CT=C'T'.

Run the two tangents to their point of intersection T, measure back from T the distances TC and TC', and at C and C' set stakes marked P. S1.

Set up the transit at P. Sı, sight on T, and then set stakes on the spiral exactly as on a simple circular curve, except that the deflection angle for each stake is computed by the formula or taken from the tables. When the stake at A (marked P. S2) has been set, move the transit to A, backsight on P. Sı, and deflect from this direction the angle necessary to bring the telescope tangent to the simple circular curve at A. This angle is equal to the angle of deviation ▲ minus the angle of deflection VCA. Run in the circular curve as usual.

When the stake at B (marked P. S2') has been set, move the transit to C', backsight on T, and stake out the second spiral in exactly the same manner as the first, using the deflection angles computed for the first spiral. When the last stake along C' B has been set, backsight on T, and continue the survey along the tangent C'R'.

EXAMPLE.

Two tangents that intersect at an angle of 80° 20' are to be connected with a 6° circular curve by two equal spirals, each 300 ft. long. The tangents intersect at Sta. 36. Lay out the two spirals and the circular curve. SOLUTION.-The unit degree of spiral a=·

Dc 6°

L 3

=2°; the

spiral offset F= .072709 aL3= .072709X2X33=3.93 ft.; CV= }

5,730

Dc

length-t cor.=150-.000127 a2L5=150-0.1=149.9. R=

5,730

955 ft. CT= length-t cor.+(R+F) tan

2

80° 20'

6

-149.9+(955+3.93) tan 40° 10′959.3 ft.

Since T is at Station 36, the station number of the P. Si is 36 (9+59.3)=26+40.7.

It will be assumed that stakes are set 50 ft. apart on the spirals and at the even stations on the circular curve. The spiral deflections are then figured as shown in example under the heading Angle of Deviation and Angle of Deflection.

0° 5'

They are:

0° 20' (A) Angles to be

to first stake,
to second stake,

to third stake,

The deviation angle A=

tangent. Vernier

aL2 = 1×2×32 = 9o. Therefore,

the central angle of circular curve-I-2A=80° 20' -29° = 62° 20'. The length of AB is therefore 62° 20′ ÷6-10.389 Sta. and the station number of B is 29+40.7+(10+38.9)

The angle between the chord CA and the tangent to the circular curve at A is ▲

=9°—3°=6°.

Transit at P. S2.-The deflection angles to the stakes on the circular curve are as follows:

to Sta. 30, .593X3° 1°47′; to Sta. 35, 16° 47'

Transit at P. Si'.-The angles to be deflected are the same as at P. S1. The station number of P. Sı' is (39+79.6)+3

=42+79.6.

The Field Work.-Run the two tangents to their intersection. Measure back from T the distances TC= TC'=959.3 ft., and set stakes marked P. Si at C and C'. Set the transit at C with the vernier at 0° 0′; sight on T and deflect the angles (A) to locate the first spiral. When the stake at A (marked P. S2) has been set, move to this point, set the vernier at 6° 0', backsight on C, turn the telescope until the vernier reads 0° 0', and from this direction deflect the angles (B) to locate the circular curve. When the stake B (marked P. S2) has been set, move the transit to C', set the vernier at 0° 0', backsight on T, and deflect the angles (A) to locate the second spiral.

SELECTION OF SPIRALS

For a given velocity of train, in miles per hour, V, and the degree of curve of the circular curve Dc, the best length of spiral, in stations is found by the following formula:

EXAMPLE.-Find the theoretically best length of spiral to connect with a 6° curve, the maximum train velocity being

ing to the least length of spiral that the engineer should endeavor to insert. The spiral may be longer than the length obtained from this table, but it should not be shorter, unless

topographical conditions make it necessary to use a shorter spiral than the minimum given in the table.

The least length corresponding to any value of a is found from the formula

Dc
L=

a

EXAMPLE. Find the least length for the spiral in the preceding example.

SOLUTION. The velocity is 40 mi. per hr.; therefore, from the table, a=2°, and L=6°÷23 sta. = 300 ft.

EARTHWORK

FIELD WORK

Cuts and Fills.-In building a railroad, cuts and fills are introduced to equalize the irregularities of the natural soil. Figs. 1 and 2 show a typical fill and cut in ordinary firm earth or gravel.

Slope Ratio. is usually made

In cuts in the hardest rock, the average slope :1; that is, horizontal to 1 vertical.

As

the soil becomes less firm the slope must be flattened until, for a soil of firm earth or gravel, a slope of 1 to 1 may be permissible, although a slope of 1:1 is commonly adopted. In very soft soil, the slope ratio is sometimes cut down even as far as 4 horizontal to 1 vertical. The standard practice