which subtracted from the radius, 40 chains, leaves 0500313 of a chain. This number, multiplied by 792, gives 39.6247896 inches as the length of the offset I K at the end of the second chain.

So that to obtain the length of an offset from any point on the tangent to the curve, we have only to subtract the square of the distance measured on the tangent from the square of the radius, and extract the square root of the remainder; the root so obtained subtracted from the radius gives the length of the offset.

Setting out a New Tangent.-At every eight or ten chains it may be advisable to recommence operations, which is done by setting out a new tangent to the curve. This may be done by joining the last point found in the curve to the next but one from it in the curve, viz., that corresponding to the offset two chains back. Then a line ranged forward parallel to this chord, and touching the curve at the middle point, will be the tangent required.

The following method will perhaps be found superior to the one just given, as no ground is lost, and affords a better position for ranging the new tangent. Let K be the point whence it is required to start with the new tangent. From A set off on AI2+I K2 A C, a distance equal to that is, the sum of the

2 A 1;

squares of the last offset and the whole distance measured on

the tangent, divided by

twice the same distance. A line from the point thus obtained, ranged forward through K, the last point found in the curve, will be the tangent required.

As the formula just given is not only useful in practice, but strictly geometrical, it may be well to give its demonstration. Join A K, and bisect it in Q; connect D Q, and produce it both ways to O and P. Now A K being bisected by the line D Q drawn from the

M

centre, the angles A Q P and P Q K are right angles

(Euclid 3, iii.); it follows that the angles IK Q and I P Qare together equal to two right angles; therefore a circle may be described about the quadrilateral I K_Q P. Let the circle be described. Now according to cor. Euclid 36, iii., K AXA Q I AXA P, but K A X A Q K A2 A I2+ I K2 A 12 + I K2

=

, therefore I A × A P—

2

multiply off by 2, and 2 IA × AP AI2+I K2; divide by

2 I A, and AP

A 12+ I K2

=

Now, as A Q and Q P are

equal to K Q and QP, each to each, and the contained angles equal, the line A P is equal to P K; consequently P K is a tangent (Euclid 37, iii.).

PROBLEM VII.

To lay out the curve by offsets from its chord, or chords, when obstructions occur on the convex side of the curve.

Let A C B be a portion or the whole of a railway curve, H A

K

B

a tangent at its commencement, CE a tangent at its middle point C, and A D, B D its radii. Take the chord AB, an even number of chains; find the successive offsets corresponding to the radius AD, and the tangent CE The last offset E A will be equal C F; from C F subtract the successive offsets, and the remainders will be respectively the offsets, which must be set off in an inverted order from A to F, and their order must be again inverted in setting them off from F to B. The operation may be continued by taking other chords, as B G.

= A B, as in Prob. 6, p. 164.

AF

An

NOTE. This method of laying out a curve may be advantageously used where a winding river, buildings, cliffs, &c., are close to or protrude in some places over the curve. The tangent CE, which is parallel to A B, is not used further than to explain the method of finding the offsets. important point in relation to this method has been hitherto entirely overlooked by writers on railway curves; that is, they do not furnish any means by which the direction of the chord A B is to be determined. This is necessary. Let the radius of the curve be 40 chains, and the chord A B be 10 chains, how is its direction to be determined? By No. 7 formula for the solution of right triangles, p. 160, the angle A D F is found to be 7° 10' 50", but the angle A D F is equal F A K. If a theodolite be fixed at A, and an angle equal to 7° 10' 50" be laid off with the line A K, it will determine the direction of the chord A B.

PROBLEM VIII.

8

Given the radius A F, or the tangential angle of a curve, and the angle B A C, made by the chord A C with the tangent at A, to find the length of the chord A C.

E

Suppose the curve A C is 4 chains long, and that we wish to find the length of the chord A C. In this case the angle B A C

100 sin. 4 t

and the radius

Given the line A B, which joins the fixed tangential points A and B, the angles B AI and A B I, and the first radius A E of a compound curve, to find the second radius B F to unite the tangents H A and K B.

BF=AE+

The point D may be determined in the field by laying off the angle I A D = (BAI + A B I), and measuring the distance A D = 2 A E sin. (BAI +ABI). BD and C B I may then be measured.

When the angle ABI is greater than B A I, that is, when the greater radius is given, the solution is the same, except that the angle D A B (A BI -B A I), and C B I is found by subtracting the supplement

BD
2 sin. C BI

B

of A BD from A B I. We shall also find C B C D—B D, and therefore

Given the tangents AI and B I, the angle of intersection BIR, and the first radius A E, to find the second radius B F. (See last figure.)

Solution.-Suppose the first curve to be run with the given radius from A to D, where its tangent D O becomes parallel to BI. Through D draw D P parallel to A I, and we have I P= D 0 A 0= Rad. tan. BIR (Prob. 3). Then in the triangle D P B we have D P=I 0=A Í—A 0—A I— Rad. tan. BIR=BI—IP=BI—Rad. tan. † B I R, and the included angle D P B=AIP=180° — † BIR. In the triangle D P B, the sides D P and P B are given, and the included angle, from which the angle C B I, and the side B D may be computed by formula 14, p. 161. The rest of the solution is the same as the last problem. The determination of the point D in the field is also the same; the angle I A D being in this case= BIR. When A B I is greater than B A I, that is, when the greater radius is given, the solution is the same, except that D P=Rad. tan. † B I R—A I, and B P= Rad. tan. BIR-BI.

PROBLEM XI.

H

Given the perpendicular distance between two parallel tangents B D, and the distance between the tangent points A B, to determine the reversing point C, and the common radius E C CF of a serpentine or reversed curve uniting the tangents H A and B K.

D

A

K

If the common radius E C=C F, and the perpendicular distance BD be given, and A B required, then

A B 2 √E CXB D'

The reversing point C is the middle point of A B.

PROBLEM XII.

Given the perpendicular distance between two parallel tangents B D, the distance between the two tangents' points A B, and the first radius E C, of a serpentine or reversed curve uniting the tangents H A and B K, to find the chords A C and C B, and the other radius C F.

Draw the perpendiculars E G and F L. Then the right triangles A B D and E A G are similar, since the angle B AD= AE C = AE G. Therefore A B: B D:: E A: A G, or AB: BD:: EC: A C; multiply means and extremes, and

ABX AC=BDXEC; multiply off by 2, and

ABXAC=2E CxBD. Divide by A Band AC

And A B-AC=C B.

2 ECXBD

AB

To find C F, the similar triangles A B D and F C L give AB: BD:: FC: CL; multiply extremes and means, and F CXB DA BXC L, but C L = C B ; therefore F CXB D =ABX CB; multiply off by 2, and F Cx2 BD=ABXCB; divide by 2 B D, and

If E C. C F and B D are given, to find A B, A C, and C B.