> = to the estimated distance B F. Setting up
at E, the angle B E F is found to be 39 00'. Whence, we
find the angle B F E= 180 - (103 30' + 39) =37 30'.
From the above principle we have sin 37 30' : sin 39 00' ::
217 ft. : side B F.

Sin 39 00' = .62932;

.62932 X 217 = 136.56244;

sin 37 30' = .60876;

side B F = 136.56244 -~ .60876 = 224.33 ft.

Whence, we find the station of Flo be 20 + 17 -f 224.33 =
22 + 41.33. Set up at F and turn an angle H F G = 71 00',
and set up at a point G where the line C D prolonged inter-
sects/ 7 G. Measure the angle F G // = 57 50', and the side
F G =180.3'. The angle F H G =180 (71 + 57 50') =
51 10'. From the same principle as before we have sin
51 10' : sin 57 50' :: 180.3' : side F H.

Sin 57 50' = .84650;
.84650 X 180.3 = 152.62395;
sin 51 10' = .77897;

side FH= 152.62395 -4- .77897 = 195.93 ft. ;
whence, we find the station of //to be 24+ 37.26.

1244. Vertical Angles. A vertical angle is an

]F an S^ e formed by two intersecting
lines lying in the same vertical
plane, one of which is horizontal.
If the lines A B and A C, Fig.
274, lying in the vertical plane
D E F G, meet at the point A,

and the line A B is horizontal, the angle C A is a. vertical

angle, and is measured by the arc B C.

1245. Intersection of Tangents. Let A B and

C D, Fig. 275, be tangents whose point of intersection is to


be determined and the angle which they make with each other
to be measured. First set up a flag or stake at B and another
at A, or some other point in the line A B. Set up the tran-
sit at C, backsighting to D. Reverse the instrument.
Have a flagman hold a rod in the line C D, at the same time
putting himself in range with the stakes at A and B. With
a little practice he can nearly determine the intersection /
of the two lines. Then drive two stakes K and L firmly
in the line C D, one on each side of the point /. Their dis-
tance from the point / to be determined by the obtuseness

^'Angle of Intersection.

of the angle AID. Carefully center these stakes, driving
a tack half its length in each- center. Stretch a cord between
these tacks. Next set up the instrument at J3, backsighting
to A. Reversing the telescope, set a flag at /, which will
be the intersection of the line A B prolonged with L D.
Drive a stake flush with the ground at /and drive a tack in
this stake where the prolongation of A B crosses the cord
connecting the stakes at K and L. The point / is the inter-
section of the tangents A B and C D. The external angle
C I M, formed by the intersecting tangents, is called the
angle of intersection.


1246. A line of railroad consists of a series of straight
lines and curves. In general, the straight lines, or, more
properly, the tangents, are first located and then they are
united by curves best fitting the ground lying between the
tangents. There are certain limits of curvature prescribed
for all roads, which must not be exceeded. These limits



will depend upon conditions to be explained later. Rail-
road curves are circular and are divided into simple, com-
pound, and reverse curves.

A simple curve has but one radius, as A B in Fig. 276,
whose radius is A C.

A compound curve,' shown in Fig. 277, is a continuous



FIG. 276. FIG. 277.

curve of two or more arcs of different radii, as C D E F,
which is composed of the arcs CD, D E, and E F, whose
respective radii are G C, H D, and K E.

A reverse curve, Fig. 278, is a continuous curve com-
posed of two arcs L M and M N of the same or of different

radii described in the opposite
directions, and having a com-
mon point M, called the point
of reversal. Reverse curves,
though common in the early
days of railroad building in the
United States, are now con-
FIG - 278> demned for roads of standard

gauge, and only admitted for narrow-gauge roads, when
cheapness of construction is the first requirement.

1247. Geometry of the Circle. Before attempting
to lay out curves, a knowledge of geometry relating to the
circle must be mastered. The following propositions are of
special importance:

1. A tangent to a circle is perpendicular to the radius
drawn through its tangent point. Thus, A E, Fig. 279, is
perpendicular to B O, and C E is perpendicular to C O.



2. Two tangents drawn to a circle from any point are
equal, and if a chord be drawn joining these points, the
angles between the chord and the tangents are equal. Thus,
B E and C E are equal, and the angles EEC and E C B
are equal.

3. An acute angle between a tangent and a chord is
equal to half the central angle subtended by the same chord ;
thus, the angle EBC=ECB= one-half B O C.

4. An acute angle subtended by a chord, and having its
vertex in the circumference of a circle, is equal to half the
central angle subtended by the same chord. Thus, the
angle E B G, whose vertex B is in the circumference and
subtended by the chord B G, is equal to half the central
angle BOG, subtended by the same chord B G.

5. Equal chords subtend equal angles at the center of a
circle and also at the circumference, if the angles are


inscribed in similar segments. Thus, if B G, G H, H K, and
K C are equal, BOG=GOHzn&GBH=HBK.

6. The angle of intersection of two tangents equals the
central angle subtended by the chord uniting the tangent
points. Thus, the angle C E F= B O C.

1248. Deflection Angles. When two lines meet in
the same plane, they are said to form an angle, and the point
of meeting is called the angular point. The rate of diver-
gence or deflection of the two lines from their common or
angular point determines the size of the angle. The unit
of angular measurement is the degree, equal to j^ part of a
circle. Two lines forming an angle of one degree with each
other will, at a distance of one hundred feet from the angular
point, deflect or diverge 1.745 feet.

In Fig. 280, the lines A B and A C, meeting at the point
A, are supposed to form an angle of 1, and the angle B A C
is measured by the arc B C, described with the radius A B,


FIG. 280.

which is 100 feet in length. The arc B C and the straight
line joining the extremities of that arc, i. e., the chord B C,
are assumed to be of equal length.

1249. Degree of Curvature. The curve from
which, as a unit or basis, all other railroad curves are
deduced, is called a one-degree curve. It is the circum-
ference of a circle whose radius is 5,730 feet, or, more exactly,
5,729.65 feet, in length. Two radii forming an angle of one
degree at the center of a one-degree curve will subtend a
chord of 100 feet at its circumference. The arc subtended
by this chord of 100 feet is assumed to be of the same length
as the chord.

In Fig. 281, let A B and A C be radii 5,729.65 feet in
length, forming an angle of 1 at the center A ; then the arc
B C subtended by these radii will be 100 feet in length. The
curve B C is called a 1 curve. If, from the point O as a



center, with a radius O B equal to 2,864.93 feet, we describe
an arc B D 100 feet in length, the radii O B and O D will

form an angle of 2 at the center O, and the curve B D is
called a 2 curve. A curve whose radius is nearly one-third
A B, or 1,910.08 feet, is a 3 curve, etc.

The degree of a curve is determined by the central
angle, which is subtended by a chord of 100 feet. Thus, if
BOG (Fig. 282) is 10 and B G is 100 feet, B G H K C is
a 10 curve.

The deflection angle of a curve is the angle formed at
any point of the curve between a tangent and a chord of 100
feet. The deflection angle is, therefore, half the degree


of the curve. Thus, if the chord B G is 100 feet, the angle
E B G is the deflection angle of the curve B G H K C, and is
half the angle BOG.

EXAMPLE. Given the deflection angle EB G = D (Fig. 282), to find
the radius B O = R.

SOLUTION. Draw OL perpendicular to B G. In the right-angled

triangle B O L, we have sin OL = -^-\ but SOL = EBG = D,

since OL being perpendicular to the chord B G it bisects the arc
BLG. But the angle Z> = \ BOG\ hence, angle BOL D. B L =
50 feet and the radius B O = R. Substituting these values in the


given equation, we have sin D -^ ; whence, R sin D = 50, and we
have the formula

*-v < 89 ->

For curves of from 1 to 10, the radius may be found by dividing
5,730 ft. (the radius of a 1 curve) by the degree of the curve. The re-
sults obtained are sufficiently accurate for all practical purposes. For
sharp curves, i.e., for those exceeding 10, the above formula, viz.,


R . =; should be used, especially if the radii are to be used, as a
sin D

basis for further calculation.

For example, the radius of a 4 curve is found by both methods as
follows: By first method, R = 5,730 ft. -f- 4 = 1,432.5 ft. By second
method, we find the deflection angle D of a 4 curve is 2. Applying

Kf\ KA

the formula, 1? = ^ we have 7? = Q349 = 1,432.67 ft.

In this case the error is only .27 foot, and may be ignored in prac-
tical work. For a 30 curve we have by first method, R = ^^ = 191


K/\ Kf\

ft. By second method, we have R = . ,, = , QOO = 193.18 ft. In

Sin ID . AUOO*

this case the error is 2.18 ft., and the error increases as the degree of
curve increases.

The radii given in the table of Radii and Deflections are calculated

by the formula R = . =;.
sin D

125O. Sub-Chords for Curves of Short Radii.

On curves of short radii, i. e., curves of 20 and upwards,
center stakes are driven at intervals of 25 feet. In Art.
1248, we stated that the standard chord and arc are as-
sumed to be of the same length. This is practically true


for curves of large radii, but for curves above 20 the excess
of length of arc over the chord constantly increases. If,
now, in Fig. 282, the chord B C is 100 feet in length, the arc
B G H K C must be greater than 100 feet ; and if the arcs
B G, G H, H K, and K C are equal, i. e., each equal to one-
quarter the arc B H C, then the equal chords B G, G H,
H K, and K C subtending these equal arcs must each be
greater than one-quarter of B C, which we assumed to be
100 feet. These greater chords must, therefore, be greater
than 25 feet. Suppose the curve B H C to be a 20 curve,
and the chord B C, 100 feet ; then the central angle B O C
is 20. As the arc B G is one-quarter of the arc B H C, the

central angle B O G is = 5. The line O L, drawn to

the middle point of the chord B G, is perpendicular to B G
and bisects the angle BOG. The deflection angle E B G
B O L = G O L. Let C designate the chord B G, R, the
radius O B and Z>, the deflection angle, E B G = B O L.
In the right-angled triangle B O Z, we have sin B O L =

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