and FB being equal, the angles A M D, D ME, EM F, and F M B are equal.

6. The angle KCB is called the angle of intersection of the tangents, and is equal to the central angle A M B subtended by the chord A B, which joins the tangent points A and B.

PROBLEM I.

The positions of two straight or tangential portions of a railway, GA and H B, being given, to determine the radius of the curve that joins them. (See figure, p. 159.)

This problem seems to admit of an indefinite number of solutions, but when local obstructions, such as rivers, roads, hills, buildings, &c., are considered, it is often limited to one solution. Besides, by the Standing Orders of Parliament, it is required that " a memorandum of the radius of every curve not exceeding one mile in length shall be noted on the parliamentary plan in furlongs and chains ;" and even in the case of curves having radii of one mile and upwards, no engineer in laying out a

CONCISE TRIGONOMETRICAL FORMULA.

Let A B C be a right triangle. If the angles be denoted by A, B, and C, and the sides by a, b, and c, as in the figure, we shall have these six formula:

α

b=√(c+a).(c—a), or √✅✅ c2—a®

α

8a, bA, B, ctan. A= cot. B=7› c = √ a2+b2

9A, a B, b, cB=90°-A, ba cot. A, c==

10A, 6B, a, cB=90°—A,a=b tan. A, c=

α

sin. A

b

cos. A

11]A, c|B, a, b|B=90°—A, a=c sin. A, b=c cos. A

railway is justified, under any circumstances, in passing beyond the limits of deviation which are limited by the Standing Orders of Parliament to 100 yards on each side of the centre line as shown on the parliamentary plan.

Solution. Let G A and H B be produced until they meet in C. Extend G A to K. Measure the angle K C B, which is equal to the central angle A M B, subtended by the curve; half this angle is equal to the angle A M C, which, subtracted from 90°, gives the angles A CM. The point A being the tangent point from which the curve is proposed to commence, measure the line A C. We have then given all the angles of the triangle AC M, and the side A C, by which the radius A M can be readily found by No. 6 formula, p. 160. By this formula =cot. A M C, or

A M

A C

Rad. or RT cot. A M C.

R

T

EXAMPLE.

=

cot. A M C; therefore

Given A M C = 22°30′, and T = 950, to find R, or radius.

Method of laying out a curve by tangential angles.-The method generally adopted at present for laying out curves on

the ground is by means of tangential angles, or, as they are sometimes called, deflection angles.

PROBLEM II.

Given the radius A MR, to find the tangential angle C A D for a chord of one chain. (See figure, p. 159.)

Solution.-Draw M L perpendicular to B F. Then the angle B MLB M FA M D C A D, the tangential angle, and B L= † B F=50 links. But in the right triangle M BL

we have (formula 1, p. 160) sin. B M L=

BL

BM

Second method of ascertaining the tangential angle.-Divide the constant 1718.9 by the radius in chains, which will give the angle in minutes.

In the present case the radius is 40 chains. Then 1718.9 40

=

= 42.972 minutes, or 42′ 58′′.

Having ascertained the tangential angle, fix the theodolite at the tangent point A, and lay off the tangential angle, C A D, the chain being extended from the tangent point A to D, in the same range as the visual axis of the telescope; the theodolite being still fixed at A, lay off the angle D A E=C AD, the chain being now extended from D, to meet the visual axis of the telescope at E, and so on to the end of the curve. Care should be taken to drive down some distinctive stump at the tangent point A, so that it may be easily found afterwards. Stumps should, of course, be driven down at the end of each chain in the curve, and at the point of intersection of the tangents at C.

It may happen some obstruction, such as a house, trees, &c., will prevent seeing from the point A further on the curve than E. If such should be the case, remove the theodolite to D, the point preceding E. Then sight first on E, and continue to

lay off the tangential angles as before until the end of the curve is reached.

PROBLEM III.

The radius A M and angle of intersection K C B being given, to find the length of the tangent A C. (See figure 159.)

Given K C B=22° 52′, and R=30.00, to find tangent A C.

When the point of intersection C of the tangents G A and

HB happens to fall in a sheet of water, a wood, or a pile of buildings, to find the angle of intersection K C B.

Set off D E and F I at right angles to G C, and J L and O P at right angles to H C, and let these perpendiculars measure equally, so that lines produced through the points E, I, L, and P, to intersect the tangents G C and H C, may clear

K

H

the obstacle in the way as at Q and R. Now measure the angle LS R, which is equal to the angle of intersection K C B. Measuring SQ and S R gives QC and R C, which added to A Q and B R, respectively give the length of the tangents A C and B C.

PROBLEM V.

Given the angle of intersection KC BA M B, and the tangential angle CAD, to find the length of the curve. (See figure, p. 159.)

The

Solution.-Half the angle of intersection, divided by the tangential angle, gives the length of the curve in chains. length thus found is not the length of the arc, but the length of

the chords. The length of the arc may be found by Rule 1, page 15, Stephenson's "Railway Construction," by Nugent (Weale's Series).

PROBLEM VI.

To lay out a curve on the ground with the chain only, by equidistant offsets from the tangents, no obstruction being on the convex side of the curve.

Let G A and H B be the tangents which are to be connected by a curve, the radius of which is given, and A the point whence the curve is proposed to commence. Produce the tangent G A in the same straight line towards C; measure, say, one chain from A to E. Now, if the length of the offset E F to the curve be known, by raising the offset the proper length, the point F in the curve becomes known.

To find the length of the offset E F. Join F D, and let fall

transpose L F2, and A D2L F2D L2; extract the square root of each side,

√ A D2 —L F 2

=

and

D L.

If D L be taken from A D, we get A L―E F, the length of the offset required. The length of the offset I K, at the end of the second chain on the tangent, can be found in a similar manner; for I KA M—A D—D M, D M being equal the /A D-K M2; and so on with regard to any other offset.

EXAMPLE.

Suppose the radius A D equal 40 chains, to find the length of the offset E F to the curve at the end of the first chain. By the above formula A D2—LF2=D L; that is, √/402—12—DL=√/1599=39·987498; this last number taken from the radius, 40 chains, leaves 012502 of a chain, the length of the offset E F. If we multiply 012502 by 792, the number of inches in a chain, we obtain 9·901584 inches =E F. The length of the offset I K is found in a similar manner. For DM=√AD2—K M2 — √ 402—22= √1596—39·9499687,

=