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gent distance of the first curve.

Whence, (.30891 + .37887) : .30891 :: 1,011 ft. : the tan-
gent distance.

Whence, tangent distance of the first curve = ..^'^^ =

. Ob* I 8

454.
8 ft.

Substituting known values in formula 91, T= R tan ^/
(see Art. 1251), we have 454.08 = 7? X .30891; whence,

R = ^oAon? = 1,469.94 ft. Dividing 5,730 ft., the radius of
. 30891

a 1° curve, by 1,469.94, the length of the required radius,
the quotient 3.899 is the degree of the required curve. Re-
ducing the decimal to minutes, we have the degree of the
required curves 3° 53.9'.

(796) This question also comes under Problem IV,
Art. 1426. We have

2 2

The distance between intersection points is 816 ft. From
Art. 1426, we have

(tan 10° 07' + tan 20° 34') : tan 10° 07' :: 816 ft. : the tangent
distance of the first curve.

Whence, we have (.1784:\$ + .:]75'21, : . 17^4:? :: SIC : tangent
distance.

Whence, tangfent distance = — z~-~-. = 202.085 ft.

.oo.}<;4

Substituting known values in formula 91, 7' = R tan \I
(see Art. 1251), we have 2(i2.085 = i^ X .17843; whence,

R - -j'^-^y = 1,473.88 ft. = the radius of a 3° 53.3' curve.

(797) This question comes under Problem V, Art.

1427. The required radius is equal to 470 ft. divided

28° 40'
by (tan \ 28° 40' + tan \ 30° 16'). , ^ 14° 20'; tan

14° 20' = .25552. ^^ }^\ = 15° 08'; tan 15° 08' = .27044.

The sum of these tangents is .52596, and we have R =

470 ft

-^-—-f = 893.04 ft. Dividing 5,730 ft., the radius of a V

. 52oy»)

curve, by the radius 893.64 ft., the quotient is the degree

of the required curve. ^ ' ' \. , = 0.412° = 0° 24.7' curve.

890.04

(798) This question also comes under Problem V, Art.

1 427. The required radius R = . ..o° k<v \ «. — , ,ioon' -

tan -i" 32 50 + tan ^ 41 20

\ = 10° 25'. tan 10° 25' = .29403. . = 20° 40'.

tan 20° 40'= .3772. .29403 + .37720 = .07183. Substituting

this value in the above equation, we have A' = '^ =

708.05 ft., the radius of a 7° 27.0' curve.

(799) This question comes under Problem VII, Art.

7 3 X 100
1429. The required distance across the stream= \ ^.^ =■

1.745

418.3 ft.

(800) See Art. 1433.

(801) See Art. 1434.

(802) See Art. 1435.

(803) The usual compensations for curvature are from
.03 ft. to .05 ft. per degree.

(804) As the elevation of grade at Sta. 20 is 118.5 ft.,

and that at Sta. 40 is 142.5 ft., the total actual rise between

those stations is the difference between 142.5 and 118.5,

which is 24 ft. Hence, the average grade between those

24
points is ,^ = 1-2 ft. per station. As the resistance owing to

the curvature is equivalent to an increase in grade of .03 ft.
per each degree, the total increase in grade owing to curva-
ture is equal to .03 ft. multiplied by 78, the total number of
degrees of curvature between Sta. 20 and Sta. 40. 78 X
.03 = 2.34 ft. This amount we add to 24 ft., the total actual
rise between the given stations, making a total theoretical
rise of 26.34 ft. Dividing 2G.34 by 20, we obtain for the
tangents on this portion of the line an ascending grade of
1.317 ft. per station. Hence, the grade between Sta. 20
and Sta. 24 + 50 is + 1.317 ft. per station. The distance
between Sta. 20 and Sta. 24+50 is 450 ft. =4.5 stations,
and the total rise between these stations is 1.317 ft. x 4.5 =
5.9265 ft., which we add to 118.5 ft., the elevation of grade
at Sta. 20, giving 124.4265 ft. for the elevation of grade at
Sta. 24 -|- 50. The first curve is 10°, which is equivalent to
a grade of .03 ft. X 10 = 0.3 ft. per station, which we sub-
tract from 1.317, the grade for tangents. The difference,
1.017, is the grade on the 10° curve, the length of which is
420 ft. =4.2 stations. Multiplying 1.017 ft., the grade on
the 10 curve, by 4.2, we have 4.2714 ft. as the total rise on
that curve, the P. T. of which is Sta. 28 + 70. Adding
4.2714 ft. to 124.4265 ft., we have 128.6979 ft., the elevation
of grade at Sta. 28 + 70. The line between Sta. 28 + 70
and Sta. 31 + 80 being tangent, has a grade of 1.317 ft.
The distance between these stations is 310 ft. = 3.1 stations,
and the total rise between the stations is 1.317 X 3. 1 =

4.0827 ft., which we add to 128.6979 ft., the elevation of
grade at Sta. 28 + 70, giving 132.7806 ft. for the elevation
of grade at Sea. 31 + 80. Here we commence an 8° curve
for 450 ft. =4.5 stations. The compensation in grade for
an 8° curve is .03 ft. X 8 = 0.24 ft. per station. Hence, the
grade for that curve is 1.317 ft. - 0.24 ft. = + 1.077 ft. per
station, and the total rise on the 8° curve is 1.077 ft. x 4.5 =
4.84G5 ft., which we add to 132.7806, the elevation of
grade at Sta. 31 + 80, giving 137.6271 ft. for the elevation
of grade at Sta. 36 + 30, the P. T. of the 8° curve. The
line between Sta. 36 + 30 and Sta. 40 is a tangent, and has
a grade of 4-1.317 ft. per station. The distance between
these stations is 370 ft. = 3.7 stations, and the total rise is
1.317 X 3.7 = 4.8729 ft., which, added to 137.6271 ft., the
elevation of grade at Sta. 36 + 30, gives 142.5 ft. for the
elevation of grade at Sta. 40.

(805) See Art. 1439.

(806) In this question, ^=+1.0 ft., ^'z=— 0.8 ft.
and « = 3. Substituting these values in formula lOl,

d= ^'~ ^ (see Art. 1440), we have a = - — ~ ) ~ ' ' =
4« 12

1 8

6 stations of the vertical curve are the following: g — a,
g — da, g — 5a, g — 7a, g — S)a, g — \\a. Substituting known
values of g and a, we have for the successive grades :

Heights of Curve
Above Starting
Stations. Point.

1. g- rt = 1.0 ft. -0.15 ft. = 0.85 ft 0.85 ft.

2. g- 3rt = 1.0 ft. —0.45 ft; = 0.55 ft. ... .1.40 ft.

3. g- 5rt= 1.0ft. -0.75 ft. = 0.25 ft 1.65 ft.

4. g- 7a = 1.0 ft. -1.05 ft. = -0.05 ft 1.60 ft.

5. ^ - 9<J= 1.0 ft. - 1.35 ft. = -0.35 ft 1.25 ft.

6. g- na= 1.0ft. -1.65 ft.= -0.65 ft 0.60 ft.

As the elevation of the grade at the starting point of the

curve (which we will call Sta. 0) is 110 ft., the elevations ot
the grades for all the stations of the curve are the following:

110.00

1 110.85

2 111.40

3 111.65

4 111.60

5 111.25

110.60

(807) In Fig. 72, A 6^ is an ascending grade of 1 per
cent., and C B is sl descending grade of 0.8 per cent,, which

,,-'- — « 6 « d • f~ g

Fig. 78.

are the grades specified in Question 806. To draw these
grade lines, first draw a horizontal line A D ^ in. long,
which will include both grade lines to a scale of 100 ft. to
the inch. Divide this line into six equal parts, with the
letters «, b, c, etc., at the points of division. Now, ^ being
300 ft. from A, the height of the original grade line above ^
is the rate of grade. ^= 1.0 ft. X 3 = 3 ft., which distance
we scale off above d to a. vertical scale of 5 ft. to the inch,
locating the point C. The grade of C B is — 0.8 ft. per
100 ft. Consequently, B, which is 300 ft. from C, is 0.8 ft.
X 3 = 2.4 ft. below C, and the height of B above A D is
equal to 3.0 ft. — 2.4 ft. = 0.6 ft. We scale off above D the
distance 0.6 ft., locating B. Joining A C and B C, we have
the original grade lines, which are to be united by a vertical
curve. Now, the elevation of the grade at Sta. is 110 ft,
The line A D has the same elevation. We have already
determined the heights of the curve above this line at the
several stations on the curve. Accordingly, we lay off
these distances, viz., at ^, corresponding to Sta. 1, 0.85 ft. ;
at r, Sta. 2, 1.4 ft. ; at d^ Sta. 3, 1.65 ft., etc., marking each

point so determined by a small circle. The curved line
joining these points is the vertical curve required.

(808) See Art. 1443.

(809; See Art. 1444.

(81 0) See Art. 144cf.

(811) See Art. 1447.

(812) See Art. 1451.

(QUESTIONS 8:i3-«72 )

(813) See Art. 1454.

(814) See Art. 1455.

(815) The slope given to embankment is 1^ horizontal
to 1 vertical, and the slope usually given to cuttings is
1 horizontal to 1 vertical. (See Art. 1457 and Figs. 380
and 381.)

(816) The height of the instrument being 127.4 feet

and the elevation of the grade 140 feet, the instrument is

below grade to an amount equal to the difference between

140 and 127.4 feet, which is 12.6 feet. The rod reading for

the right slope being 9.2 feet, the surface of the ground is

9.2 feet below the instrument, which we have already shown

to be 12.6 feet below grade. Hence, the distance which the

surface of the ground is below grade is the sum of 12.6 and

9.2 which is 21.8 feet, which we describe as a fill of 21.8

feet. Ans. The side distance, i. e., the distance at which we

must place the slope stake from the center line, is 1^ times

21.8 feet, the amount of the fill, plus \ the width of the

3 X 21 8

roadway, or 8 feet. Therefore, side distance = ' -{-8 =

A

40.7 feet. Ans.

(817) The height of instrument H. I. being 96.4 feet,
and the rod reading at the surface of the ground 4.7 feet,
the elevation of the surface of the ground is 96.4 — 4.7 =
91.7 feet. As the elevation of grade is 78.0 feet, the amount
of cutting is the difference between 91.7 and 78.0 = 13.7
feet. Ans.

(818) The elevation of the surface of the ground is the
height of instrument minus the rod reading ; i.e., 90. 4 — 8. 8 =
87.(3 feet. The elevation of grade is 78.0 feet; hence, the
cutting is 87.6 — 78.0 = 9.0 feet. The slope of the cutting is
1 foot horizontal to 1 foot vertical ; hence, from the foot of
the slope to its outer edge is 9.0 feet. Ans. To this we
add one-half the width of the roadway, or 9 feet, giving foi
the side distance 9.6 + 9 = 18.0 feet. Ans.

(819) See Art. 1459.

(820) See Art. 1460.

(821) See Art. 1461.

(822) We substitute the given quantities in formula 102
A = Ci/Jf (see Art. 1461), in which A = the area of the
culvert opening in square feet; C the variable coefficient,
and M the area of the given water shed in acres. We
accordingly have A = 1.8 -/iOO = 30 square feet. Ans.

(823) Applying rule I, Art. 1462, we obtain the dis-
tance from the center line to the face of the culvert as
follows : To the height of the side wall, viz., 4 feet, we add
the thickness of the covering flags and the height of the
parapet — each 1 foot, making the total height of the top of
parapet 4: -\- 2 = 6 feet. 28 feet, the height of the embank-
ment at the center line, minus feet = 22 feet. With this
as the height of the embankment, we calculate the side dis-
tance as in setting slope stakes 1|- times 22 feet = 33 feet.

One-half the width of roadway is — = 8 feet. 33 + 8 = 41 feet.

z

To this distance we add 18 inches, making 42 feet inches.
As the embankment is more than 10 feet in height, we add
to this side distance 1 inch for each foot in height above the
parapet, i. e., 22 inches = 1 foot 10 inches. Adding 1 foot
10 inches to 42 feet inches, we have for the total distance
from the center line to the face of the culvert 44 feet 4
inches. Ans,

We find the length of the wing walls by applying rule Ilr

Art. 1462. The height of top of the covering flags is
5 feet. 1^ times 5 feet =7.5 feet. Adding 2 feet, we have
length of the wing walls, i. e. , the distance from inside face
of the side walls to the end of the wing walls, 7.5 + 2 = 9
feet G inches. Ans.

(824) The span of a box culvert should not exceed
3 feet. When a larger opening is required, a double box cul-
vert with a division wall 2 feet in thickness is substituted,

(825) See Art. 1 462.

(826) See- Art. 1462.

(827) See Art. 1463.

(828) See Art. 1465.

(829) The thickness of the base should be -j^ of the
height. The height is 16 feet and the thickness of the base
should be t*^ of 16 feet, or 6.4 feet. Ans.

(830) See Art. 1468.

(831) See Art. 1468.

(832) Trautwine's formula for finding the depth of
keystone (see formula 103, Art. 1469), is as follows:

J .1 ri . • r 4/radius of arch +A^ span , , , ,
depth of keystone m feet=^^ j !-^— ^- \-0.2 foot.

In this example the arch being semicircular, the radius and
half-span are the same. Substituting known values in the
given formula, we have

4/T5 + 15
depth of keystone = j 1- 0.2 foot =1.57 feet. Ans.

Applying Rankine's formula (formula 104), depth of
keystone = -/. 12 radius, we have depth of keystone =
y. 12 X 15= 1.34 feet. Ans.

(833) Applying the rule given in Art. 1 469, we find

,. r J- • , . 19'-|- 12* 505 ,, ,^/^ ^
the length of radius is equal to jj = —j- = 21.04 teet.

Ans.

(834) Applying formula 105, Art. 1470, we have

thickness of abutment _ 12 feet 8 feet . ., r ^ _ ► o f
at spring line o 10 .

(835) See Art. 1472.

(836) See Art. 1471.

(837) See Art. 1472.

(838) See Art. 1473.

(839) See Art. 1473.

(840) See Art. 1474.

(841) See Art. 1475.

(842) See Art. 1475.

(843) They should be laid in the radial lines of the arch.
See Art. 1477.

(844) See Art. 1477.

(845) Until the arch is half built, the effect of the
weight of the arch upon the centering is to cause a lifting at
the crown. After that point is passed, the effect of the
weight is to cause a lifting of the haunches.

(846) 120°. Ans.

(847) See Art. 1480.

(848) According to the rule given in Art. 1480, the
thickness of the base should be ^ of the vertical height
10 feet X j*o = 4 feet. Ans.

(849) See Art. 1480.

285

(850) The friction of the backing against the wall adds
considerably to its stability.

(851) See Art. 1481.

(852) See Art. 1482.

(853) See Art. 1483 and Fig. 401.

(854) See Art. 1484.

(855) The line d c in Fig. 73 forms an angle of 33° 41'
with the horizontal d h and is the natural slope of earth.

I 5^ - i

Fig. 73.

The line df bisects the angle ode. The angle o df is called
the angle, the slope df is called the slope, and the triangular
prism o d f \s called the prism of maximum pressure.

(856) See Art. 1486.

(857) See Art. 1486.

(858) From Fig. 73, given above, we have, by applying
formula 106, Art. 1486,

perpendicular pressure;/ P =

the weight o f the tri angle of earth odfx of

the vertical height od

286

(859) First £^raviijf, i.e.,
the weight of the wall itself,
and second friction produced
by the weight of the wall upon
its foundation and by the
pressure of the backing
against the wall.

(860) The angle of wall
friction is the angle at which
a plane of masonry must be
inclined in order that dry
sand and earth may slide
freely over it.

(861) The base ^/, Fig.
74, of the triangle b d f is
12.73 feet, and the altitude
^ ^ is 16 feet. Hence, area of
b d f is 12.73 X 8 = 101.84
square feet. Taking the
weight of the backing at 120
pounds per cubic foot, we have
the weight of bdf^ 101.84 X

120 = 12,231 pounds. Multiplying this weight by -/ / =
8.56 feet, we have 12,221 X 8.50= 104,612 pounds, which,
divided by o d, 16 feet, = 6,538 pounds = the pressure of
the backing. This pressure to a scale of 4,000 pounds to
the inch equals 1.63 inches. P, the center of pressure,
is at ^ of the height of b d measured from d. At P erect a
perpendicular to the back of the wall. Lay off on this per-
pendicular the distance /*//= 1.63 in. Draw /*/, making
the angle ;//*/= 33° 41' = the angle of wall friction. At //
draw a perpendicular to P n^ intersecting the line P t \n h.
Draw h k, completing the parallelogram ;/ h k P\ h n, or its
equal k J\ will represent the friction of the backing against
the wall and the diagonal h P, which, to the same scale, =
7,857 pounds, will be the resultant of the pressure of the
backing and the friction. Produce P // to 5. The section
a b d c oi the wall is a trapezoid. Its area is 84 square feet,
which, multiplied by 154 pounds, the weight of rubble per
cubic foot, gives 12,936 pounds, as the weight of the wall,
which, to a scale of 4,000 pounds to the inch, = 3.23 inches.
Find the center of gravity g of the section a b d c, diS ex-
plained in Art. 1488. Through g draw the vertical line
g ty intersecting the prolongation of P h in /. Lay off from
/, on g i, the distance / z\ 3.23 inches = weight of the wall,
and on Is, the distance I in = 1.96 inches, the length of the
resultant P h\ complete the parallelogram I in u v. The di-
agonal / u is the resultant of the weight of the wall and the
pressure. The distance c r from the toe c to the point where
the resultant / u cuts this base is 3.6 feet, or nearly \ of the
base c d. This guarantees abundant stability to the retaining
wall.

The distance o f\s obtained as follows : c d=S feet ; bat-
ter of c a = 1 inch for each foot of length of ^ ^= 16 inches;
hence, a o = S feet — 16 inches = 6 feet 8 inches; a b =
2. 5 feet; therefore, ^/= 12.73 — (6 feet 8 inches — 2. 5 feet) =
8 56 feet.

(862) See Art. 1491.

(863) See Art. 1491 and Figs. 409, 410, and 411

(864) See Art. 1492.

(865) See Art. 1493.

,_^ , 540, the number of working minutes in a day,

^^^^^ 1:25+1

,_ ^, . 102.9 ^ o^ u- A ^115

102.9 trips. — — — = 7.3o cubic yards per man; — — -— =
14 7.o5

15. G4 cents per cubic yard for wheeling. One picker serves

5 wheelbarrows; he will accordirigly loosen 7.35 X 5 = 36.75

\$1.15
cubic yards. Cost of picking will, therefore, be " =:

3.13 cents per cubic yard. There are 25 men in the gang,
one-fifth of whom are pickers. There are, consequently,
20 wheelers, who together will wheel in one day 7.35x20 =
147 cubic yards. One foreman at \$2.00 and one water-car-
rier at 90 cents per day are required for such a gang. Their
combined wages are \$2.90. The cost of superintendence

\$2 90
and water-carrier is, therefore, ,',^ = 1.97 cents per cubic

' 147

yard. Use of tools and wheelbarrows is placed at ^ cent

per cubic yard.

Placing items of cost in order, we have

Cost of wheeling 15. G4 cents per cubic yard.

Cost of picking 3.13 cents per cubic yard.

Cost of water-carrier and super-
intendence 1.97 cents per cubic yard.

Use of tools and wheelbarrows. .50 cents per cubic yard.

21.24 cents per cubic yard.

(867) The number of carts loaded by each shoveler

420
(Art. 1494), is - - =84. 84 -^ 3 = 28 cubic yards handled

per day by each shoveler. The cost of shoveling is, there-

120
fore, — — = 4.28 cents per cubic yard. The number of cart
28

trips per day is rr—— J = 54. 54. As a cart carries ^ cubic

yard, each cart will in one day carry — - — = 18.18 cubic

o

yards. As cart and driver cost \$1.40 per day, the cost of
hauling is r-~-r^ =7.7 cents per cubic yard. Foreman and

lo. lo

water-carrier together cost \$3.25. The gang contains
12 carts, which together carry 18.18x12 = 218.2 cubic

\$3 25

yards. ' = 1.49 cents per cubic yard for water-carrier
218. 2

and superintendence. As loosening soil costs 2 cents per

cubic yard, dumping and spreading 1 cent per cubic yard,

and wear of carts and tools | cent per cubic yard, we have

the total cost per cubic yard as follows:

Loosening soil 2.00 cents per cubic yard.

Shoveling into carts 4.28 cents per cubic yard.

Hauling 7.70 cents per cubic yard.

Superintendence and water-car-
rier 1.49 cents per cubic yard.

Wear and tear of carts and tools, 0.50 cents per cubic yard.

Dumping and spreading 1.00 cents per cubic yard.

Cost for delivering on the dump, 16.97 cents per cubic yard.

(868) See Art. 1499.

(869) See Art. 1499.

(870) See Art. 1500.

(871) See Art. 1500.

(872) In carrying the steam a long distance through
iron pipes its pressure is greatly reduced by condensation,
whereas compressed air may be carried a great distance
without suffering any loss in pressure excepting that due tc
friction and leakage.

(QUESTIONS 878-946.)

(873) See Art. 1503.

(874) See Art. 1504.

(875) See Art. 1504.

(876) See Art. 1505 and Figs. 426 and 428.

(877) As 60" F. is assumed as normal temperature, and
as the temperature at the time of measuring the line is 94°,
we must, in determining the length of the line, make an
allowance for expansion due to an increase of temperature
equal to the difference between 60° and 94°, or 34°. The
allowance per foot per degree, as stated in Art. 1505, is
.0000066 ft., and for 34° the allowance per foot is. 0000066 X
34 = .0002244 ft. ; for 89.621 ft. the allowance is .0002244 X
89.621 = 0.020 ft. The normal length of the line will, there-
fore, be 89.621 + .020 = 89.641 ft. Ans.

(878) Let the line A B in Fig. 75 represent the slope
distance as measured, viz., 89.72 ft. This line will, together

with the difference of ^ c 88.986' A

elevation between the ^ ' ■

extremities A and B,

viz. , 11. 44 ft. , and the " fig. n.

required horizontal distance A C, form a right-angled tri-
angle, right-angled at C. By rule 1, Art. 754, we have

sin A = ^^41 = .12751; whence, A = 7° 20'. Again, by

o9. 7/0

A C
rule 3, Art. 754, cos 7° 20' = g^nj^; whence, .99182 =

^^ and >4 C = 89. 72 X . 99182 = 88. 986 ft. Ans.

89.72'

(879) See Art. 1508.

(880) See Art. 1511.

(881 ) See Art. 1511 and Figs. 435, 430. 445, and 440.

(882) See Art. 1513.

(883) See Art. 1514.

(884) See Art. 1517.

(885) If the elevation of the grade of the station is
103 ft. and the height of the tunnel section is 24 ft., the
elevation of the tunnel roof at that station is 162 -f 24 =
180 ft. The height of instrument is 179.3 ft., and when the
roof at the given station is at grade, the rod reading will be
186-179.3 = 0.7 ft. Ans.

(886) See Art. 1515.

(887) See Art. 1 51 7 and Fig. 448.

(888) See Art. 1520 and Fig. 452.

(889) See Art. 1523.

(890) See Art. 1525.

(891) See Art. 1526.

(892) The area of an 18-inch air pipe is 1.5' X .7854 =
1.77 sq. ft. At a velocity of 13 ft. per second, the amount
of foul air removed from the heading per second is 1.77 X
13 = 23.01 cu. ft. And as each cubic foot of foul air re-
moved is replaced by one of pure air, in 1 minute the
amount of pure air furnished is 23.01 x 60 = 1,380.6 cu. ft.
As each man requires 100 cu. ft. of pure air per minute,
there will be a supply for as many laborers as 100 is con-
tained times in 1,380.6, which is 13.8, say 14. Ans.

(893) See Art. 1528.

(894) See Art. 1533 and Fig. 455.

(895) See Art. 1535 and Fig. 457.

293

(896) See Art. 1 536 and Fig. 458.

(897) The height of embankment is 21 ft. ; the culvert
opening, 3 ft. in height, and the covering flags and parapet
each 1 ft. in height, making the top of parapet 5 ft. above
the foundation. In Fig. TO the angle B A L is 75° and
represents the skew of the culvert. Drawing the line A D
at right angles to the center line A C^ we have the angle
BAD= S)(f - 75° = 15°. If the culvert were built at right
angles to the center line, the side distance A D from the

Fig. 76.

center line to the end of the culvert would be as follows:
21 ft. - 5 ft. = 10 ft. 10 ft. X li = 24 ft. Adding 1.5 ft.,
and, in addition, 1 in. for each foot of embankment above
the parapet, i. e., 10 in., or 1.33 ft., we have 24 -|- 1.5 + 1 33 =
we have 20.83 + 8 = 34.83 ft. = A D. At D erect the per-
pendicular DB. In the right-angled triangle A D B, we

have cos A — -j— n, i- e., cos 15° = ' .' ' ; whence, .90503 =
A. B A. L)

•^^•^^^ and ^^ = 4t^, = 30.00 ft.

AB

.90593

Ans.

(898) See Art. 1541.

(899) See Art. 1541 and Fig. 402.

(900) See Art. 1542.

(901) See Art. 1544.

between the height of instrument and the elevation of the
grade for the given station ; i. e., 125.5 — 118.7 =6.8 ft.

Ans.

(903) See Art. 1547.

(904) See Art. 1548.

(905) The skew of abridge is the angle which its center
line makes with the general direction of the channel spanned
by the bridge.

(906) See Art. 1549.

(907) The length of A ^ is determined by the principles
of trigonometry stated in Art. 1243, from the following
proportion : sin 46° 55' : sin 43° 22' :: 421.532 : A B; whence,
^^=396.31 ft. Ans.

(908) See Art. 1551.

(909) See Art. 1553.

(91 0) See Art. 1554 and Figs. 465 and 466.

(91 1) The depth of the center of gravity of the water

below the surface is — = 5 ft. Applying the law for lateral

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