and this equation applies to a corrected subchord if we insert in it its uncorrected length.

Thus for a 14-foot subchord on a 3° curve the deflectionangle is 0° 12'.6.

Let us suppose that we are given a 3° Curve to the Right to locate from a P.C. at Sta. 421 + 36, I being equal to 12° 30'.

The length of the curve we find from Equation 5—since this is assumed as the standard method of measurement for railroad curves--to be 416.7 feet, therefore the P.T. will be at Sta. 42552.7; then if we intend to use 50-foot subchords, our notes will be arranged as follows:

The Index-reading at any station equals the sum of the deflections up to that station; then since the Index-reading at the P.T. is represented by the angle Aba in Fig. 20, and Aba is

I

easily proved equal to therefore the Index-reading at the

2'

P.T. must equal half the intersection-angle, thereby giving a check on the calculations.

Having the notes worked out as above, set the transit up at the P.C. as in Fig. 22, and setting the index to zero, clamp the telescope on to a back-sight on the tangent (or on to the apex

if it has been put in); then for any station the vernier must read the angle given in the index-column for that station. But suppose that when we have reached Sta. 423 + 50 we are uhable to see any farther. Then set a hub (with a tack in it) at that station and a back-sight at the P.C. Set up over the hub,

and setting the vernier back to zero, clamp the telescope on to the back-sight and turn off the remaining deflections by making the readings for the respective stations the same as those given in the Index-column. Thus:

(1) When pointing to any station, the vernier must always be set to read the Index-reading for that station.

(2) When on the tangent at any station, the vernier must always be set to read the Index-reading for that station.

By adhering to these two rules all possibility of error as regards the index-readings is avoided, and with the notes worked out as above we may locate the curve equally well from either end.

In order to find the bearing of the tangent at any station with reference to the tangent at the P.C., we have simply to multiply the index-reading at that station by two. Thus, if in the above example the tangent at the P.C. lies north and south, the bearing of the curve at Sta. 423 + 50 will be N. 6° 25'.2 E.

Usually in locating railroad curves there is no necessity to work out the deflections closer than to the nearer half-minute. In places where accurate measurement is difficult to obtain, and great exactness is wanted, as in giving centres for piers in the middle of a river, we can often do better work by using

Two Transits, one on either side of the stream, and fixing the points by intersection. (See Sec. 163.)

77. To locate a curve when the apex is inaccessible.

P.C.

FIG. 23.

-Suppose, as in Fig. 23, we have been unable to locate the apex of a proposed curve, but have connected the two tangents at a and b by the line ab.

Then in the triangle Aab we know the distance ab and the angles at a and b; therefore we have

where A

180°

40°;

(a+b). We can then find the position of the P.C. For example, suppose Aa =320 feet and I = then if we wish to connect the two tangents by a 5° curve, since the distance from A to the P.C. is given by Equation 9 (or Table II) 417.2 feet, therefore the P. C. will be situated

=

97.2 feet back on the tangent from a.

We can then locate the curve according to Sec. 76.

But suppose, instead of running a direct line ab, it is more

convenient to run a succession of courses as in Fig. 24. Then, if the position of the stations a and b has been worked out by "Lats. and Deps." we can at once find the angles at a and b and the length ab.

For instance, let

Tot. Lat. of a = 1020 N.
Tot. Lat. of b = 810 N.

Tot. Dep. of a = 560 E.
Tot. Dep. of b = 1430 E.

Then the bearing of ab will be given by the angle at a in the triangle aeb; thus

ab (1020

=

810) sec a = 895.2. Then if the bearing of the tangent at a N. 80° E., and of the tangent at b = S. 60° E., we have in the triangle Aab, a = 23° 34′ and b = 16° 26′ from which we can find the position of the P.C. as above.

If the notes have not been already worked out by Lats. and Deps. the position of b with reference to a can be most easily . calculated by taking the tangent at a as the N. and S. base. 78. To locate a curve by offsets from a tangent.-Let

ab be a tangent to the curve at a. Now the value of the tangential offset at any station is

t = R vers C.

ND where N = number of Stas. along the curve to

But C =
t, therefore

Similarly, the distance along the tangent from a to the offset t equals

Thus, for example, suppose a falls at Sta. 10 + 40, and we wish from this point to set out a 10° curve by offsets from the tangent at a; then at Sta. 11

t = R vers 6° = 3.14 feet,

and the distance along the tangent at which this offset must be set off equals

X= R sin 6° 59.95 feet.

The values of t at distances along the curves from a, 100 feet apart, are given in Table III, calculated by Equation 20. A formula that often comes in handy in the field for computing tangential offsets, and which is usually true enough when X does not exceed 150 feet, is

Tangential offsets may often be made use of when, on account of some obstacle or other, the method given in Sec. 76 cannot be used. By offsetting the tangent itself occasionally, as in Fig. 26, we can with ease run a curve past a succession of obstacles, and at the same time keep the offsets comparatively short.

FIG. 26.

Another occasion on which this method can be used to advantage is when the apex, P.C. and P.T. are inaccessible. Suppose, by way of example, that we have to locate a 10° curve in a position such as is represented in Fig. 27, the angle