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spare time on their hands. They should provide themselves
with profile paper and keep the profile platted as the work

1411. Bench Marks. Bench marks are established
at intervals of from 1,000 to 1,500 feet, according as the line
is rough or smooth. At every stream which the line crosses
the elevation of the surface of the water and of the bed of
the stream should be taken. If there are any marks indica-
ting the elevation of high water, a rod reading should be
taken at such point and a record made of it.

1412. The Topographical Party. The topograph-
ical party follows the level party, though their rate of progress
will be more uncertain than either of the parties preceding


them. Where the slopes are uniform, they need not be
taken oftener than every third or fourth station. If, how-
ever, the country is broken and rough, it may be necessary
to take them at each hundred feet, and with great

1413. Office Work. At the conclusion of each day's
work, the field notes, both transit and level, are carefully
checked, and a plat of the line is made, either by tangents
or by latitudes and departures, carefully marking the cross-
ings of streams and highways, and noting any important
point which would enable the chief engineer to readily
locate any particular section of the line. Where the coun-
try is smooth the line may be platted to a scale of 400 feet
to the inch. Rough parts of the line may not exceed a
scale of 200 feet to the inch, and where difficult country is
encountered, involving detailed topographical maps, a scale
of 100 feet to the inch is advisable. Plat the line on sheets
24" X 30* in size, numbering them in regular order, each
sheet containing a part of the line on the immediately pre-
ceding sheet, so that by matching and pinning them
together, there may be had a continuous map of the line.
So arrange the line on the different sheets that when the
paper location is made they shall contain as many curve
centers as possible.

The topographer will do his proportional share of the
work, which will consist mainly in a detailed explanation of
the notes of the day's work to the draftsman, whose princi-
pal work is the contour maps. The leveler will plat the
day's levels on the continuous profile kept in the office, the
rodman reading the notes. This profile will contain as full
information as possible, especially relating to highways and

Some engineers prefer to wait for a rainy day in which to
do the office work, but more make it an invariable rule to
plat each night the work of the day. This practice enables
the chief of party to have a complete record of his work
always ready for the inspection of the chief engineer, who



is liable to appear at any time. If he is his own cJiicf, per-
sonal interest in the work would warrant him in making
such a rule. Notes which are platted when/res/i are always
of more value than when stale, and the daily office work
unites the different parties which are separated during the
day, sustaining a common interest in the work. If the con-
tour maps are to keep pace with the survey, the draftsman
must be an expert. Each day he plats the work of the pre-
ceding day, and under the direction of the topographer,
every point is covered.

1414. Spur Lines. At certain points of the main
line, two and sometimes three different routes will present


themselves for reaching another point of that line, and re-
quire the running of spur lines to determine the most advan-
tageous route. The main line being run, the spur lines are
tied into it, designating them by different letters, as line A,
line />', etc. The comparative advantage of the different
alinements will show themselves in the plat. Their com-
parative profiles are commonly shown by platting them in
different colors. A case requiring spur lines is shown in
Fig. 358. Here the general direction of the main line is
C D F, D being the point where the spur lines A and 8
diverge from the main line, and E where they again unite.
It will be evident from an inspection of the map that
the main line is superior to both the spur lines in point
of alinement. Their comparative lengths are already
known. With the comparative profiles before the engineer,
he, knowing the nature of the ground on the different lines,
will have no difficulty in making a judicious choice of lines.
Sometimes where the merits of the different lines are nicely
balanced, it becomes necessary to locate on two lines and
base a decision upon actual estimate of cost of construction.

1415. Gradients. The preliminary survey having
been completed, a careful study of the profiles will enable
the chief engineer to establish a gradient whose maximum
will limit the train loads passing over it.

The character of the expected traffic will greatly modify
this maximum. If the road is to do a passenger business
principally, the gradient may be raised, but if the bulk of
the business is to be freight, the gradient must be placed at
the lowest possible limit which the finances of the company
and the nature of the country will permit.

Should all the heavy grades occur on a short section of the
line, it may be policy to mass them within the smallest pos-
sible limits and proportionately reduce the gradient of the
remainder of the line. In such an arrangement of grades,
an assistant engine would be used on the summit section,
and so cover the entire line without any change of train



1416. Curvature. There is no absolute rule for
limiting curvature. The approximate limit will depend
upon the topography of the country and upon the character
of the expected traffic, a freight traffic admitting a higher
and a passenger traffic requiring a lower degree of curvature.
For all ordinary traffic conditions, i. e., where freight and
passenger traffic will be about equal, the invariable rule is
use suck curves as ?vi!t best conform to the existing
topographical conditions.

Any curve up to 10 degrees will be no obstacle to a speed
of 30 miles per hour, the average speed of passenger trains.
This affords a range in curvature which will meet the
requirements of most localities.

Curvature is no blemish to a line, if it secures the great
advantages of low gradients and moderate cost. At points

where moderate curves are possible only at great cost, it is
often a wise policy to build a temporary line, using sharp
curves, and put off the expensive work until the financial
strength of the company warrants its undertaking.


An instance where a temporary line is expedient is shown
in Fig. 359. Here the track follows the windings of a stream
in a narrow valley, whose sides are steep and rough. Unless
the company is financially strong, it will be much better
policy to build the line A B C D E, using curves as high as
15, and reducing cost to a minimum, than to build the line
A F E y giving a single curve of 6, but requiring a heavy
rock cut at F, or perhaps a tunnel at that point. The line
A F E is always possible, and when the road has built up a
paying traffic and finances are easy, the cut or tunnel at F
can be made without interfering in any way with traffic, and,
in all probability, at much better prices than when the
temporary line was built.

The question of gradients being decided, a preliminary
profile is made, which will serve as a basis for a paper

1417. A Paper Location. The paper location is
substantially the one which takes permanent form in the
final or located line. It is laid down on the contour maps,
which contain all the information accumulated by the pre-
ceding surveys. The grade for each station is taken from
the preliminary profile and marked on the contour maps
opposite the corresponding station. This is readily done, as
the contours are but five feet apart, -and intermediate eleva-
tions can easily be estimated. These grade points are com-
monly marked by small red dots enclosed in small circles of
the same color, and show where the plane of the grade
would cut the surface of the ground. A piece of fine thread
is then stretched, covering as many of these points as pos-
sible, and a pencil line drawn in place of the thread. This
pencil line will locate a tangent on the map. In the same
way any number of tangents may be located.

A set of office curves will be of great assistance in fitting
curves to the tangents, the curves like the tangents fol-
lowing the grade line as nearly as possible. Having
determined the degrees of curve uniting the tangents, the
intersection angles are calculated by tangents and the



tangent distances accu-
rately scaled from the
intersection points.

1418. Field Notes
from the Paper Lo-
cation. In taking notes
from the paper location
for actual field work, the
points of curve and points
of tangent should be care-
fully referred to fixed
points in the preliminary
line (either stakes or
plugs, the latter are pref-
erable) so that if, upon
the completion of any
curve, the following tan-
gent does not take the
position prescribed for it
in the notes, it may be
swung' into that position.
It is impossible, especial-
ly in a rough country, to
make the actual measure-
ments agree with the cal-
culated measurements,
hence small inaccuracies
need cause no concern.

The platting of a paper
location is illustrated in
Fig. 360. Here the grade
of the line is determined
by the grade of the stream,
which it closely follows.
The grade averages . 5 per
cent. The preliminary
line is shown dotted, and


the located line is drawn full. Let the grade for Sta. 1 be 11.0
feet. The grade for Sta. 2 will, therefore, be 11.0 feet plus. 5
foot, or 11.5 feet. The grade for Sta. 3 will be 11. 5 + .5,
or 12 feet. By the same process we find the grade for each
of the stations given in the plat. The grade for each station
is then marked on the contour map opposite the correspond-
ing station of the preliminary line by a small dot enclosed
in a circle. Straight lines A B and C D, which are to form
tangents in the paper location, are then drawn, covering as
many of these small circles as may be, and produce, until
they intersect at E. The line A E is then produced to F,
making E F = 300 feet or any other convenient length of
radius suitable for measuring the intersection angle by its
tangent. At F erect the perpendicular F G, which will
be the tangent of the intersection angle PEG. Meas-
uring F G by scale F G = 140 feet, though by calculation
139.89 feet. Dividing 140 by 300 we have a quotient of
.4667 = tan 25. We find by trying different curves that
an 8 curve will most nearly cover the grade points between
the tangents^ B and C D. From formula 91,7^= R tan /
(Art. 1251), we find the tangent distance = 158.9 feet.
Scaling from the intersection point E on both tangents this
distance we locate the P. C. and P. T. - The station of the
P. C. we determine by scaling from the P. T. of the last
curve. The station of the P. T. is, of course, found by

1419. Paper Location Profile. A profile, called a
paper location profile, is made from elevations taken
from the contour map at each station of the paper location,
and a grade line drawn on it which should be substantially
that of the final location, and if the preliminary work has
been thoroughly done, the discrepancy will be but slight. ,

1420. Actual Location. The location party has the
same organization as the preliminary party, excepting the
topographer and his assistants. Their work is supposed to
be completed. The chief of party carries, besides the notes
of the location, which is to be run in on the groutid, the map


covering the section immediately in hand, as not infre-
quently it is necessary to slightly modify the paper location.
He will need in addition a short scale and compass carrying
a pencil point.

Where the country is open, it is good practice to locate
the tangents by offsets from the preliminary line and make
the intersections on the ground ; but if the ground is covered
with brush or timber, the paper location must be strictly
followed, and the results will generally fulfil all reasonable


1421. Problems in Location. The tangents being
fixed in the paper location, the purpose is to so fix the point
of curve, the P. C., that, the curve being run, its tangent
shall coincide with the following tangent as laid down in the
paper location. Frequently, the actual tangent fails to
coincide with the theoretical tangent, in which case it must
be swung into place. Sometimes the tangents not only fail
to coincide, but form an angle with each other, in which
case the central angle of the curve must be either increased
or diminished, as the case demands. These modifications
of the paper location give rise to the following problems,
which will cover all ordinary cases:

1422. Problem I. To change the P. C. of a curve so
that the curve shall terminate in a tangent parallel to a
given tangent and at a given distance from it :

Let A B, Fig. 3G1, be a curve terminating in the tangent
B C, and it is required to change the P. C. of the curve
from A to A', so that it shall terminate in a tangent B' C'
parallel to B C and at a fixed distance from it.

The angle B B' D = 7, the angle of intersection of the

We have sin B B' D = ^, whence B B' = ~- r
B B sin /

B B' O O' A A', the required distance to move the
P. C. of the curve either backwards or forwards, according



as the required tangent is within or without the given

7? 7~)

Substituting A A' for B B' we have A A' = -r r -


In the figure the required tangent is within the given

tangent. Let the intersection angle be 68 and the distance
B D = 40 feet. Sin 68 = .92718, whence A A' -^^U =

43.14 feet.

That is, the P. C. of the required curve must be moved
backwards 43.14 feet from the P. C. of the given curve.

1 423. Problem II. To change the point of compound
curvature, the P. C. C., so that the second curve shall ter-
minate in a tangent parallel to a given tangent, and at a
given distance from it.

Case I. When the second curve is of shorter radius than
the first curve:

Let A B D, Fig. 362, be a compound curve terminating


in the tangent D H, and let it be required to change the

P. C. C. from B to some
point E, so that the curve
shall terminate (as shown in
the figure) in a tangent FG,
parallel to and at a given
distance H G from D H. To
determine the point E (the
new P. C. C. ) the angle ELF
must be determined, and sub-
stituted for the known angle
B K D. From C, the center

MX w of the larger curve, let fall

upon F G the perpendicular

CG. From K and L let fall upon C G the perpendiculars K M
and L N. Call the longer radius C B, R\ the shorter radius
K D, r; the distance I F or its equal H G, D; the angle
B K D or its equal K C M, ;r, and the required angle ELF
or its equal L C N, y. Then,

_ (R r) cos .r + D

05 y ~ R-r

That is, the distance I F or H G measured rectangularly

between the two tangents, being added to the difference of the

radii times cos .r, and the result divided by the difference of

the radii, will give the cosine of the angle ELF, or y, to be

turned on the smaller curve.

Subtracting the angle y from the angles, will give the
angle B C E, to be added to the larger curve; and dividing
this angle by the degree of curvature in A B we find the
distance from B to E, the required P. C. C.

If E F be the second curve located, and the required
tangent lies within, i. e., D .//is the required tangent, it is
evident that instead of advancing on the curved B, we must
retreat on it to find the required P. C. C. Accordingly,
we subtract D instead of adding, as in the preceding case.

EXAMPLE. .4 B, Fig. 362, is a 4 curve to the right, located -and
compounding at B into a 7 curve, the latter being continued through
an angle of 38. At the P. T. we find that the proper tangent is 46 ft.


to the left, i.e., without the actual tangent, so that the curve will be
thrown out to meet the required tangent. How far must the 4 curve
be continued f

SOLUTION. The radius of a 4 curve is 1,432.69 ft. ; the radius of a
7 curve is 819.02 ft., hence, R- r = 613.67 ft. The cosine of 38 =
.78801. Substituting known quantities in formula 99,
(V?-r)cos.r+ D 613.67 x .78801 + 46

R-r 613.67

Hence, angle y = 30 21'. Subtracting this angle from 38 00', there
remains a difference of 7 39', which must be added to the 4 curve.
7 39' reduced to decimal form is 7.65 C -=- 4 gives 1.9125 stations =
191.25 feet, which must be- added to the 4 curve to reach the correct
P. C. C. Ans.

In the above example, it is evident that, had the required
tangent been within the given tangent, it would have been
necessary to move the P. C. C. backwards instead of advan-
cing it. This will increase the angle y of the second curve,
and, consequently, its cosine will be reduced. The distance
D will, therefore, be subtracted and the formula will read,

_ (R - r} cos x - D

1424. Problem II. Case II. When the second curve
is of longer radius than the first curve :

Let A B F, in Fig. 363, be a compound curve terminating
in the tangent F H, and it is required to change the P. C. C.
from B to some point E so that the curve shall terminate in
the tangent D G parallel to F H and at a given distance
H G from it.

Let the required tangent D G be without the given
tangent F H. Calling the perpendicular distance H G
between the tangents D\ the radius of the larger curve, R,
and the radius of the smaller curve, r; the given angle
B O F of the second curve, x, and the required angle E C G
of the second curve 7, we have, as in formula 1OO,

(R-r) cos x D

~ - -

That is, the distance H G, measured rectangularly between
the two tangents, should be subtracted as in formula 1OO.



It will be seen that the required tangent is without the
given tangent, consequently, it will be necessary to move







c* ^

- r




l tf

the P. C. C. backwards on the first curve, i. e., the angle of
the first curve must be reduced.

EXAMPLE. A B is a 6 curve compounding at B into a 3 curve
whose angle is 42 30', and whose tangent F H is 52 ft. within the
required tangent. How far backwards must the P. C. C. be moved ?

SOLUTION. The radius of a 6 curve is 955.37 ft., and the radius of
a 3 curve is 1,910.08 ft.; hence, 7? - r = 1,910.08 - 955.37 = 954.71.
Cos x cos 42 30' = . 73728. Substituting the known values in formula
954.71 X. 73728 -52

1OO, we have cos y =

whence, we find


y 46 56'. Deducting 42 30' from 46 56', we have a difference of
4 26', which must be deducted from the first curve. 4 26' in decimal
form is 4.433; 4.433 -H 6, the degree of the first curve, gives a quotient
of .739 of a full station = 73.9 ft., the distance backwards from B to the
correct P. C. C. at E. Ans.

If the required tangent were within the given tangent,
the P. C. C. would be advanced and the angle y would be



reduced. The distance D would then be added and the
formula would read

_ (R r) cos x+D
R-r '

cos y =

1425. Problem III. To avoid obstacles on a curve:
Let it be required to run a curve A DEC between the
points A and C, and suppose an obstacle lies directly in the
path of the curve. The obstacle may be avoided by tracing
a parallel curve FG HI, and from the stations on this par-
allel curve, the corresponding stations on the required


FIG. 364.

curve may be located. The process is as follows (see
Fig. 3G4):

Having determined either P. C. or P. T., erect a per-
pendicular A F to the tangent A K. Now, in the curve
FGH I it is evident that while the angle A O C remains
constant, the chords F G, G H, and HI shorten as we
approach the center O of the required curve. L,etAF=z


90 ft., and the radius A (9 = 819 ft. The chords of the
required curve being 100 ft. in length, we have the follow-
ing proportion: O A : O A A F::WO : FG, the length of
the chord of the parallel curve.

Substituting known values in the proportion, we have
819 : 729:: 100 : FG; whence, F G = 89.01ft. Set up the in-
strument at F and trace the curve F ' G H I, setting a transit
point at each station of 89.01 ft. Then set the transit at
each of these points, as at G, and turn to a tangent of the
curve as run. Then, turning a right angle, set a stake 90
ft. from 6", locating the point D. In a similar manner
locate each of the stations upon the required curve.

1426. Problem IV. Having given two angles of
intersection D B E and G F H, and the distance B F between
the points of intersection (Fig. 365), it is required to find
the radius of the easiest reverse curve which will unite

the tangents A D and F K. The angle DDE is equal to
the angle A O E, half of which is B O E. The angle G FH
is equal to the angle E CG, half of which is E C F. Then,
(tan O E+ tan E C F) : tan B O E :: BF : BE. But
E F '= B F BE. Reducing the proportion, we have

tan B O E X BF
"tan BO E + tan E C F'

Now, B E is the tangent distance T of the curve A E, and


substituting known values in formula 91, T=R tan \I
(Art. 1 251 ), we have B E = O E tan B O ; whence,


radius O E = - ., and radius C h = - ^ ^ .
tan B O E' t&nECF

EXAMPLE. Let the angle D B E, Fig. 365, = 40 00', the angle
G FH= 62 30', and the distance ^9 /^= 922 ft. Find the radius of
the reverse curve.

SOLUTION. The angle BOE=\D B E 20, and the angle ECF=
iGFtf=3l 15'. Tan 20 = .36397; tan 31 15' = .60681. The sum of
these tangents is .97078, and we have the proportion .97078 : .36397::
922 : B E; whence, we find B E= 345.68 ft. Substituting the value
of B E in the formula, T = R tan $ /, we have 345.68 = O E X .

whence, we have radius O E = = 949. 75 ft. Substituting this


value of R in the formula, R = -. ^ (Art. 1 249), we have sin D =


,, , whence, sin D = .05264 and D = 3 01', which, multiplied by 2,
949. ^

gives 6 02', the required degree of the curve A E. To show the
student that the curve E G is of the same degree as the curve A E,
we complete the calculation as follows : B F= 922 ft. and EF - 922 -
345.68 = 576.32 ft. Substituting the value of ^.Fin the formula T =

R tan 4 /, we have radius CE= ^jj%- = 949. 75 ft. Ans.

Substituting this value of R in the formula, R = - jy we have
sin D = gj^g-, whence sin D = .05264, and D = 3 01' ; this, multiplied

by 2, gives 6 02', the required degree of the curve EG, which is the
same as that of the curve A E. Ans.

1427. Problem V. To find the radius of a curve
which will be tangent to three given straight lines: Let
A B, B C, and C D, in Fig. 366, be the given lines, then the
required radius will be equal to

_ BC_ _ ._
tan E B C + tan E C B

EXAMPLE. Let ^C = 428 feet, the angle EB C= 23 20', and the
angle ECB = 25 20'. Find the radius of the curve that will be tan-
gent to A B, B C, and CD.

SOLUTION. Tan $ EB C= .20648, and tan | E C B = . 22475. The
sum of these tangents is .43123. Substituting these values in the


equation, radius = tan ^. c + tan ^ cv/ we have radius =
whence the required radius = 992.51 ft. Substituting this value of R


in formula 89, Art. 1249, we have sin Z> = j^ =. 05038, and

D = 2 53.3', which, multiplied by 2, gives 5 46.6', the required degree

of the curve. Ans. The required degree of curve may be found by
the following and simpler operation, viz.: Dividing 5,730 ft., the
approximate radius of a 1 curve, by the given radius, 992.51 ft., we
obtain a quotient of 5.773 = 5 46.38', a result amply close for practical
work. The angle of intersection C E F is equal \.oEBC+ECB =
48 40'. Having found the radius and angle of intersection, the tan-
gent distance is calculated by formula 91, T R tan /. (See Art.

1428. Problem VI. To swing a tangent so that it
will pass through a given point:

Let A B, in Fig. 367, be a curve whose tangent B X passes
through the point C, and it is required to swing the tangent
B X into the position B' X' , so that it shall pass through
the point C'. With the instrument at B measure the angle
C B C' . Divide this angle by the degree of the curve A B.

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