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Font size C D of 400 feet for laying off the angle of the first curve.
The angle of this curve is 63 10'. The tan of 63 10' is
1.97081. The radius 400 ft. X 1.97G81 = 790.7 ft., the length
of the required tangent. At D erect a perpendicular to

768

MAPPING.
NOTES FOR FIG. 1.

Station.

Deflection.

Total

Angle.

Magnetic
Course.

Calculated

Course.

40

12 00'

39

6 00'

38 + 00

18 00'

36 00'

37

12 00'

36

6 00'

35 + 00

P. C. 12 L.

34

33 + 04.9
33

10 40. 3' P.T.
10 30'

35 50'

S 36 30' E

S 36 40' E

32

7 00'

31

3 30'

30

7 14.7'

14 29.4'

29

3 44.7'

28

14 7'

27 + 93

P C 7 R

24

21

20 + 54.9
20

10 38.8' P. T.
9 00'

44 20'

S 72 30' E

S 72 D 30' E

19

18

6 00'
3 00'

17

11 31 2'

23 02 4'

16

8 31 2'

15

5 31 2'

14

2 31 2'

13 + 16

P. C. 6 R.

12

11

10

9

8

7 + 89.6

7

15 35' P. T.
12 00'

63 10'

N 63 00' E

N 63 10' E

6

8 00'

5

4 00'

4

16 00'

32 00'

3

12 00'

2

8 00'

1

4 00'

P. C. 8 R.

North

North

MAPPING.
NOTES FOR FIG. 1.

769

Remarks. June 28, 1894.

Int. Ang. =

72 00'

From intersection to intersection.

12 curve, L. R. =

478.34 ft.

Tan preceding curve = 264.8 ft.

T. =

347.5 ft.

Tan between curves = 195.1 ft.

P. C. =

35 + 00

f>{\,\ f4.

Tan 12 curve = 347.6 ft.

Length curve =
P. C. C. =

bOU tt.
41 + 00

Total, = 807.5 ft.

Def. 100 ft. =

6 00'

tan 72 00' = 3. 07768

Def. 1 ft. =

3.6'

400ft. X 3. 07768 = 1,231. 1ft.

Int. Ang. =

35 50'

From intersection to intersection.

7 curve, R. R. =

819.02 ft.

Tan preceding curve = 389.2 ft.

T. =

264.8 ft.

Tan between curves = 738.1 ft.

P. C. =

27 + 93

K-t 1 Q f -t-

Tan 7 curve = 264.8 ft.

Length curve
P. T. =

oii. y it.
33 + 04.9

Total, =1,392. 1ft.

Def. 100 ft. =

3 30'

tan 35 50' = .72211

Def. 1 ft. =

2.1'

400 ft. X .72211 = 288.8 ft.

Int. Ang. =

44 20'

From intersection to intersection.

6 curve, R. R. =

955.37 ft.

Tan preceding curve = 440.7ft.

T. =

389.2 ft.

Tan between curves = 526.4 ft.

P. C. =

13 + 16

Tan 6 curve = 389.2ft.

Length of curve =

738.9 ft.

Total, =1,356. 3 ft.

P. T. =

20 + 54.9

tan 44 20' = .977

Def. 100 ft. =

3 00'

400ft. X. 977= 390.8ft.

Def. 1 ft. =

1.8'

Int. Ang. =

63 10'

8 curve, R. R. =

716.78 ft.

tan 63" 10' = 1.97681

T. =

440.7 ft.

400ft. X 1.97681 = 790. 7 ft..

P. C. =

Length of curve =

789.6 ft.

P. T. =

7 + 89.6

Def. 100 ft. =

4 00'

Def. 1 ft. =

2.4'

770

MAPPING.
NOTES FOR FIG. 2.

Station.

Deflection.

Total
Angle.

Magnetic
Course.

Calculated
Course.

13 + 41.7
13

10 15' P. T.

9 00'

32 30'

S 79 00' E

S 79 00' E

12
11

6 00'
3 00'

10 + 00

6 00'

12 00'

9

3 00'

8 + 00

P. C. 6 R.

5

3

S 46 30' E

NOTES FOR FIG. 1 Continued.

Station.

Deflection.

Total
Angle.

Magnetic
Course.

Calculated
Course.

69 + 10.1
61 + 65. 1
61

15 40' P. T.
12 44 1'

End
31 20'

of Line.
N 39 45' E

N 39 40' E

60

8 14 1'

59

3 44.1'

58 + 17

P. C. 9 R.

55

54

53

52

51

50 + 00
49

17 30' P. T.
14 00'

63 00'

N 8 15' E

N 8 20' E

48

10 30'

47

7 00'

46
45

3 30'

14 00'

28 00'

44

10 30'

43

7 00'

42

3 30'

41 + 00

18 00' P. C. C. T L.

72 00'

N 71" 15' E

N 71 20' E

MAPPING.
NOTES FOR FIG. 2.

771

Remarks.

June 28, 1894.

Int. Ang. = 32 30'
6 curve, L. R. = 955.37 ft.

T. = 278.5 ft.
P. C. = 8 + 00
Length curve = 541.7 ft.

P. T. =13 + 41.7
Def. 100 ft. = 3 00'
Def. 1 ft. = 1.8'

From Sta. to P. C. = 800.0 ft.
Tan 6 curve = 278.5ft.

Total from P. C. to P. I. = 1,078.5 ft.

tan 32 30' = .63707
400ft. x. 63707= 254.8ft.

NOTES FOR FIG. I Continued.

June 28, 1894.

Int. Ang. = 31 20'
curve, R. R. = 637.27 ft.

T. = 178.7 ft.
P. C. = 58 + 17
Length curve = 348.1 ft.
P. T. = 61 + 65.1
Def. 100 ft. = 4 30'
Def. 1 ft. = 2.7'

Int. Ang. = 63 00'
7 curve, L. R. = 819.02 ft.

T. = 501.9 ft.
P. C. C. = 41 + 00
Length curve = 900 ft.
P. T. = 50 + 00

From intersection to intersection.
Tan preceding curve = 501.9 ft.
Tan between curves = 817.0 ft.
Tan 9 curve = 178.7ft.

Total, = 1,497.6 ft.

tan 31 20' =.60881
400ft. X. 60881 = 243.5ft.

From intersection to intersection.
Tan preceding curve = 347.6 ft.
Tan between curves = 0.0 ft,.
Tan 7 curve = 501.9 ft.

Total,

= 849.5ft.
tan 63 = 1.96261
400 ft. X 1.96261 = 785 ft.

772 MAPPING.

NOTES FOR FIG. 2Contimied.

Station.

Deflection.

Total
Angle.

Magnetic
Course.

Calculated
Course.

57 + 40
47 + 19
47
46
45

8 46.3' P'. T.
8 15'
5 30'

2 45'

End
34 00'

of Line.
N 11 15' E

N 11 20' E

44 + 00

8 13.7'

16 27.4'

43

5 28.7'

42

2 43.7'

41 + 00.8
-. 41

14 01.7'P.C.C. 5 30' L.
14 -00'

52 00'

N 45 15' E

N 45 20' E

40

10 30'

39

7 00'

38

3 30'

37 + 00

11 58.2'

23 56.4'

36

8 28.2'

35

4 58.2'

34

1 28.2'

33 + 58

P. C. 7 L.

32

30 + 36.6
30

13 C 27.8' P. T.
12 00'

48 10'

S 82 30' E

S 82 40' E

29

28

8 00'
4 00'

27 + 00

10 37.2'

21 14 4'

26

6 37.2'

25

2 37.2'

24 + 34.5

P. C. 8 L.

24

23

22 + 14.4
22

9 39 P. T.

9 00'

44 30'

S 34 20' E

S 34 30 E

21

4 30'

20 + 00

12 36'

25 12'

19

8 06'

18

3 36'

17 + 20

P. C. 9 L.

17

15

MAPPING. 773

NOTES FOR FIG. 2 Continued.

Remarks. June 28, 1894.

Int. Ang. = 34 00'

From intersection to intersection.

5\$Q',L. R. =1,042. 14 ft.

Tan preceding curve = 399.5 ft.

T. =318.6 ft.

Tan between curves = 0.0ft.

P. C. C. = 41 + 00.8

Tan 5 30' curve = 318.6 ft.

Length curve = 618. 2 ft.

Total, = 718. 1 ft.

P. T. =47 + 19

tan 34 00' = .67451

Def. 100 ft. = 2 45'

400 ft. X .67451 = 269.8 ft.

Djjf. 1 ft. =1.65'

Int. Ang. = 52 00'

From intersection to intersection.

7 curve, L. R. = 819.02 ft.

Tan preceding curve = 320.4 ft.

T. = 399.5 ft.

Tan between curves = 321.4 ft.

P. C. = 33 + 58

Tan 7 curve = 399.5 ft.

Length curve 742.8 ft.
P. C. C. = 41 + 00.8

Total, =1,041. 3 ft.

Def. 100 ft. = 3 30'

tan 52 00' = 1.27994

Def. 1 tt. =2.1'

400 ft. X 1.27994 = 512 ft.

Int. Ang. = 48 10'

From intersection to intersection.

8 curve, L. R. = 716.78 ft.

Tan preceding curve = 260.7 ft.

T. = 320.4 ft.

Tan between curves = 220. 1 ft.

P. C. =24 + 34.5

Tan 8 curve = 320.4 ft.

Length curve = 602.1 ft.
P. T. = 30 + 36.6

Total, =801. 2 ft.

Def. 100 ft. = 4 00'

tan 48 10' = 1.11713

Def. 1 ft. = 2.4'

400 ft. X 1.11713 = 446.8 ft.

Int. Ang. = 44 30'

From intersection to intersection.

9 curve, R. R. = 637.27 ft.

Tan preceding curve = 278.5 ft.

T. = 260.7 ft.

Tan between curves = 378.3 ft.

P. C. = 17 + 20

Tan 9 curve = 260.7 ft.

Length curve = 494.4 ft.
P. T. = 22 + 14.4

Total, =917. 5 ft.

Def. 100 ft. = 4 30'

tan 44 30' = .98270

Def. 1 ft. = 2.7'

400 ft. X- 98270 = 393.1 ft.

774 MAPPING.

A B, towards the right, as the curve is to the right, and
upon this perpendicular scale off the calculated tangent
790.7 ft., locating the point E. A line joining the points
C and E will give the direction of the forward tangent. On
the line C E, scale off from C the tangent distance, 440.7 ft.,
locating the point F, which -is the P. T. of the first curve.
From ^4 and Fas centers, with radii of 716.78 ft., the radius
of an 8 curve, describe arcs intersecting at G. Then, from
G as a center, with the same radius, describe a curve joining
the points A and F. The curve A F is an 8 curve and tan-
gent to the lines A A' and F E at the points A and F.

We find from the notes that the next curve is 6 R. Its
P. C. is at Sta. 13 + 16, and its central angle is 44 20'. We
find its tangent distance is 389.2 ft. We next calculate the
distance from the intersection point of the first curve to the
intersection point of the second curve. The distance is com-
posed of three parts; viz., the tangent of the preceding
curve, which is 440.7 ft. ; the intermediate tangent, i. e.,
from the P. T. of the preceding curve at Sta. 7 + 89.6 to the
P. C. of the second or 6 curve at Sta. 13 -f- 16, a distance of
526.4 ft., and the tangent of the 6 curve, which is 389.2 ft.,
making a total distance of 1,356.3 ft. Produce the line CE,
and scale off from C on that line a total distance of 1,356. 3 ft. ,
locating the point //", which is the intersection point of the
second or 6 curve. Produce C H 400 ft. to K for a radius
in laying off the central angle, 44 20' R., of the second
curve. The tangent of 44 20' is .977, which, multiplied by
400, gives 390.8 ft. At K erect a perpendicular to H K, and
upon it scale off the tangent 390.8 ft., locating the point L.
The line joining H and L gives the direction of the forward
tangent of the second curve. Next, from the intersection
point H, scale off on both back and forward tangents the
tangent distance 389.2 ft., locating the P. C. of the second
curve at M, Sta. 13 + 16, and its P. T. at N, Sta. 20 + 54.9.
Next, from M and A 7 " as centers, with a radius of 955.37 ft.,
the radius of a 6 curve, describe arcs intersecting at O.
Then, from O as a center, with the same radius, describe a
curve joining the points M and N. The curve M N is, a 6

MAPPING. 775

curve and tangent to the lines F H and H L at the points of
curve M and N.

The student w.ill draw the tangent distances and the radii
and tangents for laying off angles in dotted lines, as they
are simply construction lines. The line of survey he will
draw in a full, bold line, as shown in the plate. The inter-
section points and the points of curve and tangent are
marked by small circles, the latter being more fully described
by their station numbers. Dotted radial lines are drawn
from the center of each curve to its P. C. and P. T. On
one of these radial lines the length of the radius of the curve
is written, and the amount of the central angle written with-
in the radial lines. The student will need no further direc-
tions to enable him to plat the balance of the line and also
the notes for Example 2, a plat of which is given in Fig. 2.

1364. Office Curves and Beam Compass. Office
curves are curves of different radii, whose principal object

FIG. 836.

is to enable the engineer to readily select a curve which
shall best fit the ground lying between tangents, as shown
in the topographical map. They are commonly made of
pasteboard, each piece containing arcs of two different radii,
the degrees of curvature of which, together with the scale of
each, being distinctly written, as shown in Fig. 336. A l6
curve to a scale of 100 feet to the inch will serve for a 5
curve to a scale of 200 feet to the inch, or a 2 30' curve to
a scale of 400 feet to the inch. In the same way, a 12
curve to a scale of 100 feet to the inch will serve for a 6
curve to a scale of 200 feet to the inch, or a 3 curve to a

776 MAPPING.

scale of 400 feet to the inch. Office curves are applied
directly to the contour map upon which a grade line has
been platted, and the curves fitted to ground and tangent.
Compound curves are as readily fitted as simple curves. A
satisfactory line being decided upon, the tangent distances
are calculated and the curves struck with a compass.

When the radius is of considerable length, it is difficult to
describe a true circle with the ordinary compass and length-
ening bar.

An accurate substitute is found in the beam compass,
which consists of two upright legs; one pointed and fixed
at the center of the circle; the other leg carrying either
pencil or pen, with which the circle is described. Both legs

FIG. asr.

are clamped to a horizontal arm called a beam, which is
lengthened or shortened to suit the radius. A cut of a beam
compass is shown in Fig. 337. A B is the beam, to which is
clamped the needle point at C and the pen or pencil at D.
At E is a milled-headed screw, which gives slow movement
to the pen or pencil at D and adjusts it to the required

TOPOGRAPHICAL DRAWING.

1365. General Definition. Topographical
drawing consists of the representation of the different
features of any portion of the earth's surface. The different
features will comprise all its inequalities of surface, such
as hills, hollows, streams, lakes, valleys, and plains; the
location of towns, highways, canals, and railroads. Detailed
topographical maps give individual dwellings, boundaries of
fields, their owners, the charactei of the vegetation, etc.

MAPPING.

777

1366. Systems. There are three principal systems of
representing topographical features, viz. :

1. By level contours or horizontal sections.

2. By lines of greatest slope perpendicular to contours.

3. By shades from vertical light.

1367. Ridge Lines and Valley Lines. Ridge
lines are the lines which divide the water falling upon them
and from which it passes off on opposite sides. They are
the lines of least slope when looking along them from above
downwards, and the lines of greatest slope when looking
along them from below upwards. They can be readily
determined by the slope level. On these lines are found the
projecting or protruding bends of the contour lines.

Valley lines are the reverse of ridge lines. They are
indicated by the water courses which follow or occupy them.
They are the lines of greatest slope when looked at from
above and of least slope when looked at from below. On
these lines are found all the receding or reentering points
of the contour lines.

1368. Forms of Ground. It will be found from a
general examination of any surface that ground exists under
one of the five following conditions, viz. :

1. Sloping down on all sides, i. e. , a hill, as shown in Fig.

FIG

338, the direction in which water would flow being indicated
by the arrows.

778

MAPPING.

2. Sloping up on all sides, i. e., a hollow, as shown in
Fig. 339.

3. Sloping down on three sides, i. e., shoulder or prom-
ontory; the end of a ridge or watershed line, as shown in
Fig. 340.

FIG. 340. PIG. 341.

4. Sloping up on three sides and down on one side, i. e. , a
valley, as shown in Fig. 341.

5. Sloping up on two sides and down on two sides,

shown in Fig. 342.

1369. Clear and Intelligi-
ble Maps. No pains should be
spared in making maps clear and
intelligible. All water courses,
whether occupied or dry, should
be accurately sketched, and gaps
should be left in the contour lines
at suitable intervals, with the eleva-
tion of the contours written in them. Where time and cost
are not to be considered, the lower sides of the contours may
be hatched as though water were draining off them, and the
valleys and low places tinted with a light shade of India ink.
Sometimes the spaces between the contour lines are tinted
with India ink, increasing the tint as the depth increases.

Ground under water is commonly so represented. Begin-
ning at low water-line, the space to the depth of 6 feet is
covered with a dark shade of India ink ; from 6 feet to 12 feet

FIG. 342.

MAPPING. 779

with a lighter shade ; from 12 feet to 18 feet with still lighter,
and from 18 feet to 24 feet with lightest of all. Greater
depths are marked in fathoms.

1 37O. Uses of Contours :

2. To obtain vertical sections profiles.

3. To calculate excavation and embankment.

In both railroad and highway location, the contour map
is used in platting a grade line to which the final location
should closely approximate. The paper location is then made,
conforming as closely as possible to the grade line in the
contour map, and from that location a final profile is platted.

When the contour map is to be used as a basis for the cal-
culation of excavation and embankment, a hill or hollow is
conceived to be divided into horizontal sections. The areas
of the upper and lower bases of any section are then calcu-
lated and their average is multiplied by the altitude of the
section, which gives the content of that section.

PLATE, TITLE : CONTOURS AND SLOPES.

1 37 1 . This plate contains three examples in topograph-
ical drawing. Each example affords practice in drawing
shore lines. Fig. 1 is an example under the first system in
which the topography is represented by level contours, and
affords the student excellent practice in contour mapping.
Figs. 2 and 3 are examples under the second system in which
topography is represented by lines of greatest slope or
hatchings.

First System By Level Contours. In Fig. 1 the
situation is a steep hillside bordering upon a lake. The en
gineer before commencing the field work should examine the
ground thoroughly in order that he may intelligently choose
a method of work well suited to the situation. In the case
in hand, the surface of the water in the lake is adopted as
the datum plane.

780 MAPPING.

It is a common and excellent practice to divide the area to
be contoured into squares, the dimensions of which will de-
pend upon the area to be treated and the degree of detail
required in the work. Large areas are usually divided into
squares containing 100 ft. on a side. The division lines
serve as guides to those taking the levels. The intersections
of the division lines being 100 ft. apart, they render the loca-
tion of any intermediate point an easy task. These inter-
sections are called stations, and are usually numbered
consecutively or distinguished in some other way. In Fig.
1 the area is divided into squares 100 ft. on a side. Base
lines A X and A H are first established where distant and
well-defined targets may be set up and the lines carefully meas-
ured. The importance of an accurately measured base line,
and of a distant fixed target can not be overestimated. The
lines of division are determined by laying off lines at 90 to
these bases, and are supposed to be parallel, a difficult thing
to accomplish in rough country where short sights are fre-
quent, and impossible if the initial angle of each line is not
turned from the same backsight, and that a comparatively
distant one. The base lines being established, the lines of
division are carefully run. The vertical division lines, i. e.,
those parallel to general direction of the lake shore, are des-
ignated by the letters of the alphabet, the first being de-
scribed as line A, the next as line B, the next as line C, and
so on. The horizontal division lines are numbered consec-
utively, Commencing with the bottom line, which is num-
bered 0, the next parallel to it 1, the next 2, and so on.
The intersections of the division lines locate the succeeding
stations on each line. This greatly simplifies the keeping of
the notes, and enables the engineer to readily locate any
point and briefly describe it. Thus, the starting point of
line A is called line A, 0; the next intersection is called line
A, 1; the next line A, 2, etc. The engineer determines the
form of notes best suited to the situation. He will find a
leveling rod 20 ft. in length of great assistance when work-
ing in a locality where changes in elevation are frequent
and abrupt. The form of notes best adapted to the work in

MAPPING.

781

hand is the following, the notes being a record of the levels
which are taken at each intersection of the division lines:

LEVELS OF LINE A.

Station.

Height of
Instrument.

Elevation.

B. M.

+ 8.80

56.05

47.25

Q

7.20

48.80

1

13.70

42. 30

T. P.

19.72

36. 33

+ 4. 20

40. 53

2

5.70

34.80

3

14.80

25.70

T. P.

18.03

21.90

+ 3. 53

25. 43

4

7.10

18.30

5

9.30

16.10

T. P.

- 18.55

6.88

+ 3. 22

10.10

6

9.60

0.50

6 + 05

Edge of Lake.

1372. The contour map is made as follows: First
we draw the outlines of the given area, 1,100 ft. in length
by 750 ft. in width, to a scale of 100 ft. to the inch. These
boundaries are then divided into equal spaces of 100 ft. each,
as shown in the engraving, and the lines of division drawn,
the boundaries being drawn full and the division lines dotted.
The vertical division lines, as before stated, are designated
by the letters of the alphabet, and the horizontal lines by
numerals. *,

From the level notes we find the elevations of the stations
on lines A and B. These elevations we mark on the map at
their proper stations, and then locate the contour lines as
follows: Beginning with line A, we find that the elevation
of Station is 48.8 ft. ; that of Station 1 is 42.3 ft. ; hence,

782

MAPPING.

the difference of elevation between these stations is 48.8
42.3 = 6.5 ft. The distance between these stations is 100 ft.,

TOO

and the rate of fall between Stations and 1 is equal to r =

o. o

15.4, which is called a descending slope of 1 to 15.4. The
contours are 5 ft. apart, and, therefore, the elevations of
each contour will be some multiple of 5 ft. Contour 45 ft.
will come between Stations and 1 of line A. As the eleva-
tion of Station is 48.8 ft., we must, to reach contour 45, go

LEVELS OF LINE B.

Station.

Height of
Instrument.

Elevation.

7-4-12

10.10

Edge of Lake.

7

8 00

2 10

T P

1.24

8 86

+ 19.84

28.70

6

8 30

20 40

T. P.

1.65

27.05

+ 19 91

46 96

5

15.00

32 00

4

9.20

37.80

3

1 70

45 30

T. P.

2.21

44 75

+ 18.88

63.63

2

6.90

56 70

T P

2 24

61 39

+ 18 31

79 70

1

15 50

64 20

4.40

75.30

towards Station 1 far enough to descend an amount equal
to 48.8 45.0 = 3.8 ft. As the rate of fall is 1 in 15.4, to
fall 3.8 ft. we must go 15.4 X 3.8 = 58.5 ft., which brings us

MAPPING. 783

to contour 45. This distance we scale off to a scale of
100 ft. to the inch, marking the point where contour 45
crosses line A by a small dot. The next two lower contours
are 40 and 35. As the elevation of Station 1 is 42.3 and that
of Station 2 is 34.8, both of these contour lines will cross the
line between these stations. The total fall between Stations

1 00
1 and 2 is 42.3 34.8 = 7.5 ft., and the rate of fall is ^- =

7.5

13.3. To reach contour 40 we must fall 2.3 ft. below Station
1, and the distance will be 2.3 X 13.3 = 30.6 ft. To reach
contour 35 we must fall 5 ft. more, and the additional dis-
tance will be 5 X 13.3 66.5 ft. We accordingly locate
those contours at 30.6 ft. and at 30.6 + 66.5 = 97.1 ft. from
Station 1. The difference of elevation between Stations 2
and 3 is 34.8 25.7 = 9.1 ft., and equivalent to a de-
scending slope of 1 to 11 between them. Contour 30 will
come at 4.8 X 11 = 52.8 ft. from Station 2. The difference
of elevation between Stations 3 and 4 is 25.7 18.3 = 7.4 ft.,
which gives a descending slope of say 1 to 14. This is not
the exact rate of slope, but where decimal fractions are small
and slopes easy the fractions may be ignored, as they will not
to a perceptible degree affect the accuracy of the work. Con-
tour 25 will come at 14 X .7 = 9.8 ft. from Station 3, and
contour 20 at a point 70 ft. farther, or at say 80 ft. from
Station 3. The difference of elevation between Stations 4
and 5 is 18. 3 16.1 = 2. 2, but no contour line passes between
these points. The difference of elevation between Stations
5 and 6 is 16.1 .5 = 15.6, which gives a slope of 1 to 6.4.
This brings contour 15 at 7 ft., contour 10 at 39 ft., and
contour 5 at 71 ft. from Station 5.

1 373. The usual custom is to ivork up the notes, as it is
called, before commencing the platting of the contours,

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