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Stations.


Bearings.


28 + 15


End of line.


23 + 55


S 45 00' E


18 + 92


S 70 45' E


14 + 20


N 80 30' E


10 + 40


SS1 20' E


6 + 90


N 83 30' E


3+75


N 60 00' E





N 10 15' E



750



MAPPING.



Sta. 14 -f 20. The length of the course is, therefore, the
difference between 14 -f 20 and 10 + 40, which is 380 ft.
Call Sta. 14 + 20, K. Through K draw the meridian K L.
The bearing here is N 80 30' E. From the meridian K L,
lay off this bearing and draw a line in the direction of the
course. In a similar manner locate the remaining stations
and lay off the remaining bearings of the line. The bearing
of each course should be distinctly written above it, the
letters reading in the same direction in which the line is
measured.

The notes for the lines b and c are as follows:

NOTES FOR LINE b.



Stations.


Bearings.


Distances.


1


N 40 E


4.22 chains


2


N 65 E


6.75 chains


3


S 75i E


8.70 chains


4


S 45i E


6.60 chains


5


S 20i W


5.18 chains



NOTES FOR LINE c.



Stations.


Bearings.


Distances.


1


S 47 E


6.60 chains


2


N 20 E


8.80 chains


3


S 80 E


4.32 chains


4


S 20 E


6.54 chains


5


N 65 E


7.48 chains



1354. The regular 100-foot stationing is used in rail-
road and highway surveying, but in land surveying the
lengths of the courses are given in surveyors' chains. As
the fractional parts of chains are given decimally, the






MAPPING. 751

length of each course is readily scaled on the plat with a
decimal scale. The notes of line b are platted as follows:
The starting point is called Sta. 1, and so marked on the
plat. Call Sta. 1, A. Through A draw a meridian A B,
and from it lay off the first bearing, N 40 E. The first
course is 4.22 chains in length, which lay off to a scale of
2 chains to the inch, locating Sta. 2, which call C. Through
C draw a meridian C D, and lay off the given bearing
N 65^ E. The course with this bearing is 6.75 chains in
length, which scale off, locating Sta. 3. In similar manner
plat the remainder of line b, and also line c. Mark dis-
tinctly each course, giving its direction and length, being
careful that the figures and letters shall read in the same
direction in which the line is being run.

1355. To Lay Off an Angle by its Tangent. In

laying off an angle by its tangent, the line from which the
angle is turned is prolonged to a distance equal to the
length of the assumed radius. The length of the tangent
of the given angle is then found in terms of the assumed
radius and the tangent platted. A line joining the angular
point with the extremity of the calculated tangent will give
the direction of the required line, which is then measured to
the given scale.

Let A B, in Fig. 330, be the given line, from which an
angle of 30 15' is to be laid off to the right at the point B.'
Produce A B to C, ma- B R=400' r

kmgC= 400 feet, the
length of the assumed
radius. The tangent
of 30 15' in terms of a
radius 1, is. 58318, which, FIG> m

multiplied by 400 feet, the length of the assumed radius,,
gives 233.27 feet, the length of the required tangent. At
C erect a perpendicular to B C 233.27 feet in length, equal
to the calculated tangent. Denote the end of this tangent
by C. Join B and C. The angle C B C' = 30 15', the
given angle, and the line B C' is the required line.




752



MAPPING.



The following notes which are platted in Plate, Title:
Platting Angles II, Fig. 2, the student will plat to a scale
of 200 feet to the inch :

NOTES FOR LINE a.



Stations.


Angles.


Bearings.


25 + 00


End of Line




19 + 97


L. 40 10'


N7845'E


13 + 22


R. 32 15'


SG105'E


5 + 00


R. 43 30'


N 80 40' E







N 43 10' E



The notes for line a are platted as follows: Having ad-
justed the paper to the drawing board and drawn a merid-
ian N S, fix the starting point A; which number 0. The
first course is 500 feet in length, which plat by drawing a
meridian A B through Sta. 0, and scale off 400 ft. equal to
the length of the assumed radius A C. The bearing of
the first course is N 43 10' E. The tangent of 43 10'
is .93797, which, multiplied by 400, the length of the radius,
gives 375.19, the length of the required tangent. Erect a
perpendicular to A B at C. and on this perpendicular scale
off to the right, the tangent 375.19 ft., calling the extremity
of the tangent D. Draw A D. The angle CAD will
be 43 10'. The first course is 500 feet in length, which
scale off on the line A D at 200 ft. to the inch, locating Sta.
5 + 00 at E. The angle at Sta. 5 + 00 is 43 30' to the right.
Produce A E and scale off the radius 7^=400 ft. The
tangent of 43 30' = .94896, which, multiplied by 400, gives
379.58 ft., the length of the required tangent. Erect a per-
pendicular to E F at /% and scale off, to the right, the tan-
gent 379.58 ft., locating the point G. Draw E G. The
angle F E G is 43 30', and the line E G the required line,
the bearing of which is N 80 40' E. Produce E G to //,
making E H= 1,322 - 500 -"822 ft. in length.



MAPPING.



753



The line changes direction again at Sta. 13 + 22, where
an angle of 32 15' is turned to the right. Denote Sta.
13 + 22 by H. Produce E H 400 feet, equal to the assumed
radius, calling its extremity K. The tangent of 32 15' =
.63095, which, multiplied by 400 feet, gives 252.38 feet, the
length of the required tangent. Erect a perpendicular to
H K at K and on that perpendicular scale off, to the right,
the tangent 252.38 feet, locating the point L. Join H
and L. The angle K H L is 32 15' and the bearing of H L
is S 61 05' E.

The line changes direction again at Sta. 19 -f- 97. Call
this station M. The angle at this point is 40 10' to the left.
Produce H M 400' to N, and at N erect a perpendicular
to M N. The tangent of 40 10' is. 84407, which, multiplied
by 400, gives 337.63 feet, the length of the required tangent.
On the perpendicular to M N scale off, to the left, this tan-
gent, locating the point O. Join J/and O. The end of the
line is Sta. 25 + 00. The length of the last course is readily
found by subtracting 19 + 97 from 25 -f 00. The difference,
503 feet, is scaled off on M O, locating the point P, the end
of the line. The bearing of M Pis N 78 45' E. In a
similar manner plat the notes of line b.

NOTES FOR LINE b.



Stations.


Angles.


Bearings.


27 + 47


End of Line




20 + 97


R. 42 20'


S 34 25' E '


13 + 73


R. 49 10'


S 76 45' E


7 + 63


L. 62 15'


N5405' E







S 63 40' E



1356. To Lay Off an Angle by Latitude and
Departure. The subject of latitudes and departures was
discussed in the section on Land Surveying, and the theory
needs no explanation in connection with this subject.



754



MAPPING.



Suppose the bearing of a line is N 40 E, and its length is 300
feet. Its latitude and departure are calculated as follows:



Distances.
3 ft.
00ft.
Oft.

ft.



Latitudes.
2298
0000
0000

229.8 ft.



Departures.
1928
0000
0000

4-192.8 ft.



The student should bear in mind that north latitudes and
east departures are-)-, and south
latitudes and west departures are .
Let A, in Fig. 331, be the station
at which the bearing is taken.
Through A draw the meridian N S.
From A upwards scale the calculated
latitude 229.8 ft., marking the ex-
tremity B. At B erect a perpen-
dicular to the meridian N S, draw-
ing it from left to right, as the
bearing is east. On this perpen-
dicular scale off the calculated de-
parture 192.8 ft., locating the point
C. Join A and C. The angle BAC

is 40, equal to the given bearing, and A C is equal to the

length of the given course, viz., 300 ft.

EXAMPLE. Calculate the latitudes and departures of the following
courses, and plat them by means of total latitudes and total departures
from Sta. 1.










Latitudes.


Departures.


Stations.


Bearings.


Distances.


N +


S -


E +


W -


1


N 10i E


250 ft.


246 ft.




44.5 ft.




2


N 41f E


123 ft.


91.76 ft.




81.92 ft.




3


N 84| E


215 ft.


20.64 ft.




214.03ft.




4


S 25i E


210 ft.




189.94 ft.


89.57 ft.





MAPPING.



755



SOLUTION.






Distances.


Latitudes.


Departures.


250ft.






2 ft.


1968


0356


50ft.


4920


0890


Oft.


0000


0000



250ft.

123ft.

1 ft.

20ft.

3ft.

1^3 ft.

215ft.

2 ft.
10ft.

5ft.

215ft.

2 1 ft.

200ft.

10ft.

Oft.

2 1 ft.



246.000 ft.



0746
1492

2238



91.758ft.



0192
0096
0479

20.639ft.



1809
0904
0000

189.940ft.



44.500ft.



1332 '
1998

81.918ft.



1991
0995
4977

214.027ft.



0853
0427
0000

89.570ft.



Stations.


Total Latitudes from
Station 1.


Total Departures from
Station 1.


1


0.00 ft.


0.00 ft.


2


+ 246.00 ft.


+ 44.50 ft.


3


+ 337. 76 ft.


+ 126.42 ft.


4


+ 358.40 ft.


+ 340.45 ft.


5


+ 168.46 ft.


+ 430. 02 ft.



The platting of the courses is as follows: On the merid-
ian N S, Fig. 332, take a point which call Sta. 1. The total
latitude of Sta. 2 is -f- 246 feet, and, as it is a plus latitude,
it must be scaled off on the meridian above Sta. 1, locating



756



MAPPING.



the point^. The total departure of Sta. 2 is +44.5 feet.
This departure will therefore be to the right of the merid-
ian N S. At A, erect a perpendicular to the meridian, and
upon it scale off the total latitude 44.5 feet, locating Sta. 2.
The line joining Stas. 1 and 2, i. e., the first course, will
have a bearing of N 10 E. Its length, viz., 250 feet, we
write on the plat, the figures reading in the same direction
in which the line is being run.



Total Lat.+358.40
Total Lat.+ 337.76 g



TotalLal.+246'



Total Lat+168.46




FIG. 332.

The total latitude of Sta. 3 is +337.76 feet, which we scale
off on the meridian above Sta. 1, locating the point B, and
on a perpendicular to the meridian at B, we scale off the total
departure of Sta. 3, which is + 126.42 feet, locating Sta. 3.
The line joining Stas. 2 and 3 has a bearing of N 41f E,
and length of 123 feet. The total latitude of Sta. 4 is
+ 358. 4 feet, which we scale off on the meridian above Sta. 1,
locating the point C, where we erect a perpendicular to the
meridian and upon it scale off the total departure of Sta. 4,



MAPPING. 757

viz., + 340.45 feet, locating Sta. 4. The line joining Stas. 3
and 4 will have a bearing of N 84^ E, and a length of 215 feet.
The total latitude of Sta. 5 is + 168.46 feet, which we scale
off on the meridian locating the point Z>, where we erect a
perpendicular to the meridian and upon it scale off the total
departure of Sta. 5, viz., +430.02 ft., locating that point.
The line joining Stas. 4 and 5 has a bearing of S 25^ E,
and a length of 210 ft. This method of platting bearings or
angles is more accurate than either of the foregoing methods,
as each course is platted independently. Great care must,
however, be observed in making the additions by which total
latitudes and departures are obtained. Tables of latitudes
and departures are commonly calculated to quarter degrees.
See table of Latitudes and Departures. Where angles are
read to single minutes, a table of sines and cosines may be
used to advantage. The two following formulas should be
memorized:

Latitude = distance X cos bearing. (97.)
Departure = distance X sin bearing. (98.)

1357. In preliminary railroad work, angles are com-
monly platted by tangents, but on difficult parts of the line
where all dependence must be placed on a paper location,
latitudes and departures should be used and the line platted
to a scale that will admit of full topographical details.

For practice in platting lines by latitudes and departures,
the following examples are given. The notes of Example 1
are platted in Fig. 333, and those of Example 2 are platted
in Fig. 334.

The student should carefully study the different steps
given under Art. 1356, and illustrated in Fig. 332, before
undertaking to plat the following notes. He should cal- '
culate the latitudes and departures for each course, compa-
ring his results with those given in the text and likewise with
the total latitudes and departures for platting. These plats
he will submit for inspection, together with Plate, Title:
Platting Angles II.



758



MAPPING.



Wj

5



OO50000O5i - ICOCOi-l

O**^HrHCOo6o6,-HrH

r-ICOCO-*GOrHt-O

<?? t- i I CO O5 - CO i t

O05coioaocoo^co
OidT-Hco'cdt^co^cd

Z>- CO CO CO CO GO O CO

CO-^-^i IGOCOr- ICO

+ + + I I I I I

GO GO i I i I

S ^ g
(TO CO CO CO

O5 O5 O rH
TJH O5 i> rH

1 I O O T t

; co co

O CO 00

CO CO i-l

. CO CO 1O

O5 CO CO : CO O

id id d

* GO

CO

COt-t-Or- IO1O1O ^

"^ id "^ co ?> id "^ id i_J



IlU



rt a

-*- 1 o



H* o r ^ % ^, c
os o TH ^r ^ jr w



GO rH CO c-j
CO C/2 b-



10 o t- co cs



MAPPING.



759



CD *-*

Q >



i I O O -*' Oi i I O I
i I I I I I rH |

COOOOOCOC500CO

:+ + + + + T T 7

00 O rH t- 1 **'

01 co i> o co

01 O CO 00 C5

O rH O

o co z^

CO 05 1C

*O rH C^ O Ci *

O *> O CO O
O O t- CO CO

rH

rH

I]

r* r rrj



is "g a c
11.81

H 3^l



CO "* 5 CO J> OO OS



760



MAPPING.



A piece of drawing paper one-half the size of an ordinary
drawing plate will be large enough to contain plats of both
lines. The total latitudes and departures in both examples
are reckoned from Station 1, which in Fig. 333 is the most
westerly station, and in Fig. 334 the most easterly one.
Plat the courses of Example 1 to a scale of 2 chains to the
inch, and those of Example 2 to a scale of 200 feet to the
inch.




FIG. 333.

In Fig. 333, a magnetic meridian is drawn through Sta. 1,
which we call A. We find in the column for total latitudes
and departures from Sta. 1 (see Example 1), that the total
latitude of Sta. 2 is+ 375.9 links, which we scale off on the
meridian above A to a scale of 2 chains or 200 links to the
inch, locating the point B. The total departure of Sta. 2 is
+ 214.9 links, which we scale off at 2 chains to the inch on a



MAPPING.



761



perpendicular to the meridian at B, locating Sta. 2, which we
call C. A line joining A and C will have the given length
and bearing of the first course, viz., length 4.33 chains, and
bearing N 29f E.

The total latitude of Sta. 3 is + 461.2 links, which we scale
upon the meridian above A at 2 chains or 200 links to the
inch, locating the point D. The total departure of Sta. 3 is
-f- 724.8 links, which we scale off on a perpendicular to the




meridian at D, locating Sta. 3, which we call E. A line
joining C and E will have the given length and bearing qf
the second course, viz., length 5.17 chains, and bearing
N 80 E. In a similar manner plat the remaining courses,
bearing in mind that positive latitudes are measured on the
meridian above Sta. 1, and positive departures on perpen-
diculars to the right of the meridian, while negative latitudes



762 MAPPING.

are measured on the meridian below Sta. 1, and negative
departures, if there were any, on perpendiculars to the left
of the meridian. The notes for Example 2 are similarly
platted, excepting that the meridian passes through the
most easterly station, as all the departures from Sta. 1 are
negative. The lengths of the courses in this example are
given in feet, and are to be platted to a scale of 200 feet to
the inch. Write the bearing of each line distinctly, being
careful that the letters read in the same direction in which
the line is run. The student is expected to accompany each
drawing with a brief description of the successive steps
taken in the work.

1358. Parallel Rulers. A parallel rule is a straight
edge carried on milled rollers of equal diameter, having a
common axis. They are of great service in drawing meridian
lines. A magnetic meridian is drawn the entire width of
the sheet which is to contain the plat. The straight edge
of the rule is then made to coincide with the meridian line
and then rolled across the paper until the straight edge
passes through the point where the angle is to be measured.
A line is then drawn following the straight edge; this will
be a meridian line.

1359. The Line of Survey. The line of prelim-
inary survey is a succession of straight lines and angles,
or an angle line, as it is commonly called, while the
located line is a succession of straight lines and curves.

1360. Tangents and Curves. Though this subject
has been considered in the section on Surveying, yet some
additional matter may be of advantage in connection with
the subject of mapping.

1361. Map of Final Location. In mapping a final
location the measurements should be made from inter-
section point to intersection point, and the angles platted
either by tangents or by latitudes and departures. The
points of curve are then located by scaling the tangent
distances from the intersection points. The curve centers



MAPPING.



763



1 25+50




764



MAPPING.



are best determined by describing intersecting arcs from
the tangent points as centers, with radii equal to that of the
given curve.

Let it be required to plat by tangents the following
location notes:



Stations.


Degree of
Curve.


Intersection
Angle.


Tangent.


Magnetic
Bearing.


25 + 50


p T






N 1 00' E


20+ 10
14 _[_ 55


P. C. 5 L.
P T


27 00'


275.20 ft.


N 28 00' E


10 + 80


P C 6 R


22 30'


190.03 ft




o








N 5 30' E













The tangent distances are found by the formula



(See Art. 1251.) The first curve is 6 R. ; the intersec-
tion angle / is 22 30'. The radius of a 6 curve is 955.37
feet. See table of Radii and Chord and Tangent Deflec-
tions.

| 7=11 15'. Tan 11 15' = .19891; then 955.37 X
.19891 ft. = 190.03 ft. = T, which we place in the column
headed "tangent," opposite the intersection angle 22 30'.
The second curve is 5 L; the intersection angle is 27 00'.
The radius of a 5 curve is 1,146.28 feet. 7-13 30'.
Tan 13 30' = .24008, and 1,146.28 ft. X .24008 = 275.2 ft. =
T, which we place in tangent column opposite the
intersection angle 27 00'.

A plat of these notes is given in Fig. 335. The order of
work is the following: First we select a starting point A,
which we number 0, and through this point draw a meridian
A B with its north point at the top of the plat.

The first course has a bearing N 5 30' E. From Sta-
tion 0, scale off on the meridian 600 feet, the length of our
radius for platting angles. The bearing angle is 5 30' and



MAPPING. 765

its tangent .09629, which, multiplied by 600, the radius,
gives 57.77 feet, the length of the required tangent. Call
the extremity of the radius C. At C erect a perpendicular
to A B, and on it lay off the tangent 57.77 feet, locating the
point D. Join A and D. The angle C .A D = 5 30'. The
P. C. of the first curve is at Station 10+80. The tangent
distance, as given in the preceding table, is 190.03 feet.
Hence, the distance from the starting point to the first in-
tersection point is the sum of 1,080 and 190.03 feet, which
is 1,270.03 feet. Produce A D, making a total distance of
1,270.03 feet to the point of intersection , and 600 feet
additional for the radius by which the next angle is platted.
Call the extremity of this radius F. The intersection angle
of the first curve is 22 30'. Its tangent is .41421, which,
multiplied by 600, the gfven radius, gives 248.52 feet as the
required tangent for platting the angle. At F erect a per-
pendicular to the radius E F and scale off the tangent
F G = 248.52 feet, locating the point G. Join E and G.
The angle F G is 22 30', and the bearing of the tangent
E G is N 28 E. Next, from the point of intersection E,
scale off on the lines E D and E G the tangent distance
190.03 feet, locating the P. C. at H, Station 10 + 80, and
the P. T. at K, Station 14+55. Now, from H and K as
centers with a radius 955.37 feet = radius of 6 curve, de-
scribe arcs intersecting at the point L. Then, from L as a
center with the same radius, describe a curve joining the
points H and K. The curve H K will be a 6 curve and
will be tangent to the lines H D and K G at the points of
curve H and K. The next intersection point M is in the line
E G produced. The distance between these intersection
points is made up of three parts, viz., the tangent of pre-
ceding curve, which we know to be 190.03 feet; the inter-
mediate tangent, i. e. , the distance from the P. T. of the
first curve to the P. C. of the second curve, and the tangent
of the next curve following. The P. T. of the first curve is
at Station 14+55; the P. C. of the second curve is at
Station 20 + 10; the intermediate tangent is, therefore, the
difference between 14 + 55 and 20 + 10, which is 555 feet.



766 MAPPING.

The tangent of the second curve is 275.2 feet. Hence,
the distance from the intersection point E of the first curve
to the intersection point M of the second curve is the sum
of 190.03, 555, and 275.2 ft., which is 1,020.23 ft. Produce
E G so as to contain 1,020.23 ft., and 600 additional feet for
a radius, the extremity of which call N. The intersection
angle of the second course is 27 00' L., tan 27 = .50953.
Radius 600ft. X .50953 = 305.72 ft., the length of the re-
quired tangent. Accordingly, at N we erect a perpendicular
to the radius M N, and. on that perpendicular, scale off the
tangent 305.72 feet, locating the point O. Join M and O.
The angle A" M O is 27 00', equal to the given intersection
angle, and the bearing of the tangent M O is N 1 E.
From M on the lines M K and M O scale off the tangent
distance 275.20 feet, locating the P. C. at P, Sta. 20+10,
and the P. T. at Q, Sta. 25 + 50. Then, from P and Q as
centers with radii of 1,146.28 feet, the radius of a 5 curve,
describe arcs intersecting at R. From R as a center with
the same radius describe the curve P Q, which is a 5 curve,
and is tangent to the lines M K and M O at P and Q.
Write the bearing of each tangent in its proper place, being
careful that the bearings shall read in the same direction in
which the line is being run.



PLATE, TITLE: MAP OF RAILROAD LOCATION.

1362. This plate contains two maps of railroad loca-
tion, Figs. 1 and 2, the notes for which are given in the fol-
lowing pages. All the angles are laid off by tangents and
the notes of the alinement given in detail, all of which the
student must carefully go over and check.

The student, before commencing these drawings, should
first note that the magnetic meridian (by means of which
the direction of the first tangent of each line is determined)
is parallel to the right and left border lines of the plate.
He must also determine by measurement from the border
lines the location of the starting point of each line



MAPPING. 767

Without these precautions, the lines are liable to run off the
paper, necessitating a repetition of the work, and involving
the erasure of lines, which always soils the paper and mars
the appearance of the drawing.

He will make the drawing to a scale of 300 feet to the
inch. If his scale reads only 200 feet to the inch, he will
reduce the distances given to a scale of 300 feet to the inch,
to their equivalent to a scale of 200 feet to the inch. The
process of reduction is simple and may be readily understood
from the following: A line which measures 300 feet in
length to a scale of 300 feet to the inch will measure but
200 feet to a scale of 200 feet to the inch. Hence, in chang-
ing a scale from 300 feet to 200 feet to the inch the distances
and dimensions will scale but f of the original distances and
dimensions.

EXAMPLE. A line measures 963 feet to a scale of 300 feet to the
inch. What will it measure to a scale of 200 feet to the inch ?

SOLUTION. of 963 = 642, i.e., to a scale of 200 feet to the inch, the
line will measure 642 feet.

1 363. The order of platting the notes is as follows : First
draw a meridian as indicated by the arrow. Next, having
located the starting point A, Fig. ], which is numbered 0,
draw through that station a parallel meridian A B. We find
from the notes that the direction of the back tangent A A'
(which we will consider a part of a line of railroad already con-
structed) is due north and south, and that Sta. is the P. C.
of an 8 R. curve with a central or intersection angle of
63 10'. The tangent distance we find by the formula
T = R tan % /, is 440.7 feet. This distance we scale off on
the meridian above the point A to a scale of 300 ft. to the
inch, locating the point C, which is the intersection point of
the back and forward tangents.

Next, from Con the same meridian, we scale off the radius



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