We also find o⇒ 5 and o'= 10. The angle at the centre sub

tended by the 100-ft. chain is found by the relation

chord

r

sine angle at centre (Leg., Trig., Art. 64). In the case supposed we find the angle subtended to be 5° 44'. The number of full 48° 5° 44''

chords is equal to

which gives 8 full chords, and a remain

ing angle at the centre of 2° 8'. To find the sub-chord we have

The

peg at the end of the sub-chord should fall at the second tangent peg, if no errors have been made in chaining.

NOTE.--In employing this method of locating curves, the aligning by which the chords are produced should be done with much care, as any error in locating a stake involves much greater and increasing errors in succeeding stakes.

This is called, by engineers, "the method by offsetting from tangent and chords produced."

EXAMPLES.

1. What are the tangent and chord offsets, for a curve of 2000 feet radius; the stakes to be 100 feet apart?

Ans. From tangent, 2.5 ft.; from chord produced, 5 ft.

2. Find offsets for a one-degree curve.

Ans. Tangent, .87 ft.; chord, 1.74 ft.

360. Let it be required to run out a curve of 500 ft. radius, the stations being 25 feet apart on the curve.

As the chord of the angle i is to be 25 ft., we have (Fig. 172),,

The next tangent distance to z equals sine 3ix 500, and the offset at z equals 500-(cos 3ix 500), and so on for succeeding points.

When the offsets become too long to be readily and correctly set out, a new tangent is located thus: Prolong cd to h making dh = cd, and then bisect he (e having been already located) in f and range a line df, which line will be tangent to the curve at d. The correctness of the new tangent should always be checked by locating from it the third or fourth station counting back towards T. The same computed values that were used in locating the points already fixed may be again used in locating new points from the new tangent.

Laying off the Ordinates.

361. The methods described thus far for locating railway curves, apply to points 100 feet apart. This is sufficiently

accurate for the earthwork. In laying the track, however, stakes every ten or twelve feet are necessary. These are set by drawing the chain or tape in a straight line between the 100-ft. stakes, and measuring from it, offsets, as often as desirable, to the intermediate points of the curve.

The length of these offsets, or ordinates, is calculated in the following manner:

Let VW represent a 100-ft. chord of a railway curve, of which C is the centre. Draw the diameter HK parallel to VW, and drop the perpendicular VL. Then,

VL2 = HL × LK.

(Legendre, Bk. IV., Prop. 23, cor. 2). Since

the value of VL is readily calculated for known values of the radius.

Let NM be an ordinate, at any distance from VL, say 10 feet. Then,

whence,

NM2 = HM × MK ;

NM2 (r40) (r+40).

=

Having determined NM, subtract VL from it, and we have Nt, one of the ordinates required.

In this manner, by calculating the full ordinate to the diameter, and subtracting VL, any desired number of offsets are determined for the half chain VF. For FW, the ordinates have the same length, but are located in the inverse order.

The middle ordinate, FE, is found by subtracting VL from the radius.

EXAMPLE.

Determine the ordinates 10 feet apart on a 100-foot chord, for a two-degree curve. Radius, 2864.79 feet.

Reversed and Compound Curves.*

362. Two curves, of the same or different radii, may join each other and have a common tangent at the point of junction. If the curves lie on opposite sides of the common tangent, they form a reversed curve, and their radii may be equal or unequal; if they lie on the same side of the common tangent, they have unequal radii and form a compound curve. Thus ABC (Fig. 174) is a reversed curve, and

Taken, by permission and with slight alterations, from Henck's

Railroad Engineers."

FIG. 175.

-K

B

point, C, is in the line

AB joining the points of contact A and B.

A reversed curve is of use in certain track-work near stations, constructing turnouts, &c., though it is not, or ought not to be, used on the ordinary running track.

363. Given the perpendicular distance between two parallel tangents BD = b (Fig. 175), the distance between the two tangent points AB = a, and the first radius, EC R, of a reversed curve uniting the tangents HA and BK, to find the chords AC a' and CB a", and the second radius CF = R'.

=

=

Draw the perpendiculars EG and FL (Fig. 175). Then the right-angled triangles ABD and EAG have the angle BAD = AEG, since each is one-half AEC, and are, therefore, similar; hence,

or

AB: BD: EA: AG,

To find R', we have, from the similar triangles ABD and FBL,

ab R' ja".